diff --git "a/wiki/proofwiki/shard_0.txt" "b/wiki/proofwiki/shard_0.txt" new file mode 100644--- /dev/null +++ "b/wiki/proofwiki/shard_0.txt" @@ -0,0 +1,19371 @@ +\section{Closed Form for Triangular Numbers} +Tags: Triangular Numbers, Sums of Sequences, Closed Forms, Closed Form for Triangular Numbers + +\begin{theorem} +The [[Definition:Closed-Form Expression|closed-form expression]] for the $n$th [[Definition:Triangular Number|triangular number]] is: +:$\displaystyle T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ +\end{theorem} + +\begin{proof} +We have that: +:$\displaystyle \sum_{i \mathop = 1}^n i = 1 + 2 + \cdots + n$ +Consider $\displaystyle 2 \sum_{i \mathop = 1}^n i$. +Then: +{{begin-eqn}} +{{eqn | l = 2 \sum_{i \mathop = 1}^n i + | r = 2 \paren {1 + 2 + \dotsb + \paren {n - 1} + n} + | c = +}} +{{eqn | r = \paren {1 + 2 + \dotsb + \paren {n - 1} + n} + \paren {n + \paren {n - 1} + \dotsb + 2 + 1} + | c = +}} +{{eqn | r = \paren {1 + n} + \paren {2 + \paren {n - 1} } + \dotsb + \paren {\paren {n - 1} + 2} + \paren {n + 1} + | c = [[Integer Addition is Commutative]], [[Integer Addition is Associative]] +}} +{{eqn | r = \paren {n + 1}_1 + \paren {n + 1}_2 + \dotsb + \paren {n + 1}_n + | c = +}} +{{eqn | r = n \paren {n + 1} + | c = +}} +{{end-eqn}} +So: +{{begin-eqn}} +{{eqn | l = 2 \sum_{i \mathop = 1}^n i + | r = n \paren {n + 1} + | c = +}} +{{eqn | ll= \leadsto + | l = \sum_{i \mathop = 1}^n i + | r = \frac {n \paren {n + 1} } 2 + | c = +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +Observe that: +{{begin-eqn}} +{{eqn | l = \sum_{i \mathop = 1}^n \paren {\paren {i + 1}^2 - i^2} + | r = -\sum_{i \mathop = 1}^n \paren {i^2 - \paren {i + 1} ^2} +}} +{{eqn | r = -\paren {1 - \paren {n + 1}^2} + | c = [[Telescoping Series/Example 1|Telescoping Series]] +}} +{{eqn | r = \paren {n + 1}^2 - 1 +}} +{{end-eqn}} +Moreover, we have: +:$\paren {i + 1}^2 - i^2 = 2 i + 1$ +And also: +:$\paren {n + 1}^2 - 1 = n^2 + 2 n$ +Combining these results, we obtain: +{{begin-eqn}} +{{eqn | l = n + 2 \sum_{i \mathop = 1}^n i + | r = n^2 + 2 n +}} +{{eqn | ll= \leadsto + | l = 2 \sum_{i \mathop = 1}^n i + | r = n \left({n + 1}\right) +}} +{{eqn | ll= \leadsto + | l = \sum_{i \mathop = 1}^n i + | r = \frac {n \paren {n + 1} } 2 +}} +{{end-eqn}} +This concludes the proof. +{{qed}} +\end{proof} + +\begin{proof} +We have that: +:$\displaystyle \map T n = 1 + 2 + \dotsb + n = \sum_{i \mathop = 1}^n i$ +Thus: +{{begin-eqn}} +{{eqn | l = \map T n + | r = n + \paren {n - 1} + \paren {n - 2} + \dotsb + 2 + 1 + | c = +}} +{{eqn | r = n + \paren {n - 1} + \paren {n - 2} + \dotsb + \paren {n - \paren {n - 2} } + \paren {n - \paren {n - 1} } + | c = +}} +{{eqn | r = \underbrace {n + n + \dotsb + n}_{n \text { times} } + | c = extracting $n$ from each term +}} +{{eqn | o = + | ro= + + | r = \paren {-1} + \paren {-2} + \dotsb + \paren {-\paren {n - 2} } + \paren {-\paren {n - 1} } + | c = +}} +{{eqn | r = n^2 - \paren {1 + 2 + \dotsb + \paren {n - 1} } + | c = +}} +{{eqn | n = 1 + | ll= \leadsto + | l = \map T n + | r = n^2 - \map T {n - 1} + | c = +}} +{{end-eqn}} +Then: +{{begin-eqn}} +{{eqn | n = 2 + | l = \map T n + | r = n + \map T {n - 1} + | c = {{Defof|Triangular Number|index = 1}} +}} +{{eqn | ll= \leadsto + | l = 2 \, \map T n + | r = n^2 + n + | c = $(1) + (2)$ +}} +{{eqn | ll= \leadsto + | l = \map T n + | r = \frac {n \paren {n + 1} } 2 + | c = +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +Proof by [[Principle of Mathematical Induction|induction]]: +=== Basis for the Induction === +When $n = 1$, we have: +:$\displaystyle \sum_{i \mathop = 1}^1 i = 1$ +Also: +:$\dfrac {n \paren {n + 1} } 2 = \dfrac {1 \cdot 2} 2 = 1$ +This is our [[Definition:Basis for the Induction|base case]]. +=== Induction Hypothesis === +:$\forall k \in \N: k \ge 1: \displaystyle \sum_{i \mathop = 1}^k i = \frac {k \paren {k + 1} } 2$ +This is our [[Definition:Induction Hypothesis|induction hypothesis]]. +It is to be demonstrated that: +:$\displaystyle \sum_{i \mathop = 1}^{k + 1} i = \frac {\paren {k + 1} \paren {k + 2} } 2$ +=== Induction Step === +This is our [[Definition:Induction Step|induction step]]: +Consider $n = k + 1$. +By the properties of [[Definition:Summation|summation]]: +:$\displaystyle \sum_{i \mathop = 1}^{k + 1} i = \sum_{i \mathop = 1}^k i + k + 1$ +Using the [[Closed Form for Triangular Numbers/Proof by Induction#Induction Hypothesis|induction hypothesis]] this can be simplified to: +{{begin-eqn}} +{{eqn | l = \frac {k \paren {k + 1} } 2 + k + 1 + | r = \frac {k \paren {k + 1} + 2 k + 2} 2 + | c = +}} +{{eqn | r = \frac {k^2 + 3 k + 2} 2 + | c = +}} +{{eqn | r = \frac {\paren {k + 1} \paren {k + 2} } 2 + | c = +}} +{{eqn | r = \frac {\paren {k + 1} \paren {\paren {k + 1} + 1} } 2 + | c = +}} +{{end-eqn}} +Hence the result by [[Principle of Mathematical Induction|induction]]. +{{qed}} +\end{proof} + +\begin{proof} +From [[Binomial Coefficient with One]]: +:$\forall k \in \Z, k > 0: \dbinom k 1 = k$ +Thus: +{{begin-eqn}} +{{eqn | l = \sum_{k \mathop = 1}^n k + | r = \sum_{k \mathop = 1}^n \binom k 1 + | c = [[Binomial Coefficient with One]] +}} +{{eqn | r = \binom {n + 1} 2 + | c = [[Sum of k Choose m up to n]] +}} +{{eqn | r = \frac {\left({n+1}\right) n} 2 + | c = {{Defof|Binomial Coefficient}} +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \sum_{i \mathop = 0}^{m - 1} \paren {a + i d} + | r = m \paren {a + \frac {m - 1} 2 d} + | c = [[Sum of Arithmetic Sequence]] +}} +{{eqn | l = \sum_{i \mathop = 0}^n \paren {a + i d} + | r = \paren {n + 1} \paren {a + \frac n 2 d} + | c = Let $n = m - 1$ +}} +{{eqn | l = \sum_{i \mathop = 0}^n i + | r = \paren {n + 1} \paren {\frac n 2} + | c = Let $a = 0$ and $d = 1$ +}} +{{eqn | l = 0 + \sum_{i \mathop = 1}^n i + | r = \frac {n \paren {n + 1} } 2 + | c = +}} +{{eqn | l = \sum_{i \mathop = 1}^n i + | r = \frac {n \paren {n + 1} } 2 + | c = +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +[[Definition:Triangular Number|Triangular numbers]] are [[Definition:Polygonal Number|$k$-gonal numbers]] where $k = 3$. +From [[Closed Form for Polygonal Numbers]] we have that: +:$P \left({k, n}\right) = \dfrac n 2 \left({\left({k - 2}\right) n - k + 4}\right)$ +Hence: +{{begin-eqn}} +{{eqn | l = T_n + | r = \frac n 2 \left({\left({3 - 2}\right) n - 3 + 4}\right) + | c = [[Closed Form for Polygonal Numbers]] +}} +{{eqn | r = \frac n 2 \left({n + 1}\right) + | c = +}} +{{end-eqn}} +Hence the result. +{{qed}} +\end{proof} + +\begin{proof} +Let $\N_n^* = \set {1, 2, 3, \cdots, n}$ be the [[Definition:Initial Segment of One-Based Natural Numbers|initial segment of natural numbers]]. +Let $A = \set {\tuple {a, b}: a \le b, a, b \in \N_n^*}$ +Let $B = \set {\tuple {a, b}: a \ge b, a, b, \in \N_n^*}$ +Let $\phi: A \to B$ be the [[Definition:Mapping|mapping]]: +:$\map \phi {x, y} = \tuple {y, x}$ +By definition of [[Definition:Dual Ordering|dual ordering]], $\phi$ is a [[Definition:Bijection|bijection]]: +:$(1): \quad \size A = \size B$ +We have: +{{begin-eqn}} +{{eqn | l = A \cup B + | r = \set {\tuple {a, b}: a, b \in \N_n^*} +}} +{{eqn | l = A \cap B + | r = \set {\tuple {a, b}: a = b: a, b \in \N_n^*} +}} +{{end-eqn}} +Thus: +{{begin-eqn}} +{{eqn | l = \size A + \size B + | r = \size {A \cup B} + \size {A \cap B} + | c = [[Inclusion-Exclusion Principle]] +}} +{{eqn | r = n^2 + n + | c = [[Definition:Count|Count]] of a [[Definition:Finite Set|finite set]] +}} +{{end-eqn}} +Combined with $\left({1}\right)$ this yields: +:$\size A = \dfrac {n^2 + n} 2 = \dfrac {n \paren {n + 1} } 2$ +It remains to prove that: +:$T_n = \size A$ +{{begin-eqn}} +{{eqn | l = T_n + | r = \sum_{i \mathop = 1}^n i + | c = Definition of $T_n$ +}} +{{eqn | r = \sum_{i \mathop \in \N_n^*} i + | c = +}} +{{eqn | r = \sum_{i \mathop \in \N_n^*} \size {\set {a: a \in \N_i^*} } + | c = [[Definition:Count|Count]] of [[Definition:Finite Set|Finite Set]],{{Defof|Initial Segment of One-Based Natural Numbers}} +}} +{{eqn | r = \size {\set {\tuple {a, i} :a\in \N_i^*, i \in \N_n^*} } + | c = [[Inclusion-Exclusion Principle]], sets are [[Definition:Mutually Exclusive|mutually exclusive]] as their second argument in the ordered pair are different +}} +{{eqn | r = \size {\set {\tuple {a, b}: a \le b, a, b \in \N_n^*} } + | c = Change of Variable, {{Defof|Initial Segment of One-Based Natural Numbers}} +}} +{{eqn | r = \size A + | c = Definition of $A$ +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \sum_{j \mathop = 1}^n \paren {2 j - 1} + | r = n^2 + | c = [[Odd Number Theorem]] +}} +{{eqn | ll= \leadsto + | l = \sum_{j \mathop = 1}^n \paren {2 j - 1} + \sum_{j \mathop = 1}^n 1 + | r = n^2 + n + | c = +}} +{{eqn | ll= \leadsto + | l = \sum_{j \mathop = 1}^n \paren {2 j} + | r = n \paren {n + 1} + | c = +}} +{{eqn | ll= \leadsto + | l = \sum_{j \mathop = 1}^n j + | r = \frac {n \paren {n + 1} } 2 + | c = +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +From [[Sum of Powers of Positive Integers]]: +{{begin-eqn}} +{{eqn | l = \sum_{i \mathop = 1}^n i^p + | r = 1^p + 2^p + \cdots + n^p + | c = +}} +{{eqn | r = \frac {n^{p + 1} } {p + 1} + \sum_{k \mathop = 1}^p \frac {B_k \, p^{\underline {k - 1} } \, n^{p - k + 1} } {k!} + | c = +}} +{{end-eqn}} +where $B_k$ are the [[Definition:Bernoulli Numbers|Bernoulli numbers]]. +Setting $p = 1$: +{{begin-eqn}} +{{eqn | l = \sum_{i \mathop = 1}^n i^1 + | r = 1 + 2 + \cdots + n + | c = +}} +{{eqn | r = \frac {n^2} 2 + \frac {B_1 \, p^{\underline 0} n^1} {1!} + | c = +}} +{{eqn | r = \frac {n^2} 2 + \frac 1 2 \frac {1 \times n} 1 + | c = {{Defof|Bernoulli Numbers}}, [[Number to Power of Zero Falling is One]] +}} +{{eqn | r = \frac {n^2 + n} 2 + | c = +}} +{{end-eqn}} +Hence the result. +{{qed}} +\end{proof}<|endoftext|> +\section{Sum of Sequence of Squares} +Tags: Square Numbers, Sums of Sequences, Sum of Sequence of Squares + +\begin{theorem} +:$\displaystyle \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$ +\end{theorem} + +\begin{proof} +Proof by [[Principle of Mathematical Induction|induction]]: +For all $n \in \N$, let $\map P n$ be the [[Definition:Proposition|proposition]]: +:$\displaystyle \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$ +When $n = 0$, we see from the definition of [[Definition:Vacuous Summation|vacuous sum]] that: +:$0 = \displaystyle \sum_{i \mathop = 1}^0 i^2 = \frac {0 \paren 1 \paren 1} 6 = 0$ +and so $\map P 0$ holds. +=== Basis for the Induction === +When $n = 1$: +:$\displaystyle \sum_{i \mathop = 1}^1 i^2 = 1^2 = 1$ +Now, we have: +:$\displaystyle \frac {n \paren {n + 1} \paren {2 n + 1} } 6 = \frac {1 \paren {1 + 1} \paren {2 \times 1 + 1} } 6 = \frac 6 6 = 1$ +and $\map P 1$ is seen to hold. +This is our [[Definition:Basis for the Induction|basis for the induction]]. +=== Induction Hypothesis === +Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. +So this is our [[Definition:Induction Hypothesis|induction hypothesis]]: +:$\displaystyle \sum_{i \mathop = 1}^k i^2 = \frac {k \paren {k + 1} \paren {2 k + 1} } 6$ +Then we need to show: +:$\displaystyle \sum_{i \mathop = 1}^{k + 1} i^2 = \frac {\paren {k + 1} \paren {k + 2} \paren {2 \paren {k + 1} + 1} } 6$ +=== Induction Step === +This is our [[Definition:Induction Step|induction step]]: +Using the properties of summation, we have: +:$\displaystyle \sum_{i \mathop = 1}^{k + 1} i^2 = \sum_{i \mathop = 1}^k i^2 + \paren {k + 1}^2$ +We can now apply our [[Sum of Sequence of Squares/Proof by Induction#Induction Hypothesis|induction hypothesis]], obtaining: +{{begin-eqn}} +{{eqn | l = \sum_{i \mathop = 1}^{k + 1} i^2 + | r = \frac {k \paren {k + 1} \paren {2 k + 1} } 6 + \paren {k + 1}^2 + | c = +}} +{{eqn | r = \frac {k \paren {k + 1} \paren {2 k + 1} + 6 \paren {k + 1}^2} 6 + | c = +}} +{{eqn | r = \frac {\paren {k + 1} \paren {k \paren {2 k + 1} + 6 \paren {k + 1} } } 6 + | c = +}} +{{eqn | r = \frac {\paren {k + 1} \paren {2 k^2 + 7 k + 6} } 6 + | c = +}} +{{eqn | r = \frac {\paren {k + 1} \paren {k + 2} \paren {2 \paren {k + 1} + 1} } 6 + | c = +}} +{{end-eqn}} +So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]]. +Therefore: +: $\displaystyle \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$ +{{qed}} +\end{proof} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \sum_{i \mathop = 1}^n 3 i \left({i + 1}\right) + | r = n \left({n + 1}\right) \left({n + 2}\right) + | c = [[Sum of Sequence of Products of Consecutive Integers]] +}} +{{eqn | ll= \implies + | l = \sum_{i \mathop = 1}^n 3 i^2 + \sum_{i \mathop = 1}^n 3 i + | r = n \left({n + 1}\right) \left({n + 2}\right) +}} +{{eqn | ll= \implies + | l = \sum_{i \mathop = 1}^n 3 i^2 + | r = n \left({n + 1}\right) \left({n + 2}\right) - 3 \frac {n \left({n + 1}\right) } 2 + | c = [[Closed Form for Triangular Numbers]] +}} +{{eqn | ll= \implies + | l = \sum_{i \mathop = 1}^n i^2 + | r = \frac {n \left({n + 1}\right) \left({n + 2}\right)} 3 - \frac {n \left({n + 1}\right) } 2 + | c = +}} +{{eqn | ll= \implies + | l = \sum_{i \mathop = 1}^n i^2 + | r = \frac {2 n \left({n + 1}\right) \left({n + 2}\right) - 3 n \left({n + 1}\right) } 6 + | c = +}} +{{eqn | ll= \implies + | l = \sum_{i \mathop = 1}^n i^2 + | r = \frac {n \left({n + 1}\right) \left({2 n + 1}\right)} 6 + | c = +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +Observe that: +: $3 i \left({i + 1}\right) = i \left({i + 1}\right) \left({i + 2}\right) - i \left({i + 1}\right) \left({i - 1}\right)$ +That is: +: $(1): \quad 6 T_i = \left({i + 1}\right) \left({\left({i + 1}\right) + 1}\right) \left({\left({i + 1}\right) - 1}\right) - i \left({i + 1}\right) \left({i - 1}\right)$ +Then: +{{begin-eqn}} +{{eqn | l = n^2 + | r = \frac {n^2 + n + n^2 - n} 2 +}} +{{eqn | r = \frac {n \left({n + 1}\right) + n \left({n - 1}\right)} 2 +}} +{{eqn | r = \frac {n \left({n + 1}\right)} 2 + \frac {n \left({n - 1}\right)} 2 +}} +{{eqn | r = T_n + T_{n-1} +}} +{{end-eqn}} + +where $T_n$ is the $n$th Triangular number. +Then: +{{begin-eqn}} +{{eqn | l = \sum_{i \mathop = 1}^n i^2 + | r = 1 + \left({T_1 + T_2}\right) + \left({T_2 + T_3}\right) + \left({T_3 + T_4}\right) + \cdots + \left({T_{n - 1} + T_n}\right) +}} +{{eqn | r = 1 + 2 T_2 + 2 T_3 + 2 T_4 + \cdots + 2 T_{n - 1} + T_n +}} +{{eqn | r = 1 - T_1 - T_n + 2 \left({T_1 + T_2 + T_3 + T_4 + \cdots + T_n}\right) +}} +{{eqn | r = 2 \left({\sum_{i \mathop = 1}^n T_i} \right) - \frac {n \left({n + 1}\right)} 2 +}} +{{eqn | r = 2 \left({\frac {n \left({n + 1}\right) \left({n + 2}\right)} 6}\right) - \frac {n \left({n + 1}\right)} 2 + | c = [[Definition:Telescoping Series|Telescoping Series]] from $(1)$ above +}} +{{eqn | r = \frac {2 n \left({n^2 + 3 n + 2}\right) - \left({3 n^2 + 3 n}\right)} 6 +}} +{{eqn | r = \frac {2 n^3 + 3 n^2 + n} 6 +}} +{{eqn | r = \frac {n \left({n + 1}\right) \left({2 n + 1}\right)} 6 +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +:[[File:Sum of Sequences of Squares.jpg]] +We can observe from the above diagram that: +: $\displaystyle \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \sum_{i \mathop = 1}^n \left({\sum_{j \mathop = i}^n j}\right)$ +Therefore we have: +{{begin-eqn}} +{{eqn | l = \sum_{i \mathop = 1}^n i^2 + | r = \sum_{i \mathop = 1}^n \left({\sum_{j \mathop = i}^n j}\right) + | c = +}} +{{eqn | r = \sum_{i \mathop = 1}^n \left( {\sum_{j \mathop = 1}^n j - \sum_{j \mathop = 1}^{i - 1} j} \right) + | c = +}} +{{eqn | r = \sum_{i \mathop = 1}^n \left({\frac {n \left({n + 1}\right)} 2 - \frac {i \left({i - 1}\right)} 2}\right) + | c = +}} +{{eqn | ll= \implies + | l = 2 \sum_{i \mathop = 1}^n i^2 + | r = n^2 \left({n + 1}\right) - \sum_{i \mathop = 1}^n i^2 + \sum_{i \mathop = 1}^n i + | c = +}} +{{eqn | ll= \implies + | l = 3 \sum_{i \mathop = 1}^n i^2 + | r = n^2 \left({n + 1}\right) + \sum_{i \mathop = 1}^n i + | c = +}} +{{eqn | ll= \implies + | l = 3 \sum_{i \mathop = 1}^n i^2 + | r = n^2 \left({n + 1}\right) + \frac {n \left({n + 1}\right)} 2 + | c = [[Closed Form for Triangular Numbers]] +}} +{{eqn | ll= \implies + | l = 6 \sum_{i \mathop = 1}^n i^2 + | r = 2 n^2 \left({n + 1}\right) + n \left({n + 1}\right) + | c = +}} +{{eqn | r = n \left({n + 1}\right) \left({2 n + 1}\right) + | c = +}} +{{eqn | ll= \implies + | l = \sum_{i \mathop = 1}^n i^2 + | r = \frac {n \left({n + 1}\right) \left({2 n + 1}\right)} 6 + | c = +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \sum_{i \mathop = 1}^n \left({\left({i + 1}\right)^3 - i^3}\right) + | r = \sum_{i \mathop = 1}^n \left({i^3 + 3 i^2 + 3 i + 1 - i^3}\right) + | c = [[Binomial Theorem]] +}} +{{eqn | r = \sum_{i \mathop = 1}^n \left({3 i^2 + 3 i + 1}\right) + | c = +}} +{{eqn | r = 3 \sum_{i \mathop = 1}^n i^2 + 3 \sum_{i \mathop = 1}^n i + \sum_{i \mathop = 1}^n 1 + | c = [[Summation is Linear]] +}} +{{eqn | r = 3\sum_{i \mathop = 1}^n i^2 + 3 \frac {n \left({n + 1}\right)} 2 + n + | c = [[Closed Form for Triangular Numbers]] +}} +{{end-eqn}} +On the other hand: +{{begin-eqn}} +{{eqn | l = \sum_{i \mathop = 1}^n \left({\left({i + 1}\right)^3 - i^3}\right) + | r = \left({n + 1}\right)^3 - n^3 + n^3 - \left({n - 1}\right)^3 + \left({n - 1}\right)^3 - \cdots + 2^3 - 1^3 + | c = Definition of [[Definition:Summation|Summation]] +}} +{{eqn | r = \left({n + 1}\right)^3 - 1^3 + | c = [[Telescoping Series/Example 2|Telescoping Series: Example 2]] +}} +{{eqn | r = n^3 + 3n^2 + 3n + 1 - 1 + | c = [[Binomial Theorem]] +}} +{{eqn | r = n^3 + 3n^2 + 3n + | c = +}} +{{end-eqn}} +Therefore: +{{begin-eqn}} +{{eqn | l = 3 \sum_{i \mathop = 1}^n i^2 + 3 \frac {n \left({n + 1}\right)} 2 + n + | r = n^3 + 3 n^2 + 3 n +}} +{{eqn | ll= \implies + | l = 3 \sum_{i \mathop = 1}^n i^2 + | r = n^3 + 3 n^2 + 3 n - 3 \frac {n \left({n + 1}\right)} 2 - n +}} +{{end-eqn}} +Therefore: +{{begin-eqn}} +{{eqn | l = \sum_{i \mathop = 1}^n i^2 + | r = \frac 1 3 \left({n^3 + 3 n^2 + 3 n - 3 \frac {n \left({n + 1}\right)}2 - n}\right) +}} +{{eqn | r = \frac 1 3 \left({n^3 + 3 n^2 + 3 n - \frac {3 n^2} 2 - \frac {3 n} 2 - n}\right) +}} +{{eqn | r = \frac 1 3 \left({n^3 + \frac {3 n^2} 2 + \frac n 2}\right) +}} +{{eqn | r = \frac 1 6 n \left({2 n^2 + 3 n + 1}\right) +}} +{{eqn | r = \frac 1 6 n \left({n + 1}\right) \left({2 n + 1}\right) +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +From [[Binomial Coefficient with One]]: +:$\dbinom n 1 = n$ +From [[Binomial Coefficient with Two]]: +:$\dbinom n 2 = \dfrac {n \paren {n - 1} } 2$ +Thus: +{{begin-eqn}} +{{eqn | l = 2 \binom n 2 + \binom n 1 + | r = 2 \dfrac {n \paren {n - 1} } 2 + n + | c = +}} +{{eqn | r = n \paren {n - 1} + n + | c = +}} +{{eqn | r = n^2 + | c = +}} +{{end-eqn}} +Hence: +{{begin-eqn}} +{{eqn | l = \sum_{i \mathop = 1}^n i^2 + | r = \sum_{i \mathop = 1}^n \paren {2 \binom i 2 + \binom i 1} + | c = +}} +{{eqn | r = 2 \binom {n + 1} 3 + \binom {n + 1} 2 + | c = [[Sum of Binomial Coefficients over Upper Index]] +}} +{{eqn | r = \frac {2 \paren {n + 1} n \paren {n - 1} } {3 \times 2 \times 1} + \frac {\paren {n + 1} n} 2 + | c = {{Defof|Binomial Coefficient}} +}} +{{eqn | r = \frac {n \paren {n + 1} \paren {2 n + 1} } 6 + | c = after algebra +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +From [[Sum of Powers of Positive Integers]]: +{{begin-eqn}} +{{eqn | l = \sum_{i \mathop = 1}^n i^p + | r = 1^p + 2^p + \cdots + n^p + | c = +}} +{{eqn | r = \frac {n^{p + 1} } {p + 1} + \sum_{k \mathop = 1}^p \frac {B_k \, p^{\underline {k - 1} } \, n^{p - k + 1} } {k!} + | c = +}} +{{end-eqn}} +where $B_k$ are the [[Definition:Bernoulli Numbers|Bernoulli numbers]]. +Setting $p = 2$: +{{begin-eqn}} +{{eqn | l = \sum_{i \mathop = 1}^n i^2 + | r = 1^2 + 2^2 + \cdots + n^2 + | c = +}} +{{eqn | r = \frac {n^{2 + 1} } {2 + 1} + \frac {B_1 \, p^{\underline 0} \, n^2} {1!} + \frac {B_2 \, p^{\underline 1} \, n^1} {2!} + | c = +}} +{{eqn | r = \frac {n^3} 3 + B_1 n^2 + B_2 n + | c = {{Defof|Falling Factorial}} and simplification +}} +{{eqn | r = \frac {n^3} 3 + \frac {n^2} 2 + \frac n 6 + | c = {{Defof|Bernoulli Numbers}} +}} +{{eqn | r = \frac {n \left({n + 1}\right) \left({2 n + 1}\right)} 6 + | c = after algebra +}} +{{end-eqn}} +Hence the result. +{{qed}} +\end{proof}<|endoftext|> +\section{Union is Associative} +Tags: Set Union, Direct Proofs, Associativity, Union is Associative + +\begin{theorem} +[[Definition:Set Union|Set union]] is [[Definition:Associative Operation|associative]]: +:$A \cup \paren {B \cup C} = \paren {A \cup B} \cup C$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | o = + | r = x \in A \cup \paren {B \cup C} + | c = {{Defof|Set Union}} +}} +{{eqn | o = \leadstoandfrom + | r = x \in A \lor \paren {x \in B \lor x \in C} + | c = {{Defof|Set Union}} +}} +{{eqn | o = \leadstoandfrom + | r = \paren {x \in A \lor x \in B} \lor x \in C + | c = [[Rule of Association/Disjunction|Rule of Association: Disjunction]] +}} +{{eqn | o = \leadstoandfrom + | r = x \in \paren {A \cup B} \cup C + | c = {{Defof|Set Union}} +}} +{{end-eqn}} +Therefore: +:$x \in A \cup \paren {B \cup C}$ {{iff}} $x \in \paren {A \cup B} \cup C$ +Thus it has been shown that: +: $A \cup \paren {B \cup C} = \paren {A \cup B} \cup C$ +{{qed}} +\end{proof}<|endoftext|> +\section{Pythagoras's Theorem} +Tags: Right Triangles, Pythagoras's Theorem + +\begin{theorem} +Let $\triangle ABC$ be a [[Definition:Right Triangle|right triangle]] with $c$ as the [[Definition:Hypotenuse|hypotenuse]]. +Then: +:$a^2 + b^2 = c^2$ +\end{theorem} + +\begin{proof} +Consider the [[Definition:Triangle (Geometry)|triangle]] shown below. +:[[File:Pythagoras1-1.png|350px]] +We can take $4$ copies of this [[Definition:Triangle (Geometry)|triangle]] and form them into a [[Definition:Square (Geometry)|square]] using isometries, specifically rotations and translations. +This new figure is shown below. +:[[File:Pythagoras1-2.png|500px]] +This figure is clearly a [[Definition:Square (Geometry)|square]], since all the [[Definition:Angle|angles]] are [[Definition:Right Angle|right angles]], and the lines connecting the corners are easily seen to be straight. +Now to calculate the [[Definition:Area|area]] of this figure. +On the one hand, we can add up the [[Definition:Area|area]] of the component parts of the [[Definition:Square (Geometry)|square]]. +Specifically, we can add up the $4$ [[Definition:Triangle (Geometry)|triangles]] and the inner [[Definition:Square (Geometry)|square]]. +Thus we calculate the [[Definition:Area|area]] of the large [[Definition:Square (Geometry)|square]] to be: +:$4 \paren {\dfrac 1 2 a b} + c^2 = 2 a b + c^2$ +On the other hand, we can similarly calculate the [[Definition:Area|area]] of the large [[Definition:Square (Geometry)|square]] to be: +:$\paren {a + b}^2 = a^2 + 2 a b + b^2$ +Now these two expressions have to be equal, since they both represent the [[Definition:Area|area]] of the [[Definition:Square (Geometry)|square]]. +Thus: +:$a^2 + 2 a b + b^2 = 2 a b + c^2 \iff a^2 + b^2 = c^2$ +{{qed}} +\end{proof} + +\begin{proof} +:[[File:Pythagoras2.png|400px]] +We have: +:$\dfrac b c = \dfrac d b$ +and: +:$\dfrac a c = \dfrac e a$ +using the fact that all the [[Definition:Triangle (Geometry)|triangles]] involved are [[Definition:Similar Triangles|similar]]. +That is: +:$b^2 = c d$ +:$a^2 = c e$ +Adding, we now get: +:$a^2 + b^2 = c d + c e = c \paren {d + e} = c^2$ +{{qed}} +\end{proof} + +\begin{proof} +We start with the algebraic definitions for [[Definition:Sine|sine]] and [[Definition:Cosine|cosine]]: +:$\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$ +:$\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$ +From these, we derive the [[Sum of Squares of Sine and Cosine|proof]] that $\cos^2 x + \sin^2 x = 1$. +Then from the [[Equivalence of Definitions of Sine and Cosine]], we can use the geometric interpretation of [[Definition:Sine|sine]] and [[Definition:Cosine|cosine]]: +:[[File:SineCosine.png|400px]] +:$\sin \theta = \dfrac {\text{Opposite}} {\text{Hypotenuse}}$ +:$\cos \theta = \dfrac {\text{Adjacent}} {\text{Hypotenuse}}$ +Let $\text{Adjacent} = a, \text{Opposite} = b, \text{Hypotenuse} = c$, as in the diagram at the top of the page. +Thus: +{{begin-eqn}} +{{eqn | l = \cos^2 x + \sin^2 x + | r = 1 + | c = [[Sum of Squares of Sine and Cosine]] +}} +{{eqn | ll= \leadsto + | l = \paren {\frac a c}^2 + \paren {\frac b c}^2 + | r = 1 + | c = +}} +{{eqn | ll= \leadsto + | l = a^2 + b^2 + | r = c^2 + | c = multiplying both sides by $c^2$ +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +:[[File:Euclid-I-47.png|400px]] +Let $ABC$ be a [[Definition:Right Triangle|right triangle]] whose angle $BAC$ is a [[Definition:Right Angle|right angle]]. +[[Construction of Square on Given Straight Line|Construct squares]] $BDEC$ on $BC$, $ABFG$ on $AB$ and $ACKH$ on $AC$. +[[Construction of Parallel Line|Construct $AL$ parallel]] to $BD$ (or $CE$). +Since $\angle BAC$ and $\angle BAG$ are both [[Definition:Right Angle|right angles]], from [[Two Angles making Two Right Angles make Straight Line]] it follows that $CA$ is in a [[Definition:Straight Line|straight line]] with $AG$. +[[Two Angles making Two Right Angles make Straight Line|For the same reason]] $BA$ is in a [[Definition:Straight Line|straight line]] with $AH$. +We have that $\angle DBC = \angle FBA$, because both are [[Definition:Right Angle|right angles]]. +We add $\angle ABC$ to each one to make $\angle FBC$ and $\angle DBA$. +By [[Axiom:Euclid's Common Notions|common notion 2]], $\angle FBC = \angle DBA$. +By [[Triangle Side-Angle-Side Equality]], $\triangle ABD = \triangle FBC$. +We have that the [[Definition:Parallelogram|parallelogram]] $BDLM$ is on the same base $BD$ and between the same [[Definition:Parallel Lines|parallels]] $BD$ and $AL$ as the [[Definition:Triangle (Geometry)|triangle]] $\triangle ABD$. +So, by [[Parallelogram on Same Base as Triangle has Twice its Area]], the [[Definition:Parallelogram|parallelogram]] $BDLM$ is twice the area of $\triangle ABD$. +Similarly, we have that the [[Definition:Parallelogram|parallelogram]] $ABFG$ (which happens also to be a [[Definition:Square (Geometry)|square]]) is on the same base $FB$ and between the same [[Definition:Parallel Lines|parallels]] $FB$ and $GC$ as the [[Definition:Triangle (Geometry)|triangle]] $\triangle FBC$. +So, by [[Parallelogram on Same Base as Triangle has Twice its Area]], the [[Definition:Parallelogram|parallelogram]] $ABFG$ is twice the area of $\triangle FBC$. +So $BDLM = 2 \triangle ABD = 2 \triangle FBC = ABFG$. +By the same construction, we have that $CELM = 2 \triangle ACE = 2 \triangle KBC = ACKH$. +But $BDLM + CELM$ is the whole of the [[Definition:Square (Geometry)|square]] $BDEC$. +Therefore the area of the [[Definition:Square (Geometry)|square]] $BDEC$ is equal to the combined area of the [[Definition:Square (Geometry)|squares]] $ABFG$ and $ACKH$. +{{qed}} +\end{proof} + +\begin{proof} +:[[File:Pythagoras3.png|300px]] +The area of the big square is $c^2$. +It is also equal to $4 \dfrac {a b} 2 + \paren {a - b}^2$. +So: +{{begin-eqn}} +{{eqn | l = c^2 + | r = 4 \frac {a b} 2 + \paren {a - b}^2 + | c = +}} +{{eqn | r = 2 a b + a^2 - 2 a b + b^2 + | c = +}} +{{eqn | r = a^2 + b^2 + | c = +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +:[[File:Pythagoras4.png|800px]] +Dissect the [[Definition:Square (Geometry)|square]] on the left (which has [[Definition:Area|area]] $c^2$) as shown. +Rearrange the pieces to make the two [[Definition:Square (Geometry)|squares]] on the right, with areas $a^2$ and $b^2$. +{{qed}} +\end{proof} + +\begin{proof} +:[[File:Pythagoras5.png|700px]] +The two squares both have the same area, that is, $\paren {a + b}^2$. +The one on the left has four triangles of area $\dfrac {a b} 2$ and a square of area $c^2$. +The one on the right has four triangles of area $\dfrac {a b} 2$ and two squares: one of area $a^2$ and one of area $b^2$. +Take away the triangles from both of the big squares and you are left with $c^2 = a^2 + b^2$. +{{qed}} +\end{proof} + +\begin{proof} +:[[File:Pythagoras6.png|500px]] +We have that $CH = BS = AB = AJ$. +Hence the result follows directly from [[Pythagoras's Theorem for Parallelograms]]. +{{qed}} +\end{proof} + +\begin{proof} +:[[File:Pythagoras7.png|500px]] +Let $\triangle ABC$ be a [[Definition:Right Triangle|right triangle]] and $h_c$ the [[Definition:Altitude of Triangle|altitude]] from $c$. +We have: +:$\angle CAB \cong \angle DCB$ +:$\angle ABC \cong \angle ACD$ +Then we have: +:$\triangle ADC \sim \triangle ACB \sim \triangle CDB$ +Use the fact that if $\triangle XYZ \sim \triangle X'Y'Z'$ then: +:$\dfrac {\paren {XYZ} } {\paren {X' Y' Z'} } = \dfrac {XY^2} {X'Y'^2} = \dfrac {h_z^2} {h_{z'}^2} = \dfrac {t_z^2} {t_{z'}^2} = \ldots$ +where $\paren {XYZ}$ denotes the area of $\triangle XYZ$. +{{explain|What do $t_z$, $t_{z'}$ mean? A diagram would help this last statement to make sense here.}} +This gives us: +:$\dfrac {\paren {ADC} } {\paren {ACB} } = \dfrac {AC^2} {AB^2}$ +and +:$\dfrac {\paren {CDB} } {\paren {ACB} } = \dfrac {BC^2} {AB^2}$ +Taking the sum of these two equalities we obtain: +:$\dfrac {\paren {ADC} } {\paren {ACB} } + \dfrac {\paren {CDB} } {\paren {ACB} } = \dfrac {BC^2} {AB^2} + \dfrac {AC^2} {AB^2}$ +Thus: +:$\dfrac {\overbrace {\paren {ADC} + \paren {CDB} }^{\paren {ACB} } } {\paren {ACB} } = \dfrac {BC^2 + AC^2} {AB^2}$ +Thus: +:$AB^2 = BC^2 + AC^2$ +as desired. +{{qed}} +\end{proof}<|endoftext|> +\section{Euclid's Theorem} +Tags: Prime Numbers, Euclid's Theorem + +\begin{theorem} +For any [[Definition:Finite Set|finite set]] of [[Definition:Prime Number|prime numbers]], there exists a [[Definition:Prime Number|prime number]] not in that [[Definition:Set|set]]. +{{:Euclid:Proposition/IX/20}} +\end{theorem} + +\begin{proof} +Let $\mathbb P$ be a [[Definition:Finite Set|finite set]] of [[Definition:Prime Number|prime numbers]]. +Consider the number: +:$\displaystyle n_p = \paren {\prod_{p \mathop \in \mathbb P} p} + 1$ +Take any $p_j \in \mathbb P$. +We have that: +: $\displaystyle p_j \divides \prod_{p \mathop \in \mathbb P} p$ +Hence: +:$\displaystyle \exists q \in \Z: \prod_{p \mathop \in \mathbb P} p = q p_j$ +So: +{{begin-eqn}} +{{eqn | l = n_p + | r = q p_j + 1 + | c = [[Division Theorem]] +}} +{{eqn | ll= \leadsto + | l = n_p + | o = \perp + | r = p_j + | c = {{Defof|Coprime Integers}} +}} +{{end-eqn}} +So $p_j \nmid n_p$. +There are two possibilities: +:$(1): \quad n_p$ is [[Definition:Prime Number|prime]], which is not in $\mathbb P$. +:$(2): \quad n_p$ is [[Definition:Composite Number|composite]]. +But from [[Positive Integer Greater than 1 has Prime Divisor]], it must be divisible by ''some'' prime. +That means it is divisible by a prime which is not in $\mathbb P$. +So, in either case, there exists at least one [[Definition:Prime Number|prime]] which is not in the original set $\mathbb P$ we created. +{{qed}} +{{Euclid Note|20|IX}} +\end{proof}<|endoftext|> +\section{Limsup Squeeze Theorem} +Tags: Analysis, Direct Proofs, Named Theorems + +\begin{theorem} +Let $\left \langle {x_n} \right \rangle$ and $\left \langle {y_n} \right \rangle$ be [[Definition:Real Sequence|sequences in $\R$]]. +Let: +: $(1): \quad \forall n \ge n_0: \left|{x_n}\right| \le y_n$ +: $(2): \quad \displaystyle \limsup_{n \mathop \to \infty} \left({y_n}\right) = 0$, where $\limsup$ denotes the [[Definition:Limit Superior|limit superior]]. +Then: +: $\displaystyle \lim_{n \mathop \to \infty} x_n = 0$ +\end{theorem}<|endoftext|> +\section{Square Root of Prime is Irrational} +Tags: Irrationality Proofs, Prime Numbers, Square Roots, Square Root of Prime is Irrational + +\begin{theorem} +The [[Definition:Square Root|square root]] of any [[Definition:Prime Number|prime number]] is [[Definition:Irrational Number|irrational]]. +\end{theorem} + +\begin{proof} +Let $p$ be [[Definition:Prime Number|prime]]. +{{AimForCont}} that $\sqrt p$ is [[Definition:Rational Number|rational]]. +Then there exist [[Definition:Natural Numbers|natural numbers]] $m$ and $n$ such that: +{{begin-eqn}} +{{eqn | l = \sqrt p + | r = \frac m n + | c = +}} +{{eqn | ll= \leadsto + | l = p + | r = \frac {m^2} {n^2} + | c = +}} +{{eqn | ll= \leadsto + | l = n^2 p + | r = m^2 + | c = +}} +{{end-eqn}} +Any [[Definition:Prime Number|prime]] in the [[Definition:Prime Decomposition|prime factorizations]] of $n^2$ and $m^2$ must occur an [[Definition:Even Integer|even number]] of times because they are [[Definition:Square Number|squares]]. +Thus, $p$ must occur in the [[Definition:Prime Decomposition|prime factorization]] of $n^2 p$ an [[Definition:Odd Integer|odd number]] of times. +Therefore, $p$ occurs as a factor of $m^2$ an [[Definition:Odd Integer|odd number]] of times, a [[Proof by Contradiction|contradiction]]. +So $\sqrt p$ must be [[Definition:Irrational Number|irrational]]. +{{qed}} +\end{proof} + +\begin{proof} +Let $p \in \Z$ be a [[Definition:Prime Number|prime number]]. +Consider the [[Definition:Polynomial over Ring in One Variable|polynomial]]: +:$\map P x = x^2 - p$ +over the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] $\Q \sqbrk X$ over the [[Definition:Rational Number|rational numbers]]. +From [[Difference of Two Squares]]: +:$x^2 - p = \paren {x + \sqrt p} \paren {x + \sqrt p}$ +Because $p$ is [[Definition:Prime Number|prime]], $\sqrt p$ is not an [[Definition:Integer|integer]]. +From [[Polynomial which is Irreducible over Integers is Irreducible over Rationals]] it follows that $\paren {x + \sqrt p}$ and $\paren {x + \sqrt p}$ do not have [[Definition:Rational Number|rational]] [[Definition:Polynomial Coefficient|coefficients]]. +That is: +:$\sqrt p$ is not [[Definition:Rational Number|rational]]. +Hence the result. +{{qed}} +\end{proof}<|endoftext|> +\section{Derivative of Exponential Function} +Tags: Derivatives involving Exponential Function, Exponential Function, Derivative of Exponential Function + +\begin{theorem} +Let $\exp$ be the [[Definition:Exponential Function|exponential function]]. +Then: +:$\map {\dfrac \d {\d x} } {\exp x} = \exp x$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \map {\frac \d {\d x} } {\exp x} + | r = \lim_{h \mathop \to 0} \frac {\map \exp {x + h} - \exp x} h + | c = {{Defof|Derivative}} +}} +{{eqn | r = \lim_{h \mathop \to 0} \frac {\exp x \cdot \exp h - \exp x} h + | c = [[Exponential of Sum]] +}} +{{eqn | r = \lim_{h \mathop \to 0} \frac {\exp x \paren {\exp h - 1} } h +}} +{{eqn | r = \exp x \paren {\lim_{h \mathop \to 0} \frac {\exp h - 1} h} + | c = [[Multiple Rule for Limits of Functions]], as $\exp x$ is constant +}} +{{eqn | r = \exp x + | c = [[Derivative of Exponential at Zero]] +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +We use the fact that the [[Definition:Real Exponential Function|exponential function]] is the [[Definition:Inverse Mapping|inverse]] of the [[Definition:Natural Logarithm|natural logarithm function]]: +:$y = e^x \iff x = \ln y$ +{{begin-eqn}} +{{eqn | l = \dfrac {\d x} {\d y} + | r = \dfrac 1 y + | c = [[Derivative of Natural Logarithm Function]] +}} +{{eqn | ll= \leadsto + | l = \dfrac {\d y} {\d x} + | r = \dfrac 1 {1 / y} + | c = [[Derivative of Inverse Function]] +}} +{{eqn | r = y +}} +{{eqn | r = e^x +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \map {\frac \d {\d x} } {\ln e^x} + | r = \map {\frac \d {\d x} } x + | c = [[Exponential of Natural Logarithm]] +}} +{{eqn | ll= \leadsto + | l = \frac 1 {e^x} \map {\frac \d {\d x} } {e^x} + | r = 1 + | c = [[Chain Rule for Derivatives]], [[Derivative of Natural Logarithm Function]], [[Derivative of Identity Function]] +}} +{{eqn | ll= \leadsto + | l = \map {\frac \d {\d x} } {e^x} + | r = e^x + | c = multiply both sides by $e^x$ +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +This proof assumes the [[Definition:Exponential Function/Real/Sum of Series|series definition of $\exp$]]. +That is, let: +:$\displaystyle \exp x = \sum_{k \mathop = 0}^\infty \frac{x^k}{k!}$ +From [[Series of Power over Factorial Converges]], the [[Definition:Interval of Convergence|interval of convergence]] of $\exp$ is the entirety of $\R$. +So we may apply [[Differentiation of Power Series]] to $\exp$ for all $x \in \R$. +Thus we have: +{{begin-eqn}} +{{eqn | l = \frac \d {\d x} \exp x + | r = \frac \d {\d x} \sum_{k \mathop = 0}^\infty \frac {x^k} {k!} +}} +{{eqn | r = \sum_{k \mathop = 1}^\infty \frac k {k!} x^{k - 1} + | c = [[Differentiation of Power Series]], with $n = 1$ +}} +{{eqn | r = \sum_{k \mathop = 1}^\infty \frac {x^{k - 1} } {\paren {k - 1}!} +}} +{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {x^k} {k!} +}} +{{eqn | r = \exp x +}} +{{end-eqn}} +Hence the result. +{{qed}} +\end{proof} + +\begin{proof} +This proof assumes the [[Definition:Exponential Function/Real/Limit of Sequence|limit definition of $\exp$]]. +So let: +:$\forall n \in \N: \forall x \in \R: \map {f_n} x = \paren {1 + \dfrac x n}^n$ +Let $x_0 \in \R$. +Consider $I := \closedint {x_0 - 1} {x_0 + 1}$. +Let: +:$N = \ceiling {\max \set {\size {x_0 - 1}, \size {x_0 + 1} } }$ +where $\ceiling {\, \cdot \,}$ denotes the [[Definition:Ceiling Function|ceiling function]]. +From [[Closed Real Interval is Compact in Metric Space]], $I$ is [[Definition:Compact (Real Analysis)|compact]]. +From [[Chain Rule for Derivatives]]: +:$\dfrac \d {\d x} \map {f_n} x = \dfrac n {n + x} \map {f_n} x$ +=== [[Derivative of Exponential Function/Proof 5/Lemma|Lemma]] === +{{:Derivative of Exponential Function/Proof 5/Lemma}}{{qed|lemma}} +From the [[Derivative of Exponential Function/Proof 5/Lemma|lemma]]: +:$\forall x \in I: \sequence {\dfrac \d {\d x} \map {f_{n + N} } x}$ is [[Definition:Increasing Real Sequence|increasing]] +Hence, from [[Dini's Theorem]], $\sequence {\dfrac \d {\d x} f_{n + N} }$ is [[Definition:Uniform Convergence|uniformly convergent]] on $I$. +Therefore, for $x \in I$: +{{begin-eqn}} +{{eqn | l = \frac \d {\d x} \exp x + | r = \frac \d {\d x} \lim_{n \mathop \to \infty} \map {f_n} x +}} +{{eqn | r = \frac \d {\d x} \lim_{n \mathop \to \infty} \map {f_{n + N} } x + | c = [[Tail of Convergent Sequence]] +}} +{{eqn | r = \lim_{n \mathop \to \infty} \frac \d {\d x} \map {f_{n + N} } x + | c = [[Derivative of Uniformly Convergent Sequence of Differentiable Functions]] +}} +{{eqn | r = \lim_{n \mathop \to \infty} \frac n {n + x} \map {f_n} x + | c = from above +}} +{{eqn | r = \lim_{n \mathop \to \infty} \map {f_n} x + | c = [[Combination Theorem for Sequences]] +}} +{{eqn | r = \exp x +}} +{{end-eqn}} +In particular: +:$\dfrac \d {\d x} \exp x_0 = \exp x_0$ +{{qed}} +\end{proof}<|endoftext|> +\section{Derivative of Sine Function} +Tags: Derivatives of Trigonometric Functions, Sine Function, Derivative of Sine Function + +\begin{theorem} +:$\map {\dfrac \d {\d x} } {\sin x} = \cos x$ +\end{theorem} + +\begin{proof} +From the definition of the [[Definition:Sine|sine function]], we have: +:$\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}$ +From [[Radius of Convergence of Power Series over Factorial]], this series converges for all $x$. +From [[Power Series is Differentiable on Interval of Convergence]]: +{{begin-eqn}} +{{eqn | l = \map {\frac \d {\d x} } {\sin x} + | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} + | c = +}} +{{eqn | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} + | c = +}} +{{end-eqn}} +The result follows from the definition of the [[Definition:Cosine|cosine function]]. +{{qed}} +\end{proof} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \map {\frac \d {\d x} } {\sin x} + | r = \lim_{h \mathop \to 0} \frac {\map \sin {x + h} - \sin x} h + | c = {{Defof|Derivative of Real Function at Point}} +}} +{{eqn | r = \lim_{h \mathop \to 0} \frac {\sin x \cos h + \sin h \cos x - \sin x} h + | c = [[Sine of Sum]] +}} +{{eqn | r = \lim_{h \mathop \to 0} \frac {\sin x \paren {\cos h - 1} + \sin h \cos x} h + | c = collecting terms containing $\map \sin x$ and factoring +}} +{{eqn | r = \lim_{h \mathop \to 0} \frac {\sin x \paren {\cos h - 1} } h + \lim_{h \mathop \to 0} \frac {\sin h \cos x} h + | c = [[Sum Rule for Limits of Functions]] +}} +{{eqn | r = \map \sin x \times 0 + 1 \times \cos x + | c = [[Limit of Sine of X over X]] and [[Limit of (Cosine (X) - 1) over X]] +}} +{{eqn | r = \cos x +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \dfrac \d {\d x} \sin x + | r = \dfrac \d {\d x} \map \cos {\frac \pi 2 - x} + | c = [[Cosine of Complement equals Sine]] +}} +{{eqn | r = \map \sin {\frac \pi 2 - x} + | c = [[Derivative of Cosine Function]] and [[Chain Rule for Derivatives]] +}} +{{eqn | r = \cos x + | c = [[Sine of Complement equals Cosine]] +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \map {\frac \d {d x} } {\sin x} + | r = \lim_{h \mathop \to 0} \frac {\map \sin {x + h} - \sin x} h + | c = {{Defof|Derivative of Real Function at Point}} +}} +{{eqn | r = \lim_{h \mathop \to 0} \frac {\map \sin {\paren {x + \frac h 2} + \frac h 2} - \map \sin {\paren {x + \frac h 2} - \frac h 2} } h +}} +{{eqn | r = \lim_{h \mathop \to 0} \frac {2 \map \cos {x + \frac h 2} \map \sin {\frac h 2} } h + | c = [[Simpson's Formula for Cosine by Sine]] +}} +{{eqn | r = \lim_{h \mathop \to 0} \map \cos {x + \frac h 2} \lim_{h \mathop \to 0} \frac {\map \sin {\frac h 2} } {\frac h 2} + | c = [[Multiple Rule for Limits of Functions]] and [[Product Rule for Limits of Functions]] +}} +{{eqn | r = \cos x \times 1 + | c = [[Cosine Function is Continuous]] and [[Limit of Sine of X over X]] +}} +{{eqn | r = \cos x +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \map \arcsin x + | r = \int_0^x \frac {\d x} {\sqrt {1 - x^2} } + | c = [[Arcsin as an Integral]] +}} +{{eqn | ll= \leadsto + | l = \dfrac {\map \d {\map \arcsin y} } {\d y} + | r = \dfrac {\map \d {\displaystyle \int_0^y \dfrac 1 {\sqrt {1 - y^2} } \rd y} } {\d y} +}} +{{eqn | r = \dfrac 1 {\sqrt {1 - y^2} } + | c = [[Fundamental Theorem of Calculus/First Part/Corollary|Corollary to Fundamental Theorem of Calculus: First Part]] +}} +{{end-eqn}} +Note that we get the same answer as [[Derivative of Arcsine Function]]. +By definition of [[Definition:Real Arcsine|real $\arcsin$ function]], $\arcsin$ is [[Definition:Bijection|bijective]] on its [[Definition:Domain of Mapping|domain]] $\closedint {-1} 1$. +Thus its [[Definition:Inverse of Mapping|inverse]] is itself a [[Definition:Mapping|mapping]]. +From [[Inverse of Inverse of Bijection]], its [[Definition:Inverse Mapping|inverse]] is the [[Definition:Real Sine Function|$\sin$ function]]. +{{begin-eqn}} +{{eqn | l = \dfrac {\map \d {\sin \theta} } {\d \theta} + | r = 1 / \dfrac 1 {\sqrt {1 - \sin^2 \theta} } + | c = [[Derivative of Inverse Function]] +}} +{{eqn | r = \pm \sqrt {1 - \sin^2 \theta} + | c = [[Definition:Positive Real Number|Positive]] in [[Definition:First Quadrant|Quadrant $\text I$]] and [[Definition:First Quadrant|Quadrant $\text {IV}$]], [[Definition:Negative Real Number|Negative]] in [[Definition:Second Quadrant|Quadrant $\text {II}$]] and [[Definition:Third Quadrant|Quadrant $\text {III}$]] +}} +{{eqn | l = \dfrac {\map \d {\sin \theta} } {\d \theta} + | r = \cos \theta +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Cauchy-Bunyakovsky-Schwarz Inequality} +Tags: Algebra, Linear Algebra, Analysis, Integral Calculus, Inequalities, Cauchy-Bunyakovsky-Schwarz Inequality + +\begin{theorem} +=== [[Cauchy-Bunyakovsky-Schwarz Inequality/Inner Product Spaces|Semi-Inner Product Spaces]] === +{{:Cauchy-Bunyakovsky-Schwarz Inequality/Inner Product Spaces}} +\end{theorem}<|endoftext|> +\section{0.999...=1} +Tags: Number Theory, Direct Proofs, 0.999...=1 + +\begin{theorem} +:$0.999 \ldots = 1$ +\end{theorem} + +\begin{proof} +By [[Sum of Infinite Geometric Sequence]]: +:$0.999 \ldots = \dfrac a {1 - r}$ +where $a = \dfrac 9 {10}$ and $r = \dfrac 1 {10}$. +Since our ratio is less than $1$, then we know that $\displaystyle \sum_{n \mathop = 0}^\infty \frac 9 {10} \paren {\frac 1 {10} }^n$ must converge to: +:$\dfrac a {1 - r} = \dfrac {\frac 9 {10} } {1 - \frac 1 {10} } = \dfrac {\frac 9 {10} } {\frac 9 {10} } = 1$ +{{qed}} +\end{proof} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = 0.333 \ldots + | r = 1 / 3 +}} +{{eqn | ll= \leadsto + | l = 3 \paren {0.333 \ldots} + | r = 3 \paren {1 / 3} +}} +{{eqn | ll= \leadsto + | l = 0.999 \ldots + | r = 3 / 3 +}} +{{eqn | r = 1 +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +Let $c = 0.999 \ldots$ +Then: +{{begin-eqn}} +{{eqn | l = c + | r = 0.999 \ldots +}} +{{eqn | ll= \leadsto + | l = 10 c + | r = \paren {9.999 \ldots} + | c = multiplying $c$ by $10$ +}} +{{eqn | ll= \leadsto + | l = 10 c - c + | r = \paren {9.999 \ldots} - \paren {0.999 \ldots} + | c = subtracting $c$ from each side +}} +{{eqn | ll= \leadsto + | l = 9 c + | r = 9 +}} +{{eqn | ll= \leadsto + | l = c + | r = 1 +}} +{{end-eqn}} +It follows that: +:$0.999 \ldots = 1$ +{{qed}} +\end{proof} + +\begin{proof} +We begin with the knowledge that: +{{begin-eqn}} +{{eqn | l = \frac 9 9 + | r = \frac 1 1 = 1 +}} +{{end-eqn}} +Now we divide $9$ by $9$ using the standard process of long division, only instead of stating that $90$ divided by $9$ is $10$, we say that it is "$9$ remainder $9$," yielding the following result: +:$ +\require{enclose} +\phantom{0..}0.9999\ldots\\ +9\enclose{longdiv}{9.0000\ldots}\\ +\phantom{0}\underline{\phantom{0}8.1}\\ +\phantom{000.}90\\ +\phantom{0.}\underline{\phantom{00}81}\\ +\phantom{0000.}90\\ +\phantom{0.}\underline{\phantom{000}81}\\ +\phantom{00000.}9\ldots\\ +$ +Thus, we are compelled to believe that: +:$0.999\ldots = \dfrac 9 9 = 1$ +{{qed}} +\end{proof} + +\begin{proof} +{{begin-eqn}} +{{eqn | n = 1 + | l = 0 . \underset n {\underbrace {999 \cdots 9} } + | r = 1 - 0.1^n + | c = +}} +{{eqn | l = 0.999 \cdots + | r = \eqclass {\sequence {0.9, \, 0.99, \, 0.999, \, \cdots} } {} + | c = {{Defof|Real Numbers}} +}} +{{eqn | r = \eqclass {\sequence {1 - 0.1^1, \, 1 - 0.1^2, \, 1 - 0.1^3, \, \cdots} } {} + | c = from $(1)$ +}} +{{eqn | r = \eqclass {\sequence {1, \, 1, \, 1, \, \cdots} } {} - \eqclass {\sequence {0.1^1, \, 0.1^2, \, 0.1^3, \, \cdots} } {} + | c = +}} +{{eqn | r = 1 - 0 + | c = [[Sequence of Powers of Number less than One]], {{Defof|Real Numbers}} +}} +{{eqn | r = 1 + | c = +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Schur-Zassenhaus Theorem} +Tags: Group Theory, Hall Subgroups + +\begin{theorem} +Let $G$ be a [[Definition:Finite Group|finite group]] and $N$ be a [[Definition:Normal Subgroup|normal subgroup]] in $G$. +Let $N$ be a [[Definition:Hall Subgroup|Hall subgroup]] of $G$. +Then there exists $H$, a [[Definition:Complement of Subgroup|complement]] of $N$, such that $G$ is the [[Definition:Inner Semidirect Product|semidirect product]] of $N$ and $H$. +\end{theorem} + +\begin{proof} +The proof proceeds by [[Principle of Mathematical Induction|induction]]. +By definition, $N$ is a [[Definition:Hall Subgroup|Hall subgroup]] {{iff}} the [[Definition:Index of Subgroup|index]] and [[Definition:Order of Structure|order]] of $N$ in $G$ are [[Definition:Coprime Integers|relatively prime numbers]]. +Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$. +We induct on $\order G$, where $\order G$ is the [[Definition:Order of Structure|order]] of $G$. +We may assume that $N \ne \set e$. +Let $p$ be a [[Definition:Prime Number|prime number]] dividing $\order N$. +Let $\Syl p N$ be the [[Definition:Set|set]] of [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] of $N$. +By the [[First Sylow Theorem]]: +:$\Syl p N \ne \O$ +Let: +: $P \in \Syl p N$ +: $G_0$ be the [[Definition:Normalizer|normalizer]] in $G$ of $P$ +: $N_0 = N \cap G_0$. +By [[Frattini's Argument]]: +:$G = G_0 N$ +By the [[Second Isomorphism Theorem for Groups]] and thence [[Lagrange's Theorem (Group Theory)]], it follows that: +: $N_0$ is a [[Definition:Hall Subgroup|Hall subgroup]] of $G_0$ +: $\index {G_0} {N_0} = \index G H$ +Suppose $G_0 < G$. +Then by [[Principle of Mathematical Induction|induction]] applied to $N_0$ in $G_0$, we find that $G_0$ contains a [[Definition:Complement of Subgroup|complement]] $H \in N_0$. +We have that: +:$\order H = \index {G_0} {N_0}$ +and so $H$ is also a [[Definition:Complement of Subgroup|complement]] to $N$ in $G$. +So we may assume that $P$ is [[Definition:Normal Subgroup|normal]] in $G$ (that is: $G_0 < G$). +{{explain|What is the significance of $G_0 < G$ here? It has already been stated above. What is its purpose at this point in the argument?}} +Let $Z \paren P$ be the [[Definition:Center of Group|center]] of $P$. +By: +:[[Center is Characteristic Subgroup]] +:$P$ is [[Definition:Normal Subgroup|normal]] in $G$ +:[[Characteristic Subgroup of Normal Subgroup is Normal]] +$Z \paren P$ is also [[Definition:Normal Subgroup|normal]] in $G$. +Let $Z \paren P = N$. +Then there exists a [[Definition:Long Exact Sequence|long exact sequence]] of [[Definition:Cohomology Groups|cohomology groups]]: +:$0 \to H^1 \paren {G / N, P^N} \to H^1 \paren {G, P} \to H^1 \paren {N, P} \to H^2 \paren {G / N, P} \to H^2 \paren {G, P}$ +which splits as desired. +{{explain|The link leads to "exact sequence". An explanation is needed as to what a "long exact sequence" is.}} +{{explain|The definition of $P^N$ in this context}} +{{explain|The definition of "splits" in this context}} +Otherwise: +:$Z \paren P \ne N$ +In this case $N / Z \paren P$ is a [[Definition:Normal Subgroup|normal]] [[Definition:Hall Subgroup|(Hall) subgroup]] of $G / Z \paren P$. +By [[Principle of Mathematical Induction|induction]]: +:$N / Z \paren P$ has a [[Definition:Complement of Subgroup|complement]] $H / Z \paren P$ in $E // Z \paren P$. +{{explain|The meaning of $E // Z \paren P$ and definition of $E$ in this context.}} +{{Explain|Although it is stated that this proof is by induction, it is unclear what the base case, induction hypothesis and induction step actually are.}} +Let $G_1$ be the [[Definition:Preimage of Subset under Relation|preimage]] of $H // Z \paren P$ in $G$ (under the [[Definition:Equivalence Relation|equivalence relation]]). +{{explain|Under what equivalence relation?}} +Then: +:$\order {G_1} = \order {K / Z \paren P} \times \order {Z \paren P} = \order {G / N} \times \order {Z \paren P}$ +{{explain|The definition of $K$ in $\order {G_1} {{=}} \order {K / Z \paren P} \times \order {Z \paren P} {{=}} \order {G / N}\ \times \order {Z \paren P}$.}} +Therefore, $Z \paren P$ is [[Definition:Normal Subgroup|normal]] [[Definition:Hall Subgroup|Hall subgroup]] of $G_1$. +By [[Principle of Mathematical Induction|induction]], $Z \paren P$ has a [[Definition:Complement of Subgroup|complement]] in $G_1$ and is also a [[Definition:Complement of Subgroup|complement]] of $N$ in $G$. +{{Explain|Again, although it is stated that this proof is by induction, it is unclear what the base case, induction hypothesis and induction step actually are.}} +{{qed}} +\end{proof}<|endoftext|> +\section{Sum of Sequence of Cubes} +Tags: Sums of Sequences, Cube Numbers, Sum of Sequence of Cubes + +\begin{theorem} +:$\displaystyle \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$ +\end{theorem} + +\begin{proof} +First, from [[Closed Form for Triangular Numbers]]: +:$\displaystyle \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ +So: +:$\displaystyle \paren {\sum_{i \mathop = 1}^n i}^2 = \dfrac {n^2 \paren {n + 1}^2} 4$ +Next we use [[Principle of Mathematical Induction|induction]] on $n$ to show that: +:$\displaystyle \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$ +The proof proceeds by [[Principle of Mathematical Induction|induction]]. +For all $n \in \Z_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]: +:$\displaystyle \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$ +=== Basis for the Induction === +$\map P 1$ is the case: +:$1^3 = \dfrac {1 \paren {1 + 1}^2} 4$ +{{begin-eqn}} +{{eqn | l = \sum_{i \mathop = 1}^1 i^3 + | r = 1^3 + | c = +}} +{{eqn | r = \dfrac {1^2 \paren {1 + 1}^2} 4 + | c = +}} +{{end-eqn}} +Thus $\map P 1$ is seen to hold. +This is the [[Definition:Basis for the Induction|basis for the induction]]. +=== Induction Hypothesis === +Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. +So this is the [[Definition:Induction Hypothesis|induction hypothesis]]: +:$\displaystyle \sum_{i \mathop = 1}^k i^3 = \dfrac {k^2 \paren {k + 1}^2} 4$ +from which it is to be shown that: +:$\displaystyle \sum_{i \mathop = 1}^{k + 1} i^3 = \dfrac {\paren {k + 1}^2 \paren {k + 2}^2} 4$ +=== Induction Step === +This is the [[Definition:Induction Step|induction step]]: +{{begin-eqn}} +{{eqn | l = \sum_{i \mathop = 1}^{k + 1} i^3 + | r = \sum_{i \mathop = 1}^k i^3 + \paren {k + 1}^3 + | c = +}} +{{eqn | r = \frac {k^2 \paren {k + 1}^2} 4 + \paren {k + 1}^3 + | c = [[Sum of Sequence of Cubes/Proof by Induction#Induction Hypothesis|Induction Hypothesis]] +}} +{{eqn | r = \frac {k^4 + 2 k^3 + k^2} 4 + \frac {4 k^3 + 12 k^2 + 12 k + 4} 4 + | c = +}} +{{eqn | r = \frac {k^4 + 6 k^3 + 13 k^2 + 12 k + 4} 4 + | c = +}} +{{eqn | r = \frac {\paren {k + 1}^2 \paren {k + 2}^2} 4 + | c = +}} +{{end-eqn}} +So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]]. +Therefore: +:$\forall n \in \Z_{>0}: \displaystyle \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$ +\end{proof} + +\begin{proof} +By [[Nicomachus's Theorem]], we have: +:$\forall n \in \N_{>0}: n^3 = \left({n^2 - n + 1}\right) + \paren {n^2 - n + 3} + \ldots + \paren {n^2 + n - 1}$ +Also by [[Nicomachus's Theorem]], we have that the first term for $\paren {n + 1}^3$ is $2$ greater than the last term for $n^3$. +So if we add them all up together, we get: +{{begin-eqn}} +{{eqn | l = \sum_{i \mathop = 1}^n i^3 + | r = \sum_{\stackrel {1 \mathop \le i \mathop \le n^2 \mathop + n \mathop - 1} {i \text { odd} } } i + | c = ... the sum of all the odd numbers up to $n^2 + n - 1$ +}} +{{eqn | r = \sum_{i \mathop = 1}^{\frac {n^2 \mathop + n} 2} 2 i - 1 + | c = ... that is, the first $\dfrac {n^2 + n} 2$ odd numbers +}} +{{eqn | r = \paren {\frac {n^2 + n} 2}^2 + | c = [[Odd Number Theorem]] +}} +{{end-eqn}} +Hence the result. +{{qed}} +\end{proof} + +\begin{proof} +From [[Closed Form for Triangular Numbers]]: +: $(1): \quad \displaystyle \map A n := \sum_{i \mathop = 1}^n i = \frac{n \paren {n + 1} } 2$ +From [[Sum of Sequence of Squares]]: +: $(2): \quad \displaystyle \map B n := \sum_{i \mathop = 1}^n i^2 = \frac{n \paren {n + 1} \paren {2 n + 1} } 6$ +Let $\displaystyle \map S n = \sum_{i \mathop = 1}^n i^3$. +Then: +{{begin-eqn}} +{{eqn | l = \map S n + | r = n^3 + \paren {n - 1}^3 + \paren {n - 2}^3 + \cdots + 2^3 + 1^3 + | c = +}} +{{eqn | r = \sum_{k \mathop = 0}^{n \mathop - 1} \paren {n - k}^3 + | c = +}} +{{eqn | r = \sum_{k \mathop = 0}^{n \mathop - 1} \paren {n^3 - 3 n^2 k + 3 n k^2 - k^3} + | c = +}} +{{eqn | r = n^4 - 3 n^2 \cdot \map A {n - 1} + 3 n \cdot \map B {n - 1} - \map S {n - 1} + | c = +}} +{{eqn | n = 3 + | ll= \leadsto + | l = \map S n + | r = n^4 - 3 n^2 \cdot \frac {n \paren {n - 1} } 2 + 3n \cdot \frac{n \paren {n - 1} \paren {2 n - 1} } 6 - \map S {n - 1} + | c = substituting from $(1)$ and $(2)$ +}} +{{eqn | n = 4 + | l = \map S n + | r = n^3 + \map S {n - 1} + | c = recursive definition +}} +{{eqn | ll= \leadsto + | l = 2 \map S n + | r = n^4 + n^3 - 3 n^2 \cdot \frac {n \paren {n - 1} } 2 + 3 n \cdot \frac {n \paren {n - 1} \paren {2 n - 1} } 6 + | c = adding $(3)$ and $(4)$ +}} +{{eqn | ll= \leadsto + | r = \frac {n^2 \paren {n + 1}^2} 2 + | c = simplification +}} +{{eqn | ll= \leadsto + | l = \map S n + | r = \frac {n^2 \paren {n + 1}^2} 4 + | c = +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \sum_{i \mathop = 1}^n i^3 + | r = \frac {n^2 \paren {n + 1}^2} 4 + | c = [[Sum of Sequence of Cubes]] +}} +{{eqn | r = \paren {\frac {n \paren {n + 1} } 2}^2 + | c = +}} +{{eqn | r = {T_n}^2 + | c = [[Closed Form for Triangular Numbers]] +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +The proof proceeds by [[Principle of Mathematical Induction|induction]]. +For all $n \in \Z_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]: +:$\displaystyle \sum_{i \mathop = 1}^n i^3 = {T_n}^2$ +=== Basis for the Induction === +$\map P 1$ is the case: +{{begin-eqn}} +{{eqn | l = \sum_{i \mathop = 1}^1 i^3 + | r = 1^3 + | c = +}} +{{eqn | r = 1 + | c = +}} +{{eqn | r = \paren {\frac {1 \paren {1 + 1} } 2}^2 + | c = +}} +{{eqn | r = {T_1}^2 + | c = [[Closed Form for Triangular Numbers]] +}} +{{end-eqn}} +Thus $\map P 1$ is seen to hold. +This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]]. +=== Induction Hypothesis === +Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. +So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]: +:$\displaystyle \sum_{i \mathop = 1}^k i^3 = {T_k}^2$ +from which it is to be shown that: +:$\displaystyle \sum_{i \mathop = 1}^{k + 1} i^3 = {T_{k + 1} }^2$ +=== Induction Step === +This is the [[Principle of Mathematical Induction#Induction Step|induction step]]: +{{begin-eqn}} +{{eqn | l = \sum_{i \mathop = 1}^{k + 1} i^3 + | r = \sum_{i \mathop = 1}^k i^3 + \paren {k + 1}^3 + | c = +}} +{{eqn | r = {T_k}^2 + \paren {k + 1}^3 + | c = [[Square of Triangular Number equals Sum of Sequence of Cubes/Proof 2#Induction Hypothesis|Induction Hypothesis]] +}} +{{eqn | r = {T_k}^2 + \paren { {T_{k + 1} }^2 - {T_k}^2} + | c = [[Cube Number as Difference between Squares of Triangular Numbers]] +}} +{{eqn | r = {T_{k + 1} }^2 + | c = +}} +{{end-eqn}} +So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]]. +Therefore: +:$\forall n \in \Z_{> 0}: \displaystyle \sum_{i \mathop = 1}^n i^3 = {T_n}^2$ +{{qed}} +\end{proof} + +\begin{proof} +From [[Sum of Powers of Positive Integers]]: +{{begin-eqn}} +{{eqn | l = \sum_{i \mathop = 1}^n i^p + | r = 1^p + 2^p + \cdots + n^p + | c = +}} +{{eqn | r = \frac {n^{p + 1} } {p + 1} + \sum_{k \mathop = 1}^p \frac {B_k \, p^{\underline {k - 1} } \, n^{p - k + 1} } {k!} + | c = +}} +{{end-eqn}} +where $B_k$ are the [[Definition:Bernoulli Numbers|Bernoulli numbers]]. +Setting $p = 3$: +{{begin-eqn}} +{{eqn | l = \sum_{i \mathop = 1}^n i^3 + | r = \frac {n^{3 + 1} } {3 + 1} + \sum_{k \mathop = 1}^3 \frac {B_k \, 3^{\underline {k - 1} } \, n^{3 - k + 1} } {k!} + | c = +}} +{{eqn | r = \frac {n^4} 4 + \frac {B_1 \, 3^{\underline 0} \, n^3} {1!} + \frac {B_2 \, 3^{\underline 1} \, n^2} {2!} + \frac {B_3 \, 3^{\underline 2} \, n^1} {3!} + | c = +}} +{{eqn | r = \frac {n^4} 4 + \frac 1 2 \frac {n^3} {1!} + \frac 1 6 \frac {3 n^2} {2!} + 0 \frac {3 \times 2 n} {3!} + | c = {{Defof|Bernoulli Numbers}} and {{Defof|Falling Factorial}} +}} +{{eqn | r = \frac {n^4} 4 + \frac {n^3} 2 + \frac {n^2} 4 + | c = simplifying +}} +{{eqn | r = \frac {n^2 \left({n + 1}\right)^2} 4 + | c = after algebra +}} +{{end-eqn}} +\end{proof} + +\begin{proof} +We aim to demonstrate that the "Sum of Cubes" is the "Square of the Sum" using simple Multiplication Tables. +On the right hand side of the equation, the "Square of the Sum" will be represented by a two dimensional array. +The sum of all numbers contained inside the two dimensional (n x n) array is the "Square of the Sum". +For example: +:$ \paren {1 + 2}^2 = 1 + 2 + 2 + 4$ +:$\begin{array}{r|cccccccccc} +\paren {1 + 2}^2 & 1 & 2 \\ +\hline +1 & 1 & 2 \\ +2 & 2 & 4 \\ +\end{array}$ +:$ \paren {1 + 2 + 3}^2 = 1 + 2 + 3 + 2 + 4 + 6 + 3 + 6 + 9$ +:$\begin{array}{r|cccccccccc} +\paren {1 + 2 + 3}^2 & 1 & 2 & 3 \\ +\hline +1 & 1 & 2 & 3 \\ +2 & 2 & 4 & 6 \\ +3 & 3 & 6 & 9 \\ +\end{array}$ +:$ \cdots $ +:$ \paren {1 + 2 + 3 + \cdots + N}^2 = $ +:$\begin{array}{r|cccccccccc} +\paren {\sum_{i \mathop = 1}^n i}^2 & 1 & 2 & 3 & 4 & 5 & 6 & \cdots & N \\ +\hline +1 & 1 & 2 & 3 & 4 & 5 & 6 & \cdots & N \\ +2 & 2 & 4 & 6 & 8 & 10 & 12 & \cdots & 2N \\ +3 & 3 & 6 & 9 & 12 & 15 & 18 & \cdots & 3N \\ +4 & 4 & 8 & 12 & 16 & 20 & 24 & \cdots & 4N \\ +5 & 5 & 10 & 15 & 20 & 25 & 30 & \cdots & 5N \\ +6 & 6 & 12 & 18 & 24 & 30 & 36 & \cdots & 6N \\ +\cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ +N & N & 2N & 3N & 4N & 5N & 6N & \cdots & N^2 \\ +\end{array}$ +From [[Closed Form for Triangular Numbers]]: +:$\displaystyle \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$ +So the sum of the numbers in the two dimensional n x n array (i.e. "Square of the Sum") is: +:$\displaystyle \paren {\sum_{i \mathop = 1}^n i}^2 = \dfrac {n^2 \paren {n + 1}^2} 4$ +So where is the "Sum of the Cubes"? +Consider the sum of the set of all numbers in each outermost square edge above of length $1$, $2$, $3$, $\ldots$. +We have: +{{begin-eqn}} +{{eqn | l = 1 + | r = \paren 2 \paren 1 \paren 1 - 1^2 = 1^3 + | c = +}} +{{eqn | l = 2 + 4 + 2 + | r = \paren 2 \paren 2 \paren {1 + 2} - 2^2 = 2^3 + | c = +}} +{{eqn | l = 3 + 6 + 9 + 6 + 3 + | r = \paren 2 \paren 3 \paren {1 + 2 + 3} - 3^2 = 3^3 + | c = +}} +{{eqn | o = \cdots +}} +{{eqn | l = n + 2 n + 3 n + \cdots + \paren {n - 1} \paren n + n^2 + \paren {n - 1} \paren n + \cdots + 2 n + n + | r = \paren 2 \paren n \paren {1 + 2 + \cdots + n} - n^2 + | c = +}} +{{eqn | r = 2 n \sum_{i \mathop = 1}^n i - n^2 + | c = +}} +{{eqn | r = 2 n \frac {n \paren {n + 1} } 2 - n^2 + | c = [[Closed Form for Triangular Numbers]] +}} +{{eqn | r = n^3 + n^2 - n^2 + | c = +}} +{{eqn | r = n^3 + | c = +}} +{{end-eqn}} +Therefore, +The "Square of the Sum" (i.e. the sum of all of the numbers in an n x n multiplication table) is identical to +The "Sum of the Cubes" (i.e. the sum of the numbers on each incremental outer square-edge is a cube, sum up the outer edges), or: +:$\forall n \in \Z_{>0}: \displaystyle \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$ +{{qed}} +\end{proof}<|endoftext|> +\section{Law of Cosines} +Tags: Triangles, Cosine Function, Named Theorems, Law of Cosines + +\begin{theorem} +Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]] whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$. +Then: +:$c^2 = a^2 + b^2 - 2 a b \cos C$ +\end{theorem} + +\begin{proof} +Let $\triangle ABC$ be embedded in a [[Definition:Cartesian Coordinate System|Cartesian coordinate system]] by identifying: +:$C := \tuple {0, 0}$ +:$B := \tuple {a, 0}$ +:[[File:CosineRuleCartesian.png|400px]] +Thus by definition of [[Definition:Sine of Angle|sine]] and [[Definition:Cosine of Angle|cosine]]: +:$A = \tuple {b \cos C, b \sin C}$ +By the [[Distance Formula]]: +:$c = \sqrt {\paren {b \cos C - a}^2 + \paren {b \sin C - 0}^2}$ +Hence: +{{begin-eqn}} +{{eqn | l = c^2 + | r = \paren {b \cos C - a}^2 + \paren {b \sin C - 0}^2 + | c = squaring both sides of [[Distance Formula]] +}} +{{eqn | r = b^2 \cos^2 C - 2 a b \cos C + a^2 + b^2 \sin^2 C + | c = [[Square of Difference]] +}} +{{eqn | r = a^2 + b^2 \paren {\sin^2 C + \cos^2 C} - 2 a b \cos C + | c = [[Real Multiplication Distributes over Addition]] +}} +{{eqn | r = a^2 + b^2 - 2 a b \cos C + | c = [[Sum of Squares of Sine and Cosine]] +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]]. +=== Case 1: $AC$ greater than $AB$ === +Using $AC$ as the radius, we construct a [[Definition:Circle|circle]] whose [[Definition:Center of Circle|center]] is $A$. +Now we extend: +: $CB$ to $D$ +: $AB$ to $F$ +: $BA$ to $G$ +: $CA$ to $E$. +$D$ is joined with $E$, thus: +:[[File:CosineRule.png|300px]] +Using the [[Intersecting Chord Theorem]] we have: +: $GB \cdot BF = CB \cdot BD$ +$AF$ is a radius, so $AF = AC = b = GA$ and thus: +: $GB = GA + AB = b + c$ +: $BF = AF - AB = b - c$ +Thus: +{{begin-eqn}} +{{eqn | l = \left({b + c}\right) \left({b - c}\right) + | r = a \cdot BD +}} +{{eqn | ll= \implies + | l = \frac {b^2 - c^2} a + | r = BD +}} +{{end-eqn}} +Next: +{{begin-eqn}} +{{eqn | l = CD + | r = CB + BD +}} +{{eqn | r = a + \frac {b^2 - c^2} a +}} +{{eqn | r = \frac {a^2 + b^2 - c^2} a +}} +{{end-eqn}} +As $CA$ is a radius, $CE$ is a diameter. +By [[Thales' Theorem]], it follows that $\angle CDE$ is a [[Definition:Right Angle|right angle]]. +Then using the definition of [[Definition:Cosine of Angle|cosine]], we have +{{begin-eqn}} +{{eqn | l = \cos C + | r = \frac {CD} {CE} +}} +{{eqn | r = \frac {\left({\dfrac{a^2 + b^2 - c^2} a}\right)} {2 b} +}} +{{eqn | r = \frac {a^2 + b^2 - c^2} {2 a b} +}} +{{eqn | ll= \implies + | l = c^2 + | r = a^2 + b^2 - 2 a b \cos C +}} +{{end-eqn}} +{{qed|lemma}} +=== Case 2: $AC$ less than $AB$ === +When $AC$ is less than $AB$, the point $B$ lies outside the circle and so the diagram needs to be modified accordingly: +:[[File:CosineRule2.png|300px]] +{{expand|Draw a diagram for the case where $\angle ACB$ is a right angle and where it is a convex angle to show that the formula will be the same.}} +Now we extend: +: $BA$ to $G$ +: $CA$ to $E$. +Then we construct: +:$D$ as the point at which $CB$ [[Definition:Intersection (Geometry)|intersects]] the [[Definition:Circle|circle]] +:$F$ as the point at which $AB$ [[Definition:Intersection (Geometry)|intersects]] the [[Definition:Circle|circle]]. +Finally $D$ is joined to $E$. +Using the [[Secant Secant Theorem]] we have: +: $GB \cdot BF = CB \cdot BD$ +$AF$ is a radius, so $AF = AC = b = GA$ and thus: +: $GB = GA + AB = b + c$ +: $BF = AB - AF = b - c$ +Thus: +{{begin-eqn}} +{{eqn | l = \left({b + c}\right) \left({b - c}\right) + | r = CB \cdot BD + | c = [[Secant Secant Theorem]] +}} +{{eqn | l = \left({b + c}\right) \left({b - c}\right) + | r = a \cdot BD +}} +{{eqn | ll= \implies + | l = \frac {b^2 - c^2} a + | r = BD +}} +{{end-eqn}} +Next: +{{begin-eqn}} +{{eqn | l = CD + | r = CB - BD +}} +{{eqn | r = a - \frac {b^2 - c^2} a +}} +{{eqn | r = \frac {a^2 - b^2 + c^2} a +}} +{{end-eqn}} +As $CA$ is a radius, $CE$ is a diameter. +By [[Thales' Theorem]], it follows that $\angle CDE$ is a [[Definition:Right Angle|right angle]]. +Then using the definition of [[Definition:Cosine of Angle|cosine]], we have +{{begin-eqn}} +{{eqn | l = \cos C + | r = \frac {CD} {CE} +}} +{{eqn | r = \frac {\left({\dfrac {a^2 - b^2 + c^2} a}\right)} {2 b} +}} +{{eqn | r = \frac {a^2 - b^2 + c^2} {2 a b} + | c = +}} +{{eqn | ll= \implies + | l = c^2 + | r = a^2 + b^2 - 2 a b \cos C +}} +{{end-eqn}} +{{qed|lemma}} +=== Case 3: $AC = AB$ === +When $AC = AB$ the points $B$, $D$ and $F$ coincide on the [[Definition:Circumference of Circle|circumference]] of the [[Definition:Circle|circle]]: +:[[File:CosineRule3.png|300px]] +We extend: +: $BA$ to $G$ +: $CA$ to $E$ +and immediately: +: $GB = CB$ +{{finish}} +{{qed}} +\end{proof} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \sin c \sin a \cos B + | r = \cos b - \cos c \cos a + | c = [[Spherical Law of Cosines]] +}} +{{eqn | r = \cos b - \cos c \paren {\cos b \cos c + \sin b \sin c \cos A} + | c = [[Spherical Law of Cosines]] +}} +{{eqn | r = \cos b \paren {1 - \cos^2 c} - \sin b \sin c \cos c \cos A + | c = rearranging +}} +{{eqn | r = \sin^2 c \cos b - \sin b \sin c \cos c \cos A + | c = [[Sum of Squares of Sine and Cosine]] +}} +{{eqn | ll= \leadsto + | l = \sin a \cos B + | r = \sin c \cos b - \sin b \cos c \cos A + | c = simplifying +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | l = \sin a \sin b \cos C + | r = \cos c - \cos a \cos b + | c = [[Spherical Law of Cosines]] +}} +{{eqn | r = \cos c - \cos b \paren {\cos b \cos c + \sin b \sin c \cos A} + | c = [[Spherical Law of Cosines]] +}} +{{eqn | r = \cos c \paren {1 - \cos^2 b} - \sin b \sin c \cos b \cos A + | c = rearranging +}} +{{eqn | r = \sin^2 b \cos c - \sin b \sin c \cos b \cos A + | c = [[Sum of Squares of Sine and Cosine]] +}} +{{eqn | ll= \leadsto + | l = \sin a \cos C + | r = \cos b \sin c - \sin c \cos b \cos A + | c = simplifying +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +:[[File:Spherical-Cosine-Formula-Analog.png|500px]] +Suppose $c$ is less than $\dfrac \pi 2$. +Let $BA$ be [[Definition:Production|produced]] to $D$ so that $BD = \dfrac \pi 2$. +Then: +:$AD = \dfrac \pi 2 - c$ +and: +:$\angle CAD = pi - A$ +Let $C$ and $D$ be joined by an [[Definition:Arc of Circle|arc]] of a [[Definition:Great Circle|great circle]], denoted $x$. +From the [[Definition:Spherical Triangle|triangle]] $\sphericalangle DAC$, using the [[Spherical Law of Cosines]]: +{{begin-eqn}} +{{eqn | l = \cos x + | r = \map \cos {\dfrac \pi 2 - c} \cos b + \map \sin {\dfrac \pi 2 - c} \sin b \, \map \cos {\pi - A} + | c = +}} +{{eqn | r = \sin c \cos b - \cos c \sin b \cos A + | c = +}} +{{end-eqn}} +From the [[Definition:Spherical Triangle|triangle]] $\sphericalangle DBC$, using the [[Spherical Law of Cosines]]: +{{begin-eqn}} +{{eqn | l = \cos x + | r = \cos \dfrac \pi 2 \cos a + \sin \pi 2 \sin a \cos B + | c = +}} +{{eqn | r = \sin a \cos B + | c = +}} +{{end-eqn}} +Hence: +:$\sin a \cos B = \sin c \cos b - \cos c \sin b \cos A$ +The case where $c > \dfrac \pi 2$ is worked similarly, but by making $D$ the [[Definition:Point|point]] between $A$ and $B$ such that $BD$ is $\dfrac \pi 2$. +{{qed}} +\end{proof} + +\begin{proof} +:[[File:Spherical-Cosine-Formula.png|500px]] +Let $A$, $B$ and $C$ be the [[Definition:Vertex of Polygon|vertices]] of a [[Definition:Spherical Triangle|spherical triangle]] on the surface of a [[Definition:Sphere (Geometry)|sphere]] $S$. +By definition of a [[Definition:Spherical Triangle|spherical triangle]], $AB$, $BC$ and $AC$ are [[Definition:Arc of Circle|arcs]] of [[Definition:Great Circle|great circles]] on $S$. +By definition of a [[Definition:Great Circle|great circle]], the [[Definition:Center of Circle|center]] of each of these [[Definition:Great Circle|great circles]] is $O$. +Let $AD$ be the [[Definition:Tangent Line|tangent]] to the [[Definition:Great Circle|great circle]] $AB$. +Let $AE$ be the [[Definition:Tangent Line|tangent]] to the [[Definition:Great Circle|great circle]] $AC$. +Thus the [[Definition:Radius of Sphere|radius]] $OA$ of $S$ is [[Definition:Perpendicular|perpendicular]] to $AD$ and $AE$. +By construction, $AD$ lies in the same [[Definition:Plane|plane]] as $AB$. +Thus when $OB$ is [[Definition:Production|produced]], it will [[Definition:Intersection (Geometry)|intersect]] $AD$ at $D$, say. +Similarly, $OC$ can be [[Definition:Production|produced]] to [[Definition:Intersection (Geometry)|intersect]] $AE$ at $E$, say. +The [[Definition:Spherical Angle|spherical angle]] $\sphericalangle BAC$ is defined as the [[Definition:Angle|angle]] between the [[Definition:Tangent Line|tangents]] $AD$ and $AE$. +Thus: +:$\sphericalangle BAC = \angle DAE$ +or, denoting that [[Definition:Spherical Angle|spherical angle]] $\sphericalangle BAC$ as $A$: +:$A = \angle DAE$ +In the [[Definition:Triangle (Geometry)|(plane) triangle]] $OAD$, we have that $\angle OAD$ is a [[Definition:Right Angle|right angle]]. +We also have that $\angle AOD = \angle AOB$ is equal to $c$, by definition of the [[Definition:Length of Side of Spherical Triangle|length of a side of a spherical triangle]]. +Thus: +{{begin-eqn}} +{{eqn | l = AD + | r = OA \tan c + | c = +}} +{{eqn | l = OD + | r = OA \sec c + | c = +}} +{{end-eqn}} +and by similar analysis of $\triangle OAE$, we have: +{{begin-eqn}} +{{eqn | l = AE + | r = OA \tan b + | c = +}} +{{eqn | l = OE + | r = OA \sec b + | c = +}} +{{end-eqn}} +From consideration of $\triangle DAE$: +{{begin-eqn}} +{{eqn | l = DE^2 + | r = AD^2 + AE^2 - 2 AD \cdot AE \cos \angle DAE + | c = [[Law of Cosines]] +}} +{{eqn | n = 1 + | r = OA^2 \paren {\tan^2 c + \tan^2 b - 2 \tan b \tan c \cos A} + | c = +}} +{{end-eqn}} +From consideration of $\triangle DOE$: +{{begin-eqn}} +{{eqn | l = DE^2 + | r = OD^2 + OE^2 - 2 OD \cdot OE \cos \angle DOE + | c = [[Law of Cosines]] +}} +{{eqn | n = 2 + | r = OA^2 \paren {\sec^2 c + \sec^2 b - 2 \sec b \sec c \cos a} + | c = as $\angle DOE = \angle BOC$ +}} +{{end-eqn}} +Thus: +{{begin-eqn}} +{{eqn | l = \sec^2 c + \sec^2 b - 2 \sec b \sec c \cos a + | r = \tan^2 c + \tan^2 b - 2 \tan b \tan c \cos A + | c = from $(1)$ and $(2)$ +}} +{{eqn | ll= \leadsto + | l = \paren {1 + \tan^2 c} + \paren {1 + \tan^2 b} - 2 \sec b \sec c \cos a + | r = \tan^2 c + \tan^2 b - 2 \tan b \tan c \cos A + | c = [[Difference of Squares of Secant and Tangent]] +}} +{{eqn | ll= \leadsto + | l = 1 - \sec b \sec c \cos a + | r = \tan b \tan c \cos A + | c = simplifying +}} +{{eqn | ll= \leadsto + | l = \cos b \cos c - \cos a + | r = \sin b \sin c \cos A + | c = multiplying both sides by $\cos b \cos c$ +}} +{{end-eqn}} +and the result follows. +{{qed}} +\end{proof} + +\begin{proof} +:[[File:Spherical-Cosine-Formula-2.png|500px]] +Let $A$, $B$ and $C$ be the [[Definition:Vertex of Polygon|vertices]] of a [[Definition:Spherical Triangle|spherical triangle]] on the surface of a [[Definition:Sphere (Geometry)|sphere]] $S$. +By definition of a [[Definition:Spherical Triangle|spherical triangle]], $AB$, $BC$ and $AC$ are [[Definition:Arc of Circle|arcs]] of [[Definition:Great Circle|great circles]] on $S$. +By definition of a [[Definition:Great Circle|great circle]], the [[Definition:Center of Circle|center]] of each of these [[Definition:Great Circle|great circles]] is $O$. +Let $O$ be joined to each of $A$, $B$ and $C$. +Let $P$ be an arbitrary [[Definition:Point|point]] on $OC$. +Construct $PQ$ [[Definition:Perpendicular|perpendicular]] to $OA$ meeting $OA$ at $Q$. +Construct $PR$ [[Definition:Perpendicular|perpendicular]] to $OB$ meeting $OB$ at $R$. +In the [[Definition:Plane|plane]] $OAB$: +:construct $QS$ [[Definition:Perpendicular|perpendicular]] to $OA$ +:construct $RS$ [[Definition:Perpendicular|perpendicular]] to $OB$ +where $S$ is the [[Definition:Point|point]] where $QS$ and $RS$ [[Definition:Intersection (Geometry)|intersect]]. +Let $OS$ and $PS$ be joined. +Let [[Definition:Tangent Line|tangents]] be constructed at $A$ to the [[Definition:Arc of Circle|arcs]] of the [[Definition:Great Circle|great circles]] $AC$ and $AB$. +These [[Definition:Tangent Line|tangents]] [[Definition:Containment of Angle|contain]] the [[Definition:Spherical Angle|spherical angle]] $A$. +But by construction, $QS$ and $QP$ are [[Definition:Parallel Lines|parallel]] to these [[Definition:Tangent Line|tangents]] +Hence $\angle PQS = \sphericalangle A$. +Similarly, $\angle PRS = \sphericalangle B$. +Also we have: +{{begin-eqn}} +{{eqn | l = \angle COB + | r = a +}} +{{eqn | l = \angle COA + | r = b +}} +{{eqn | l = \angle AOB + | r = c +}} +{{end-eqn}} +It is to be proved that $PS$ is [[Definition:Line Perpendicular to Plane|perpendicular]] to the [[Definition:Plane|plane]] $AOB$. +By construction, $OQ$ is [[Definition:Perpendicular|perpendicular]] to both $PQ$ and $QS$. +Thus $OQ$ is [[Definition:Line Perpendicular to Plane|perpendicular]] to the [[Definition:Plane|plane]] $PQS$. +Similarly, $OR$ is [[Definition:Line Perpendicular to Plane|perpendicular]] to the [[Definition:Plane|plane]] $PRS$. +Thus $PS$ is [[Definition:Perpendicular|perpendicular]] to both $OQ$ and $OR$. +Thus $PS$ is [[Definition:Perpendicular|perpendicular]] to every [[Definition:Straight Line|line]] in the [[Definition:Plane|plane]] of $OQ$ and $OR$. +That is, $PS$ is [[Definition:Line Perpendicular to Plane|perpendicular]] to the [[Definition:Plane|plane]] $OAB$. +In particular, $PS$ is [[Definition:Perpendicular|perpendicular]] to $OS$, $SQ$ and $SR$ +It follows that $\triangle PQS$ and $\triangle PRS$ are [[Definition:Right Triangle|right triangles]]. +From the [[Definition:Right Triangle|right triangles]] $\triangle OQP$ and $\triangle ORP$, we have: +{{begin-eqn}} +{{eqn | n = 1 + | l = PQ + | r = OP \sin b +}} +{{eqn | n = 2 + | l = PR + | r = OP \sin a +}} +{{eqn | n = 3 + | l = OQ + | r = OP \cos b +}} +{{eqn | n = 4 + | l = OR + | r = OP \cos a +}} +{{end-eqn}} +Let us denote the [[Definition:Angle|angle]] $\angle SOQ$ by $x$. +Then: +:$\angle ROS = c - x$ +We have that: +{{begin-eqn}} +{{eqn | l = OS + | r = OQ \sec x +}} +{{eqn | l = OS + | r = OR \, \map \sec {c - x} +}} +{{eqn | ll= \leadsto + | l = OR \cos x + | r = OQ \, \map \cos {c - x} + | c = +}} +{{eqn | ll= \leadsto + | l = OP \cos a \cos x + | r = OP \cos b \, \map \cos {c - x} + | c = from $(3)$ and $(4)$ +}} +{{eqn | ll= \leadsto + | l = \cos a \cos x + | r = \cos b \paren {\cos c \cos x - \sin c \sin x} + | c = [[Cosine of Difference]] +}} +{{eqn | n = 5 + | ll= \leadsto + | l = \cos a + | r = \cos b \cos c + \cos b \sin c \tan x + | c = dividing both sides by $\cos x$ and multiplying out +}} +{{end-eqn}} +But we also have: +{{begin-eqn}} +{{eqn | l = \tan x + | r = \dfrac {QS} {OQ} + | c = +}} +{{eqn | r = \dfrac {PQ \cos A} {OQ} + | c = +}} +{{eqn | r = \tan b \cos A + | c = +}} +{{eqn | ll= \leadsto + | l = \cos a + | r = \cos b \cos c + \cos b \sin c \tan b \cos A + | c = substituting for $\tan x$ from $(5)$ +}} +{{eqn | r = \cos b \cos c + \sin b \sin c \cos A + | c = +}} +{{end-eqn}} +Hence the result. +{{qed}} +\end{proof}<|endoftext|> +\section{Euler's Formula} +Tags: Exponential Function, Trigonometric Functions, Euler's Formula + +\begin{theorem} +Let $z \in \C$ be a [[Definition:Complex Number|complex number]]. +Then: +:$e^{i z} = \cos z + i \sin z$ +\end{theorem} + +\begin{proof} +As [[Sine Function is Absolutely Convergent]] and [[Cosine Function is Absolutely Convergent]], we have: +{{begin-eqn}} +{{eqn | l = \cos z + i \sin z + | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {z^{2 n} } {\paren {2 n}!} + i \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {z^{2 n + 1} } {\paren {2 n + 1}!} + | c = {{Defof|Complex Cosine Function}} and {{Defof|Complex Sine Function}} +}} +{{eqn | r = \sum_{n \mathop = 0}^\infty \paren {\paren {-1}^n \dfrac {z^{2 n} } {\paren {2 n}!} + i \paren {-1}^n \dfrac {z^{2 n + 1} } {\paren {2 n + 1}!} } + | c = [[Sum of Absolutely Convergent Series]] +}} +{{eqn | r = \sum_{n \mathop = 0}^\infty \paren {\dfrac {\paren {i z}^{2 n} } {\paren {2 n}!} + \dfrac {\paren {i z}^{2 n + 1} } {\paren {2 n + 1}!} } + | c = {{Defof|Imaginary Unit}} +}} +{{eqn | r = \sum_{n \mathop = 0}^\infty \dfrac {\paren {i z}^n} {n!} +}} +{{eqn | r = e^{i z} + | c = {{Defof|Exponential Function/Complex|subdef = Sum of Series|Complex Exponential Function}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Lagrange's Theorem (Group Theory)} +Tags: Order of Groups, Index of Subgroups, Subgroups, Lagrange's Theorem (Group Theory) + +\begin{theorem} +Let $G$ be a [[Definition:Finite Group|finite group]]. +Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$. +Then: +: $\order H$ [[Definition:Divisor of Integer|divides]] $\order G$ +where $\order G$ and $\order H$ are the [[Definition:Order of Structure|order]] of $G$ and $H$ respectively. +In fact: +:$\index G H = \dfrac {\order G} {\order H}$ +where $\index G H$ is the [[Definition:Index of Subgroup|index]] of $H$ in $G$. +When $G$ is an [[Definition:Infinite Group|infinite group]], we can still interpret this theorem sensibly: +:A [[Definition:Subgroup|subgroup]] of [[Definition:Finite Index|finite index]] in an [[Definition:Infinite Group|infinite group]] is itself an [[Definition:Infinite Group|infinite group]]. +:A [[Definition:Finite Group|finite]] [[Definition:Subgroup|subgroup]] of an [[Definition:Infinite Group|infinite group]] has [[Definition:Infinite Index|infinite index]]. +\end{theorem} + +\begin{proof} +Let $G$ be a [[Definition:Group|group]]. +Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$. +From [[Cosets are Equivalent]], a [[Definition:Left Coset|left coset]] $y H$ has the same number of [[Definition:Element|elements]] as $H$, namely $\order H$. +Since [[Definition:Left Coset|left cosets]] are [[Congruence Class Modulo Subgroup is Coset|identical or disjoint]], each [[Definition:Element|element]] of $G$ belongs to exactly one [[Definition:Left Coset|left coset]]. +From the definition of [[Definition:Index of Subgroup|index of subgroup]], there are $\index G H$ [[Definition:Left Coset|left cosets]], and therefore: +:$\order G = \index G H \order H$ +Let $G$ be of [[Definition:Finite Group|finite order]]. +All three numbers are finite, and the result follows. +Now let $G$ be of [[Definition:Infinite Group|infinite order]]. +If $\index G H = n$ is [[Definition:Finite Index|finite]], then $\order G = n \order H$ and so $H$ is of [[Definition:Infinite Group|infinite order]]. +If $H$ is of [[Definition:Finite Group|finite order]] such that $\order H = n$, then $\order G = \index G H \times n$ and so $\index G H$ is [[Definition:Infinite Index|infinite]]. +{{qed}} +\end{proof} + +\begin{proof} +Follows directly from the [[Orbit-Stabilizer Theorem]] applied to [[Group Action on Coset Space]]. +{{Qed}} +\end{proof} + +\begin{proof} +Let $G$ be [[Definition:Finite Group|finite]]. +Consider the [[Definition:Mapping|mapping]] $\phi: G \to G / H^l$, defined as: +:$\phi: G \to G / H^l: \map \phi x = x H^l$ +where $G / H^l$ is the [[Definition:Left Coset Space|left coset space of $G$ modulo $H$]]. +For every $y H \in G / H^l$, there exists a corresponding $y \in G$, so $\phi$ is a [[Definition:Surjection|surjection]]. +From [[Cardinality of Surjection]] it follows that $G / H^l$ is [[Definition:Finite Set|finite]]. +From [[Cosets are Equivalent]], $G / H^l$ has the same number of [[Definition:Element|elements]] as $H$. +We have that the [[Left Coset Space forms Partition|$G / H^l$ is a partition of $G$]]. +It follows from [[Number of Elements in Partition]] that $\index G H = \dfrac {\order G} {\order H}$ +{{Qed}} +\end{proof}<|endoftext|> +\section{ProofWiki:Sandbox} +Tags: Euler Numbers, Binomial Coefficients, Sum of Euler Numbers by Binomial Coefficients Vanishes + +\begin{theorem} +$\forall n \in \Z_{>0}: \displaystyle \sum_{k \mathop = 0}^{n} \binom {2n} {2k} E_{2n - 2k } = 0$ +where $E_k$ denotes the $k$th [[Definition:Euler Numbers|Euler number]]. +\end{theorem} + +\begin{proof} +Take the definition of [[Definition:Euler Numbers/Generating Function|Euler numbers]]: +{{begin-eqn}} +{{eqn | l = \sum_{n \mathop = 0}^\infty \frac {E_n x^n} {n!} + | r = \frac {2e^x} {e^{2x} + 1} + | c = +}} +{{eqn | r = \paren {\frac {2e^x } {e^{2x} + 1 } } \paren {\frac {e^{-x } } {e^{-x } } } + | c = Multiply by [[1]] +}} +{{eqn | r = \paren {\frac 2 {e^{x} + e^{-x} } } + | c = +}} +{{end-eqn}} +From the definition of the [[Definition:Exponential Function/Real/Sum of Series|exponential function]]: +{{begin-eqn}} +{{eqn | l = e^x + | r = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!} + | c = +}} +{{eqn | r = 1 + x + \frac {x^2} {2!} + \frac {x^3} {3!} + \frac {x^4} {4!} + \cdots + | c = +}} +{{eqn | l = e^{-x} + | r = \sum_{n \mathop = 0}^\infty \frac {\paren {-x }^n } {n!} + | c = +}} +{{eqn | r = 1 - x + \frac {x^2} {2!} - \frac {x^3} {3!} + \frac {x^4} {4!} - \cdots + | c = +}} +{{eqn | l = \paren {\frac {e^x + e^{-x } } 2 } + | r = \paren {\sum_{n \mathop = 0}^\infty \frac {x^{2n } } {\paren {2n }!} } + | c = +}} +{{eqn | r = 1 + \frac {x^2} {2!} + \frac {x^4} {4!} + \cdots + | c = [[Definition:Odd Integer|odd]] terms cancel in the [[Definition:Sum|sum]]. +}} +{{end-eqn}} +Thus: +{{begin-eqn}} +{{eqn | l = 1 + | r = \paren {\frac 2 {e^x + e^{-x } } } \paren {\frac {e^x + e^{-x } } 2 } + | c = +}} +{{eqn | r = \paren {\sum_{n \mathop = 0}^\infty \frac {E_n x^n} {n!} } \paren {\sum_{n \mathop = 0}^\infty \frac {x^{2n } } {\paren {2n }!} } + | c = +}} +{{end-eqn}} +By [[Product of Absolutely Convergent Series]], we will let: +{{begin-eqn}} +{{eqn | l = a_n + | r = \frac {E_n x^n} {n!} + | c = +}} +{{eqn | l = b_n + | r = \frac {x^{2n} } {\paren {2n }!} + | c = +}} +{{end-eqn}} +Then: +{{begin-eqn}} +{{eqn | l = \sum_{n \mathop = 0}^\infty c_n + | r = \paren { \displaystyle \sum_{n \mathop = 0}^\infty a_n } \paren {\displaystyle \sum_{n \mathop = 0}^\infty b_n } + | rrr = =1 + | c = +}} +{{eqn | l = c_n + | r = \sum_{k \mathop = 0}^n a_k b_{n - k} + | c = +}} +{{eqn | l = c_0 + | r = \frac {E_0 x^0} {0!} \frac {x^{0} } {0!} + | rrr = = 1 + | c = $c_0 = \paren {a_0 } \paren {b_{0 - 0 } } = \paren {a_0 } \paren {b_0 }$ +}} +{{eqn | lll = \leadsto + | l = \sum_{n \mathop = 1}^\infty c_n + | r = \paren { \displaystyle \sum_{n \mathop = 0}^\infty a_n } \paren {\displaystyle \sum_{n \mathop = 0}^\infty b_n } - a_0 b_0 + | rrr = =0 + | c = [[Definition:Subtraction|Subtract]] [[1]] from both sides of the [[Definition:Equation|equation]]. +}} +{{end-eqn}} +$\forall n \in \Z_{\gt 0}$, term by term $c_n$ is equal to: +{{begin-eqn}} +{{eqn | l = c_1 + | r = \frac {E_0 x^0} {0!} \frac {x^2 } {2!} + \frac {E_1 x^1} {1!} \frac {x^0 } {0!} + | rrr = = \frac {x^2 } {2! } E_0 + | c = $= a_0 b_1 + a_1 b_0$ +}} +{{eqn | l = c_2 + | r = \frac {E_0 x^0} {0!} \frac {x^4 } {4!} + \frac {E_1 x^1} {1!} \frac {x^2 } {2!} + \frac {E_2 x^2} {2!} \frac {x^0 } {0!} + | rrr = = \frac {x^4 } {4! } E_0 + \frac {x^2 } {2! } E_2 + | c = $= a_0 b_2 + a_1 b_1 + a_2 b_0$ +}} +{{eqn | l = c_3 + | r = \frac {E_0 x^0} {0!} \frac {x^6 } {6!} + \frac {E_1 x^1} {1!} \frac {x^4 } {4!} + \frac {E_2 x^2} {2!} \frac {x^2 } {2!} + \frac {E_3 x^3} {3!} \frac {x^0 } {0!} + | rrr = = \frac {x^6 } {6! } E_0 + \frac {x^4 } {2! 2! } E_2 + | c = $= a_0 b_3 + a_1 b_2 + a_2 b_1 + a_3 b_0$ +}} +{{eqn | l = c_4 + | r = \frac {E_0 x^0} {0!} \frac {x^8 } {8!} + \frac {E_1 x^1} {1!} \frac {x^6 } {6!} + \frac {E_2 x^2} {2!} \frac {x^4 } {4!} + \frac {E_3 x^3} {3!} \frac {x^2 } {2!} + \frac {E_4 x^4} {4!} \frac {x^0 } {0!} + | rrr = = \frac {x^8 } {0! 8! } E_0 + \frac {x^6 } {2! 4! } E_2 + \frac {5 x^4 } {4! 0! } E_4 + | c = $= a_0 b_4 + a_1 b_3 + a_2 b_2 + a_3 b_1 + a_4 b_0$ +}} +{{eqn | l = \cdots + | r = \cdots +}} +{{eqn | l = c_n + | r = \frac {E_0 x^0} {0!} \frac {x^{2n} } {\paren {2n }!} + \frac {E_1 x^1} {1!} \frac {x^{2n - 2} } {\paren {2n - 2 }!} + \frac {E_2 x^2} {2!} \frac {x^{2n - 4} } {\paren {2n - 4 }!} + \cdots + \frac {E_n x^n} {n!} \frac {x^0 } {0!} + | c = +}} +{{end-eqn}} +Grouping like [[Definition:Even Integer|even]] terms produces: +{{begin-eqn}} +{{eqn | l = \paren {\frac 1 {0! 2!} } E_0 + \paren {\frac 1 {2! 0!} } E_2 + | r = 0 + | c = $x^2$ term +}} +{{eqn | l = \paren {\frac 1 {0! 4!} } E_0 + \paren {\frac 1 {2! 2!} } E_2 + \paren {\frac 1 {4! 0!} } E_4 + | r = 0 + | c = $x^4$ term +}} +{{end-eqn}} +Multiplying $c_n$ through by $n!$ gives: +{{begin-eqn}} +{{eqn | l = n! c_n + | r = \frac {E_0 x^0} {0!} \frac {n! x^n } {n!} + \frac {E_1 x^1} {1!} \frac {n! x^{n-1} } {\paren {n - 1 }!} + \cdots + \frac {E_n x^n} {n!} \frac {n! x^{0} } {0!} + | rrr = = 0 + | c = +}} +{{eqn | r = x^n \paren {\frac {n! } {0! n!} E_0 + \frac {n! } {1! \paren {n - 1 }!} E_1 + \cdots + \frac {n! } {n! 0!} E_n } + | rrr = = 0 + | c = factoring out $x^n$ +}} +{{end-eqn}} +But those coefficients are the [[Definition:Binomial Coefficient|binomial coefficients]]: +{{begin-eqn}} +{{eqn | l = n! c_n + | r = \dbinom n 0 E_0 + \dbinom n 1 E_1 + \dbinom n 2 E_2 + \cdots + \dbinom n n E_n + | rrr = = 0 + | c = +}} +{{end-eqn}} +Hence the result. +{{qed}} +{{begin-eqn}} +{{eqn | l = \sum_{k \mathop = 0}^{n} \dbinom {2 n} {2 k} E_{2 n - 2 k} + | r = \binom {2 n} 0 E_{2 n} + \binom {2 n} 2 E_{2 n - 2} + \binom {2 n} 4 E_{2 n - 4} + \binom {2 n} 6 E_{2 n - 6} + \cdots + 1 + | rrr = = 0 + | c = +}} +{{end-eqn}} +\end{proof}<|endoftext|> +\section{Carathéodory's Theorem (Analysis)} +Tags: Direct Proofs, Real Analysis, Differential Calculus + +\begin{theorem} +Let $I \subseteq \R$. +Let $c \in I$ be an [[Definition:Interior Point|interior point]] of $I$. +{{Disambiguate|Definition:Interior Point}} +{{explain|In this case, there appears not to be a definition for "interior point" which appropriately captures the gist of this. It is clearly a point inside a real interval, but the concept has not yet been defined.}} +Let $f : I \to \R$ be a [[Definition:Real Function|real function]]. +Then $f$ is [[Definition:Differentiable Real Function at Point|differentiable]] at $c$ {{iff}}: +:There exists a [[Definition:Real Function|real function]] $\varphi : I \to \R$ that is [[Definition:Continuous Real Function at Point|continuous]] at $c$ and satisfies: +:$(1): \quad \forall x \in I: f \left({x}\right) - f \left({c}\right) = \varphi \left({x}\right) \left({x - c}\right)$ +:$(2): \quad \varphi \left({c}\right) = f' \left({c}\right)$ +\end{theorem} + +\begin{proof} +=== Necessary Condition === +Suppose $f$ is [[Definition:Differentiable Real Function at Point|differentiable]] at $c$. +Then by definition $f' \left({c}\right)$ exists. +So we can define $\varphi$ by: +:$\varphi \left({x}\right) = \begin{cases} +\dfrac {f \left({x}\right) - f \left({c}\right)} {x - c} & : x \ne c, x \in I \\ +f' \left({c}\right) & : x = c +\end{cases}$ +Condition $(2)$, that $\varphi$ is [[Definition:Continuous Real Function at Point|continuous]] at $c$, is satisfied, since: +{{begin-eqn}} +{{eqn | l = \varphi \left({c}\right) + | r = f' \left({c}\right) + | c = Definition of $\varphi$ +}} +{{eqn | r = \lim_{x \to c} \frac {f \left({x}\right) - f \left({c}\right)} {x - c} + | c = Definition of [[Definition:Derivative|derivative]] +}} +{{eqn | r = \lim_{x \to c} \varphi \left({x}\right) + | c = Definition of $\varphi$ +}} +{{end-eqn}} +Finally, condition $(1)$ is vacuous for $x = c$. +For $x \ne c$, it follows from the definition of $\varphi$ by dividing both sides of $(1)$ by $x - c$. +{{qed|lemma}} +=== Sufficient Condition === +Suppose a $\varphi$ as in the theorem statement exists. +Then for $x \ne c$, we have that: +:$\varphi \left({x}\right) = \dfrac {f \left({x}\right) - f \left({c}\right)} {x - c}$ +Since $\varphi$ is [[Definition:Continuous Real Function at Point|continuous]] at $c$: +:$\displaystyle \varphi \left({c}\right) = \lim_{x \to c} \varphi \left({x}\right) = \lim_{x \to c} \frac {f \left({x}\right) - f \left({c}\right)} {x - c}$ +That is, $f$ is [[Definition:Differentiable Real Function at Point|differentiable]] at $c$, and $f' \left({c}\right) = \varphi \left({c}\right)$. +{{qed}} +{{Namedfor|Constantin Carathéodory|cat = Carathéodory}} +[[Category:Direct Proofs]] +[[Category:Real Analysis]] +[[Category:Differential Calculus]] +kuvgsqeze4423l7w2tclrn6y4dqellz +\end{proof}<|endoftext|> +\section{One-Step Subgroup Test} +Tags: Named Theorems, Subgroups + +\begin{theorem} +Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. +Let $H$ be a [[Definition:Subset|subset]] of $G$. +Then $\struct {H, \circ}$ is a [[Definition:Subgroup|subgroup]] of $\struct {G, \circ}$ {{iff}}: +: $(1): \quad H \ne \O$, that is, $H$ is [[Definition:Non-Empty Set|non-empty]] +: $(2): \quad \forall a, b \in H: a \circ b^{-1} \in H$. +\end{theorem} + +\begin{proof} +=== Necessary Condition === +Let $H$ be a [[Definition:Subset|subset]] of $G$ that fulfils the conditions given. +It is noted that the fact that $H$ is [[Definition:Non-Empty Set|non-empty]] is one of the conditions. +It is also noted that the [[Definition:Group Operation|group operation]] of $\struct {H, \circ}$ is the same as that for $\struct {G, \circ}$, that is, $\circ$. +So it remains to show that $\struct {H, \circ}$ is a [[Definition:Group|group]]. +We check the four [[Definition:Group Axioms|group axioms]]: +'''$\text G 1$: Associativity''': +From [[Subset Product within Semigroup is Associative]], [[Definition:Associative Algebraic Structure|associativity]] is inherited by $\struct {H, \circ}$ from $\struct {G, \circ}$. +'''$\text G 2$: Identity''' +Let $e$ be the [[Definition:Identity Element|identity]] of $\struct {G, \circ}$. +Since $H$ is [[Definition:Non-Empty Set|non-empty]], $\exists x \in H$. +If we take $a = x$ and $b = x$, then $a \circ b^{-1} = x \circ x^{-1} = e \in H$, where $e$ is the [[Definition:Identity Element|identity element]]. +'''$\text G 3$: Inverses''' +If we take $a = e$ and $b = x$, then $a \circ b^{-1} = e \circ x^{-1} = x^{-1} \in H$. +Thus every element of $H$ has an [[Definition:Inverse Element|inverse]] also in $H$. +'''$\text G 0$: Closure''' +Let $x, y \in H$. +Then $y^{-1} \in H$, so we may take $a = x$ and $b = y^{-1}$. +So: +:$a \circ b^{-1} = x \circ \paren {y^{-1} }^{-1} = x \circ y \in H$ +Thus, $H$ is [[Definition:Closed Algebraic Structure|closed]]. +Therefore, $\struct {H, \circ}$ satisfies all the [[Definition:Group Axioms|group axioms]], and is therefore a [[Definition:Group|group]]. +Therefore $\struct {H, \circ}$ is a [[Definition:Subgroup|subgroup]] of $\struct {G, \circ}$. +{{qed|lemma}} +=== Sufficient Condition === +Now suppose $\struct {H, \circ}$ is a [[Definition:Subgroup|subgroup]] of $\struct {G, \circ}$. +: $(1): \quad H \le G \implies H \ne \O$ from the fact that $H$ is a [[Definition:Group|group]] and therefore [[Group is not Empty|can not be empty]]. +: $(2): \quad$ As $\struct {H, \circ}$ is a [[Definition:Group|group]], it is [[Definition:Closed Algebraic Structure|closed]] and every element has an [[Definition:Inverse Element|inverse]]. So it follows that $\forall a, b \in H: a \circ b^{-1} \in H$. +{{qed}} +\end{proof}<|endoftext|> +\section{Two-Step Subgroup Test} +Tags: Named Theorems, Subgroups + +\begin{theorem} +Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. +Let $H$ be a [[Definition:Subset|subset]] of $G$. +Then $\struct {H, \circ}$ is a [[Definition:Subgroup|subgroup]] of $\struct {G, \circ}$ {{iff}}: +:$(1): \quad H \ne \O$, that is, $H$ is [[Definition:Non-Empty Set|non-empty]] +:$(2): \quad a, b \in H \implies a \circ b \in H$ +:$(3): \quad a \in H \implies a^{-1} \in H$. +That is, $\struct {H, \circ}$ is a [[Definition:Subgroup|subgroup]] of $\struct {G, \circ}$ {{iff}} $\struct {H, \circ}$ is a $H$ be a [[Definition:Non-Empty Set|nonempty]] [[Definition:Subset|subset]] of $G$ which is: +:[[Definition:Closed Algebraic Structure|closed]] under its [[Definition:Group Operation|operation]] +and: +:[[Definition:Closed under Inversion|closed under inversion]]. +\end{theorem} + +\begin{proof} +=== Necessary Condition === +Let $H$ be a [[Definition:Subset|subset]] of $G$ that fulfils the conditions given. +It is noted that the fact that $H$ is [[Definition:Non-Empty Set|nonempty]] is one of the conditions. +It is also noted that the [[Definition:Group Operation|group operation]] of $\struct {H, \circ}$ is the same as that for $\struct {G, \circ}$, that is, $\circ$. +So it remains to show that $\struct {H, \circ}$ is a [[Definition:Group|group]]. +We check the four [[Definition:Group Axioms|group axioms]]: +==== $\text G 0$: Closure ==== +The [[Definition:Closed Algebraic Structure|closure]] condition is given by condition $(2)$. +{{qed|lemma}} +==== $\text G 1$: Associativity ==== +From [[Closed Substructure of Semigroup is Semigroup]], [[Definition:Associative Operation|associativity]] is inherited by $\struct {H, \circ}$ from $\struct {G, \circ}$. +{{qed|lemma}} +==== $\text G 2$: Identity ==== +Let $e$ be the [[Definition:Identity Element|identity]] of $\struct {G, \circ}$. +From condition $(1)$, $H$ is [[Definition:Non-Empty Set|non-empty]]. +Therefore $\exists x \in H$. +From condition $(3)$, $\struct {H, \circ}$ is [[Definition:Closed under Inversion|closed under inversion]]. +Therefore $x^{-1} \in H$. +Since $\struct {H, \circ}$ is [[Definition:Closed Algebraic Structure|closed]] under $\circ$, $x \circ x^{-1} = e = x^{-1} \circ x \in H$. +{{qed|lemma}} +==== $\text G 3$: Inverses ==== +From condition $(3)$, every element of $H$ has an [[Definition:Inverse Element|inverse]]. +{{qed|lemma}} +So $\struct {H, \circ}$ satisfies all the [[Definition:Group Axioms|group axioms]], and is therefore a [[Definition:Group|group]]. +So by definition $\struct {H, \circ}$ is a [[Definition:Subgroup|subgroup]] of $\struct {G, \circ}$. +{{qed|lemma}} +=== Sufficient Condition === +Now suppose $\struct {H, \circ}$ is a [[Definition:Subgroup|subgroup]] of $\struct {G, \circ}$. +:$(1): \quad H \le G \implies H \ne \O$ from the fact that $H$ is a [[Definition:Group|group]] and therefore [[Group is not Empty|can not be empty]]. +:$(2): \quad a, b \in H \implies a \circ b \in H$ follows from {{GroupAxiom|0}} as applied to the [[Definition:Group|group]] $\struct {H, \circ}$. +:$(3): \quad a \in H \implies a^{-1} \in H$ follows from {{GroupAxiom|3}} as applied to the [[Definition:Group|group]] $\struct {H, \circ}$. +{{qed}} +\end{proof}<|endoftext|> +\section{Fundamental Theorem of Arithmetic} +Tags: Prime Numbers, Fundamental Theorems, Factorization, Prime Decompositions + +\begin{theorem} +For every [[Definition:Integer|integer]] $n$ such that $n > 1$, $n$ can be expressed as the [[Definition:Integer Multiplication|product]] of one or more [[Definition:Prime Number|primes]], uniquely up to the order in which they appear. +\end{theorem} + +\begin{proof} +In [[Integer is Expressible as Product of Primes]] it is proved that every [[Definition:Integer|integer]] $n$ such that $n > 1$, $n$ can be expressed as the [[Definition:Integer Multiplication|product]] of one or more [[Definition:Prime Number|primes]]. +In [[Prime Decomposition of Integer is Unique]], it is proved that this [[Definition:Prime Decomposition|prime decomposition]] is [[Definition:Unique|unique]] up to the order of the [[Definition:Factor (Algebra)|factors]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Euclid's Lemma} +Tags: Euclid's Lemma, Divisors, Coprime Integers + +\begin{theorem} +Let $a, b, c \in \Z$. +Let $a \divides b c$, where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]]. +Let $a \perp b$, where $\perp$ denotes [[Definition:Coprime Integers|relative primeness]]. +Then $a \divides c$. +\end{theorem} + +\begin{proof} +We have that the [[Integers are Euclidean Domain|integers form a Euclidean domain]]. +Then from [[Irreducible Elements of Ring of Integers]] we have that the [[Definition:Irreducible Element of Ring|irreducible elements]] of $\Z$ are the [[Definition:Prime Number|primes]] and their [[Definition:Negative|negatives]]. +The result then follows directly from [[Euclid's Lemma for Irreducible Elements]]. +{{qed}} +\end{proof} + +\begin{proof} +Let $p \divides a b$. +Suppose $p \nmid a$. +Then from the definition of [[Definition:Prime Number|prime]]: +:$p \perp a$ +where $\perp$ indicates that $p$ and $a$ are [[Definition:Coprime Integers|coprime]]. +Thus from [[Euclid's Lemma]] it follows that: +:$p \divides b$ +Similarly, if $p \nmid b$ it follows that $p \divides a$. +So: +:$p \divides a b \implies p \divides a$ or $p \divides b$ +as we needed to show. +{{qed}} +\end{proof} + +\begin{proof} +Follows directly from [[Integers are Euclidean Domain]]. +{{qed}} +\end{proof} + +\begin{proof} +Let $a, b, c \in \Z$. +We have that $a \perp b$. +That is: +:$\gcd \set {a, b} = 1$ +where $\gcd$ denotes [[Definition:Greatest Common Divisor of Integers|greatest common divisor]]. +From [[Bézout's Lemma]], we may write: +:$a x + b y = 1$ +for some $x, y \in \Z$. +Upon multiplication by $c$, we see that: +:$c = c \paren {a x + b y} = c a x + c b y$ +Now note that $c a x + c b y$ is an [[Definition:Integer Combination|integer combination]] of $a c$ and $b c$. +So, since: +:$a \divides a c$ +and: +:$a \divides b c$ +it follows from [[Common Divisor Divides Integer Combination]] that: +:$a \divides \paren {c a x + c b y}$ +However: +: $c a x + c b y = c \paren {a x + b y} = c \cdot 1 = c$ +Therefore: +: $a \divides c$ +{{qed}} +\end{proof} + +\begin{proof} +Let $p \divides a b$. +Suppose $p \nmid a$. +Then by {{EuclidPropLink|book = VII|prop = 29|title = Prime not Divisor implies Coprime}}: +:$p \perp a$ +As $p \divides a b$, it follows by definition of [[Definition:Divisor of Integer|divisor]]: +:$\exists e \in \Z: e p = a b$ +So by {{EuclidPropLink|book = VII|prop = 19|title = Relation of Ratios to Products}}: +:$p : a = b : e$ +But as $p \perp a$, by: +:{{EuclidPropLink|book = VII|prop = 21|title = Coprime Numbers form Fraction in Lowest Terms}} +and: +:{{EuclidPropLink|book = VII|prop = 20|title = Ratios of Fractions in Lowest Terms}} +it follows that: +:$p \divides b$ +Similarly, if $p \perp b$ then $p \divides a$. +Hence the result. +{{qed}} +\end{proof}<|endoftext|> +\section{Fermat's Little Theorem} +Tags: Number Theory, Fermat's Little Theorem + +\begin{theorem} +Let $p$ be a [[Definition:Prime Number|prime number]]. +Let $n \in \Z_{>0}$ be a [[Definition:Positive Integer|positive integer]] such that $p$ is not a [[Definition:Divisor of Integer|divisor]] of $n$. +Then: +:$n^{p - 1} \equiv 1 \pmod p$ +\end{theorem} + +\begin{proof} +Consider the sequence of integers $n, 2n, 3n, \dots, (p-1)n$. +Note that none of these integers are [[Definition:Congruence (Number Theory)|congruent modulo $p$]] to the others. +If this were the case, we would have $a n \equiv b n \pmod p$ for some $1 \le a < b \le p-1$. +Then as $\gcd \left\{{n, p}\right\} = 1$, and we can [[Cancellability of Congruences|cancel the $n$]], we get $a \equiv b \pmod p$ and so $a = b$. +Also, since $p \nmid n$ and $p \nmid c$, for any $1 \le c \le p-1$, then by [[Euclid's Lemma]] $p \nmid c n$ for any such $c n$, which means $c n \not \equiv 0 \pmod p$. +Thus, each integer in the sequence can be reduced modulo $p$ to exactly one of $1, 2, 3, \ldots, p-1$. +So $\left\{{1, 2, 3, \ldots, p-1}\right\}$ is the set of [[Definition:Reduced Residue System|Reduced Residue System]] modulo $p$. +So, upon taking the product of these congruences, we see that $n \times 2n \times 3n \times \dots \times (p-1)n \equiv 1 \times 2 \times 3 \times \cdots \times (p-1) \pmod p$. +This simplifies to $n^{p-1} \times (p-1)! \equiv (p-1)! \pmod p$. +Since $p \nmid (p-1)!$, we can [[Cancellability of Congruences|cancel $(p-1)!$ from both sides]], leaving us with $n^{p-1}\equiv 1 \pmod p$. +{{qed}} +\end{proof} + +\begin{proof} +By [[Prime not Divisor implies Coprime]]: +: $p \nmid n \implies p \perp n$ +and [[Euler's Theorem]] can be applied. +Thus: +: $n^{\map \phi p} \equiv 1 \pmod p$ +But from [[Euler Phi Function of Prime Power]]: +: $\map \phi p = p \paren {1 - \dfrac 1 p} = p - 1$ +and the result follows. +{{qed}} +\end{proof} + +\begin{proof} +Let $\struct {\Z'_p, \times}$ denote the [[Definition:Multiplicative Group of Reduced Residues|multiplicative group of reduced residues modulo $p$]]. +From [[Reduced Residue System under Multiplication forms Abelian Group/Corollary|the corollary to Reduced Residue System under Multiplication forms Abelian Group]], $\struct {\Z'_p, \times}$ forms a [[Definition:Group|group]] of [[Definition:Order of Structure|order]] $p - 1$ under [[Definition:Modulo Multiplication|modulo multiplication]]. +By [[Element to Power of Group Order is Identity]], we have: +: $n^{p - 1} \equiv 1 \pmod p$ +{{qed}} +\end{proof} + +\begin{proof} +[[Second Principle of Mathematical Induction|Proof by induction]] over $n$. +Induction base: +:$1^p \equiv 1 \pmod p$ +Induction step: +Assume $n^p \equiv n \pmod p$ +{{begin-eqn}} +{{eqn | l = \paren {n + 1}^p + | r = \sum_{k \mathop = 0}^p {p \choose k} n^{p - k} \cdot 1^k + | c = [[Binomial Theorem]] +}} +{{eqn | l = \forall k: 0 < k < p: {p \choose k} + | o = \equiv + | r = 0 \pmod p + | c = [[Binomial Coefficient of Prime]] +}} +{{end-eqn}} +and so: +{{begin-eqn}} +{{eqn | l = \sum_{k \mathop = 0}^p {p \choose k} n^{p - k} + | o = \equiv + | r = n^p + n^0 \pmod p +}} +{{eqn | o = \equiv + | r = n^p + 1 \pmod p +}} +{{eqn | o = \equiv + | r = n + 1 \pmod p + | c = by induction step +}} +{{end-eqn}} +Dividing by $n$: +:$\forall n: n^p \equiv n \pmod p \implies n^{p - 1} \equiv 1 \pmod p$ +{{qed}} +\end{proof}<|endoftext|> +\section{De Morgan's Laws (Set Theory)} +Tags: Set Difference, Set Union, Set Intersection, Relative Complement, Set Complement, De Morgan's Laws + +\begin{theorem} +{{:De Morgan's Laws (Set Theory)/Set Difference}} +\end{theorem}<|endoftext|> +\section{De Moivre's Formula} +Tags: Complex Analysis, De Moivre's Formula + +\begin{theorem} +Let $z \in \C$ be a [[Definition:Complex Number|complex number]] expressed in [[Definition:Polar Form of Complex Number|complex form]]: +:$z = r \paren {\cos x + i \sin x}$ +Then: +:$\forall \omega \in \C: \paren {r \paren {\cos x + i \sin x} }^\omega = r^\omega \paren {\map \cos {\omega x} + i \, \map \sin {\omega x} }$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \paren {r \paren {\cos x + i \sin x} }^\omega + | r = \paren {r e^{i x} }^\omega + | c = [[Euler's Formula]] +}} +{{eqn | r = r^\omega e^{i \omega x} + | c = [[Power of Power]] +}} +{{eqn | r = r^\omega \paren {\map \cos {\omega x} + i \, \map \sin {\omega x} } + | c = [[Euler's Formula]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{First Isomorphism Theorem} +Tags: Named Theorems, Isomorphism Theorems + +\begin{theorem} +=== [[First Isomorphism Theorem/Groups|Groups]] === +{{:First Isomorphism Theorem/Groups}} +=== [[First Isomorphism Theorem/Rings|Rings]] === +{{:First Isomorphism Theorem/Rings}} +\end{theorem}<|endoftext|> +\section{Power Rule for Derivatives} +Tags: Derivatives, Power Rule for Derivatives + +\begin{theorem} +Let $n \in \R$. +Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as $\map f x = x^n$. +Then: +:$\map {f'} x = n x^{n - 1}$ +everywhere that $\map f x = x^n$ is defined. +When $x = 0$ and $n = 0$, $\map {f'} x$ is undefined. +\end{theorem} + +\begin{proof} +This can be done in sections. +=== [[Power Rule for Derivatives/Natural Number Index/Proof by Binomial Theorem|Proof for Natural Number Index]] === +{{:Power Rule for Derivatives/Natural Number Index/Proof by Binomial Theorem}} +=== [[Power Rule for Derivatives/Integer Index|Proof for Integer Index]] === +{{:Power Rule for Derivatives/Integer Index}} +=== [[Power Rule for Derivatives/Fractional Index/Proof 1|Proof for Fractional Index]] === +{{:Power Rule for Derivatives/Fractional Index/Proof 1}} +=== [[Power Rule for Derivatives/Rational Index|Proof for Rational Index]] === +{{:Power Rule for Derivatives/Rational Index}} +=== [[Power Rule for Derivatives/Real Number Index/Proof 1|Proof for Real Number Index]] === +{{:Power Rule for Derivatives/Real Number Index/Proof 1}} +\end{proof}<|endoftext|> +\section{Basel Problem} +Tags: Named Theorems, Riemann Zeta Function, Formulas for Pi, Basel Problem, Riemann Zeta Function at Even Integers + +\begin{theorem} +:$\displaystyle \map \zeta 2 = \sum_{n \mathop = 1}^\infty {\frac 1 {n^2} } = \frac {\pi^2} 6$ +where $\zeta$ denotes the [[Definition:Riemann Zeta Function|Riemann zeta function]]. +\end{theorem} + +\begin{proof} +By [[Riemann Zeta Function as a Multiple Integral]], +:$\displaystyle \map \zeta 2 = \int_0^1 \int_0^1 \frac 1 {1 - x y} \rd A$ +Let $\tuple {u, v} = \tuple {\dfrac {x + y} 2, \dfrac{y - x} 2}$ so that: +:$\tuple {x, y} = \tuple {u - v, u + v}$ +Let: +:$\size J = \size {\dfrac {\partial \tuple {x, y} } {\partial \tuple {u, v} } } = 2$ +Then, by [[Change of Variables Theorem (Multivariable Calculus)]]: +:$\map \zeta 2 = \displaystyle 2 \iint \limits_S \frac {\d u \rd v} {1 - u^2 + v^2}$ +where $S$ is the square defined by the coordinates: +:$\tuple {0, 0}, \ \tuple {\dfrac 1 2, -\dfrac 1 2}, \ \tuple {1, 0}, \ \tuple {\dfrac 1 2, \dfrac 1 2}$ +Exploiting the symmetry of the square and the function over the $u$-axis, we have: +:$\map \zeta 2 = \displaystyle 4 \paren {\int_0^{\frac 1 2} \! \int_0^u \frac {\d v \rd u} {1 - u^2 + v^2} + \int_{\frac 1 2}^1 \! \int_0^{1 - u} \frac {\d v \rd u} {1 - u^2 + v^2} }$ +Factoring $1 - u^2$ gives us: +:$\map \zeta 2 = \displaystyle 4 \paren {\int_0^{\frac 1 2} \! \int_0^u \frac 1 {1 - u^2} \frac {\d v \rd u} {\frac {v^2} {1 - u^2} + 1} + \int_{\frac 1 2}^1 \! \int_0^{1 - u} \frac 1 {1 - u^2} \frac {\d v \rd u} {\frac {v^2} {1 - u^2} + 1} }$ +and letting: +:$s = \dfrac v {\sqrt {1 - u^2} }, \rd s = \dfrac 1 {\sqrt {1 - u^2} }$ +allows us to make a [[Integration by Substitution|substitution into each integral]], giving: +:$\map \zeta 2 = \displaystyle 4 \paren {\int_0^{\frac 1 2} \frac 1 {\sqrt {1 - u^2} } \map \arctan {\frac u {\sqrt {1 - u^2} } } \rd u + \int_{\frac 1 2}^1 \frac 1 {\sqrt {1 - u^2} } \map \arctan {\frac {1 - u} {\sqrt {1 - u^2} } } \rd u}$ +Consider the [[Definition:Right Triangle|right triangle]] with [[Definition:Side of Polygon|sides]] $1$, $x$ and $\sqrt {1 - x^2}$. +Applying [[Pythagoras's Theorem]]: +:$\arcsin x = \arctan \dfrac x {\sqrt {1 - x^2} }$ +Let: +{{begin-eqn}} +{{eqn | l = \theta + | r = \map \arctan {\dfrac{1 - u} {\sqrt{1 - u^2} } } + | c = +}} +{{eqn | ll= \leadsto + | l = \tan^2 \theta + | r = \frac {1 - u} {1 + u} + | c = +}} +{{eqn | ll= \leadsto + | l = \sec^2 \theta + | r = \frac 2 {1 + u} + | c = [[Difference of Squares of Secant and Tangent]] +}} +{{eqn | ll= \leadsto + | l = \cos 2 \theta + | r = u + | c = [[Double Angle Formulas/Cosine/Corollary 1|Double Angle Formula for Cosine: Corollary 1]] +}} +{{eqn | ll= \leadsto + | l = \theta + | r = \dfrac 1 2 \arccos u + | c = taking [[Definition:Arccosine|arccosines]] of both sides +}} +{{eqn | r = \frac \pi 4 - \frac {\arcsin u} 2 + | c = [[Sine of Complement equals Cosine]] +}} +{{end-eqn}} +This allows us to convert the [[Definition:Arctangent|arctangents]] from the integrals into [[Definition:Arcsine|arcsines]]: +:$\map \zeta 2 = \displaystyle 4 \paren {\int_0^{\frac 1 2} {\frac {\arcsin u} {\sqrt {1 - u^2} } \rd u} + \int_{\frac 1 2}^1 {\frac 1 {\sqrt {1 - u^2} } \paren {\frac \pi 4 - \frac {\arcsin u} 2} \rd u} }$ +Substituting: +:$s = \arcsin u$, $\rd s = \dfrac 1 {\sqrt{1 - u^2} }$ +into the [[Definition:Arcsine|arcsines]], and splitting the second integral: +:$\map \zeta 2 = 4 \paren {\dfrac {\pi^2} {72} + \dfrac {\pi^2} {36} } = \dfrac {\pi^2} 6$ +{{qed}} +\end{proof} + +\begin{proof} +Let: +:$\displaystyle P_k = x \prod_{n \mathop = 1}^k \paren {1 - \frac {x^2} {n^2 \pi^2} }$ +We note that: +{{begin-eqn}} +{{eqn | l = P_k - P_{k - 1} + | r = \paren {-\frac {x^3} {k^2 \pi^2} } \prod_{n \mathop = 1}^{k - 1} \paren {1 - \frac {x^2} {n^2 \pi^2} } + | c = +}} +{{eqn | r = -\frac {x^3} {k^2 \pi ^2} + \map O {x^5} + | c = [[Definition:Big-O Notation|Big-O Notation]] +}} +{{end-eqn}} +'''Examples:''' +{{begin-eqn}} +{{eqn | l = P_2 - P_1 + | r = x \paren {1 - \frac {x^2} {\pi^2} } \paren {\paren {1 - \frac {x^2} {4 \pi^2} } - 1 } + | c = +}} +{{eqn | r = -\frac {x^3} {4 \pi ^2} + \frac {x^5} {4 \pi ^4} + | c = +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | l = P_3 - P_2 + | r = x \paren {1 - \frac {x^2} {\pi^2} } \paren {1 - \frac {x^2} {4 \pi^2} } \paren {\paren {1 - \frac {x^2} {9 \pi^2} } - 1 } + | c = +}} +{{eqn | r = -\frac {x^3} {9 \pi ^2} + \frac {5 x^5} {36 \pi ^4} - \frac {x^7} {36 \pi ^4} + | c = +}} +{{end-eqn}} +By [[Definition:Telescoping Series|Telescoping Series]] we find that the [[Definition:Coefficient|coefficient]] of $x^3$ in $P_k$ is given by: +:$(1): \quad \displaystyle -\frac 1 {\pi^2} \sum_{i \mathop = 1}^k \frac 1 {i^2}$ +We also have the following two representations of the [[Definition:Sine|Sine of x]]: +{{begin-eqn}} +{{eqn | l = \sin x + | r = x \prod_{n \mathop = 1}^\infty \paren {1 - \dfrac {x^2} {n^2 \pi^2} } + | c = [[Euler Formula for Sine Function]] +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | l = \sin x + | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \dfrac {x^3} {3!} + \dfrac {x^5} {5!} - \dfrac {x^7} {7!} + \cdots + | c = [[Power Series Expansion for Sine Function]] +}} +{{end-eqn}} +So by taking the [[Definition:Limit of Function|limit]] as $k \to \infty$ in $(1)$ and equating with the coefficient of $x^3$ in the [[Power Series Expansion for Sine Function]], we can deduce: +:$\displaystyle -\frac 1 {\pi^2} \sum_{i \mathop = 1}^{\infty} \frac 1 {i^2} = -\frac 1 {3!}$ +hence: +:$\displaystyle \sum_{i \mathop = 1}^{\infty} \frac 1 {i^2} = \frac {\pi^2} 6$ +{{qed}} +\end{proof} + +\begin{proof} +Let $x \in \openint 0 {\dfrac \pi 2}$ and let $n$ be a [[Definition:Positive Integer|non-negative integer]]. +{{begin-eqn}} +{{eqn | l = \frac {\cos \paren {2 n + 1} x + i \sin \paren {2 n + 1} x} {\sin^{2 n + 1} x} + | r = \frac {\paren {\cos x + i \sin x}^{2 n + 1} } {\sin^{2 n + 1} x} + | c = [[De Moivre's Formula]] +}} +{{eqn | r = \paren {\cot x + i}^{2 n + 1} + | c = +}} +{{eqn | r = \sum_{r \mathop = 0}^{2 n + 1} \binom {2 n + 1} r i^r \cot^{2 n + 1 - r} x + | c = [[Binomial Theorem]] +}} +{{eqn | ll= \leadsto + | l = \frac {\sin \paren {2 n + 1} x} {\sin^{2 n + 1} x} + | r = \sum_{r \mathop = 0}^n \binom {2 n + 1} {2 r + 1} \paren {-1}^r \cot^{2 \paren {n - r} } x + | c = equating [[Definition:Imaginary Part|imaginary parts]] +}} +{{end-eqn}} +Let $x_k = \dfrac {k \pi} {2 n + 1}$ for $k = 1, 2, \ldots, n$. +Then: +:$\sin \paren {2 n + 1} x_k = 0$ +So we have: +:$\displaystyle \sum_{r \mathop = 0}^n \binom {2 n + 1} {2 r + 1} \paren {-1}^r \cot^{2 \paren {n - r} } x_k = 0$ +for $k = 1, 2, \ldots, n$. +The numbers $x_k$ are all distinct and in the [[Definition:Open Real Interval|interval]] $\openint 0 {\dfrac \pi 2}$. +By [[Shape of Cotangent Function]], $\cot x$ is positive and [[Definition:Injective|injective]] in the [[Definition:Open Real Interval|interval]] $\openint 0 {\dfrac \pi 2}$. +Therefore $\cot^2 x$ is also [[Definition:Injective|injective]] in this interval. +Hence the numbers $c_k = \cot^2 x_k$ are distinct for $k = 1, 2, \ldots, n$. +These numbers are the $n$ distinct roots of the $n$th [[Definition:Degree of Polynomial|degree]] polynomial: +:$\displaystyle \map f c := \sum_{r \mathop = 0}^n \binom {2 n + 1} {2 r + 1} (-1)^r c^{n - r}$ +By [[Viète's Formulas]], we can calculate the sum of the roots: +:$\displaystyle \sum_{k \mathop = 1}^n \cot^2 x_k = \frac {\binom {2 n + 1} 3} {\binom {2 n + 1} 1} = \frac {2 n \paren {2 n - 1} } 6$ +From the [[Difference of Squares of Cosecant and Cotangent]] $\cot^2 x = \csc^2 x - 1$ we can similarly deduce: +:$\displaystyle \sum_{k \mathop = 1}^n \csc^2 x_k = \frac {2 n \paren {2 n - 1} } 6 + n = \frac {2 n \paren {2 n + 2} } 6$ +By [[Shape of Sine Function]], $\sin x$ is positive in the [[Definition:Open Real Interval|interval]] $\openint 0 {\dfrac \pi 2}$. +So from the [[Tangent Inequality]] $\sin x < x < \tan x$ for $x \in \openint 0 {\dfrac \pi 2}$, we can deduce that $\cot^2 x < 1 / x^2 < \csc^2 x$ in the same interval. +Summing this inequality from $x_1$ to $x_n$ gives: +{{begin-eqn}} +{{eqn | l = \frac {2 n \paren {2 n - 1} } 6 + | o = < + | r = \sum_{k \mathop = 1}^n \paren {\frac {2 n + 1} {k \pi} }^2 + | c = +}} +{{eqn | o = < + | r = \frac {2 n \paren {2 n + 2} } 6 + | c = +}} +{{end-eqn}} +or equivalently: +{{begin-eqn}} +{{eqn | l = \frac {\pi^2} 6 \frac {2 n \paren {2 n - 1} } {\paren {2 n + 1}^2} + | o = < + | r = \sum_{k \mathop = 1}^n \frac 1 {k^2} + | c = +}} +{{eqn | o = < + | r = \frac {\pi^2} 6 \frac {2 n \paren {2 n + 2} } {\paren {2 n + 1}^2} + | c = +}} +{{end-eqn}} +By [[Combination Theorem for Limits of Functions]], the {{LHS}} and {{RHS}} both tend to $\dfrac {\pi^2} 6$ as $n$ tends to infinity. +Therefore by the [[Squeeze Theorem]]: +:$\map \zeta 2 = \displaystyle \sum_{k \mathop = 1}^\infty \frac 1 {k^2} = \frac {\pi^2} 6$ +{{qed}} +\end{proof} + +\begin{proof} +From [[Sum of Reciprocals of Squares of Odd Integers]], +:$\displaystyle \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2n+1}^2} = \frac {\pi^2} 8$ +Note that: +{{begin-eqn}} +{{eqn | l = \sum_{n \mathop = 1}^\infty \frac 1 {n^2} + | r = \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n}^2} + \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} + | c = +}} +{{eqn | r = \frac 1 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2} + \frac {\pi^2} 8 + | c = +}} +{{eqn | ll= \leadsto + | l = \sum_{n \mathop = 1}^\infty \frac 1 {n^2} + | r = \frac {\pi^2} 6 +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +Let the [[Definition:Real Function|function]] $f_n$ be defined as: +:$(1): \quad f_n \left({x}\right) := \dfrac 1 2 + \cos x + \cos 2 x + \cdots + \cos n x$ +By [[Sum of Cosines of Multiples of Angle]]: +:$(2): \quad f_n \left({x}\right) = \dfrac {\sin \left({\left({2 n + 1}\right) x / 2}\right)} {2 \sin \left({x / 2}\right)}$ +Let $E_n$ be defined as: +{{begin-eqn}} +{{eqn | l = E_n + | r = \int_0^\pi x f_n \left({x}\right) \mathrm d x + | c = +}} +{{eqn | n = 3 + | r = \frac {\pi^2} 4 + \sum_{k \mathop = 1}^n \left({\frac {\left({-1}\right)^k - 1} {k^2} }\right) + | c = [[Primitive of x by Cosine of a x|Primitive of $x \cos a x$]] and algebra +}} +{{end-eqn}} +The terms for [[Definition:Even Integer|even]] $k$ on the {{{RHS}} are zero, so: +:$\displaystyle (4): \quad \dfrac 1 2 E_{2 n - 1} = \dfrac {\pi^2} 8 - \sum_{k \mathop = 1}^n \dfrac 1 {\left({2 k - 1}\right)^2}$ +It remains to be shown that: +:$\displaystyle \lim_{n \mathop \to \infty} E_{2 n - 1} = 0$ +which will establish: +:$\displaystyle (5): \quad \dfrac {\pi^2} 8 = \sum_{k \mathop = 1}^n \dfrac 1 {\left({2 k - 1}\right)^2}$ +Using $(2)$, let $g \left({x}\right)$ be the [[Definition:Real Function|function]] defined as: +:$g \left({x}\right) := \dfrac {\mathrm d} {\mathrm d x} \left({\dfrac {x / 2} {\sin \left({x /2}\right)} }\right)$ +Using [[Integration by Parts]], we obtain: +:$\displaystyle (6): \quad E_{2 n - 1} = \dfrac 1 {4 n - 1} \left({2 + 2 \int_0^\pi g \left({x}\right) \cos \dfrac {\left({4 n - 1}\right) x} 2 \, \mathrm d x}\right)$ +during which we use [[Limit of Sine of X over X/Corollary|Limit of $\dfrac {\sin x} x$: Corollary]]: +:$\displaystyle \lim_{x \mathop \to 0} \dfrac {x / 2} {\sin \left({x / 2}\right)} = 1$ +We have that $g \left({x}\right)$ is [[Definition:Increasing Mapping|increasing]] on the [[Definition:Closed Real Interval|interval]] of integration. +Therefore $g \left({x}\right)$ is [[Definition:Bounded Mapping|bounded]] on the [[Definition:Closed Real Interval|interval]] $\left[{ 0 \,.\,.\, \pi}\right]$ by $g \left({\pi}\right) = \dfrac \pi 2$. +Hence $(5)$ has been established as being true. +Now we divide the [[Definition:Strictly Positive Integer|(strictly) positive integers]] into [[Definition:Even Integer|even]] and [[Definition:Odd Integer|odd]], and use $(5)$ to obtain: +{{begin-eqn}} +{{eqn | l = \sum_{k \mathop = 1}^\infty \frac 1 {k^2} + | r = \sum_{k \mathop = 1}^\infty \frac 1 {\left({2 k}\right)^2} + \sum_{k \mathop = 1}^\infty \frac 1 {\left({2 k - 1}\right)^2} + | c = +}} +{{eqn | r = \frac 1 4 \sum_{k \mathop = 1}^\infty \frac 1 {k^2} + \frac {\pi^2} 8 + | c = +}} +{{eqn | ll= \leadsto + | l = \frac 3 4 \sum_{k \mathop = 1}^\infty \frac 1 {k^2} + | r = \frac {\pi^2} 8 + | c = +}} +{{eqn | ll= \leadsto + | l = \sum_{k \mathop = 1}^\infty \frac 1 {k^2} + | r = \frac 4 3 \frac {\pi^2} 8 + | c = +}} +{{eqn | r = \frac {\pi^2} 6 + | c = +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \zeta \left({2}\right) + | r = \left({-1}\right)^2 \dfrac {B_2 2^1 \pi^2} {2!} + | c = [[Riemann Zeta Function at Even Integers]] +}} +{{eqn | r = \left({-1}\right)^2 \left({\dfrac 1 6}\right) \dfrac {2^1 \pi^2} {2!} + | c = {{Defof|Sequence of Bernoulli Numbers}} +}} +{{eqn | r = \left({\dfrac 1 6}\right) \left({\dfrac 2 2}\right) \pi^2 + | c = {{Defof|Factorial}} +}} +{{eqn | r = \dfrac {\pi^2} 6 + | c = simplifying +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +By [[Fourier Series of x squared|Fourier Series of $x^2$]], for $x \in \openint {-\pi} \pi$: +:$\displaystyle x^2 = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\paren {-1}^n \frac 4 {n^2} \cos n x}$ +Letting $x \to \pi$ [[Definition:Limit from Left|from the left]]: +{{begin-eqn}} +{{eqn | l = \pi^2 + | r = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\paren {-1}^n \frac 4 {n^2} \cos \pi x} +}} +{{eqn | l = \pi^2 + | r = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\paren {-1}^n \paren {-1}^n \frac 4 {n^2} } + | c = [[Cosine of Multiple of Pi|Cosine of Multiple of $\pi$]] +}} +{{eqn | l = \pi^2 + | r = \frac {\pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2} +}} +{{eqn | ll= \leadsto + | l = \frac {2 \pi^2} 3 + | r = 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2} +}} +{{eqn | ll= \leadsto + | l = \sum_{n \mathop = 1}^\infty \frac 1 {n^2} + | r = \frac {\pi^2} 6 +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +By [[Fourier Series/Identity Function over Minus Pi to Pi|Fourier Series of Identity Function over $-\pi$ to $\pi$]], for $x \in \openint {-\pi} \pi$: +:$\displaystyle x \sim 2 \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \map \sin {n x}$ +Hence: +{{begin-eqn}} +{{eqn | l = \frac 1 \pi \int_{-\pi}^\pi x^2 \rd x + | r = \sum_{n \mathop = 1}^\infty \paren {\frac{2 \paren {-1}^{n + 1} } n}^2 + | c = [[Parseval's Theorem]] +}} +{{eqn | ll= \leadsto + | l = \frac 2 \pi \int_0^\pi x^2 \rd x + | r = \sum_{n \mathop = 1}^\infty \frac 4 {n^2} + | c = [[Definite Integral of Even Function]] +}} +{{eqn | ll=\leadsto + | l = \frac {2 \pi^2} 3 + | r = \sum_{n \mathop = 1}^\infty \frac 4 {n^2} +}} +{{eqn | ll=\leadsto + | l = \sum_{n \mathop = 1}^\infty \frac 1 {n^2} + | r = \frac {\pi^2} 6 +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +Let $\map f x$ be the [[Definition:Real Function|real function]] defined on $\openint 0 {2 \pi}$ as: +:$\map f x = \begin {cases} +\paren {x - \pi}^2 & : 0 < x \le \pi \\ +\pi^2 & : \pi < x < 2 \pi \end {cases}$ +From [[Fourier Series/Square of x minus pi, Square of pi|Fourier Series: Square of x minus pi, Square of pi]], its [[Definition:Fourier Series over Range 2 Pi|Fourier series]] can be expressed as: +:$\map f x \sim \displaystyle \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\frac {2 \cos n x} {n^2} + \paren {\frac {\paren {-1}^n \pi} n + \frac {2 \paren {\paren {-1}^n - 1} } {\pi n^3} } \sin n x}$ +Setting $x = 0$: +{{begin-eqn}} +{{eqn | l = \map f 0 + | r = \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\frac {2 \cos 0} {n^2} + \paren {\frac {\paren {-1}^n \pi} n + \frac {2 \paren {\paren {-1}^n - 1} } {\pi n^3} } \sin 0} + | c = +}} +{{eqn | ll= \leadsto + | l = \paren {0 - \pi}^2 + | r = \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\frac {2 \cos 0} {n^2} } + | c = [[Sine of Zero is Zero]] +}} +{{eqn | ll= \leadsto + | l = \pi^2 + | r = \frac {2 \pi^2} 3 + 2 \sum_{n \mathop = 1}^\infty \frac 1 {n^2} + | c = [[Cosine of Zero is One]] +}} +{{eqn | ll= \leadsto + | l = \frac {\pi^2} 2 - \frac {\pi^2} 3 + | r = \sum_{n \mathop = 1}^\infty \frac 1 {n^2} + | c = +}} +{{eqn | ll= \leadsto + | l = \frac {\pi^2} 6 + | r = \sum_{n \mathop = 1}^\infty \frac 1 {n^2} + | c = +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +From [[Mittag-Leffler Expansion for Hyperbolic Cotangent Function]], we have: +:$\displaystyle \frac 1 {2 z} \paren {\pi \map \coth {\pi z} - \frac 1 z} = \sum_{n \mathop = 1}^\infty \frac 1 {z^2 + n^2}$ +We can write: +{{begin-eqn}} +{{eqn | l = \frac 1 {2 z} \paren {\pi \map \coth {\pi z} - \frac 1 z} + | r = \frac 1 {2 z} \paren {\frac {\pi \paren {e^{\pi z} + e^{-\pi z} } } {e^{\pi z} - e^{-\pi z} } - \frac 1 z} + | c = {{Defof|Hyperbolic Cotangent}} +}} +{{eqn | r = \frac 1 {2 z} \paren {\frac {\pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1} {z \paren {e^{2 \pi z} - 1} } } +}} +{{end-eqn}} +We therefore have: +:$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2} = \lim_{z \mathop \to 0} \paren {\frac {\pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1} {2 z^2 \paren {e^{2 \pi z} - 1} } }$ +We have at $z = 0$: +{{begin-eqn}} +{{eqn | l = \pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1 + | r = \pi \times 0 \times \paren {e^0 + 1} - e^0 + 1 +}} +{{eqn | r = 1 - e^0 +}} +{{eqn | r = 0 + | c = [[Exponential of Zero]] +}} +{{end-eqn}} +and: +:$2 z^2 \paren {e^{2 \pi z} + 1} = 0$ +So by [[L'Hopital's Rule]]: +{{begin-eqn}} +{{eqn | l = \lim_{z \mathop \to 0} \paren {\frac {\pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1} {2 z^2 \paren {e^{2 \pi z} - 1} } } + | r = \lim_{z \mathop \to 0} \paren {\frac {\pi \paren {e^{2 \pi z} + 1} + 2 \pi^2 z e^{2 \pi z} - 2 \pi e^{2 \pi z} } {4 z \paren {e^{2 \pi z} - 1} + 4 \pi z^2 e^{2 \pi z} } } + | c = [[Product Rule]], [[Derivative of Exponential Function]], [[Derivative of Power]] +}} +{{eqn | r = \lim_{z \mathop \to 0} \paren {\frac {2 \pi^2 e^{2 \pi z} + 2 \pi^2 e^{2 \pi z} + 4 \pi^3 z e^{2 \pi z} - 4 \pi^2 e^{2 \pi z} } {4 \paren {e^{2 \pi z} - 1} + 8 \pi z e^{2 \pi z} + 8 \pi z e^{2 \pi z} + 8 \pi^2 z^2 e^{2 \pi z} } } + | c = [[L'Hopital's Rule]] +}} +{{eqn | r = 4 \pi^3 \lim_{z \mathop \to 0} \paren {\frac {z e^{2 \pi z} } {-4 + e^{2 \pi z} \paren {8 \pi^2 z^2 + 16 \pi z + 4} } } +}} +{{eqn | r = 4 \pi^3 \lim_{z \mathop \to 0} \paren {\frac {e^{2 \pi z} + 2 \pi z e^{2 \pi z} } {2 \pi e^{2 \pi z} \paren {8 \pi^2 z^2 + 16 \pi z + 4} + e^{2 \pi z} \paren {16 \pi z + 16 \pi} } } + | c = [[L'Hopital's Rule]] +}} +{{eqn | r = 4 \pi^3 \lim_{z \mathop \to 0} \paren {\frac {1 + 2 \pi z} {16 \pi^3 z^2 + 32 \pi^2 z + 8 \pi + 16 \pi z + 16 \pi} } +}} +{{eqn | r = \frac {4 \pi^3} {24 \pi} +}} +{{eqn | r = \frac {\pi^2} 6 +}} +{{end-eqn}} +giving: +:$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2} = \frac {\pi^2} 6$ +{{qed}} +\end{proof} + +\begin{proof} +From the [[Euler Formula for Sine Function]], we have: +:$\displaystyle \sin x = x \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} }$ +Therefore, if we divide by x, we get +:$\displaystyle \dfrac {\sin x} {x} = \dfrac {x} {x} \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} } = \paren {1 - \frac {x^2} {1^2 \pi^2} } x \paren {1 - \frac {x^2} {2^2 \pi^2} } x \paren {1 - \frac {x^2} {3^2 \pi^2} } x \cdots$ +:$\displaystyle $ +From the [[Power Series Expansion for Sine Function]], we have: +:$\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n + 1} } {\left({2 n + 1}\right)!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \frac {x^7} {7!} + \cdots$ +Therefore, if we divide by x, we get +:$\displaystyle \dfrac {\sin x} {x} = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n} } {\left({2 n + 1}\right)!} = 1 - \frac {x^2} {3!} + \frac {x^4} {5!} - \frac {x^6} {7!} + \cdots$ +:$\displaystyle $ +Equating the infinite product with the infinite sum, we have: +:$\displaystyle \dfrac {\sin x} {x} = \dfrac {x} {x} \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} } = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n} } {\left({2 n + 1}\right)!}$ +:$\displaystyle $ +:$\displaystyle \dfrac {\sin x} {x} = \paren {1 - \frac {x^2} {1 \pi^2} } x \paren {1 - \frac {x^2} {4 \pi^2} } x \paren {1 - \frac {x^2} {9 \pi^2} } x \cdots = 1 - \frac {x^2} {3!} + \frac {x^4} {5!} - \frac {x^6} {7!} + \cdots$ +:$\displaystyle $ +From this, we can see the following relationships: +:$\displaystyle - \frac {x^2} {3!} = - \frac {x^2} { \pi^2} \paren {1 + \frac {1} {4} + \frac {1} {9} + \frac {1} {16} \cdots} ($"Choose 1" - used to calculate $ \map \zeta 2)$ +:$\displaystyle $ +:$\displaystyle + \frac {x^4} {5!} = +\frac {x^4} { \pi^4} +\paren {\paren {1} \paren {\frac {1} {4}} + \paren {1} \paren {\frac {1} {9}} + \paren {1} \paren {\frac {1} {16}} + \cdots ++ \paren {\frac {1} {4}} \paren {\frac {1} {9}} + \paren {\frac {1} {4}} \paren {\frac {1} {16}} + \cdots ++ \paren {\frac {1} {9}} \paren {\frac {1} {16}} + \cdots} ($"Choose 2" - used to calculate $ \map \zeta 4)$ +:$\displaystyle $ +:$\displaystyle - \frac {x^6} {7!} = - \frac {x^6} { \pi^6} +\paren {\paren {1} \paren {\frac {1} {4}} \paren {\frac {1} {9}} + \paren {1} \paren {\frac {1} {4}} \paren {\frac {1} {16}} + \cdots + +\paren {1} \paren {\frac {1} {9}} \paren {\frac {1} {16}} + \cdots + + \paren {\frac {1} {4}} \paren {\frac {1} {9}} \paren {\frac {1} {16}} + \cdots} ($"Choose 3" - used to calculate $\map \zeta 6)$ +:$\displaystyle $ +The equations above can be used to calculate the zeta function at even integer arguments. +Notice that +"Choose $1$" is the summation of each term chosen once. +"Choose $2$" is the summation of the product of every non-repeating combination of two terms. +"Choose $3$" is the summation of the product of every non-repeating combination of three terms. +:$\displaystyle $ +For example, from "Choose $1$" above: +:$\displaystyle \map \zeta 2 = \dfrac { \pi^2} {3!}$ +{{qed}} +\end{proof}<|endoftext|> +\section{1+2+...+n+(n-1)+...+1 = n^2} +Tags: Sums of Sequences, Square Numbers, 1+2+...+n+(n-1)+...+1 = n^2 + +\begin{theorem} +:$\forall n \in \N: 1 + 2 + \cdots + n + \paren {n - 1} + \cdots + 1 = n^2$ +\end{theorem} + +\begin{proof} +Proof by [[Principle of Mathematical Induction|induction]]: +=== Basis for the Induction === +$n = 1$ holds trivially. +Just to make sure, we try $n = 2$: +:$1 + 2 + 1 = 4$ +Likewise $n^2 = 2^2 = 4$. +So shown for [[Definition:Basis for the Induction|basis for the induction]]. +=== Induction Hypothesis === +This is our [[Definition:Induction Hypothesis|induction hypothesis]]: +:$1 + 2 + \cdots + k + \paren {k - 1} + \cdots + 1 = k^2$ +Now we need to show true for $n = k + 1$: +:$1 + 2 + \cdots + \paren {k + 1} + k + \paren {k - 1} + \cdots + 1 = \paren {k + 1}^2$ +=== Induction Step === +This is our [[Definition:Induction Step|induction step]]: +{{begin-eqn}} +{{eqn | o = + | r = 1 + 2 + \cdots + \paren {k + 1} + k + \paren {k - 1} + \cdots + 1 + | c = +}} +{{eqn | r = \paren {1 + 2 + \cdots + k + \paren {k - 1} + \cdots + 1} + k + \paren {k + 1} + | c = +}} +{{eqn | r = k^2 + k + \paren {k + 1} + | c = from [[1+2+...+n+(n-1)+...+1 = n^2/Proof 3#Induction Hypothesis|induction hypothesis]] +}} +{{eqn | r = k^2 + 2k + 1 + | c = +}} +{{eqn | r = \paren {k + 1}^2 + | c = +}} +{{end-eqn}} +The result follows by [[Principle of Mathematical Induction|induction]]. +{{qed}} +\end{proof} + +\begin{proof} +{{begin-eqn}} +{{eqn | o = + | r = 1 + 2 + \cdots + \paren {n - 1} + n + \paren {n - 1} + \cdots + 1 + | c = +}} +{{eqn | r = 1 + 2 + \cdots + \paren {n - 1} + \paren {n - 1} + \cdots + 1 + n + | c = +}} +{{eqn | r = 2 \paren {1 + 2 + \cdots + \paren {n - 1} } + n + | c = +}} +{{eqn | r = 2 \paren {\frac {\paren {n - 1} n} 2} + n + | c = [[Closed Form for Triangular Numbers]] +}} +{{eqn | r = \paren {n - 1} n + n + | c = +}} +{{eqn | r = n^2 - n + n + | c = +}} +{{eqn | r = n^2 + | c = +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +{{begin-eqn}} +{{eqn | o = + | r = 1 + 2 + \cdots + \paren {n - 1} + n + \paren {n - 1} + \cdots + 1 + | c = +}} +{{eqn | r = \paren {1 + 2 + \cdots + \paren {n - 1} } + \paren {1 + 2 + \cdots + \paren {n - 1} + n} + | c = +}} +{{eqn | r = \frac {\paren {n - 1} n} 2 + \frac {n \paren {n + 1} } 2 + | c = [[Closed Form for Triangular Numbers]] +}} +{{eqn | r = \frac {n^2 - n + n^2 + n} 2 + | c = +}} +{{eqn | r = n^2 + | c = +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +Let $T_n = 1 + 2 + \cdots + n + \left({n - 1}\right) + \cdots + 1$. +We have $T_1 = 1$ +and +{{begin-eqn}} +{{eqn | l = T_n - T_{n-1} + | r = \left({1 + 2 + \cdots + n + \left({n - 1}\right) + \cdots + 1 }\right) + | c = Definition of $T_n$ +}} +{{eqn | o = + | ro= - + | r = \left({1 + 2 + \cdots + \left({n - 1}\right) + \left({n - 2}\right) + \cdots + 1}\right) + | c = +}} +{{eqn | r = \left({\left({1 + 2 + \cdots + n}\right) - \left({1 + 2 + \cdots + \left({n - 1}\right)}\right)}\right) + | c = [[Integer Addition is Associative]] +}} +{{eqn | o = + | ro= + + | r = \left({\left({\left({n - 1}\right) + \left({n - 2}\right) + \cdots + 1}\right) - \left({\left({n - 2}\right) + \left({n - 3}\right) + \cdots + 1}\right)}\right) + | c = [[Integer Addition is Commutative]] +}} +{{eqn | r = n + \left({n - 1}\right) + | c = simplifying +}} +{{eqn | r = 2 n - 1 + | c = +}} +{{end-eqn}} +Thus we have: +{{begin-eqn}} +{{eqn | l = T_n + | r = \left({T_n - T_{n-1} }\right) + \left({T_{n-1} - T_{n-2} }\right) + \cdots + \left({T_2 - T_1}\right) + T_1 +}} +{{eqn | r = \left({2 n - 1}\right) + \left({2 \left({n - 1}\right) - 1}\right) + \cdots + \left({2 \times 2 - 1}\right) + 1 +}} +{{eqn | r = \sum_{k \mathop = 1}^n 2 k - 1 +}} +{{eqn | r = n^2 + | c = [[Odd Number Theorem]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Necessary Conditions for Existence of Skolem Sequence} +Tags: Combinatorics + +\begin{theorem} +A [[Definition:Skolem Sequence|Skolem sequence]] of order $n$ can only exist if $n \equiv 0, 1 \pmod 4$. +\end{theorem} + +\begin{proof} +Let $S$ be a [[Definition:Skolem Sequence|Skolem sequence]] of order $n$. +Let $a_i$ and $b_i$ be the positions of the first and second occurrences respectively of the integer $i$ in $S$, where $1 \le i \le n$. +We can thus conclude that: +:$b_i - a_i = i$ +for each $i$ from 1 to $n$. +Summing both sides of this equation we obtain: +:$\displaystyle \sum_i^n b_i - \sum_i^n a_i = \sum_i^n i = \frac {n \paren {n + 1} } 2$ +Now the $a_i$ and the $b_i$ represent the positions in the sequence from $1$ to $2 n$. +Hence: +:$\displaystyle \sum_i^n b_i + \sum_i^n a_i = \frac {2 n \paren {2 n + 1} } 2 = n \paren {2 n + 1}$ +Summing the previous two equations we obtain the identity: +:$\displaystyle \sum_i^n b_i = \frac {n \paren {5 n + 3} } 4$ +The {{LHS}} of this last equality is a sum of positions and thus must be an [[Definition:Integer|integer]]. +We conclude that the [[Definition:RHS|right hand side]] must also be an [[Definition:Integer|integer]]. +This occurs exactly when: +$n \equiv 0, 1 \pmod 4$ +{{qed}} +\end{proof}<|endoftext|> +\section{Divisibility by 9} +Tags: Divisibility Tests, 9, Divisibility by 9 + +\begin{theorem} +A number expressed in [[Definition:Decimal Notation|decimal notation]] is [[Definition:Divisor of Integer|divisible]] by $9$ {{iff}} the sum of its [[Definition:Digit|digits]] is [[Definition:Divisor of Integer|divisible]] by $9$. +That is: +:$N = \sqbrk {a_0 a_1 a_2 \ldots a_n}_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is [[Definition:Divisor of Integer|divisible]] by $9$ +{{iff}}: +:$a_0 + a_1 + \ldots + a_n$ is [[Definition:Divisor of Integer|divisible]] by $9$. +\end{theorem} + +\begin{proof} +Let $N$ be [[Definition:Divisor of Integer|divisible]] by $9$. +Then: +{{begin-eqn}} +{{eqn | l = N + | o = \equiv + | r = 0 \pmod 9 +}} +{{eqn | ll= \leadstoandfrom + | l = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n + | o = \equiv + | r = 0 \pmod 9 +}} +{{eqn | ll= \leadstoandfrom + | l = a_0 + a_1 1 + a_2 1^2 + \cdots + a_n 1^n + | o = \equiv + | r = 0 \pmod 9 + | c = as $10 \equiv 1 \pmod 9$ +}} +{{eqn | ll= \leadstoandfrom + | l = a_0 + a_1 + \cdots + a_n + | o = \equiv + | r = 0 \pmod 9 +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +This is a special case of [[Congruence of Sum of Digits to Base Less 1]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Existence of Rational Powers of Irrational Numbers} +Tags: Number Theory, Roots of Numbers, Existence of Rational Powers of Irrational Numbers + +\begin{theorem} +There exist [[Definition:Irrational Number|irrational numbers]] $a$ and $b$ such that $a^b$ is [[Definition:Rational Number|rational]]. +\end{theorem} + +\begin{proof} +We have that: +: [[Square Root of 2 is Irrational|$\sqrt 2$ is irrational]]. +: $2$ is trivially rational, as $2 = \dfrac 2 1$. +Consider the number $q = \sqrt 2^{\sqrt 2}$, which is [[Definition:Irrational Number|irrational]] by the [[Gelfond-Schneider Theorem]]. +Thus: +: $q^{\sqrt 2} = \left({\sqrt 2^{\sqrt 2}}\right)^{\sqrt 2} = \sqrt 2 ^{\left({\sqrt 2}\right) \left({\sqrt 2}\right)} = \sqrt 2^2 = 2$ +is [[Definition:Rational Number|rational]]. +So $a = q = \sqrt 2^{\sqrt 2}$ and $b = \sqrt 2$ are the desired [[Definition:Irrational Number|irrationals]]. +{{qed}} +\end{proof} + +\begin{proof} +Given that $2$ is [[Definition:Rational Number|rational]] and [[Square Root of 2 is Irrational|$\sqrt 2$ is irrational]], consider the number $q = \sqrt 2^{\sqrt 2}$. +We consider the two cases. +:$(1): \quad$ If $q$ is [[Definition:Rational Number|rational]] then $a = \sqrt 2$ and $b = \sqrt 2$ are the desired [[Definition:Irrational Number|irrational numbers]]. +:$(2): \quad$ If $q$ is [[Definition:Irrational Number|irrational]] then $q^{\sqrt 2} = \paren {\sqrt 2^{\sqrt 2} }^{\sqrt 2} = \sqrt 2^{\paren {\sqrt 2} \paren {\sqrt 2} } = \sqrt 2^2 = 2$ is [[Definition:Rational Number|rational]], so $a = q = \sqrt2^{\sqrt 2}$ and $b = \sqrt 2$ are the desired [[Definition:Irrational Number|irrational numbers]]. +{{qed}} +\end{proof} + +\begin{proof} +Consider the equation: +:$\left({\sqrt 2}\right)^{\log_{\sqrt 2} 3} = 3$ +We have that [[Square Root of 2 is Irrational|$\sqrt 2$ is irrational]] and $3$ is (trivially) [[Definition:Rational Number|rational]]. +It remains to be proved $\log_{\sqrt 2} 3$ is [[Definition:Irrational Number|irrational]]. +We have: +{{begin-eqn}} +{{eqn | l = \log_{\sqrt 2} 3 + | r = \frac{\log_2 3}{\log_2 \sqrt 2} + | c = [[Change of Base of Logarithm]] +}} +{{eqn | r = \frac{\log_2 3}{1/2} + | c = [[Logarithms of Powers]]: $\sqrt 2 = 2^{1/2}$ and $\log_2 2 = 1$ +}} +{{eqn | r = 2 \, \log_2 3 + | c = +}} +{{end-eqn}} +From [[Irrationality of Logarithm]], $\log_2 3$ is irrational. +Therefore $2 \, \log_2 3$ is [[Definition:Irrational Number|irrational]]. +Thus there exist two [[Definition:Irrational Number|irrational numbers]] $a = \sqrt 2$ and $b = \log_{\sqrt 2} 3$ such that $a^b$ is [[Definition:Rational Number|rational]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Intersection is Associative} +Tags: Direct Proofs, Set Intersection, Associativity, Intersection is Associative + +\begin{theorem} +[[Definition:Set Intersection|Set intersection]] is [[Definition:Associative Operation|associative]]: +:$A \cap \paren {B \cap C} = \paren {A \cap B} \cap C$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | o = + | r = x \in A \cap \paren {B \cap C} +}} +{{eqn | o = \leadstoandfrom + | r = x \in A \land \paren {x \in B \land x \in C} + | c = {{Defof|Set Intersection}} +}} +{{eqn | o = \leadstoandfrom + | r = \paren {x \in A \land x \in B} \land x \in C + | c = [[Rule of Association/Conjunction|Rule of Association: Conjunction]] +}} +{{eqn | o = \leadstoandfrom + | r = x \in \paren {A \cap B} \cap C + | c = {{Defof|Set Intersection}} +}} +{{end-eqn}} +Therefore: +:$x \in A \cap \paren {B \cap C}$ {{iff}} $x \in \paren {A \cap B} \cap C$ +Thus it has been shown that: +:$A \cap \paren {B \cap C}\ = \paren {A \cap B} \cap C$ +{{qed}} +\end{proof}<|endoftext|> +\section{Center of Symmetric Group is Trivial} +Tags: Symmetric Groups, Center of Symmetric Group is Trivial, Centers of Groups + +\begin{theorem} +Let $n \in \N$ be a [[Definition:Natural Number|natural number]]. +Let $S_n$ denote the [[Definition:Symmetric Group|symmetric group]] of [[Definition:Order of Structure|order]] $n$. +Let $n \ge 3$. +Then the [[Definition:Center of Group|center]] $\map Z {S_n}$ of $S_n$ is [[Definition:Trivial Group|trivial]]. +\end{theorem} + +\begin{proof} +From its definition, the [[Definition:Identity Element|identity]] (here denoted by $e$) of a [[Definition:Group|group]] $G$ [[Definition:Commute|commutes]] with all [[Definition:Element|elements]] of $G$. +So by definition of [[Definition:Center of Group|center]]: +:$e \in \map Z {S_n}$ +By definition of [[Definition:Center of Group|center]]: +:$\map Z {S_n} = \set {\tau \in S_n: \forall \sigma \in S_n: \tau \sigma = \sigma \tau}$ +Let $\pi, \rho \in S_n$ be [[Definition:Permutation|permutations]] of $\N_n$. +Let us choose an arbitrary $\pi \in S_n: \pi \ne e, \map \pi i = j, i \ne j$. +Since $n \ge 3$, we can find $\rho \in S_n$ which interchanges $j$ and $k$ (where $k \ne i, j$) and [[Definition:Fixed Element of Permutation|fixes]] everything else. +It follows that $\rho^{-1}$ does the same thing, and in particular both $\rho$ and $\rho^{-1}$ [[Definition:Fixed Element of Permutation|fix]] $i$. +So: +{{begin-eqn}} +{{eqn | l = \map {\rho \pi \rho^{-1} } i + | r = \map {\rho \pi} i + | c = +}} +{{eqn | r = \map \rho j + | c = +}} +{{eqn | r = k + | c = +}} +{{end-eqn}} +So: +:$\map {\rho \pi \rho^{-1} } i = k \ne j = \map \pi i$ +From [[Conjugate of Commuting Elements]], if $\rho$ and $\pi$ were to [[Definition:Commute|commute]], $\rho \pi \rho^{-1} = \pi$. +But they don't. +Whatever $\pi \in S_n$ is, you can ''always'' find a $\rho$ such that $\rho \pi \rho^{-1} \ne \pi$. +So no non-identity elements of $S_n$ [[Definition:Commute|commute]] with all elements of $S_n$. +Hence: +:$\map Z {S_n} = \set e$ +{{Qed}} +\end{proof} + +\begin{proof} +Let us choose an arbitrary $\pi \in S_n: \pi \ne e, \pi \left({i}\right) = j, i \ne j$. +Then $\pi \left({j}\right) \ne j$ since [[Definition:Permutation|permutations]] are [[Definition:Injection|injective]]. +Since $n \ge 3$, we can find $ k \ne j, k \ne \pi \left({j}\right)$ and $\rho \in S_n$ which interchanges $j$ and $k$ and [[Definition:Fixed Element of Permutation|fixes]] everything else. +Let $\pi \left({j}\right) = m$. Then $m \ne j, m \ne k$ so $\rho$ [[Definition:Fixed Element of Permutation|fixes]] $m$. +Then $ \rho \pi \left({j}\right) = \rho \left({m}\right) = m = \pi \left({j}\right)$ since $\rho$ [[Definition:Fixed Element of Permutation|fixes]] $m$. +Now $k = \rho \left({j}\right)$ by definition of $\rho$. +So $\pi \left({k}\right) = \pi \rho \left({j}\right)$. +But $\pi \left({j}\right) \ne \pi \left({k}\right)$ since [[Definition:Permutation|permutations]] are [[Definition:Injection|injective]]. +Thus $\rho \pi \left({j}\right) \ne \pi \rho \left({j}\right)$. +So arbitrary $\pi \ne e$ is not in the [[Definition:Center of Group|center]] since there exists a $\rho$ with which $\pi$ does not [[Definition:Commute|commute]]. +Thus only $e$ is in the [[Definition:Center of Group|center]], which, by definition, is [[Definition:Trivial Group|trivial]]. +{{Qed}} +\end{proof}<|endoftext|> +\section{Square Root of 2 is Irrational} +Tags: Square Root of 2, Irrationality Proofs, Square Root of 2 is Irrational + +\begin{theorem} +:$\sqrt 2$ is [[Definition:Irrational Number|irrational]]. +\end{theorem} + +\begin{proof} +First we note that, from [[Parity of Integer equals Parity of its Square]], if an [[Definition:Integer|integer]] is [[Definition:Even Integer|even]], its [[Definition:Square Root|square root]], if an [[Definition:Integer|integer]], is also [[Definition:Even Integer|even]]. +Thus it follows that: +:$(1): \quad 2 \divides p^2 \implies 2 \divides p$ +where $2 \divides p$ indicates that $2$ is a [[Definition:Divisor of Integer|divisor]] of $p$. +{{AimForCont}} that $\sqrt 2$ is [[Definition:Rational Number|rational]]. +So: +:$\sqrt 2 = \dfrac p q$ +for some $p, q \in \Z$, and: +:$\gcd \set {p, q} = 1$ +where $\gcd$ denotes the [[Definition:Greatest Common Divisor|greatest common divisor]]. +Squaring both sides yields: +:$2 = \dfrac {p^2} {q^2} \iff p^2 = 2 q^2$ +Therefore from $(1)$: +:$2 \divides p^2 \implies 2 \divides p$ +That is, $p$ is an [[Definition:Even Integer|even integer]]. +So $p = 2 k$ for some $k \in \Z$. +Thus: +:$2 q^2 = p^2 = \paren {2 k}^2 = 4 k^2 \implies q^2 = 2 k^2$ +so by the same reasoning: +:$2 \divides q^2 \implies 2 \divides q$ +This contradicts our assumption that $\gcd \set {p, q} = 1$, since $2 \divides p, q$. +Therefore, from [[Proof by Contradiction]], $\sqrt 2$ cannot be [[Definition:Rational Number|rational]]. +{{qed}} +\end{proof} + +\begin{proof} +Special case of [[Square Root of Prime is Irrational]]. +{{qed}} +\end{proof} + +\begin{proof} +{{AimForCont}} that $\sqrt 2$ is [[Definition:Rational Number|rational]]. +Then $\sqrt 2 = \dfrac p q$ for some $p, q \in \Z_{>0}$ +Consider the quantity $\paren {\sqrt 2 - 1}$: +{{begin-eqn}} +{{eqn | ll = 1 + | l = < + | o = \sqrt 2 + | r = < + | rr = 2 + | c = [[Ordering of Squares in Reals]] +}} +{{eqn | lll = \leadsto + | ll = 0 + | l = < + | o = \sqrt 2 - 1 + | r = < + | rr = 1 +}} +{{end-eqn}} +Now, observe that for any $n \in \Z_{>0}$: +{{begin-eqn}} +{{eqn | l = \paren {\sqrt 2 - 1}^n + | r = \sum_{k \mathop = 0}^n \binom n k \paren {\sqrt 2}^k \paren {-1}^{n-k} + | c = [[Binomial Theorem]] +}} +{{eqn | r = \sum_{\substack {0 \mathop \le k \mathop \le n \\ k \, \text{even} } } \binom n k 2^{k/2} \paren {-1}^{n - k} + \sqrt 2 \sum_{\substack {0 \mathop \le k \mathop \le n \\ k \, \text{odd} } } \binom n k 2^{\paren {k - 1}/2} \paren {-1}^{n-k} +}} +{{eqn | r = a_n + b_n \sqrt 2 + | c = for some [[Definition:Integer|integers]] $a_n, b_n$ +}} +{{eqn | r = a_n + b_n \paren {\frac p q} + | c = recalling the assumption that $\sqrt 2 = \dfrac p q$ +}} +{{eqn | r = \frac {a_n q + b_n p} q +}} +{{eqn | o = \ge + | r = \frac 1 q + | c = as the [[Definition:Numerator|numerator]] is an integer and $\sqrt 2 - 1 > 0$ +}} +{{end-eqn}} +By [[Sequence of Powers of Number less than One]]: +:$\displaystyle \lim_{n \mathop \to \infty} \paren {\sqrt 2 - 1}^n = 0$ +where $\lim$ denotes [[Definition:Limit of Real Sequence|limit]]. +Recall the definition of $a_n$ and $b_n$. +By [[Lower and Upper Bounds for Sequences]]: +:$0 = \displaystyle \lim_{n \mathop \to \infty} \frac {a_n q + b_n p} q \ge \frac 1 q$ +which is a [[Proof by Contradiction|contradiction]]. +{{qed}} +\end{proof} + +\begin{proof} +{{AimForCont}} that $\sqrt 2$ is [[Definition:Rational Number|rational]]. +Let $n$ be the smallest [[Definition:Positive Integer|positive integer]] such that: +:$\sqrt 2 = \dfrac m n$ +for some $m \in \Z_{>0}$ +Then: +:$m = n \sqrt2 > n$ +so: +:$(1): \quad m - n > 0$ +We also have: +:$m = n \sqrt2 < 2 n$ +so: +:$m < 2 n$ +and therefore: +:$(2): \quad m - n < n$ +Finally, we have +{{begin-eqn}} +{{eqn | l = m^2 + | r = 2 n^2 +}} +{{eqn | ll= \leadsto + | l = m^2 - n m + | r = 2 n^2 - n m +}} +{{eqn | ll= \leadsto + | l = m \paren {m - n} + | r = \paren {2 n - m} n +}} +{{eqn | ll= \leadsto + | l = \dfrac {2 n - m} {m - n} + | r = \dfrac m n +}} +{{end-eqn}} +It follows that: +:$\dfrac {2 n - m} {m - n} = \sqrt2$ +By $(1)$, the [[Definition:Denominator|denominator]] of $\dfrac {2 n - m} {m - n}$ is [[Definition:Positive Real Number|positive]]. +By $(2)$, the denominator of $\dfrac {2 n - m} {m - n}$ is less than $n$. +We have thus written $\sqrt2$ as a [[Definition:Fraction|fraction ]] with a smaller [[Definition:Denominator|denominator]] than $n$, which is a [[Proof by Contradiction|contradiction]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Empty Set is Unique} +Tags: Empty Set, Empty Set is Unique + +\begin{theorem} +The [[Definition:Empty Set|empty set]] is [[Definition:Unique|unique]]. +\end{theorem} + +\begin{proof} +Let $\O$ and $\O'$ both be [[Definition:Empty Set|empty sets]]. +From [[Empty Set is Subset of All Sets]], $\O \subseteq \O'$, because $\O$ is [[Definition:Empty Set|empty]]. +Likewise, we have $\O' \subseteq \O$, since $\O'$ is [[Definition:Empty Set|empty]]. +Together, by the definition of [[Definition:Set Equality/Definition 2|set equality]], this implies that $\O = \O'$. +Thus there is only one [[Definition:Empty Set|empty set]]. +{{Qed}} +\end{proof} + +\begin{proof} +Let $A$ and $B$ both be [[Definition:Empty Set|empty sets]]. +Thus: +:$\forall x: \neg \paren {x \in A}$ +and: +:$\forall x: \neg \paren {x \in B}$ +Hence: +:$x \notin A \iff x \notin B$ +and so: +:$x \in A \iff x \in B$ +[[Definition:Vacuous Truth|vacuously]]. +From [[Axiom:Axiom of Extension|Axiom of Extension]]: +:$\forall x: \paren {x \in A \iff x \in B} \iff A = B$ +Hence the result. +{{Qed}} +\end{proof}<|endoftext|> +\section{Principle of Non-Contradiction} +Tags: Proof Rules, Contradiction, Principle of Non-Contradiction + +\begin{theorem} +The '''Principle of Non-Contradiction''' is a [[Definition:Valid Argument|valid]] deduction [[Definition:Sequent|sequent]] in [[Definition:Propositional Logic|propositional logic]]. +\end{theorem}<|endoftext|> +\section{Rule of Substitution} +Tags: Rule of Substitution + +\begin{theorem} +Let $S$ be a [[Definition:Sequent|sequent]] that has been [[Definition:Proof|proved]]. +Then a [[Definition:Proof|proof]] can be found for any [[Definition:Substitution Instance|substitution instance]] of $S$. +\end{theorem} + +\begin{proof} +This is apparent from inspection of the [[Definition:Proof Rule|proof rules]] themselves. +The rules concern only the broad structure of the [[Definition:Propositional Formula|propositional formulas]] involved, and this structure is unaffected by substitution. +By performing the substitutions systematically throughout the given [[Definition:Sequent|sequent]], all applications of [[Definition:Proof Rule|proof rules]] remain correct applications in the [[Definition:Sequent|sequent]]. +{{Handwaving}} +\end{proof}<|endoftext|> +\section{Rule of Sequent Introduction} +Tags: Propositional Logic, Rule of Sequent Introduction + +\begin{theorem} +Let the statements $P_1, P_2, \ldots, P_n$ be [[Definition:Conclusion|conclusions]] in a [[Definition:Proof|proof]], on various [[Definition:Assumption|assumptions]]. +Let $P_1, P_2, \ldots, P_n \vdash Q$ be a [[Definition:Substitution Instance|substitution instance]] of a [[Definition:Sequent|sequent]] for which we already have a proof. +{{explain|Question the use of "substitution instance": can we not "just" allow for $P_1, \ldots, P_n$ to be "just" statements?}} +Then we may introduce, at any stage of a proof (citing '''SI'''), either: +:The [[Definition:Conclusion|conclusion]] $Q$ of the [[Definition:Sequent|sequent]] already proved +or: +:A [[Definition:Substitution Instance|substitution instance]] of such a [[Definition:Conclusion|conclusion]], together with a reference to the [[Definition:Sequent|sequent]] that is being cited. +This conclusion depend upon the [[Definition:Pool of Assumptions|pool of assumptions]] upon which $P_1, P_2, \ldots, P_n \vdash Q$ rests. +\end{theorem} + +\begin{proof} +[[Definition:By Hypothesis|By hypothesis]] and [[Definition:Substitution Instance|substitution instance]] we have a proof, using primitive rules, of: +: $P_1, P_2, \ldots, P_n \vdash Q$ +By the [[Extended Rule of Implication]], we have: +: $\vdash P_1 \implies \left({P_2 \implies \left({P_3 \implies \left({\ldots \implies \left({P_n \implies Q}\right) \ldots }\right)}\right)}\right)$ +{{Qed}} +\end{proof}<|endoftext|> +\section{Law of Identity} +Tags: Propositional Logic, Law of Identity + +\begin{theorem} +Every [[Definition:Proposition|proposition]] entails itself: +\end{theorem}<|endoftext|> +\section{Rule of Idempotence} +Tags: Conjunction, Disjunction + +\begin{theorem} +The '''rule of idempotence''' is two-fold: +\end{theorem}<|endoftext|> +\section{Hypothetical Syllogism} +Tags: Implication + +\begin{theorem} +The '''(rule of the) hypothetical syllogism''' is a [[Definition:Valid Argument|valid]] deduction [[Definition:Sequent|sequent]] in [[Definition:Propositional Logic|propositional logic]]: +:If we can conclude that $p$ implies $q$, and if we can also conclude that $q$ implies $r$, then we may infer that $p$ implies $r$. +\end{theorem}<|endoftext|> +\section{Extended Rule of Implication} +Tags: Implication, Conjunction + +\begin{theorem} +Any [[Definition:Sequent|sequent]] can be expressed as a [[Definition:Theorem of Logic|theorem]]. +That is: +:$P_1, P_2, P_3, \ldots, P_n \vdash Q$ +means the same thing as: +:$\vdash P_1 \implies \paren {P_2 \implies \paren {P_3 \implies \paren {\ldots \implies \paren {P_n \implies Q} \ldots} } }$ +The latter expression is known as the [[Definition:Corresponding Conditional|corresponding conditional]] of the former. +Thus every [[Definition:Sequent|sequent]] containing the [[Definition:Symbol|symbol]] $\vdash$ can, if so desired, be expressed in the form of a [[Definition:Theorem of Logic|theorem]] which has $\implies$. +\end{theorem} + +\begin{proof} +From the [[Rule of Conjunction/Proof Rule|Rule of Conjunction]], we note the following. +Any sequent: +:$P_1, P_2 \vdash Q$ +can be expressed as: +:$P_1 \land p_2 \vdash Q$ +Also, from the [[Rule of Simplification/Proof Rule|Rule of Simplification]], any [[Definition:Sequent|sequent]]: +:$P_1 \land P_2 \vdash Q$ +can be expressed as: +:$P_1, P_2 \vdash Q$ +Consider the expression: +:$P_1, P_2, P_3, \ldots, P_{n - 1}, P_n \vdash Q$ +By repeated application of the above, we can arrive at: +:$P_1 \land \paren {P_2 \land \paren {P_3 \land \paren {\ldots \land \paren {P_{n - 1} \land P_n} \ldots} } } \vdash Q$ +For convenience, the [[Rule of Substitution]] can be used to substitute $R_1$ for: +:$P_2 \land \paren {P_3 \land \paren {\ldots \land \paren {P_{n - 1} \land P_n} \ldots} }$ +to get: +:$P_1 \land R_1 \vdash Q$ +From the [[Rule of Implication]]: +:$\vdash \paren {P_1 \land R_1} \implies Q$ +Using the [[Rule of Exportation]], we then get: +:$\vdash P_1 \implies \paren {R_1 \implies Q}$ +Substituting back for $R_1$: +:$\vdash P_1 \implies \paren {\paren {P_2 \land \paren {P_3 \land \paren {\ldots \land \paren {P_{n - 1} \land P_n} \ldots} } } \implies Q}$ +Making a [[Rule of Substitution|substitution]] of convenience again, substitute $R_2$ for: +:$P_3 \land \paren {\ldots \land \paren {P_{n - 1} \land P_n} \ldots}$ +Take the expression: +:$\paren {P_2 \land R_2} \implies Q$ +and express it as: +:$P_2 \land R_2 \vdash Q$ +and use the [[Rule of Exportation]] to get: +:$P_2 \implies \paren {R_2 \implies Q}$ +which, substituting back for $R_2$, gives: +:$P_2 \implies \paren {\paren {P_3 \land \paren {\ldots \land \paren {P_{n - 1} \land P_n} \ldots} } \right) \implies Q}$ +Similarly: +:$\paren {P_3 \land \paren {\ldots \land \paren {P_{n - 1} \land P_n} \ldots} } \implies Q$ +converts to: +:$P_3 \implies \paren {\paren {\ldots \land \paren {P_{n - 1} \land P_n} \ldots} \implies Q}$ +The pattern becomes apparent. +Eventually: +:$\paren {P_{n - 1} \land P_n} \implies Q$ +which converts to: +:$P_{n - 1} \implies \paren {P_n \implies Q}$ +Substituting these back into our original expression: +:$\vdash P_1 \implies \paren {P_2 \implies \paren {P_3 \implies \paren {\ldots \implies \paren {P_n \implies Q} \ldots} } }$ +{{qed}} +\end{proof}<|endoftext|> +\section{Equivalences are Interderivable} +Tags: Biconditional, Equivalences are Interderivable + +\begin{theorem} +If two [[Definition:Propositional Formula|propositional formulas]] are [[Definition:Interderivable|interderivable]], they are [[Definition:Biconditional|equivalent]]: +:$\paren {p \dashv \vdash q} \dashv \vdash \paren {p \iff q}$ +\end{theorem} + +\begin{proof} +The result follows directly from the [[Definition:Truth Table|truth table]] for the [[Definition:Biconditional|biconditional]]: +$\begin{array}{|cc||ccc|} \hline +p & q & p & \iff & q \\ +\hline +F & F & F & T & F \\ +F & T & F & F & T \\ +T & F & F & F & F \\ +T & T & F & T & T \\ +\hline +\end{array}$ +By inspection, it is seen that $\mathcal M \left({p \iff q}\right) = T$ precisely when $\mathcal M \left({p}\right) = \mathcal M \left({q}\right)$. +{{qed}} +\end{proof} + +\begin{proof} +Let $v$ be an arbitrary [[Definition:Boolean Interpretation|interpretation]]. +Then by definition of [[Definition:Interderivable|interderivable]]: +:$\map v {p \iff q}$ {{iff}} $\map v p = \map v q$ +Since $v$ is arbitrary, $\map v p = \map v q$ holds in all [[Definition:Boolean Interpretation|interpretations]]. +That is: +:$p \dashv \vdash q$ +{{qed}} +\end{proof}<|endoftext|> +\section{Union is Commutative} +Tags: Set Union, Commutativity, Union is Commutative + +\begin{theorem} +[[Definition:Set Union|Set union]] is [[Definition:Commutative Operation|commutative]]: +:$S \cup T = T \cup S$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = x + | o = \in + | r = \paren {S \cup T} + | c = +}} +{{eqn | ll= \leadstoandfrom + | l = x \in S + | o = \lor + | r = x \in T + | c = {{Defof|Set Union}} +}} +{{eqn | ll= \leadstoandfrom + | l = x \in T + | o = \lor + | r = x \in S + | c = [[Disjunction is Commutative]] +}} +{{eqn | ll= \leadstoandfrom + | l = x + | o = \in + | r = \paren {T \cup S} + | c = {{Defof|Set Union}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Intersection is Commutative} +Tags: Set Intersection, Commutativity, Intersection is Commutative + +\begin{theorem} +[[Definition:Set Intersection|Set intersection]] is [[Definition:Commutative Operation|commutative]]: +:$S \cap T = T \cap S$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = x + | o = \in + | r = \paren {S \cap T} +}} +{{eqn | ll= \leadstoandfrom + | l = x \in S + | o = \land + | r = x \in T + | c = {{Defof|Set Intersection}} +}} +{{eqn | ll= \leadstoandfrom + | l = x \in T + | o = \land + | r = x \in S + | c = [[Conjunction is Commutative]] +}} +{{eqn | ll= \leadstoandfrom + | l = x + | o = \in + | r = \paren {T \cap S} + | c = {{Defof|Set Intersection}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Russell's Paradox} +Tags: Naive Set Theory, Subsets, Antinomies, Russell's Paradox + +\begin{theorem} +The [[Axiom:Comprehension Principle|comprehension principle]] leads to a [[Definition:Contradiction|contradiction]]. +\end{theorem} + +\begin{proof} +[[Definition:Set|Sets]] have [[Definition:Element|elements]]. +Some of those elements may themselves be sets. +So, given two sets $S$ and $T$, we can ask the question: Is $S$ an element of $T$? The answer will either be ''yes'' or ''no''. +In particular, given any set $S$, we can ask the question: Is $S$ an element of $S$? Again, the answer will either be ''yes'' or ''no''. +Thus, $\map P S = S \in S$ is a [[Definition:Property|property]] on which we can use the [[Axiom:Comprehension Principle|comprehension principle]] to build this set: +:$T = \set {S: S \in S}$ +which is the set of all sets which contain themselves. +Or we can apply the [[Axiom:Comprehension Principle|comprehension principle]] to build this set: +:$R = \set {S: S \notin S}$ +($R$ for {{AuthorRef|Bertrand Russell|Russell}}, of course.) +We ask the question: Is $R$ itself an element of $R$? +There are two possible answers: ''yes'' or ''no''. +If $R \in R$, then $R$ must satisfy the property that $R \notin R$, so from that [[Definition:Contradiction|contradiction]] we know that $R \in R$ does not hold. +So the only other answer, $R \notin R$, must hold instead. But now we see that $R$ satisfies the conditions of the property that $R \in R$, so we can see that $R \notin R$ does not hold either. +Thus we have generated a [[Definition:Contradiction|contradiction]] from the [[Axiom:Comprehension Principle|comprehension principle]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Destructive Dilemma/Formulation 1} +Tags: Destructive Dilemma + +\begin{theorem} +: $p \implies q, r \implies s \vdash \neg q \lor \neg s \implies \neg p \lor \neg r$ +\end{theorem}<|endoftext|> +\section{De Morgan's Laws (Logic)} +Tags: Conjunction, Disjunction, De Morgan's Laws (Logic) + +\begin{theorem} +{{:De Morgan's Laws (Logic)/Disjunction of Negations}} +\end{theorem}<|endoftext|> +\section{Solution to Quadratic Equation} +Tags: Quadratic Equations + +\begin{theorem} +The [[Definition:Quadratic Equation|quadratic equation]] of the form $a x^2 + b x + c = 0$ has solutions: +:$x = \dfrac {-b \pm \sqrt {b^2 - 4 a c} } {2 a}$ +\end{theorem} + +\begin{proof} +Let $a x^2 + b x + c = 0$. Then: +{{begin-eqn}} +{{eqn | l = 4 a^2 x^2 + 4 a b x + 4 a c + | r = 0 + | c = multiplying through by $4 a$ +}} +{{eqn | ll= \leadsto + | l = \paren {2 a x + b}^2 - b^2 + 4 a c + | r = 0 + | c = [[Completing the Square]] +}} +{{eqn | ll= \leadsto + | l = \paren {2 a x + b}^2 + | r = b^2 - 4 a c +}} +{{eqn | ll= \leadsto + | l = x + | r = \frac {-b \pm \sqrt {b^2 - 4 a c} } {2 a} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Normal Subgroup Test} +Tags: Normal Subgroups + +\begin{theorem} +Let $G$ be a [[Definition:Group|group]] and $H \le G$. +Then $H$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$ {{iff}}: +:$\forall x \in G: x H x^{-1} \subseteq H$. +\end{theorem} + +\begin{proof} +Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$. +Suppose $H$ is [[Definition:Normal Subgroup|normal]] in $G$. +Then $\forall x \in G, a \in H: \exists b \in H: x a = b x$. +Thus, $x a x^{-1} = b \in H$ implying $x H x^{-1} \subseteq H$. +Conversely, suppose $\forall x \in G: x H x^{-1} \subseteq H$. +Then for $g \in G$, we have $g H g^{-1} \subseteq H$, which implies $g H \subseteq H g$. +Also, for $g^{-1} \in G$, we have $g^{-1} H (g^{-1})^{-1} = g^{-1} H g \subseteq H$ which implies $H g \subseteq g H$. +Therefore, $g H = H g$ meaning $H \lhd G$. +{{qed}} +[[Category:Normal Subgroups]] +htkkjglfl8j117gvu9foubxcy7ss1b0 +\end{proof}<|endoftext|> +\section{Equality of Ordered Pairs} +Tags: Cartesian Product, Axiomatic Set Theory, Equality of Ordered Tuples, Equality + +\begin{theorem} +Two [[Definition:Ordered Pair|ordered pairs]] are [[Definition:Equality|equal]] {{iff}} corresponding [[Definition:Coordinate System/Coordinate/Element of Ordered Pair|coordinates]] are equal: +:$\tuple {a, b} = \tuple {c, d} \iff a = c \land b = d$ +\end{theorem} + +\begin{proof} +=== [[Equality of Ordered Pairs/Necessary Condition|Necessary Condition]] === +{{:Equality of Ordered Pairs/Necessary Condition}}{{qed|lemma}} +=== [[Equality of Ordered Pairs/Sufficient Condition|Sufficient Condition]] === +{{:Equality of Ordered Pairs/Sufficient Condition}}{{qed}} +\end{proof}<|endoftext|> +\section{Identity is Unique} +Tags: Identity Elements, Identity is Unique, Algebraic Structures + +\begin{theorem} +Let $\left({S, \circ}\right)$ be an [[Definition:Algebraic Structure|algebraic structure]] that has an [[Definition:Identity Element|identity element]] $e \in S$. +Then $e$ is [[Definition:Unique|unique]]. +\end{theorem} + +\begin{proof} +Suppose $e_1$ and $e_2$ are both [[Definition:Identity Element|identity elements]] of $\struct {S, \circ}$. +Then by the definition of [[Definition:Identity Element|identity element]]: +:$\forall s \in S: s \circ e_1 = s = e_2 \circ s$ +Then: +:$e_1 = e_2 \circ e_1 = e_2$ +So: +:$e_1 = e_2$ +and there is only one [[Definition:Identity Element|identity element]] after all. +{{qed}} +\end{proof} + +\begin{proof} +Let $e_S$ be an identity of $\left({S, \circ}\right)$. +Then by [[Definition:Identity Element|definition]], $e_S$ is both a [[Definition:Left Identity|left identity]] and a [[Definition:Right Identity|right identity]]. +By [[More than one Left Identity then no Right Identity]], if there is more than one of either, there cannot be one of the other. +So there can be only one of each. +By [[Left and Right Identity are the Same]], they are one and the same thing. +{{qed}} +\end{proof}<|endoftext|> +\section{Cancellation Laws} +Tags: Group Theory, Cancellability, Cancellation Laws + +\begin{theorem} +Let $G$ be a [[Definition:Group|group]]. +Let $a, b, c \in G$. +Then the following hold: +;Right cancellation law +:$b a = c a \implies b = c$ +;Left cancellation law +:$a b = a c \implies b = c$ +\end{theorem} + +\begin{proof} +Let $a, b, c \in G$ and let $a^{-1}$ be the [[Definition:Inverse Element|inverse]] of $a$. +Suppose $b a = c a$. +Then: +{{begin-eqn}} +{{eqn | l = \paren {b a} a^{-1} + | r = \paren {c a} a^{-1} + | c = +}} +{{eqn | ll= \leadsto + | l = b \paren {a a^{-1} } + | r = c \paren {a a^{-1} } + | c = {{Defof|Associative Operation}} +}} +{{eqn | ll= \leadsto + | l = b e + | r = c e + | c = {{Defof|Inverse Element}} +}} +{{eqn | ll= \leadsto + | l = b + | r = c + | c = {{Defof|Identity Element}} +}} +{{end-eqn}} +Thus, the '''right cancellation law''' holds. +The proof of the '''left cancellation law''' is analogous. +{{qed}} +\end{proof} + +\begin{proof} +From its [[Definition:Group|definition]], a group is a [[Definition:Monoid|monoid]], all of whose elements have [[Definition:Inverse Element|inverses]] and thus are [[Definition:Invertible Element|invertible]]. +From [[Invertible Element of Monoid is Cancellable]], it follows that all its elements are therefore [[Definition:Cancellable Element|cancellable]]. +{{qed}} +\end{proof} + +\begin{proof} +Suppose $x = b a = c a$. +By [[Group has Latin Square Property]], there exists [[Definition:Exactly One|exactly one]] $y \in G$ such that $x = y a$. +That is, $x = b a = c a \implies b = c$. +Similarly, suppose $x = a b = a c$. +Again by [[Group has Latin Square Property]], there exists [[Definition:Exactly One|exactly one]] $y \in G$ such that $x = a y$. +That is, $a b = a c \implies b = c$. +{{qed}} +\end{proof}<|endoftext|> +\section{Center of Group is Normal Subgroup} +Tags: Normal Subgroups, Centers of Groups, Center of Group is Normal Subgroup + +\begin{theorem} +The [[Definition:Center of Group|center]] $\map Z G$ of any [[Definition:Group|group]] $G$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$ which is [[Definition:Abelian Group|abelian]]. +\end{theorem} + +\begin{proof} +Recall that [[Center of Group is Abelian Subgroup]]. +Since $g x = x g$ for each $g \in G$ and $x \in \map Z G$: +:$g \map Z G = \map Z G g$ +Thus: +:$\map Z G \lhd G$ +{{qed}} +\end{proof} + +\begin{proof} +We have: +:$\forall a \in G: x \in \map Z G^a \iff a x a^{-1} = x a a^{-1} = x \in \map Z G$ +Therefore: +:$\forall a \in G: \map Z G^a = \map Z G$ +and $\map Z G$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$. +{{qed}} +\end{proof}<|endoftext|> +\section{Composition of Relations is Associative} +Tags: Composite Relations + +\begin{theorem} +The [[Definition:Composition of Relations|composition of relations]] is an [[Definition:Associative Operation|associative]] [[Definition:Binary Operation|binary operation]]: +:$\paren {\RR_3 \circ \RR_2} \circ \RR_1 = \RR_3 \circ \paren {\RR_2 \circ \RR_1}$ +\end{theorem} + +\begin{proof} +First, note that from the definition of [[Definition:Composition of Relations|composition of relations]], the following must be the case before the above expression is even to be defined: +:$\Dom {\RR_2} = \Cdm {\RR_1}$ +:$\Dom {\RR_3} = \Cdm {\RR_2}$ +The two [[Definition:Composition of Relations|composite relations]] can be seen to have the same [[Definition:Domain of Relation|domain]], thus: +{{begin-eqn}} +{{eqn | l = \Dom {\paren {\RR_3 \circ \RR_2} \circ \RR_1} + | r = \Dom {\RR_1} + | c = [[Domain of Composite Relation]] +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | l = \Dom {\RR_3 \circ \paren {\RR_2 \circ \RR_1} } + | r = \Dom {\RR_2 \circ \RR_1} + | c = [[Domain of Composite Relation]] +}} +{{eqn | r = \Dom {\RR_1} + | c = [[Domain of Composite Relation]] +}} +{{end-eqn}} +... and also the same [[Definition:Codomain of Relation|codomain]], thus: +{{begin-eqn}} +{{eqn | l = \Cdm {\paren {\RR_3 \circ \RR_2} \circ \RR_1} + | r = \Cdm {\RR_3 \circ \RR_2} + | c = [[Codomain of Composite Relation]] +}} +{{eqn | r = \Cdm {\RR_3} + | c = [[Codomain of Composite Relation]] +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | l = \Cdm {\RR_3 \circ \paren {\RR_2 \circ \RR_1} } + | r = \Cdm {\RR_3} + | c = [[Codomain of Composite Relation]] +}} +{{end-eqn}} +So they are equal {{iff}} they have the same value at each point in their common [[Definition:Domain of Relation|domain]], which this shows: +{{begin-eqn}} +{{eqn | ll= \forall x \in \Dom {\RR_1}: + | l = \map {\paren {\paren {\RR_3 \circ \RR_2} \circ \RR_1} } x + | r = \map {\paren {\RR_3 \circ \RR_2} } {\map {\RR_1} x} + | c = {{Defof|Composition of Relations}} +}} +{{eqn | r = \map {\RR_3} {\map {\RR_2} {\map {\RR_1} x} } + | c = {{Defof|Composition of Relations}} +}} +{{eqn | r = \map {\RR_3} {\map {\paren {\RR_2 \circ \RR_1} } x} + | c = {{Defof|Composition of Relations}} +}} +{{eqn | r = \map {\paren {\RR_3 \circ \paren {\RR_2 \circ \RR_1} } } x + | c = {{Defof|Composition of Relations}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Intersection of Subsemigroups} +Tags: Set Intersection, Subsemigroups + +\begin{theorem} +Let $\left({S, \circ}\right)$ be a [[Definition:Semigroup|semigroup]]. +Let $\left({T_1, \circ}\right)$ and $\left({T_2, \circ}\right)$ be [[Definition:Subsemigroup|subsemigroups]] of $\left({S, \circ}\right)$. +Then the [[Definition:Set Intersection|intersection]] of $\left({T_1, \circ}\right)$ and $\left({T_2, \circ}\right)$ is itself a [[Definition:Subsemigroup|subsemigroup]] of that $\left({S, \circ}\right)$. +If $\left({T, \circ}\right)$ is that [[Definition:Set Intersection|intersection]] of $\left({T_1, \circ}\right)$ and $\left({T_2, \circ}\right)$, it follows that $\left({T, \circ}\right)$ is also a [[Definition:Subsemigroup|subsemigroup]] of both $\left({T_1, \circ}\right)$ and $\left({T_2, \circ}\right)$. +\end{theorem} + +\begin{proof} +Let $T = T_1 \cap T_2$ where $T_1, T_2$ are [[Definition:Subsemigroup|subsemigroups]] of $\left({S, \circ}\right)$. Then: +{{begin-eqn}} +{{eqn | o = + | r = a, b \in T +}} +{{eqn | o = \implies + | r = a, b \in T_1 \land a, b \in T_2 + | c = {{Defof|Set Intersection}} +}} +{{eqn | o = \implies + | r = a \circ b \in T_1 \land a \circ b \in T_2 + | c = [[Definition:Subsemigroup|Subsemigroups are closed]] +}} +{{eqn | o = \implies + | r = a \circ b \in T + | c = {{Defof|Set Intersection}} +}} +{{end-eqn}} +Thus $\left({T, \circ}\right)$ is [[Definition:Closed Algebraic Structure|closed]], and is therefore a semigroup from the [[Subsemigroup Closure Test]]. +{{Qed}} +The other results follow from this and [[Intersection is Subset]]. +[[Category:Set Intersection]] +[[Category:Subsemigroups]] +ju0qecld0oxti00l2zeoc6hopffz2k8 +\end{proof}<|endoftext|> +\section{Cancellable Elements of Semigroup form Subsemigroup} +Tags: Subsemigroups, Cancellability + +\begin{theorem} +Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]]. +Let $C$ be the set of [[Definition:Cancellable Element|cancellable elements]] of $\struct {S, \circ}$. +Then $\struct {C, \circ}$ is a [[Definition:Subsemigroup|subsemigroup]] of $\struct {S, \circ}$. +\end{theorem} + +\begin{proof} +Now let $C$ be the set of [[Definition:Cancellable Element|cancellable elements]] of $\struct {S, \circ}$. +Let $x, y \in C$. +Then $x$ and $y$ are both [[Definition:Left Cancellable Element|left cancellable]] and [[Definition:Right Cancellable Element|right cancellable]]. +Thus by [[Left Cancellable Elements of Semigroup form Subsemigroup]]: +:$x \circ y$ is [[Definition:Left Cancellable Element|left cancellable]] +and by [[Right Cancellable Elements of Semigroup form Subsemigroup]]: +:$x \circ y$ is [[Definition:Right Cancellable Element|right cancellable]]. +Thus $x \circ y$ is both [[Definition:Left Cancellable Element|left cancellable]] and [[Definition:Right Cancellable Element|right cancellable]], and therefore [[Definition:Cancellable Element|cancellable]]. +Thus $x \circ y \in C$. +Thus $\struct {C, \circ}$ is [[Definition:Closed Algebraic Structure|closed]] and is therefore by the [[Subsemigroup Closure Test]] a [[Definition:Subsemigroup|subsemigroup]] of $\struct {S, \circ}$. +{{qed}} +\end{proof}<|endoftext|> +\section{Subgroup of Cyclic Group is Cyclic} +Tags: Cyclic Groups, Subgroups, Subgroup of Cyclic Group is Cyclic + +\begin{theorem} +Let $G$ be a [[Definition:Cyclic Group|cyclic group]]. +Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$. +Then $H$ is [[Definition:Cyclic Group|cyclic]]. +\end{theorem} + +\begin{proof} +Let $G$ be a [[Definition:Cyclic Group|cyclic group]] [[Definition:Generator of Cyclic Group|generated]] by $a$. +Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$. +If $H = \set e$, then $H$ is a [[Definition:Cyclic Group|cyclic group]] [[Definition:Generated Subgroup|subgroup generated]] by $e$. +Let $H \ne \set e$. +By definition of [[Definition:Cyclic Group|cyclic group]], every [[Definition:Element|element]] of $G$ has the form $a^n$. +Then as $H$ is a [[Definition:Subgroup|subgroup]] of $G$, $a^n \in H$ for some $n \in \Z$. +Let $m$ be the smallest [[Definition:Strictly Positive Integer|(strictly) positive integer]] such that $a^m \in H$. +Consider an arbitrary [[Definition:Element|element]] $b$ of $H$. +As $H$ is a [[Definition:Subgroup|subgroup]] of $G$, $b = a^n$ for some $n$. +By the [[Division Theorem]], it is possible to find [[Definition:Integer|integers]] $q$ and $r$ such that $n = m q + r$ with $0 \le r < m$. +It follows that: +:$a^n = a^{m q + r} = \paren {a^m}^q a^r$ +and hence: +:$a^r = \paren {a^m}^{-q} a^n$ +Since $a^m \in H$ so is its [[Definition:Inverse Element|inverse]] $\paren {a^m}^{-1}$ +By {{GroupAxiom|0}}, so are all [[Definition:Power of Group Element|powers]] of its [[Definition:Inverse Element|inverse]]. +Now $a^n$ and $\paren {a^m}^{-q}$ are both in $H$, thus so is their [[Definition:Product Element|product]] $a^r$, again by [[Definition:Group Axioms|Group Axiom $G 0$: Closure]]. +However: +: $m$ was the smallest [[Definition:Strictly Positive Integer|(strictly) positive integer]] such that $a^m \in H$ +and: +:$0 \le r < m$ +Therefore it follows that: +: $r = 0$ +Therefore: +:$n = q m$ +and: +:$b = a^n = \paren {a^m}^q$. +We conclude that any arbitrary [[Definition:Element|element]] $b = a^n$ of $H$ is a [[Definition:Power of Group Element|power]] of $a^m$. +So, by definition, $H = \gen {a^m}$ is [[Definition:Cyclic Group|cyclic]]. +{{qed}} +\end{proof} + +\begin{proof} +Let $G$ be a [[Definition:Cyclic Group|cyclic group]] [[Definition:Generator of Cyclic Group|generated]] by $a$. +=== Finite Group === +Let $G$ be [[Definition:Finite Group|finite]]. +By [[Bijection from Divisors to Subgroups of Cyclic Group]] there are exactly as many [[Definition:Subgroup|subgroups]] of $G$ as [[Definition:Divisor of Integer|divisors]] of the [[Definition:Order of Structure|order]] of $G$. +As each one of these is [[Definition:Cyclic Group|cyclic]] by [[Subgroup of Finite Cyclic Group is Determined by Order]], the result follows. +{{qed|lemma}} +=== Infinite Group === +By [[Morphism from Integers to Group]], an [[Definition:Infinite Cyclic Group|infinite cyclic group]] is [[Definition:Group Isomorphism|isomorphic]] to $\left({\Z, +}\right)$. +So all we need to do is show that any [[Definition:Subgroup|subgroup]] of $\left({\Z, +}\right)$ is [[Definition:Cyclic Group|cyclic]]. +Suppose $H$ is a [[Definition:Subgroup|subgroup]] of $\left({\Z, +}\right)$. +From [[Subgroup of Integers is Ideal]] and [[Ring of Integers is Principal Ideal Domain]], we have that: +:$\exists m \in \Z_{> 0}: H = \left({m}\right)$ +where $\left({m}\right)$ is the [[Definition:Principal Ideal of Ring|principal ideal]] of $\left({\Z, +, \times}\right)$ generated by $m$. +But $m$ is also a [[Definition:Generator of Subgroup|generator]] of the [[Definition:Generated Subgroup|subgroup]] $\left({m}\right)$ of $\left({\Z, +}\right)$, as: +:$n \in \Z: n \circ m = n \cdot m \in \gen m$ +Hence the result. +{{qed}} +\end{proof} + +\begin{proof} +Let $G$ be a [[Definition:Cyclic Group|cyclic group]] [[Definition:Generator of Cyclic Group|generated]] by $a$. +Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$. +By [[Cyclic Group is Abelian]], $G$ is [[Definition:Abelian Group|abelian]]. +By [[Subgroup of Abelian Group is Normal]], $H$ is [[Definition:Normal Subgroup|normal]] in $G$. +Let $G / H$ be the [[Definition:Quotient Group|quotient group]] of $G$ by $H$. +Let $q_H: G \to G / H$ be the [[Definition:Quotient Group Epimorphism|quotient epimorphism]] from $G$ to $G / H$: +:$\forall x \in G: \map {q_H} x = x H$ +Then from [[Quotient Group Epimorphism is Epimorphism]], $H$ is the [[Definition:Kernel of Group Homomorphism|kernel]] of $q_H$. +From [[Kernel of Homomorphism on Cyclic Group]], $H$ is a [[Definition:Cyclic Group|cyclic group]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Empty Set is Subset of All Sets} +Tags: Subsets, Empty Set, Empty Set is Subset of All Sets + +\begin{theorem} +The [[Definition:Empty Set|empty set]] $\O$ is a [[Definition:Subset|subset]] of every [[Definition:Set|set]] (including itself). +That is: +:$\forall S: \O \subseteq S$ +\end{theorem} + +\begin{proof} +By the definition of [[Definition:Subset|subset]], $\O \subseteq S$ means: +:$\forall x: \paren {x \in \O \implies x \in S}$ +By the definition of the [[Definition:Empty Set|empty set]]: +:$\forall x: \neg \paren {x \in \O}$ +Thus $\O \subseteq S$ is [[Definition:Vacuous Truth|vacuously true]]. +{{qed}} +\end{proof} + +\begin{proof} +$S \subseteq T$ means: +:'''every [[Definition:Element|element]] of $S$ is also in $T$''' +or, equivalently: +:'''every [[Definition:Element|element]] that is ''not'' in $T$ is not in $S$ either.''' +Thus: +{{begin-eqn}} +{{eqn | o = + | r = S \subseteq T + | c = +}} +{{eqn | o = \leadstoandfrom + | r = \forall x \in S: x \in T + | c = {{Defof|Subset}} +}} +{{eqn | o = \leadstoandfrom + | r = \neg \paren {\exists x \in S: \neg \paren {x \in T} } + | c = [[De Morgan's Laws (Predicate Logic)]] +}} +{{end-eqn}} +which means '''there is no [[Definition:Element|element]] in $S$ which is not also in $T$'''. +There are no [[Definition:Element|elements]] of $\O$, from the [[Definition:Empty Set|definition of the empty set]]. +Therefore $\O$ has no [[Definition:Element|elements]] that are not also in any other [[Definition:Set|set]]. +Thus, from the above, all [[Definition:Element|elements]] of $\O$ are all ([[Definition:Vacuous Truth|vacuously]]) in every other [[Definition:Set|set]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Relation Reflexivity} +Tags: Reflexive Relations + +\begin{theorem} +Every [[Definition:Relation|relation]] has exactly one of these properties: it is either: +:[[Definition:Reflexive Relation|reflexive]], +:[[Definition:Antireflexive Relation|antireflexive]] or +:[[Definition:Non-Reflexive Relation|non-reflexive]]. +\end{theorem} + +\begin{proof} +A [[Definition:Reflexive Relation|reflexive relation]] can not be [[Definition:Antireflexive Relation|antireflexive]], and vice versa: +:$\tuple {x, x} \in \RR \iff \neg \paren {\tuple {x, x} \notin \RR}$ +By the definition of [[Definition:Non-Reflexive Relation|non-reflexive]], a [[Definition:Reflexive Relation|reflexive relation]] can not also be [[Definition:Non-Reflexive Relation|non-reflexive]]. +So a [[Definition:Reflexive Relation|reflexive relation]] is neither [[Definition:Antireflexive Relation|antireflexive]] nor [[Definition:Non-Reflexive Relation|non-reflexive]]. +An [[Definition:Antireflexive Relation|antireflexive relation]] can be neither [[Definition:Reflexive Relation|reflexive]] (see above) nor [[Definition:Non-Reflexive Relation|non-reflexive]] (by the definition of [[Definition:Non-Reflexive Relation|non-reflexive]]). +By its own definition, if a [[Definition:Relation|relation]] is neither [[Definition:Reflexive Relation|reflexive]] nor [[Definition:Antireflexive Relation|antireflexive]], then it is [[Definition:Non-Reflexive Relation|non-reflexive]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Union is Idempotent} +Tags: Set Union, Idempotence + +\begin{theorem} +[[Definition:Set Union|Set union]] is [[Definition:Idempotent Operation|idempotent]]: +:$S \cup S = S$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = x + | o = \in + | r = S \cup S + | c = +}} +{{eqn | ll= \leadstoandfrom + | l = x \in S + | o = \lor + | r = x \in S + | c = {{Defof|Set Union}} +}} +{{eqn | ll= \leadstoandfrom + | l = x + | o = \in + | r = S + | c = [[Rule of Idempotence/Disjunction|Rule of Idempotence: Disjunction]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Intersection is Idempotent} +Tags: Set Intersection, Idempotence + +\begin{theorem} +[[Definition:Set Intersection|Set intersection]] is [[Definition:Idempotent Operation|idempotent]]: +:$S \cap S = S$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = x + | o = \in + | r = S \cap S + | c = +}} +{{eqn | ll= \leadstoandfrom + | l = x \in S + | o = \land + | r = x \in S + | c = {{Defof|Set Intersection}} +}} +{{eqn | ll= \leadstoandfrom + | l = x + | o = \in + | r = S + | c = [[Rule of Idempotence/Conjunction|Rule of Idempotence: Conjunction]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Set is Subset of Union} +Tags: Set Union, Subsets + +\begin{theorem} +The [[Definition:Set Union|union]] of two [[Definition:Set|sets]] is a [[Definition:Superset|superset]] of each: +:$S \subseteq S \cup T$ +:$T \subseteq S \cup T$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = x \in S + | o = \leadsto + | r = x \in S \lor x \in T + | c = [[Rule of Addition]] +}} +{{eqn | o = \leadsto + | r = x \in S \cup T + | c = {{Defof|Set Union}} +}} +{{eqn | o = \leadsto + | r = S \subseteq S \cup T + | c = {{Defof|Subset}} +}} +{{end-eqn}} +Similarly for $T$. +{{qed}} +\end{proof}<|endoftext|> +\section{Set is Subset of Itself} +Tags: Subsets + +\begin{theorem} +Every [[Definition:Set|set]] is a [[Definition:Subset|subset]] of itself: +:$\forall S: S \subseteq S$ +Thus, by definition, the [[Definition:Relation|relation]] '''is a subset of''' is [[Definition:Reflexive Relation|reflexive]]. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | ll= \forall x: + | l = \leftparen {x \in S} + | o = \implies + | r = \rightparen {x \in S} + | c = [[Law of Identity]]: + | cc= a [[Definition:Statement|statement]] implies itself +}} +{{eqn | lll=\leadsto + | l = S + | o = \subseteq + | r = S + | c = {{Defof|Subset}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Subset of Set with Propositional Function} +Tags: Subsets, Mapping Theory + +\begin{theorem} +Let $S$ be a [[Definition:Set|set]]. +Let $P: S \to \set {\text{true}, \text{false} }$ be a [[Definition:Propositional Function|propositional function]] on $S$. +Then: +:$\set {x \in S: \map P x} \subseteq S$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = s + | o = \in + | r = \set {x \in S: \map P x} + | c = +}} +{{eqn | ll= \leadsto + | l = s + | o = \in + | r = \set {x \in S \land \map P x} + | c = +}} +{{eqn | ll= \leadsto + | l = s + | o = \in + | r = \set {x \in S} \land \map P s + | c = {{Defof|Set Definition by Predicate}} +}} +{{eqn | ll= \leadsto + | l = s + | o = \in + | r = \set {x \in S} + | c = [[Rule of Simplification]] +}} +{{eqn | ll= \leadsto + | l = s + | o = \in + | r = S + | c = {{Defof|Element}} +}} +{{eqn | ll= \leadsto + | l = \set {x \in S: \map P x} + | o = \subseteq + | r = S + | c = {{Defof|Subset}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Universal Generalisation} +Tags: Predicate Logic + +\begin{theorem} +Let $\mathbf a$ be any arbitrarily selected object in the [[Definition:Universe of Discourse|universe of discourse]]. +Then: +:$\map P {\mathbf a} \vdash \forall x: \map P x$ +In [[Definition:Natural Language|natural language]]: +:''Suppose $P$ is true of any arbitrarily selected $\mathbf a$'' in the [[Definition:Universe of Discourse|universe of discourse]].'' +:''Then $P$ is true of everything in the [[Definition:Universe of Discourse|universe of discourse]].'' +\end{theorem} + +\begin{proof} +We can express $\forall x$ using its [[Definition:Propositional Expansion|propositional expansion]]: +:$\map P {\mathbf X_1} \land \map P {\mathbf X_2} \land \map P {\mathbf X_3} \land \ldots$ +where $\mathbf X_1, \mathbf X_2, \mathbf X_3 \ldots{}$ is the complete set of the [[Definition:Object|objects]] in the [[Definition:Universe of Discourse|universe of discourse]]. +The fact that any object we care to choose has the property in question means that they ''all'' must have this property. +The result then follows by generalising the [[Rule of Conjunction]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Existential Instantiation} +Tags: Predicate Logic + +\begin{theorem} +:$\exists x: P \left({x}\right), P \left({\mathbf a}\right) \implies y \vdash y$ +Suppose we have the following: +: From our [[Definition:Universe of Discourse|universe of discourse]], ''any'' arbitrarily selected [[Definition:Object|object]] $\mathbf a$ which has the [[Definition:Property|property]] $P$ implies a conclusion $y$ +: $\mathbf a$ is not [[Definition:Free Variable|free]] in $y$ +: It is known that there ''does'' actually exists an object that has $P$. +Then we may infer $y$. +This is called the '''Rule of Existential Instantiation''' and often appears in a proof with its abbreviation '''EI'''. +When using this rule of existential instantiation: +:$\exists x: P \left({x}\right), P \left({\mathbf a}\right) \implies y \vdash y$ +the instance of $P \left({\mathbf a}\right)$ is referred to as the '''typical disjunct'''. +\end{theorem} + +\begin{proof} +This is an extension of [[Proof by Cases]]. +The [[Definition:Propositional Expansion|propositional expansion]] of $\exists x: P \left({x}\right)$ is: +:$P \left({\mathbf X_1}\right) \lor P \left({\mathbf X_2}\right) \lor P \left({\mathbf X_3}\right) \lor \ldots$ +We know that any arbitrarily selected $\mathbf a$ with the property $P$ implies $y$. +From this we can infer that ''all'' such $\mathbf a$ which have that property imply $y$. +This is equivalent to the step in [[Proof by Cases]] in which we need to prove that both disjuncts lead to the same conclusion. +The fact that we only need one of them in fact to ''be'' true is quite enough to draw the conclusion that $y$ is true. +In this context, we are assured by the statement $\exists x: P \left({x}\right)$ that at least one such disjunct in the above propositional expansion is true. +Thus the conclusion follows, and the result is proved. +{{qed}} +\end{proof}<|endoftext|> +\section{Singleton of Element is Subset} +Tags: Subsets, Singletons, Singleton of Element is Subset + +\begin{theorem} +Let $S$ be a [[Definition:Set|set]]. +Let $\set x$ be the [[Definition:Singleton|singleton of $x$]]. +Then: +:$x \in S \iff \set x \subseteq S$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | o = + | r = \set x \subseteq A + | c = +}} +{{eqn | o = \leadstoandfrom + | r = \forall y: \paren {y \in \set x \implies y \in A} + | c = {{Defof|Subset}} +}} +{{eqn | o = \leadstoandfrom + | r = \forall y: \paren {y = x \implies y \in A} + | c = {{Defof|Singleton}} +}} +{{eqn | o = \leadstoandfrom + | r = x \in A + | c = [[Equality implies Substitution]] +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +=== Necessary Condition === +Let $x \in S$. +We have: +:$\set x = \set {y \in S: y = x}$ +From [[Subset of Set with Propositional Function]]: +:$\set {x \in S: \map P x} \subseteq S$ +Hence: +:$\set x \subseteq S$ +{{qed|lemma}} +=== Sufficient Condition === +Let $\set x \subseteq S$. +From the definition of a [[Definition:Subset|subset]]: +:$x \in \set x \implies x \in S$ +{{qed}} +\end{proof}<|endoftext|> +\section{Subset Relation is Transitive} +Tags: Subsets + +\begin{theorem} +The [[Definition:Relation|relation]] "is a [[Definition:Subset|subset]] of" is [[Definition:Transitive Relation|transitive]]: +:$\paren {R \subseteq S} \land \paren {S \subseteq T} \implies R \subseteq T$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | o = + | r = \paren {R \subseteq S} \land \paren {S \subseteq T} + | c = +}} +{{eqn | o = \leadsto + | r = \paren {x \in R \implies x \in S} \land \paren {x \in S \implies x \in T} + | c = {{Defof|Subset}} +}} +{{eqn | o = \leadsto + | r = \paren {x \in R \implies x \in T} + | c = [[Hypothetical Syllogism]] +}} +{{eqn | o = \leadsto + | r = R \subseteq T + | c = {{Defof|Subset}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Set Inequality} +Tags: Set Theory, Subsets + +\begin{theorem} +:$S \ne T \iff \left({S \nsubseteq T}\right) \lor \left({T \nsubseteq S}\right)$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l=S \ne T + | o=\iff + | r=\neg \left({S = T}\right) + | c= +}} +{{eqn | o=\iff + | r=\neg \left({\left({S \subseteq T}\right) \land \left({T \subseteq S}\right)}\right) + | c=Definition of [[Definition:Set Equality/Definition 2|Set Equality]] +}} +{{eqn | o=\iff + | r=\neg \left({S \subseteq T}\right) \lor \neg \left({T \subseteq S}\right) + | c=[[De Morgan's Laws (Logic)/Disjunction of Negations|De Morgan's Laws: Disjunction of Negations]] +}} +{{eqn | o=\iff + | r=\left({S \nsubseteq T}\right) \lor \left({T \nsubseteq S}\right) + | c= +}} +{{end-eqn}} +{{qed}} +[[Category:Set Theory]] +[[Category:Subsets]] +ayqpjtr3s2pd3goy1x3ti823vrizx12 +\end{proof}<|endoftext|> +\section{Set Equals Itself} +Tags: Set Equality + +\begin{theorem} +All [[Definition:Set|sets]] are [[Definition:Set Equality|equal]] to themselves: +:$\forall S: S = S$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = S \subseteq S + | o \land + | r = S \supseteq S + | c = [[Set is Subset of Itself]] +}} +{{eqn | ll= \leadsto + | l = S + | r = S + | c = {{Defof|Set Equality}} +}} +{{end-eqn}} +{{qed}} +[[Category:Set Equality]] +l4v6a9jr0a9dowr9jyr9l6oq8u5cubj +\end{proof}<|endoftext|> +\section{Union is Smallest Superset} +Tags: Set Union, Subsets + +\begin{theorem} +Let $S_1$ and $S_2$ be [[Definition:Set|sets]]. +Then $S_1 \cup S_2$ is the [[Definition:Smallest Set by Set Inclusion|smallest set]] containing both $S_1$ and $S_2$. +That is: +:$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \iff \paren {S_1 \cup S_2} \subseteq T$ +\end{theorem} + +\begin{proof} +=== Necessary Condition === +From [[Union of Subsets is Subset]]: +:$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \implies \paren {S_1 \cup S_2} \subseteq T$ +{{qed|lemma}} +=== Sufficient Condition === +Let $\paren {S_1 \cup S_2} \subseteq T$: +{{begin-eqn}} +{{eqn | l = S_1 + | o = \subseteq + | r = S_1 \cup S_2 + | c = [[Set is Subset of Union]] +}} +{{eqn | l = \paren {S_1 \cup S_2} + | o = \subseteq + | r = T + | c = [[Definition:By Hypothesis|by hypothesis]] +}} +{{eqn | ll= \leadsto + | l = S_1 + | o = \subseteq + | r = T + | c = [[Subset Relation is Transitive]] +}} +{{end-eqn}} +Similarly for $S_2$: +{{begin-eqn}} +{{eqn | l = S_2 + | o = \subseteq + | r = S_1 \cup S_2 + | c = [[Set is Subset of Union]] +}} +{{eqn | l = \paren {S_1 \cup S_2} + | o = \subseteq + | r = T + | c = [[Definition:By Hypothesis|by hypothesis]] +}} +{{eqn | ll= \leadsto + | l = S_2 + | o = \subseteq + | r = T + | c = [[Subset Relation is Transitive]] +}} +{{end-eqn}} +That is: +:$\paren {S_1 \cup S_2} \subseteq T \implies \paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T}$ +{{qed|lemma}} +Thus from the definition of [[Definition:Biconditional|equivalence]]: +:$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \iff \paren {S_1 \cup S_2} \subseteq T$ +{{qed}} +\end{proof}<|endoftext|> +\section{Union with Empty Set} +Tags: Subsets, Empty Set, Set Union, Union with Empty Set + +\begin{theorem} +The [[Definition:Set Union|union]] of any [[Definition:Set|set]] with the [[Definition:Empty Set|empty set]] is the [[Definition:Set|set]] itself: +:$S \cup \O = S$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = S + | o = \subseteq + | r = S + | c = [[Set is Subset of Itself]] +}} +{{eqn | l = \O + | o = \subseteq + | r = S + | c = [[Empty Set is Subset of All Sets]] +}} +{{eqn | ll= \leadsto + | l = S \cup \O + | o = \subseteq + | r = S + | c = [[Union is Smallest Superset]] +}} +{{eqn | ll= \leadsto + | l = S + | o = \subseteq + | r = S \cup \O + | c = [[Set is Subset of Union]] +}} +{{eqn | ll= \leadsto + | l = S \cup \O + | r = S + | c = Definition of [[Definition:Set Equality/Definition 2|Set Equality]] +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +From [[Empty Set is Subset of All Sets]]: +:$\O \subseteq S$ +From [[Union with Superset is Superset]]: +:$S \cup \O = S$ +{{qed}} +\end{proof}<|endoftext|> +\section{Intersection is Subset} +Tags: Set Intersection, Subsets + +\begin{theorem} +The [[Definition:Set Intersection|intersection]] of two [[Definition:Set|sets]] is a [[Definition:Subset|subset]] of each: +:$S \cap T \subseteq S$ +:$S \cap T \subseteq T$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = x + | o = \in + | r = S \cap T +}} +{{eqn | ll= \leadsto + | l = x + | o = \in + | r = S \land x \in T + | c = {{Defof|Set Intersection}} +}} +{{eqn | ll= \leadsto + | l = x + | o = \in + | r = S + | c = [[Rule of Simplification]] +}} +{{eqn | ll= \leadsto + | l = S \cap T + | o = \subseteq + | r = S + | c = {{Defof|Subset}} +}} +{{end-eqn}} +Similarly for $T$. +{{qed}} +\end{proof}<|endoftext|> +\section{Intersection with Empty Set} +Tags: Set Intersection, Empty Set + +\begin{theorem} +The [[Definition:Set Intersection|intersection]] of any [[Definition:Set|set]] with the [[Definition:Empty Set|empty set]] is itself the [[Definition:Empty Set|empty set]]: +:$S \cap \O = \O$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = S \cap \O + | o = \subseteq + | r = \O + | c = [[Intersection is Subset]] +}} +{{eqn | l = \O + | o = \subseteq + | r = S \cap \O + | c = [[Empty Set is Subset of All Sets]] +}} +{{eqn | ll= \leadsto + | l = S \cap \O + | r = \O + | c = {{Defof|Set Equality|index = 2}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Intersection is Largest Subset} +Tags: Set Intersection, Subsets + +\begin{theorem} +Let $T_1$ and $T_2$ be [[Definition:Set|sets]]. +Then $T_1 \cap T_2$ is the [[Definition:Greatest Set by Set Inclusion|largest set]] contained in both $T_1$ and $T_2$. +That is: +:$S \subseteq T_1 \land S \subseteq T_2 \iff S \subseteq T_1 \cap T_2$ +\end{theorem} + +\begin{proof} +=== Sufficient Condition === +From [[Set is Subset of Intersection of Supersets]] we have that: +:$S \subseteq T_1 \land S \subseteq T_2 \implies S \subseteq T_1 \cap T_2$ +{{qed|lemma}} +=== Necessary Condition === +Let: +:$S \subseteq T_1 \cap T_2$ +From [[Intersection is Subset]] we have $T_1 \cap T_2 \subseteq T_1$ and $T_1 \cap T_2\subseteq T_2$. +From [[Subset Relation is Transitive]], it follows directly that $S \subseteq T_1$ and $S \subseteq T_2$. +So $S \subseteq T_1 \cap T_2 \implies S \subseteq T_1 \land S \subseteq T_2$. +{{qed|lemma}} +From the above, we have: +:$S \subseteq T_1 \land S \subseteq T_2 \implies S \subseteq T_1 \cap T_2$ +:$S \subseteq T_1 \cap T_2 \implies S \subseteq T_1 \land S \subseteq T_2$ +Thus $S \subseteq T_1 \land S \subseteq T_2 \iff S \subseteq T_1 \cap T_2$ from the definition of [[Definition:Biconditional|equivalence]]. +{{Qed}} +\end{proof}<|endoftext|> +\section{Intersection is Subset of Union} +Tags: Set Intersection, Set Union + +\begin{theorem} +The [[Definition:Set Intersection|intersection]] of two [[Definition:Set|sets]] is a [[Definition:Subset|subset]] of their [[Definition:Set Union|union]]: +:$S \cap T \subseteq S \cup T$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = S \cap T + | o = \subseteq + | r = S + | c = [[Intersection is Subset]] +}} +{{eqn | l = S + | o = \subseteq + | r = S \cup T + | c = [[Set is Subset of Union]] +}} +{{eqn | ll= \leadsto + | l = S \cap T + | o = \subseteq + | r = S \cup T + | c = [[Subset Relation is Transitive]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Absorption Laws (Set Theory)/Union with Intersection} +Tags: Absorption Laws + +\begin{theorem} +:$S \cup \paren {S \cap T} = S$ +\end{theorem}<|endoftext|> +\section{Absorption Laws (Set Theory)/Intersection with Union} +Tags: Absorption Laws + +\begin{theorem} +:$S \cap \paren {S \cup T} = S$ +\end{theorem}<|endoftext|> +\section{Union Distributes over Intersection} +Tags: Set Union, Set Intersection, Distributive Laws, Distributive Operations, Union Distributes over Intersection + +\begin{theorem} +[[Definition:Set Union|Set union]] is [[Definition:Distributive Operation|distributive]] over [[Definition:Set Intersection|set intersection]]: +:$R \cup \paren {S \cap T} = \paren {R \cup S} \cap \paren {R \cup T}$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | o = + | r = x \in R \cup \paren {S \cap T} +}} +{{eqn | o = \leadstoandfrom + | r = x \in R \lor \paren {x \in S \land x \in T} + | c = {{Defof|Set Union}} and {{Defof|Set Intersection}} +}} +{{eqn | o = \leadstoandfrom + | r = \paren {x \in R \lor x \in S} \land \paren {x \in R \lor x \in T} + | c = [[Disjunction is Left Distributive over Conjunction]] +}} +{{eqn | o = \leadstoandfrom + | r = x \in \paren {R \cup S} \cap \paren {R \cup T} + | c = {{Defof|Set Union}} and {{Defof|Set Intersection}} +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +From [[Intersection Distributes over Union]]: +:$R \cap \paren {S \cup T} = \paren {R \cap S} \cup \paren {R \cap T}$ +From the [[Duality Principle for Sets]], exchanging $\cup$ for $\cap$ throughout, and vice versa, reveals the result: +:$R \cup \paren {S \cap T} = \paren {R \cup S} \cap \paren {R \cup T}$ +{{qed}} +\end{proof}<|endoftext|> +\section{Intersection Distributes over Union} +Tags: Set Intersection, Set Union, Distributive Laws, Distributive Operations, Intersection Distributes over Union + +\begin{theorem} +[[Definition:Set Intersection|Set intersection]] is [[Definition:Distributive Operation|distributive]] over [[Definition:Set Union|set union]]: +:$R \cap \paren {S \cup T} = \paren {R \cap S} \cup \paren {R \cap T}$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | o = + | r = x \in R \cap \paren {S \cup T} +}} +{{eqn | o = \leadstoandfrom + | r = x \in R \land \paren {x \in S \lor x \in T} + | c = {{Defof|Set Union}} and {{Defof|Set Intersection}} +}} +{{eqn | o = \leadstoandfrom + | r = \paren {x \in R \land x \in S} \lor \paren {x \in R \land x \in T} + | c = [[Conjunction is Left Distributive over Disjunction]] +}} +{{eqn | o = \leadstoandfrom + | r = x \in \paren {R \cap S} \cup \paren {R \cap T} + | c = {{Defof|Set Union}} and {{Defof|Set Intersection}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Set Difference is Subset} +Tags: Set Difference, Subsets, Set Difference is Subset + +\begin{theorem} +:$S \setminus T \subseteq S$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = x \in S \setminus T + | o = \leadsto + | r = x \in S \land x \notin T + | c = {{Defof|Set Difference}} +}} +{{eqn | o = \leadsto + | r = x \in S + | c = [[Rule of Simplification]] +}} +{{end-eqn}} +The result follows from the definition of [[Definition:Subset|subset]]. +{{Qed}} +\end{proof} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = S \setminus T + | r = S \cap \complement_S \left({T}\right) + | c = [[Set Difference as Intersection with Relative Complement]] +}} +{{eqn | o = \subseteq + | r = S + | c = [[Intersection is Subset]] +}} +{{end-eqn}} +{{Qed}} +\end{proof}<|endoftext|> +\section{Set Difference with Empty Set is Self} +Tags: Set Difference, Empty Set + +\begin{theorem} +The [[Definition:Set Difference|set difference]] between a [[Definition:Set|set]] and the [[Definition:Empty Set|empty set]] is the [[Definition:Set|set]] itself: +:$S \setminus \O = S$ +\end{theorem} + +\begin{proof} +From [[Set Difference is Subset]]: +:$S \setminus \O \subseteq S$ +From the definition of [[Definition:Empty Set|the empty set]]: +:$\forall x \in S: x \notin \O$ +Let $x \in S$. +Thus: +{{begin-eqn}} +{{eqn | l = x \in S + | o = \leadsto + | r = x \in S \land x \notin \O + | c = [[Rule of Conjunction]] +}} +{{eqn | o = \leadsto + | r = x \in S \setminus \O + | c = {{Defof|Set Difference}} +}} +{{eqn | o = \leadsto + | r = S \subseteq S \setminus \O + | c = {{Defof|Subset}} +}} +{{end-eqn}} +Thus we have: +:$S \setminus \O \subseteq S$ +and: +:$S \subseteq S \setminus \O$ +So by definition of [[Definition:Set Equality/Definition 2|set equality]]: +:$S \setminus \O = S$ +{{qed}} +\end{proof}<|endoftext|> +\section{Set Difference with Self is Empty Set} +Tags: Set Difference, Empty Set + +\begin{theorem} +The [[Definition:Set Difference|set difference]] of a [[Definition:Set|set]] with itself is the [[Definition:Empty Set|empty set]]: +:$S \setminus S = \O$ +\end{theorem} + +\begin{proof} +From [[Set is Subset of Itself]]: +:$S \subseteq S$ +From [[Set Difference with Superset is Empty Set]] we have: +:$S \subseteq T \iff S \setminus T = \O$ +Hence the result. +{{qed}} +\end{proof}<|endoftext|> +\section{Set Difference Equals First Set iff Empty Intersection} +Tags: Set Difference, Set Intersection, Empty Set + +\begin{theorem} +:$S \setminus T = S \iff S \cap T = \O$ +\end{theorem} + +\begin{proof} +Assume $S, T \subseteq \Bbb U$ where $\Bbb U$ is a [[Definition:Universe (Set Theory)|universal set]]. +{{begin-eqn}} +{{eqn | l = S \setminus T + | r = S +}} +{{eqn | ll= \leadstoandfrom + | l = S \cap \map \complement T + | r = S + | c = [[Set Difference as Intersection with Complement]] +}} +{{eqn | ll= \leadstoandfrom + | l = S + | o = \subseteq + | r = \map \complement T + | c = [[Intersection with Subset is Subset]] +}} +{{eqn | ll= \leadstoandfrom + | l = S \cap \map \complement {\map \complement T} + | r = \O + | c = [[Intersection with Complement is Empty iff Subset]] +}} +{{eqn | ll= \leadstoandfrom + | l = S \cap T + | r = \O + | c = [[Complement of Complement]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Equal Set Differences iff Equal Intersections} +Tags: Set Difference, Set Intersection, Equal Set Differences iff Equal Intersections + +\begin{theorem} +:$R \setminus S = R \setminus T \iff R \cap S = R \cap T$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = R \setminus S + | r = R \setminus T +}} +{{eqn | ll= \leadstoandfrom + | l = \set {x \in R: x \notin S} + | r = \set {x \in R: x \notin T} + | c = {{Defof|Set Difference}} +}} +{{eqn | ll= \leadstoandfrom + | l = \forall x \in R: \quad x \notin S + | o = \iff + | r = x \notin T + | c = +}} +{{eqn | ll= \leadstoandfrom + | l = \forall x \in R: \quad x \in S + | o = \iff + | r = x \in T + | c = +}} +{{eqn | ll= \leadstoandfrom + | l = \set {\paren {x \in R} \land \paren {x \in S} } + | r = \set {\paren {x \in R} \land \paren {x \in T} } + | c = +}} +{{eqn | ll= \leadstoandfrom + | l = R \cap S + | r = R \cap T + | c = {{Defof|Set Intersection}} +}} +{{end-eqn}} +{{Qed}} +\end{proof} + +\begin{proof} +From [[Set Difference and Intersection form Partition]]: +: $\paren {R \setminus S} \cup \paren {R \cap S} = R = \paren {R \setminus T} \cup \paren {R \cap T}$ +: $\paren {R \cap S} \cap \paren {R \setminus S} = \O = \paren {R \cap T} \cap \paren {R \setminus T}$ +whatever $R, S, T$ might be. +Let $R \setminus S = R \setminus T$. +Then: +{{begin-eqn}} +{{eqn | l = \paren {\paren {R \setminus S} \cup \paren {R \cap S} } \setminus \paren {R \setminus S} + | r = \paren {\paren {R \setminus T} \cup \paren {R \cap T} } \setminus \paren {R \setminus T} +}} +{{eqn | ll= \leadsto + | l = \paren {R \cap S} \setminus \paren {R \setminus S} + | r = \paren {R \cap T} \setminus \paren {R \setminus T} + | c = [[Set Difference with Union is Set Difference]] +}} +{{end-eqn}} +Now, we have from [[Set Difference with Disjoint Set]]: +:$S \cap T = \O \iff S \setminus T = S$ +and so: +:$\paren {R \cap S} \setminus \paren {R \setminus S} = R \cap S$ +and: +:$\paren {R \cap T} \setminus \paren {R \setminus T} = R \cap T$ +So: +:$R \cap S = R \cap T$ +We can use exactly the same reasoning if we assume $R \cap S = R \cap T$: +{{begin-eqn}} +{{eqn | l = \paren {\paren {R \setminus S} \cup \paren {R \cap S} } \setminus \paren {R \cap S} + | r = \paren {\paren {R \setminus T} \cup \paren {R \cap T} } \setminus \paren {R \cap T} +}} +{{eqn | ll= \leadsto + | l = \paren {R \setminus S} \setminus \paren {R \cap S} + | r = \paren {R \setminus T} \setminus \paren {R \cap T} + | c = [[Set Difference with Union is Set Difference]] +}} +{{end-eqn}} +and then because of [[Set Difference with Disjoint Set]] as above: +:$\paren {R \setminus S} \setminus \paren {R \cap S} = R \setminus S$ +and: +:$\paren {R \setminus T} \setminus \paren {R \cap T} = R \setminus T$ +So: +:$R \setminus S = R \setminus T$ +{{qed}} +\end{proof}<|endoftext|> +\section{Euler's Number is Irrational} +Tags: Irrationality Proofs, Euler's Number + +\begin{theorem} +[[Definition:Euler's Number|Euler's number]] $e$ is [[Definition:Irrational Number|irrational]]. +\end{theorem}<|endoftext|> +\section{Results Concerning Set Difference with Union} +Tags: Union, Set Union, Set Difference, Set Union + +\begin{theorem} +Let: +* $S \setminus T$ denote [[Definition:Set Difference|set difference]] +* $S \cup T$ denote [[Definition:Set Union|set union]] +* $S \cap T$ denote [[Definition:Set Intersection|set intersection]]. +\end{theorem}<|endoftext|> +\section{Set Difference with Intersection} +Tags: Set Difference, Set Intersection, Empty Set + +\begin{theorem} +Let $S$ and $T$ be [[Definition:Set|sets]]. +\end{theorem} + +\begin{proof} +Consider $R, S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the [[Definition:Universe (Set Theory)|universe]]. +Then: +{{begin-eqn}} +{{eqn | l = \left({R \setminus S}\right) \cap T + | r = \left({R \cap \complement \left({S}\right)}\right) \cap T + | c = [[Set Difference as Intersection with Complement]] +}} +{{eqn | r = \left({R \cap T}\right) \cap \complement \left({S}\right) + | c = [[Intersection is Commutative]] and [[Intersection is Associative]] +}} +{{eqn | r = \left({R \cap T}\right) \setminus S + | c = [[Set Difference as Intersection with Complement]] +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +{{begin-eqn}} +{{eqn | o = + | r = x \in \paren {R \setminus S} \cap T + | c = +}} +{{eqn | ll= \leadstoandfrom + | o = + | r = \paren {x \in R \land x \notin S} \land x \in T + | c = {{Defof|Set Intersection}} and {{Defof|Set Difference}} +}} +{{eqn | ll= \leadstoandfrom + | o = + | r = \paren {x \in R \land x \in T} \land x \notin S + | c = [[Rule of Commutation]] and [[Rule of Association]] +}} +{{eqn | ll= \leadstoandfrom + | o = + | r = x \in \paren {R \cap T} \setminus S + | c = {{Defof|Set Intersection}} and {{Defof|Set Difference}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Results Concerning Set Difference with Intersection} +Tags: Set Difference, Set Intersection + +\begin{theorem} +Let: +:$S \setminus T$ denote [[Definition:Set Difference|set difference]] +:$S \cap T$ denote [[Definition:Set Intersection|set intersection]]. +\end{theorem}<|endoftext|> +\section{Pascal's Rule} +Tags: Binomial Coefficients, Pascal's Rule + +\begin{theorem} +For [[Definition:Positive Integer|positive integers]] $n, k$ with $1 \le k \le n$: +:$\dbinom n {k - 1} + \dbinom n k = \dbinom {n + 1} k$ +This is also valid for the [[Definition:Binomial Coefficient/Real Numbers|real number definition]]: +:$\forall r \in \R, k \in \Z: \dbinom r {k - 1} + \dbinom r k = \dbinom {r + 1} k$ +\end{theorem} + +\begin{proof} +Let $n, k \in \N$ with $1 \le k \le n$. +{{begin-eqn}} +{{eqn | l = \binom n k + \binom n {k - 1} + | r = \frac {n!} {k! \, \paren {n - k}!} + \frac {n!} {\paren {k - 1}! \, \paren {n - \paren {k - 1} }!} + | c = {{Defof|Binomial Coefficient}} +}} +{{eqn | r = \frac {n! \, \paren {n - \paren {k - 1} } } {k! \, \paren {n - k}! \, \paren {n - \paren {k - 1} } } + \frac {n! \, k} {\paren {k - 1}! \, \paren {n - \paren {k - 1} }! \ k} + | c = +}} +{{eqn | r = \frac {n! \, \paren {n - k + 1} } {k! \, \paren {n - k + 1}!} + \frac {n! \, k} {k! \, \paren {n - k + 1}!} + | c = +}} +{{eqn | r = \frac {n! \, \paren {n - k + 1} + n! \, k} {k! \, \paren {n - k + 1}!} + | c = +}} +{{eqn | r = \frac {n! \, \paren {n - k + 1 + k} } {k! \, \paren {n - k + 1}!} + | c = +}} +{{eqn | r = \frac {n! \, \paren {n + 1} } {k! \, \paren {n - k + 1}!} + | c = +}} +{{eqn | r = \frac {\paren {n + 1}!} {k! \, \paren {n + 1 - k}!} + | c = +}} +{{eqn | r = \binom {n + 1} k + | c = {{Defof|Binomial Coefficient}} +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +Suppose you were a member of a club with $n + 1$ members (including you). +Suppose it were time to elect a committee of $k$ members from that club. +From [[Cardinality of Set of Subsets]], there are $\dbinom {n + 1} k$ ways to select the members to form this committee. +Now, you yourself may or may not be elected a member of this committee. +Suppose that, after the election, you are not a member of this committee. +Then, from [[Cardinality of Set of Subsets]], there are $\dbinom n k$ ways to select the members to form such a committee. +Now suppose you ''are'' a member of the committee. Apart from you, there are $k - 1$ such members. +Again, from [[Cardinality of Set of Subsets]], there are $\dbinom n {k - 1}$ ways of selecting the ''other'' $k - 1$ members so as to form such a committee. +In total, then, there are $\dbinom n k + \dbinom n {k - 1}$ possible committees. +Hence the result. +{{qed}} +\end{proof}<|endoftext|> +\section{Set Difference Union Intersection} +Tags: Set Difference, Set Union, Set Intersection, Set Difference Union Intersection + +\begin{theorem} +:$S = \paren {S \setminus T} \cup \paren {S \cap T}$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \paren {S \setminus T} \cup \paren {S \cap T} + | r = \paren {\paren {S \setminus T} \cup S} \cap \paren {\paren {S \setminus T} \cup T} + | c = [[Union Distributes over Intersection]] +}} +{{eqn | r = S \cap \paren {\paren {S \setminus T} \cup T} + | c = [[Set Difference Union First Set is First Set]] +}} +{{eqn | r = S \cap \paren {S \cup T} + | c = [[Set Difference Union Second Set is Union]] +}} +{{eqn | r = S + | c = [[Intersection Absorbs Union]] +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \left({S \setminus T}\right) \cup \left({S \cap T}\right) + | r = S \setminus \left({T \setminus T}\right) + | c = [[Set Difference with Set Difference is Union of Set Difference with Intersection]] +}} +{{eqn | r = S \setminus \varnothing + | c = [[Set Difference with Self is Empty Set]] +}} +{{eqn | r = S + | c = [[Set Difference with Empty Set is Self]] +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +By [[Set Difference is Subset]]: +:$S \setminus T \subseteq S$ +By [[Intersection is Subset]]: +:$S \cap T \subseteq S$ +Hence from [[Union is Smallest Superset]]: +:$\paren {S \setminus T} \cup \paren {S \cap T} \subseteq S$ +Let $s \in S$. +Either: +:$s \in T$, in which case $s \in S \cap T$ by definition of [[Definition:Set Intersection|set intersection]] +or +:$s \notin T$, in which case $s \in S \setminus T$ by definition of [[Definition:Set Difference|set difference]]. +That is, by definition of [[Definition:Set Union|set union]]: +:$s \in \paren {S \setminus T} \cup \paren {S \cap T}$ +and so by definition of [[Definition:Subset|subset]]: +:$S \subseteq \paren {S \setminus T} \cup \paren {S \cap T}$ +Hence the result by definition of [[Definition:Set Equality/Definition 2|set equality]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Absorption Laws (Logic)} +Tags: Conjunction, Disjunction + +\begin{theorem} +For any two [[Definition:Proposition|propositions]] $p$ and $q$, we have: +\end{theorem}<|endoftext|> +\section{Set Difference with Set Difference} +Tags: Set Intersection, Set Difference + +\begin{theorem} +:$S \setminus \paren {S \setminus T} = S \cap T = T \setminus \paren {T \setminus S}$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = S \setminus \paren {S \setminus T} + | r = \paren {S \setminus S} \cup \paren {S \cap T} + | c = [[Set Difference with Set Difference is Union of Set Difference with Intersection]] +}} +{{eqn | r = \O \cup \paren {S \cap T} + | c = [[Set Difference with Self is Empty Set]] +}} +{{eqn | r = S \cap T + | c = [[Union with Empty Set]] +}} +{{end-eqn}} +Interchanging $S$ and $T$: +{{begin-eqn}} +{{eqn | l = T \setminus \paren {T \setminus S} + | r = T \cap S +}} +{{eqn | r = S \cap T + | c = [[Intersection is Commutative]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Set Difference is Anticommutative} +Tags: Set Difference, Empty Set + +\begin{theorem} +[[Definition:Set Difference|Set difference]] is an [[Definition:Anticommutative|anticommutative]] operation: +:$S = T \iff S \setminus T = T \setminus S = \varnothing$ +\end{theorem} + +\begin{proof} +From [[Set Difference with Superset is Empty Set]] we have: +: $S \subseteq T \iff S \setminus T = \varnothing$ +: $T \subseteq S \iff T \setminus S = \varnothing$ +The result follows from definition of [[Definition:Set Equality/Definition 2|set equality]]: +:$S = T \iff \left({S \subseteq T}\right) \land \left({T \subseteq S}\right)$ +{{qed}} +\end{proof}<|endoftext|> +\section{Dirichlet's Test for Uniform Convergence} +Tags: Direct Proofs, Analysis, Convergence + +\begin{theorem} +Let $D$ be a [[Definition:Set|set]]. +Let $\struct {V, \norm {\,\cdot\,} }$ be a [[Definition:Normed Vector Space|normed vector space]]. +Let $a_i, b_i$ be [[Definition:Mapping|mappings]] from $D \to M$. +Let the following conditions be satisfied: +:$(1): \quad$ The [[Definition:Sequence of Partial Sums|sequence of partial sums]] of $\displaystyle \sum_{n \mathop = 1}^\infty \map {a_n} x$ be [[Definition:Bounded Sequence|bounded]] on $D$ +:$(2): \quad \sequence {\map {b_n} x}$ be [[Definition:Monotone Sequence|monotonic]] for each $x \in D$ +:$(3): \quad \map {b_n} x \to 0$ [[Definition:Uniform Convergence|converge uniformly]] on $D$. +Then: +:$\displaystyle \sum_{n \mathop = 1}^\infty \map {a_n} x \, \map {b_n} x$ [[Definition:Uniform Convergence|converges uniformly]] on $D$. +\end{theorem} + +\begin{proof} +Suppose $\map {b_n} x \ge \map {b_{n + 1} } x$ for each $x \in D$. +All we need to show is that: +:$\displaystyle \sum_{n \mathop = 1}^\infty \size {\map {b_n} x - \map {b_{n + 1} } x}$ +[[Definition:Uniform Convergence|converges uniformly]] on $D$. +{{explain|Why is the above sufficient? On the surface of it, no reference has been made to $\map {a_n} x$. Its connection to the above statement needs to be clarified.}} +To do this we show that the [[Definition:Cauchy Criterion for Sequences|Cauchy criterion]] holds. +Assign $\epsilon < 0$. +Then by definition of [[Definition:Uniform Convergence|uniform convergence]]: +:$\displaystyle \exists N \in \N: \forall x \in D: \forall n \ge N: \size {\map {b_n} x} < \frac \epsilon 2$ +Let $x \in D$ and $n > m \ge N$. +Then: +{{begin-eqn}} +{{eqn | l = \sum_{k \mathop = m + 1}^n \size {\map {b_k} x - \map {b_{k + 1} } x} + | r = \sum_{k \mathop = m + 1}^n \paren {\map {b_k} x - \map {b_{k + 1} } x} +}} +{{eqn | r = \map {b_{m + 1} } x - \map {b_{n + 1} } x +}} +{{eqn | r = \size {\map {b_{m + 1} } x - \map {b_{n + 1} } x} +}} +{{eqn | o = \le + | r = \size {\map {b_{m + 1} } x + \map {b_{n + 1} } x} +}} +{{eqn | o = < + | r = \frac \epsilon 2 + \frac \epsilon 2 +}} +{{eqn | r = \epsilon +}} +{{end-eqn}} +{{qed}} +{{Namedfor|Johann Peter Gustav Lejeune Dirichlet|cat = Dirichlet}} +\end{proof}<|endoftext|> +\section{Relative Complement of Empty Set} +Tags: Relative Complement, Empty Set + +\begin{theorem} +The [[Definition:Relative Complement|relative complement]] of the [[Definition:Empty Set|empty set]] is the [[Definition:Set|set]] itself: +:$\relcomp S \O = S$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \relcomp S \O + | r = S \setminus \O + | c = {{Defof|Relative Complement}} +}} +{{eqn | r = S + | c = [[Set Difference with Empty Set is Self]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Relative Complement with Self is Empty Set} +Tags: Relative Complement, Empty Set + +\begin{theorem} +The [[Definition:Relative Complement|relative complement]] of a [[Definition:Set|set]] in itself is the [[Definition:Empty Set|empty set]]: +:$\relcomp S S = \O$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \relcomp S S + | r = S \setminus S + | c = {{Defof|Relative Complement}} +}} +{{eqn | r = \O + | c = [[Set Difference with Self is Empty Set]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Relative Complement of Relative Complement} +Tags: Relative Complement, Relative Complement of Relative Complement + +\begin{theorem} +:$\relcomp S {\relcomp S T} = T$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \relcomp S {\relcomp S T} + | r = S \setminus \paren {S \setminus T} + | c = {{Defof|Relative Complement}} +}} +{{eqn | r = S \cap T + | c = [[Set Difference with Set Difference]] +}} +{{end-eqn}} +The definition of the [[Definition:Relative Complement|relative complement]] requires that: +:$T \subseteq S$ +But from [[Intersection with Subset is Subset]]: +:$T \subseteq S \iff T \cap S = T$ +Thus: +:$\relcomp S {\relcomp S T} = T$ +follows directly. +{{qed}} +\end{proof} + +\begin{proof} +By the definition of [[Definition:Relative Complement|relative complement]]: +:$\complement_S \left({\complement_S \left({T}\right)}\right) = S \setminus \left({S \setminus T}\right)$ +Let $t \in T$. +Then by the definition of [[Definition:Set Difference|set difference]]: +:$t \notin S \setminus T$ +Since $T \in T$ and $T \subseteq S$, by the definition of [[Definition:subset|subset]]: +:$t \in S$ +Thus: +:$t \in \left({S \setminus \left({S \setminus T}\right)}\right)$ +Suppose instead that: +:$t \in \left({S \setminus \left({S \setminus T}\right)}\right)$ +Then: +:$t \in S$ +and: +:$\neg \left({t \in \left({S \setminus T}\right)}\right)$ +Thus: +:$\neg \left({\left({t \in S}\right) \land \neg \left({t \in T}\right)}\right)$ +By [[Implication Equivalent to Negation of Conjunction with Negative/Formulation 1/Reverse Implication|Implication Equivalent to Negation of Conjunction with Negative]]: +:$t \in S \implies t \in T$ +By [[Modus Ponendo Ponens]]: +:$t \in T$ +{{qed}} +\end{proof}<|endoftext|> +\section{Intersection with Relative Complement is Empty} +Tags: Relative Complement, Set Intersection + +\begin{theorem} +The [[Definition:Set Intersection|intersection]] of a [[Definition:Set|set]] and its [[Definition:Relative Complement|relative complement]] is the [[Definition:Empty Set|empty set]]: +:$T \cap \relcomp S T = \O$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = T \cap \relcomp S T + | r = \paren {S \setminus T} \cap T + | c = {{Defof|Relative Complement}} +}} +{{eqn | r = \O + | c = [[Set Difference Intersection with Second Set is Empty Set]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Union with Relative Complement} +Tags: Relative Complement, Set Union + +\begin{theorem} +The [[Definition:Set Union|union]] of a [[Definition:Set|set]] $T$ and its [[Definition:Relative Complement|relative complement]] in $S$ is the set $S$: +:$\relcomp S T \cup T = S$ +\end{theorem} + +\begin{proof} +=== Step 1 === +By the definition of [[Definition:Relative Complement|relative complement]], we have that: +:$\relcomp S T \subseteq S$ +and: +:$T \subseteq S$ +Hence by [[Union is Smallest Superset]]: +:$\relcomp S T \cup T\subseteq S$ +{{qed|lemma}} +=== Step 2 === +Let $x \in S$. +By [[Law of Excluded Middle]], one of the following two applies: +:$(1): \quad x \in T$ +:$(2): \quad x \notin T$ +If $(2)$, then by definition of [[Definition:Relative Complement|relative complement]]: +:$x \in S \setminus T = \relcomp S T$ +So: +:$x \in T \lor x \in \relcomp S T$ +By definition of [[Definition:Set Union|set union]]: +:$x \in \relcomp S T \cup T$ +Thus: +:$x \in S \implies x \in \relcomp S T \cup T$ +By definition of [[Definition:Subset|subset]] it follows that: +:$S \subseteq \relcomp S T \cup T$ +{{qed|lemma}} +=== Step 3 === +From: +:$\relcomp S T \cup T\subseteq S$ +and: +:$S \subseteq \relcomp S T \cup T$ +it follows from definition of [[Definition:Set Equality/Definition 2|set equality]] that: +:$S = \relcomp S T \cup T$ +{{qed}} +{{LEM}} +\end{proof}<|endoftext|> +\section{Set Difference as Intersection with Relative Complement} +Tags: Set Difference, Relative Complement, Set Intersection + +\begin{theorem} +Let $A, B \subseteq S$. +Then the [[Definition:Set Difference|set difference]] between $A$ and $B$ can be expressed as the [[Definition:Set Intersection|intersection]] with the [[Definition:Relative Complement|relative complement]] with respect to $S$: +:$A \setminus B = A \cap \relcomp S B$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = A \setminus B + | r = \set {x: x \in A \land x \notin B} + | c = {{Defof|Set Difference}} +}} +{{eqn | r = \set {x: \paren {x \in A \land x \in X} \land x \notin B} + | c = {{Defof|Subset}}, [[Modus Ponens]] and [[Rule of Conjunction]] +}} +{{eqn | r = \set {x: x \in A \land \paren {x \in X \land x \notin B} } + | c = [[Conjunction is Associative]] +}} +{{eqn | r = \set {x: x \in A \land x \in \relcomp S B} + | c = {{Defof|Relative Complement}} +}} +{{eqn | r = A \cap \relcomp S B + | c = {{Defof|Set Intersection}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Equivalence of Axiom Schemata for Groups} +Tags: Group Theory + +\begin{theorem} +In the definition of a [[Definition:Group|group]], the [[Definition:Group Axioms|axioms]] for the existence of an [[Definition:Identity Element|identity element]] and for [[Definition:Closed Algebraic Structure|closure]] under taking [[Definition:Inverse Element|inverses]] can be replaced by the following two axioms: +: Given a group $G$, there exists at least one [[Definition:Element|element]] $e \in G$ such that $e$ is a [[Definition:Left Identity|'''left identity''']]; +: For any element $g$ in a group $G$, there exists at least one [[Definition:Left Inverse Element|'''left inverse''']] of $g$. +Alternatively, we can also replace the aforementioned axioms with the following two: +: Given a group $G$, there exists at least one [[Definition:Element|element]] $e \in G$ such that $e$ is a [[Definition:Right Identity|'''right identity''']]; +: For any element $g$ in a group $G$, there exists at least one [[Definition:Right Inverse Element|'''right inverse''']] of $g$. +Thus we can formulate the group axioms as either of the following: +\end{theorem} + +\begin{proof} +Suppose we define a group $G$ in the usual way, but make the first pair of axiom replacements listed above: +: the existence of a [[Definition:Left Identity|left identity]] +: every element has a [[Definition:Left Inverse Element|left inverse]]. +Let $e \in G$ be a [[Definition:Left Identity|left identity]] and $g \in G$. +Then, from [[Left Inverse for All is Right Inverse]], each [[Definition:Left Inverse Element|left inverse]] is also a [[Definition:Right Inverse Element|right inverse]] with respect to the [[Definition:Left Identity|left identity]]. +Also from [[Left Identity while exists Left Inverse for All is Identity]] we have that the [[Definition:Left Identity|left identity]] is also a [[Definition:Right Identity|right identity]]. +Also we have that such an [[Identity is Unique]], so this element can rightly be called [[Definition:Identity Element|'''the''' identity]]. +So we have that: +: $G$ has an [[Definition:Identity Element|identity]]; +: each element of $G$ has an element that is both a [[Definition:Left Inverse Element|left inverse]] and a [[Definition:Right Inverse Element|right inverse]] with respect to this [[Definition:Identity Element|identity]]. +Therefore, the validity of the two axiom replacements is proved. +{{qed}} +The proof of the alternate pair of replacements (existence of a [[Definition:Right Identity|right identity]] and closure under taking [[Definition:Right Inverse Element|right inverses]]) is similar. +\end{proof}<|endoftext|> +\section{Group has Latin Square Property} +Tags: Group Theory, Group has Latin Square Property + +\begin{theorem} +Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. +Then $G$ satisfies the [[Definition:Latin Square Property|Latin square property]]. +That is, for all $a, b \in G$, there exists a [[Definition:Unique|unique]] $g \in G$ such that $a \circ g = b$. +Similarly, there exists a [[Definition:Unique|unique]] $h \in G$ such that $h \circ a = b$. +\end{theorem} + +\begin{proof} +We shall prove that this is true for the first equation: +{{begin-eqn}} +{{eqn | l = a \circ g + | r = b +}} +{{eqn | ll= \leadstoandfrom + | l = a^{-1} \circ \paren {a \circ g} + | r = a^{-1} \circ b + | c = $\circ$ is a [[Cancellation Laws|Cancellable Binary Operation]] +}} +{{eqn | ll= \leadstoandfrom + | l = \paren {a^{-1} \circ a} \circ g + | r = a^{-1} \circ b + | c = {{GroupAxiom|1}} +}} +{{eqn | ll= \leadstoandfrom + | l = e \circ g + | r = a^{-1} \circ b + | c = {{GroupAxiom|3}} +}} +{{eqn | ll= \leadstoandfrom + | l = g + | r = a^{-1} \circ b + | c = {{GroupAxiom|2}} +}} +{{end-eqn}} +Because the statements: +:$a \circ g = b$ +and +:$g = a^{-1} \circ b$ +are [[Definition:Logically Equivalent|equivalent]], we may conclude that $g$ is indeed the only solution of the equation. +The proof that the unique solution of $h$ is $b \circ a^{-1}$ in the second equation follows similarly. +{{qed}} +\end{proof} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = g + | r = a^{-1} \circ b +}} +{{eqn | ll= \leadsto + | l = a \circ g + | r = a \circ \paren {a^{-1} \circ b} +}} +{{eqn | ll= \leadsto + | l = a \circ g + | r = \paren {a \circ a^{-1} } \circ b + | c = {{GroupAxiom|1}} +}} +{{eqn | ll= \leadsto + | l = a \circ g + | r = e \circ b + | c = {{GroupAxiom|3}} +}} +{{eqn | ll= \leadsto + | l = a \circ g + | r = b + | c = {{GroupAxiom|2}} +}} +{{end-eqn}} +Thus, such a $g$ exists. +Suppose $g, g' \in G$ where $a \circ g = b = a \circ g'$. +Then: +{{begin-eqn}} +{{eqn | l = g + | r = e \circ g + | c = {{GroupAxiom|2}} +}} +{{eqn | r = \paren {a^{-1} \circ a} \circ g + | c = {{GroupAxiom|3}} +}} +{{eqn | r = a^{-1} \circ \paren {a \circ g} + | c = {{GroupAxiom|1}} +}} +{{eqn | r = a^{-1} \circ b + | c = Substitution for $a \circ g$ +}} +{{eqn | r = a^{-1} \circ \paren {a \circ g'} + | c = Substitution for $a \circ g'$ +}} +{{eqn | r = \paren {a^{-1} \circ a} \circ g' + | c = {{GroupAxiom|1}} +}} +{{eqn | r = e \circ g' + | c = {{GroupAxiom|3}} +}} +{{eqn | r = g' + | c = {{GroupAxiom|2}} +}} +{{end-eqn}} +Thus uniqueness holds. +To prove the second part of the theorem, let $h = b \circ a^{-1}$. +The remainder of the proof follows a similar procedure to the above. +{{qed}} +\end{proof} + +\begin{proof} +Suppose that $\exists x, y \in G: a \circ x = b = a \circ y$. +{{begin-eqn}} +{{eqn | l = a \circ x + | r = a \circ y + | c = +}} +{{eqn | ll= \leadsto + | l = a^{-1} \circ \paren {a \circ x} + | r = a^{-1} \circ \paren {a \circ y} + | c = {{GroupAxiom|3}} +}} +{{eqn | ll= \leadsto + | l = \paren {a^{-1} \circ a} \circ x + | r = \paren {a^{-1} \circ a} \circ y + | c = {{GroupAxiom|1}} +}} +{{eqn | ll= \leadsto + | l = e \circ x + | r = e \circ y + | c = {{GroupAxiom|3}} +}} +{{eqn | ll= \leadsto + | l = x + | r = y + | c = {{GroupAxiom|2}} +}} +{{end-eqn}} +So such an element, if it exists, is [[Definition:Unique|unique]]. +Now it is demonstrated that $g = a^{-1} b$ satisfies the requirement for $a \circ g = b$ +Since $a \in G$, it follows by [[Definition:Group Axioms|group axiom $G3$]]: existence of [[Definition:Inverse Element|inverses]] that $a^{-1} \in G$. +{{begin-eqn}} +{{eqn | l = a + | o = \in + | r = G +}} +{{eqn | ll= \leadsto + | l = a^{-1} + | o = \in + | r = G + | c = {{GroupAxiom|3}} +}} +{{eqn | ll= \leadsto + | l = a^{-1} \circ b + | o = \in + | r = G + | c = {{GroupAxiom|0}} +}} +{{end-eqn}} +Then: +{{begin-eqn}} +{{eqn | l = a \circ g + | r = a \circ \paren {a^{-1} \circ b} +}} +{{eqn | r = \paren {a \circ a^{-1} } \circ b + | c = {{GroupAxiom|1}} +}} +{{eqn | r = e \circ b + | c = {{GroupAxiom|3}} +}} +{{eqn | r = b + | c = {{GroupAxiom|2}} +}} +{{end-eqn}} +Thus, such a $g$ exists. +The properties of $h$ are proved similarly. +{{qed}} +\end{proof} + +\begin{proof} +We shall prove that this is true for the first equation: +{{begin-eqn}} +{{eqn | l = b + | r = a \circ g +}} +{{eqn | ll= \leadsto + | l = a^{-1} \circ b + | r = a^{-1} \circ \paren {a \circ g} + | c = {{GroupAxiom|3}} +}} +{{eqn | r = \paren {a^{-1} \circ a} \circ g + | c = {{GroupAxiom|1}} +}} +{{eqn | r = e \circ g + | c = {{GroupAxiom|3}} +}} +{{eqn | r = g + | c = {{GroupAxiom|2}} +}} +{{end-eqn}} +Conversely: +{{begin-eqn}} +{{eqn | l = g + | r = a^{-1} \circ b +}} +{{eqn | ll= \leadsto + | l = a \circ g + | r = a \circ \paren {a^{-1} \circ b} + | c = +}} +{{eqn | r = \paren {a \circ a^{-1} } \circ b + | c = {{GroupAxiom|1}} +}} +{{eqn | r = e \circ b + | c = {{GroupAxiom|3}} +}} +{{eqn | r = b + | c = {{GroupAxiom|2}} +}} +{{end-eqn}} +The proof that the unique solution of $h$ is $b \circ a^{-1}$ in the second equation follows similarly. +{{qed}} +\end{proof}<|endoftext|> +\section{Symmetric Difference is Commutative} +Tags: Symmetric Difference, Commutativity + +\begin{theorem} +[[Definition:Symmetric Difference|Symmetric difference]] is [[Definition:Commutative Operation|commutative]]: +:$S * T = T * S$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = S * T + | r = \paren {S \setminus T} \cup \paren {T \setminus S} + | c = {{Defof|Symmetric Difference}} +}} +{{eqn | r = \paren {T \setminus S} \cup \paren {S \setminus T} + | c = [[Union is Commutative]] +}} +{{eqn | r = T * S + | c = {{Defof|Symmetric Difference}} +}} +{{end-eqn}} +{{Qed}} +\end{proof}<|endoftext|> +\section{Equivalence of Definitions of Symmetric Difference} +Tags: Symmetric Difference, Set Difference, Set Intersection, Set Union, Equivalence of Definitions of Symmetric Difference + +\begin{theorem} +Let $S$ and $T$ be [[Definition:Set|sets]]. +{{TFAE|def = Symmetric Difference|view = symmetric difference $S * T$ between $S$ and $T$}} +\end{theorem} + +\begin{proof} +=== [[Equivalence of Definitions of Symmetric Difference/(1) iff (2)|$(1)$ iff $(2)$]] === +{{:Equivalence of Definitions of Symmetric Difference/(1) iff (2)}} +{{qed|lemma}} +=== [[Equivalence of Definitions of Symmetric Difference/(1) iff (3)|$(1)$ iff $(3)$]] === +{{:Equivalence of Definitions of Symmetric Difference/(1) iff (3)}} +{{qed|lemma}} +=== [[Equivalence of Definitions of Symmetric Difference/(2) iff (4)|$(2)$ iff $(4)$]] === +{{:Equivalence of Definitions of Symmetric Difference/(2) iff (4)}} +{{qed|lemma}} +=== [[Equivalence of Definitions of Symmetric Difference/(2) iff (5)|$(2)$ iff $(5)$]] === +{{:Equivalence of Definitions of Symmetric Difference/(2) iff (5)}} +{{qed|lemma}} +=== [[Equivalence of Definitions of Symmetric Difference/(3) iff (5)|$(3)$ iff $(5)$]] === +{{:Equivalence of Definitions of Symmetric Difference/(3) iff (5)}} +{{qed}} +[[Category:Symmetric Difference]] +[[Category:Set Difference]] +[[Category:Set Intersection]] +[[Category:Set Union]] +[[Category:Equivalence of Definitions of Symmetric Difference]] +o4wdikguwfpiw83dtlgtwpjxn0au5v2 +\end{proof}<|endoftext|> +\section{Symmetric Difference of Equal Sets} +Tags: Symmetric Difference, Empty Set, Set Equality + +\begin{theorem} +The [[Definition:Symmetric Difference|symmetric difference]] of two [[Definition:Set Equality|equal sets]] is the [[Definition:Empty Set|empty set]]: +:$S = T \iff S * T = \O$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = S = T + | o = \leadstoandfrom + | r = S \subseteq T \land T \subseteq S + | c = {{Defof|Set Equality}} +}} +{{eqn | o = \leadstoandfrom + | r = \paren {S \setminus T = \O} \land \paren {T \setminus S = \O} + | c = [[Set Difference with Superset is Empty Set]] +}} +{{eqn | o = \leadstoandfrom + | r = \paren {S \setminus T} \cup \paren {T \setminus S} = \O + | c = [[Union is Empty iff Sets are Empty]] +}} +{{eqn | o = \leadstoandfrom + | r = S * T = \O + | c = {{Defof|Symmetric Difference}} +}} +{{end-eqn}} +{{Qed}} +\end{proof}<|endoftext|> +\section{Union is Empty iff Sets are Empty} +Tags: Set Union, Empty Set, Union is Empty iff Sets are Empty + +\begin{theorem} +:$S \cup T = \O \iff S = \O \land T = \O$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l=S \cup T = \varnothing + | o=\iff + | r=\neg \exists x: x \in \left ({S \cup T}\right) + | c=Definition of [[Definition:Empty Set|Empty Set]] +}} +{{eqn | o=\iff + | r=\forall x: \neg \left ({x \in \left ({S \cup T}\right)}\right) + | c=[[De Morgan's Laws (Predicate Logic)]] +}} +{{eqn | o=\iff + | r=\forall x: \neg \left ({x \in S \lor x \in T}\right) + | c=Definition of [[Definition:Set Union|Set Union]] +}} +{{eqn | o=\iff + | r=\forall x: x \notin S \land x \notin T + | c=[[De Morgan's Laws (Logic)/Conjunction of Negations|De Morgan's Laws: Conjunction of Negations]] +}} +{{eqn | o=\iff + | r=S = \varnothing \land T = \varnothing + | c=Definition of [[Definition:Empty Set|Empty Set]] +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +Let $S \cup T = \varnothing$. +We have: +{{begin-eqn}} +{{eqn | l=S + | o=\subseteq + | r=S \cup T + | c=[[Set is Subset of Union]] +}} +{{eqn | ll=\implies + | l=S + | o=\subseteq + | r=\varnothing + | c=[[Definition:By Hypothesis|by hypothesis]] +}} +{{end-eqn}} +From [[Empty Set is Subset of All Sets]]: +: $\varnothing \subseteq S$ +So it follows by definition of [[Definition:Set Equality/Definition 2|set equality]] that $S = \varnothing$. +Similarly for $T$. +{{qed}} +\end{proof}<|endoftext|> +\section{Symmetric Difference with Empty Set} +Tags: Symmetric Difference, Empty Set + +\begin{theorem} +:$S * \O = S$ +where $*$ denotes the [[Definition:Symmetric Difference|symmetric difference]]. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = S * \O + | r = \paren {S \cup \O} \setminus \paren {S \cap \O} + | c = {{Defof|Symmetric Difference|index = 2}} +}} +{{eqn | r = S \setminus \paren {S \cap \O} + | c = [[Union with Empty Set]] +}} +{{eqn | r = S \setminus \O + | c = [[Intersection with Empty Set]] +}} +{{eqn | r = S + | c = [[Set Difference with Empty Set is Self]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Intersection Distributes over Symmetric Difference} +Tags: Set Intersection, Symmetric Difference, Distributive Operations + +\begin{theorem} +[[Definition:Set Intersection|Intersection]] is [[Definition:Distributive Operation|distributive]] over [[Definition:Symmetric Difference|symmetric difference]]: +:$\paren {R * S} \cap T = \paren {R \cap T} * \paren {S \cap T}$ +:$T \cap \paren {R * S} = \paren {T \cap R} * \paren {T \cap S}$ +\end{theorem} + +\begin{proof} +From [[Set Intersection Distributes over Set Difference]], we have: +:$\paren {R \setminus S} \cap T = \paren {R \cap T} \setminus \paren {S \cap T}$ +So: +{{begin-eqn}} +{{eqn | l = \paren {R \cap T} * \paren {S \cap T} + | r = \paren {\paren {R \cap T} \setminus \paren {S \cap T} } \cup \paren {\paren {S \cap T} \setminus \paren {R \cap T} } + | c = {{Defof|Symmetric Difference}} +}} +{{eqn | r = \paren {\paren {R \setminus S} \cap T} \cup \paren {\paren {S \setminus R} \cap T} + | c = [[Set Intersection Distributes over Set Difference]] +}} +{{eqn | r = \paren {\paren {R \setminus S} \cup \paren {S \setminus R} } \cap T + | c = [[Intersection Distributes over Union]] +}} +{{eqn | r = \paren {R * S} \cap T + | c = {{Defof|Symmetric Difference}} +}} +{{end-eqn}} +The second part of the proof is a direct consequence of the fact that [[Intersection is Commutative]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Symmetric Difference of Unions} +Tags: Symmetric Difference, Set Union + +\begin{theorem} +Let $R$, $S$ and $T$ be [[Definition:Set|sets]]. +Then: +:$\paren {R \cup T} * \paren {S \cup T} = \paren {R * S} \setminus T$ +where: +:$*$ denotes the [[Definition:Symmetric Difference|symmetric difference]] +:$\setminus$ denotes [[Definition:Set Difference|set difference]] +:$\cup$ denotes [[Definition:Set Union|set union]] +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \paren {R \cup T} * \paren {S \cup T} + | r = \paren {\paren {R \cup T} \setminus \paren {S \cup T} } \cup \paren {\paren {S \cup T} \setminus \paren {R \cup T} } + | c = {{Defof|Symmetric Difference|index = 1}} +}} +{{eqn | r = \paren {\paren {R \setminus S} \setminus T} \cup \paren {\paren {S \setminus R} \setminus T} + | c = [[Set Difference with Union]] +}} +{{eqn | r = \paren {\paren {R \setminus S} \cup \paren {S \setminus R} } \setminus T + | c = [[Set Difference is Right Distributive over Union]] +}} +{{eqn | r = \paren {R * S} \setminus T + | c = {{Defof|Symmetric Difference|index = 1}} +}} +{{end-eqn}} +{{Qed}} +[[Category:Symmetric Difference]] +[[Category:Set Union]] +2awpfjbdrubswknh7xks2pus30vn6r2 +\end{proof}<|endoftext|> +\section{Intersection with Universe} +Tags: Universe, Set Intersection + +\begin{theorem} +The [[Definition:Set Intersection|intersection]] of a [[Definition:Set|set]] with the [[Definition:Universe (Set Theory)|universe]] is the set itself: +:$\mathbb U \cap S = S$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = S + | o = \subseteq + | r = \mathbb U + | c = {{Defof|Universe (Set Theory)}} +}} +{{eqn | ll= \leadstoandfrom + | l = \mathbb U \cap S + | r = S + | c = [[Intersection with Subset is Subset]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Union with Universe} +Tags: Universe, Set Union + +\begin{theorem} +The [[Definition:Set Union|union]] of a [[Definition:Set|set]] with the [[Definition:Universe (Set Theory)|universe]] is the [[Definition:Universe (Set Theory)|universe]]: +:$\mathbb U \cup S = \mathbb U$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = S + | o = \subseteq + | r = \mathbb U + | c = {{Defof|Universe (Set Theory)}} +}} +{{eqn | ll= \leadstoandfrom + | l = \mathbb U \cup S + | r = \mathbb U + | c = [[Union with Superset is Superset]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Complement of Empty Set is Universe} +Tags: Set Complement, Empty Set, Universe + +\begin{theorem} +The [[Definition:Set Complement|complement]] of the [[Definition:Empty Set|empty set]] is the [[Definition:Universe (Set Theory)|universe]]: +:$\map \complement \O = \mathbb U$ +\end{theorem} + +\begin{proof} +Substitute $\mathbb U$ for $S$ in $\relcomp S \O = S$ from [[Relative Complement of Empty Set]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Complement of Universe is Empty Set} +Tags: Set Complement, Universe, Empty Set + +\begin{theorem} +The [[Definition:Set Complement|complement]] of the [[Definition:Universe (Set Theory)|universe]] is the [[Definition:Empty Set|empty set]]: +:$\map \complement {\mathbb U} = \O$ +\end{theorem} + +\begin{proof} +Substitute $\mathbb U$ for $S$ in $\relcomp S S = \O$ from [[Relative Complement with Self is Empty Set]]. +{{Qed}} +\end{proof}<|endoftext|> +\section{Complement of Complement} +Tags: Set Complement + +\begin{theorem} +The [[Definition:Set Complement|complement]] of the [[Definition:Set Complement|complement]] of a [[Definition:Set|set]] is the set itself: +:$\map \complement {\map \complement S} = S$ +\end{theorem} + +\begin{proof} +Substitute $\mathbb U$ for $S$ and $S$ for $T$ in $\relcomp S {\relcomp S T} = T$ from [[Relative Complement of Relative Complement]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Intersection with Complement} +Tags: Set Complement, Set Intersection, Empty Set + +\begin{theorem} +The [[Definition:Set Intersection|intersection]] of a [[Definition:Set|set]] and its [[Definition:Set Complement|complement]] is the [[Definition:Empty Set|empty set]]: +:$S \cap \map \complement S = \O$ +\end{theorem} + +\begin{proof} +Substitute $\mathbb U$ for $S$ and $S$ for $T$ in $T \cap \relcomp S T = \O$ from [[Intersection with Relative Complement is Empty]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Union with Complement} +Tags: Set Union, Set Complement, Universe + +\begin{theorem} +The [[Definition:Set Union|union]] of a [[Definition:Set|set]] and its [[Definition:Set Complement|complement]] is the [[Definition:Universe (Set Theory)|universe]]: +:$S \cup \map \complement S = \mathbb U$ +\end{theorem} + +\begin{proof} +Substitute $\mathbb U$ for $S$ and $S$ for $T$ in $T \cup \relcomp S T = S$ from [[Union with Relative Complement]]. +{{qed}} +{{LEM|Union with Relative Complement}} +\end{proof}<|endoftext|> +\section{Set with Complement forms Partition} +Tags: Set Complement, Universe + +\begin{theorem} +Let $\varnothing \subset S \subset \mathbb U$. +Then $S$ and its [[Definition:Set Complement|complement]] $\complement \left({S}\right)$ form a [[Definition:Partition (Set Theory)|partition]] of the [[Definition:Universe (Set Theory)|universal set]] $\mathbb U$. +\end{theorem} + +\begin{proof} +Follows directly from [[Set with Relative Complement forms Partition]]: +If $\varnothing \subset T \subset S$, then $\left\{{T, \complement_S \left({T}\right)}\right\}$ is a [[Definition:Partition (Set Theory)|partition]] of $S$. +{{Qed}} +[[Category:Set Complement]] +[[Category:Universe]] +osay7sq601e55jilomfupzcdydrwx3i +\end{proof}<|endoftext|> +\section{Set Difference as Intersection with Complement} +Tags: Set Complement, Set Difference, Set Intersection + +\begin{theorem} +[[Definition:Set Difference|Set difference]] can be expressed as the [[Definition:Set Intersection|intersection]] with the [[Definition:Set Complement|set complement]]: +:$A \setminus B = A \cap \map \complement B$ +\end{theorem} + +\begin{proof} +This follows directly from [[Set Difference as Intersection with Relative Complement]]: +:$A \setminus B = A \cap \relcomp S B$ +Let $S = \Bbb U$. +Since $A, B \subseteq \Bbb U$ by definition of [[Definition:Universe (Set Theory)|the universe]], the result follows. +{{qed}} +\end{proof}<|endoftext|> +\section{Set Difference of Complements} +Tags: Set Difference, Set Complement + +\begin{theorem} +:$\complement \left({S}\right) \setminus \complement \left({T}\right) = T \setminus S$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l=\complement \left({S}\right) \setminus \complement \left({T}\right) + | r=\left\{ {x: x \in \complement \left({S}\right) \land x \notin \complement \left({T}\right)}\right\} + | c=[[Definition:Set Difference|Definition of Set Difference]] +}} +{{eqn | r=\left\{ {x: x \notin S \land x \in T}\right\} + | c=[[Definition:Set Complement|Definition of Complement]] +}} +{{eqn | r=\left\{ {x: x \in T \land x \notin S}\right\} + | c=[[Rule of Commutation]] +}} +{{eqn | r=T \setminus S + | c=[[Definition:Set Difference|Definition of Set Difference]] +}} +{{end-eqn}} +{{qed}} +[[Category:Set Difference]] +[[Category:Set Complement]] +o7i9en0eps0ze8bfgdd7olycgzkpgca +\end{proof}<|endoftext|> +\section{Symmetric Difference of Complements} +Tags: Symmetric Difference, Set Complement + +\begin{theorem} +The [[Definition:Symmetric Difference|symmetric difference]] of two [[Definition:Set|sets]] equals the [[Definition:Symmetric Difference|symmetric difference]] of their [[Definition:Set Complement|complements]]: +:$\map \complement S * \map \complement T = S * T$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \map \complement S * \map \complement T + | r = \paren {\map \complement S \setminus \map \complement T} \cup \paren {\map \complement T \setminus \map \complement S} + | c = {{Defof|Symmetric Difference}} +}} +{{eqn | r = \paren {T \setminus S} \cup \paren {S \setminus T} + | c = [[Set Difference of Complements]] +}} +{{eqn | r = S * T + | c = {{Defof|Symmetric Difference}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Symmetric Difference with Universe} +Tags: Symmetric Difference, Universe, Set Complement + +\begin{theorem} +:$\mathbb U * S = \map \complement S$ +where: +:$\mathbb U$ denotes the [[Definition:Universe (Set Theory)|universe]] +:$*$ denotes [[Definition:Symmetric Difference|symmetric difference]]. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \mathbb U * S + | r = \mathbb U \cup S \setminus \mathbb U \cap S + | c = {{Defof|Symmetric Difference|index = 2}} +}} +{{eqn | r = \mathbb U \cup S \setminus S + | c = [[Intersection with Universe]] +}} +{{eqn | r = \mathbb U \setminus S + | c = [[Union with Universe]] +}} +{{eqn | r = \map \complement S + | c = {{Defof|Set Complement}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Symmetric Difference with Self is Empty Set} +Tags: Symmetric Difference, Empty Set + +\begin{theorem} +The [[Definition:Symmetric Difference|symmetric difference]] of a [[Definition:Set|set]] with itself is the [[Definition:Empty Set|empty set]]: +:$S * S = \O$ +\end{theorem} + +\begin{proof} +This follows directly from [[Symmetric Difference of Equal Sets]]: +:$S * T = \O \iff S = T$ +substituting $S$ for $T$. +{{Qed}} +\end{proof}<|endoftext|> +\section{Symmetric Difference with Complement} +Tags: Symmetric Difference, Set Complement, Universe + +\begin{theorem} +The [[Definition:Symmetric Difference|symmetric difference]] of a [[Definition:Set|set]] with its [[Definition:Set Complement|complement]] is the [[Definition:Universe (Set Theory)|universe]]: +:$S * \relcomp {} S = \mathbb U$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = S * \relcomp {} S + | r = \paren {S \cup \relcomp {} S} \setminus \paren {S \cap \relcomp {} S} + | c = {{Defof|Symmetric Difference|index = 2}} +}} +{{eqn | r = \paren {S \cup \relcomp {} S} \setminus \O + | c = [[Intersection with Complement]] +}} +{{eqn | r = \mathbb U \setminus \O + | c = [[Union with Complement]] +}} +{{eqn | r = \mathbb U + | c = [[Set Difference with Empty Set is Self]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Symmetric Difference is Associative} +Tags: Symmetric Difference, Symmetric Difference is Associative + +\begin{theorem} +[[Definition:Symmetric Difference|Symmetric difference]] is [[Definition:Associative Operation|associative]]: +:$R * \paren {S * T} = \paren {R * S} * T$ +\end{theorem} + +\begin{proof} +We can directly expand the expressions for $R * \paren {S * T}$ and $\paren {R * S} * T$, and see that they come to the same thing. +Expanding the {{RHS}}: +{{begin-eqn}} +{{eqn | o = + | r = \paren {R * S} * T + | c = +}} +{{eqn | r = \paren {\paren {\paren {R \cap \overline S} \cup \paren {\overline R \cap S} } \cap \overline T} \cup \paren {\overline {R * S} \cap T} + | c = {{Defof|Symmetric Difference|index = 3}} +}} +{{eqn | r = \paren {\paren {\paren {R \cap \overline S} \cup \paren {\overline R \cap S} } \cap \overline T} \cup \paren {\overline {\paren {\paren {R \cup S} \cap \paren {\overline R \cup \overline S} } } \cap T} + | c = {{Defof|Symmetric Difference|index = 4}} +}} +{{eqn | r = \paren {R \cap \overline S \cap \overline T} \cup \paren {\overline R \cap S \cap \overline T} \cup \paren {\paren {\overline {\paren {R \cup S} } \cup \overline {\paren {\overline R \cup \overline S} } } \cap T} + | c = [[Intersection Distributes over Union]] and [[De Morgan's Laws (Set Theory)/Set Complement/Complement of Intersection|De Morgan]] +}} +{{eqn | r = \paren {R \cap \overline S \cap \overline T} \cup \paren {\overline R \cap S \cap \overline T} \cup \paren {\paren {\paren {\overline R \cap \overline S} \cup \paren {R \cap S} } \cap T} + | c = [[De Morgan's Laws (Set Theory)/Set Complement/Complement of Union|De Morgan]] +}} +{{eqn | r = \paren {R \cap \overline S \cap \overline T} \cup \paren {\overline R \cap S \cap \overline T} \cup \paren {\overline R \cap \overline S \cap T} \cup \paren {R \cap S \cap T} + | c = [[Intersection Distributes over Union]] +}} +{{end-eqn}} +Expanding the {{LHS}}: +{{begin-eqn}} +{{eqn | o = + | r = R * \paren {S * T} + | c = +}} +{{eqn | r = \paren {R \cap \overline {S * T} } \cup \paren {\overline R \cap \paren {\paren {S \cap \overline T} \cup \paren {\overline S \cap T} } } + | c = {{Defof|Symmetric Difference|index = 3}} +}} +{{eqn | r = \paren {R \cap \overline {\paren {\paren {S \cup T} \cap \paren {\overline S \cup \overline T} } } } \cup \paren {\overline R \cap \paren {\paren {S \cap \overline T} \cup \paren {\overline S \cap T} } } + | c = {{Defof|Symmetric Difference|index = 4}} +}} +{{eqn | r = \paren {R \cap \paren {\overline {\paren {S \cup T} } \cup \overline {\paren {\overline S \cup \overline T} } } } \cup \paren {\overline R \cap S \cap \overline T} \cup \paren {\overline R \cap \overline S \cap T} + | c = [[Intersection Distributes over Union]] and [[De Morgan's Laws (Set Theory)/Set Complement/Complement of Intersection|De Morgan]] +}} +{{eqn | r = \paren {R \cap \paren {\paren {\overline S \cap \overline T} \cup \paren {S \cap T} } } \cup \paren {\overline R \cap S \cap \overline T} \cup \paren {\overline R \cap \overline S \cap T} + | c = [[De Morgan's Laws (Set Theory)/Set Complement/Complement of Union|De Morgan]] +}} +{{eqn | r = \paren {R \cap \overline S \cap \overline T} \cup \paren {R \cap S \cap T} \cup \paren {\overline R \cap S \cap \overline T} \cup \paren {\overline R \cap \overline S \cap T} + | c = [[Intersection Distributes over Union]] +}} +{{end-eqn}} +From [[Union is Commutative]] it is seen that the {{LHS}} and {{RHS}} are the same, and the result is proved. +{{qed}} +\end{proof} + +\begin{proof} +Expanding the {{RHS}}: +{{begin-eqn}} +{{eqn | o = + | r = \paren {R * S} * T + | c = +}} +{{eqn | r = \paren {\paren {\paren {R \cup S} \cap \paren {\overline R \cup \overline S} } \cup T} \cap \paren {\overline {R * S} \cup \overline T} + | c = {{Defof|Symmetric Difference|index = 4}} +}} +{{eqn | r = \paren {\paren {\paren {R \cup S} \cap \paren {\overline R \cup \overline S} } \cup T} \cap \paren {\overline {\paren {\paren {R \cap \overline S} \cup \paren {\overline R \cap S} } } \cup \overline T} + | c = {{Defof|Symmetric Difference|index = 3}} +}} +{{eqn | r = \paren {R \cup S \cup T} \cap \paren {\overline R \cup \overline S \cup T} \cap \paren {\paren {\overline {\paren {R \cap \overline S} } \cap \overline {\paren {\overline R \cap S} } } \cup \overline T} + | c = [[Union Distributes over Intersection]] and [[De Morgan's Laws (Set Theory)/Set Complement/Complement of Union|De Morgan]] +}} +{{eqn | r = \paren {R \cup S \cup T} \cap \paren {\overline R \cup \overline S \cup T} \cap \paren {\paren {\paren {\overline R \cup S} \cap \paren {R \cup \overline S} } \cup \overline T} + | c = [[De Morgan's Laws (Set Theory)/Set Complement/Complement of Intersection|De Morgan]] +}} +{{eqn | r = \paren {R \cup S \cup T} \cap \paren {\overline R \cup \overline S \cup T} \cap \paren {\overline R \cup S \cup \overline T} \cap \paren {R \cup \overline S \cup \overline T} + | c = [[Union Distributes over Intersection]] +}} +{{end-eqn}} +Expanding the {{LHS}}: +{{begin-eqn}} +{{eqn | o = + | r = R * \paren {S * T} + | c = +}} +{{eqn | r = \paren {R \cup \paren {\paren {S \cup T} \cap \paren {\overline S \cup \overline T} } } \cap \paren {\overline R \cup \overline {S * T} } + | c = {{Defof|Symmetric Difference|index = 4}} +}} +{{eqn | r = \paren {R \cup \paren {\paren {S \cup T} \cap \paren {\overline S \cup \overline T} } } \cap \paren {\overline R \cup \overline {\paren {\paren {S \cap \overline T} \cup \paren {\overline S \cap T} } } } + | c = {{Defof|Symmetric Difference|index = 3}} +}} +{{eqn | r = \paren {R \cup S \cup T} \cap \paren {R \cup \overline S \cup \overline T} \cap \paren {\overline R \cup \paren {\overline {\paren {S \cap \overline T} } \cap \overline {\paren {\overline S \cap T} } } } + | c = [[Union Distributes over Intersection]] and [[De Morgan's Laws (Set Theory)/Set Complement/Complement of Union|De Morgan]] +}} +{{eqn | r = \paren {R \cup S \cup T} \cap \paren {R \cup \overline S \cup \overline T} \cap \paren {\overline R \cup \paren {\paren {\overline S \cup T} \cap \paren {S \cup \overline T} } } + | c = [[De Morgan's Laws (Set Theory)/Set Complement/Complement of Intersection|De Morgan]] +}} +{{eqn | r = \paren {R \cup S \cup T} \cap \paren {R \cup \overline S \cup \overline T} \cap \paren {\overline R \cup \overline S \cup T} \cap \paren {\overline R \cup S \cup \overline T} + | c = [[Union Distributes over Intersection]] +}} +{{end-eqn}} +From [[Intersection is Commutative]], it can be seen that the {{LHS}} and {{RHS}} are the same, and the result is proved. +{{qed}} +\end{proof}<|endoftext|> +\section{Cartesian Product is Empty iff Factor is Empty} +Tags: Cartesian Product, Empty Set + +\begin{theorem} +:$S \times T = \O \iff S = \O \lor T = \O$ +Thus: +:$S \times \O = \O = \O \times T$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = S \times T + | o = \ne + | r = \O +}} +{{eqn | ll= \leadstoandfrom + | l = \exists \tuple {s, t} + | o = \in + | r = S \times T + | c = {{Defof|Empty Set}} +}} +{{eqn | ll= \leadstoandfrom + | l = \exists s \in S + | o = \land + | r = \exists t \in T + | c = {{Defof|Cartesian Product}} +}} +{{eqn | ll= \leadstoandfrom + | l = S \ne \O + | o = \land + | r = T \ne \O + | c = {{Defof|Empty Set}} +}} +{{eqn | ll= \leadstoandfrom + | l = \neg \leftparen {S = \O} + | o = \lor + | r = \rightparen {T = \O} + | c = [[De Morgan's Laws (Logic)/Conjunction of Negations|De Morgan's Laws: Conjunction of Negations]] +}} +{{end-eqn}} +So by the [[Rule of Transposition]]: +:$S = \O \lor T = \O \iff S \times T = \O$ +{{Qed}} +\end{proof}<|endoftext|> +\section{Cartesian Product is Anticommutative} +Tags: Cartesian Product + +\begin{theorem} +Let $S, T \ne \O$. +Then: +:$S \times T = T \times S \implies S = T$ +\end{theorem} + +\begin{proof} +Suppose $S \times T = T \times S$. +Then: +{{begin-eqn}} +{{eqn | l = x \in S + | o = \land + | r = y \in T + | c = +}} +{{eqn | ll= \leadstoandfrom + | l = \tuple {x, y} + | o = \in + | r = S \times T + | c = {{Defof|Cartesian Product}} +}} +{{eqn | ll= \leadstoandfrom + | l = \tuple {x, y} + | o = \in + | r = T \times S + | c = [[Definition:By Hypothesis|by hypothesis]] +}} +{{eqn | ll= \leadstoandfrom + | l = x \in T + | o = \land + | r = y \in S + | c = {{Defof|Cartesian Product}} +}} +{{end-eqn}} +Thus it can be seen from the definition of [[Definition:Set Equality|set equality]] that $S \times T = T \times S \implies S = T$. +Note that if $S = \O$ or $T = \O$ then, from [[Cartesian Product is Empty iff Factor is Empty]], $S \times T = T \times S = \O$ whatever $S$ and $T$ are, and the result does not hold. +{{qed}} +\end{proof}<|endoftext|> +\section{Cartesian Product of Subsets} +Tags: Cartesian Product, Subsets + +\begin{theorem} +Let $A, B, S, T$ be [[Definition:Set|sets]] such that $A \subseteq B$ and $S \subseteq T$. +Then: +:$A \times S \subseteq B \times T$ +In addition, if $A, S \ne \O$, then: +:$A \times S \subseteq B \times T \iff A \subseteq B \land S \subseteq T$ +\end{theorem} + +\begin{proof} +First we show that $A \subseteq B \land S \subseteq T \implies A \times S \subseteq B \times T$. +First, let $A = \O$ or $S = \O$. +Then from [[Cartesian Product is Empty iff Factor is Empty]]: +:$A \times S = \O \subseteq B \times T$ +so the result holds. +Next, let $A, S \ne \O$. +Then from [[Cartesian Product is Empty iff Factor is Empty]]: +:$A \times S \ne \O$ +and we can use the following argument: +{{begin-eqn}} +{{eqn | o = + | r = \tuple {x, y} \in A \times S +}} +{{eqn | o = \leadsto + | r = x \in A, y \in S + | c = {{Defof|Cartesian Product}} +}} +{{eqn | o = \leadsto + | r = x \in B, y \in T + | c = {{Defof|Subset}} +}} +{{eqn | o = \leadsto + | r = \tuple {x, y} \in B \times T + | c = {{Defof|Cartesian Product}} +}} +{{end-eqn}} +Thus $A \times S \subseteq B \times T$ as we were to prove. +Now we show that if $A, S \ne \O$, then: +:$A \times S \subseteq B \times T \implies A \subseteq B \land S \subseteq T$ +So suppose that $A \times S \subseteq B \times T$. +First note that if $A = \O$, then $A \times S = \O \subseteq B \times T$, whatever $S$ is, so it is not necessarily the case that $S \subseteq T$. +Similarly if $S = \O$; it is not necessarily the case that $A \subseteq B$. +So that explains the restriction $A, S \ne \O$. +Now, as $A, S \ne \O$, $\exists x \in A, y \in S$. +Thus: +{{begin-eqn}} +{{eqn | o = + | r = x \in A, y \in S +}} +{{eqn | o = \leadsto + | r = \tuple {x, y} \in A \times S + | c = {{Defof|Cartesian Product}} +}} +{{eqn | o = \leadsto + | r = \tuple {x, y} \in B \times T + | c = {{Defof|Subset}} +}} +{{eqn | o = \leadsto + | r = x \in B, y \in T + | c = {{Defof|Cartesian Product}} +}} +{{end-eqn}} +So when $A, S \ne \O$, we have: +:$A \subseteq S \land B \subseteq T \implies A \times S \subseteq B \times T$ +:$A \times S \subseteq B \times T \implies A \subseteq B \land S \subseteq T$ +from which: +:$A \times S \subseteq B \times T \iff A \subseteq B \land S \subseteq T$ +{{qed}} +\end{proof}<|endoftext|> +\section{Cartesian Product of Intersections} +Tags: Cartesian Product, Set Intersection, Cartesian Product of Intersections + +\begin{theorem} +:$\paren {S_1 \cap S_2} \times \paren {T_1 \cap T_2} = \paren {S_1 \times T_1} \cap \paren {S_2 \times T_2}$ +where $S_1, S_2, T_1, T_2$ are [[Definition:Set|sets]]. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | o = + | r = \tuple {x, y} \in \paren {S_1 \cap S_2} \times \paren {T_1 \cap T_2} +}} +{{eqn | o = \leadstoandfrom + | r = \paren {x \in S_1 \land x \in S_2} \land \paren {y \in T_1 \land y \in T_2} + | c = {{Defof|Cartesian Product}} and {{Defof|Set Intersection}} +}} +{{eqn | o = \leadstoandfrom + | r = \paren {x \in S_1 \land y \in T_1} \land \paren {x \in S_2 \land y \in T_2} + | c = [[Rule of Commutation]], [[Rule of Association]] +}} +{{eqn | o = \leadstoandfrom + | r = \tuple {x, y} \in S_1 \times T_1 \land \tuple {x, y} \in S_2 \times T_2 + | c = {{Defof|Cartesian Product}} +}} +{{eqn | o = \leadstoandfrom + | r = \tuple {x, y} \in \paren {S_1 \times T_1} \cap \paren {S_2 \times T_2} + | c = {{Defof|Set Intersection}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Cartesian Product Distributes over Union} +Tags: Cartesian Product, Set Union, Distributive Operations + +\begin{theorem} +[[Definition:Cartesian Product|Cartesian product]] is [[Definition:Distributive Operation|distributive]] over [[Definition:Set Union|union]]: +:$A \times \paren {B \cup C} = \paren {A \times B} \cup \paren {A \times C}$ +:$\paren {B \cup C} \times A = \paren {B \times A} \cup \paren {C \times A}$ +\end{theorem} + +\begin{proof} +Take the result [[Cartesian Product of Unions]]: +:$\paren {S_1 \cup S_2} \times \paren {T_1 \cup T_2} = \paren {S_1 \times T_1} \cup \paren {S_2 \times T_2} \cup \paren {S_1 \times T_2} \cup \paren {S_2 \times T_1}$ +Put $S_1 = S_2 = A, T_1 = B, T_2 = C$: +{{begin-eqn}} +{{eqn | o = + | r = A \times \paren {B \cup C} + | c = +}} +{{eqn | r = \paren {A \cup A} \times \paren {B \cup C} + | c = [[Union is Idempotent]] +}} +{{eqn | r = \paren {A \times B} \cup \paren {A \times C} \cup \paren {A \times C} \cup \paren {A \times B} + | c = [[Cartesian Product of Unions]] +}} +{{eqn | r = \paren {A \times B} \cup \paren {A \times C} + | c = [[Union is Idempotent]] +}} +{{end-eqn}} +Thus: +:$A \times \paren {B \cup C} = \paren {A \times B} \cup \paren {A \times C}$ +The other result is proved similarly. +{{qed}} +\end{proof}<|endoftext|> +\section{Cartesian Product of Unions} +Tags: Cartesian Product, Set Union + +\begin{theorem} +:$\paren {S_1 \cup S_2} \times \paren {T_1 \cup T_2} = \paren {S_1 \times T_1} \cup \paren {S_2 \times T_2} \cup \paren {S_1 \times T_2} \cup \paren {S_2 \times T_1}$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | o = + | r = \tuple {x, y} \in \paren {S_1 \cup S_2} \times \paren {T_1 \cup T_2} +}} +{{eqn | ll= \leadstoandfrom + | o = + | r = \paren {x \in S_1 \lor x \in S_2} +}} +{{eqn | o = \land + | r = \paren {y \in T_1 \lor y \in T_2} + | c = {{Defof|Cartesian Product}} and {{Defof|Set Union}} +}} +{{eqn | ll= \leadstoandfrom + | o = + | r = \paren {\paren {x \in S_1 \lor x \in S_2} \land y \in T_1} +}} +{{eqn | o = \lor + | r = \paren {\paren {x \in S_1 \lor x \in S_2} \land y \in T_2} + | c = [[Rule of Distribution]] +}} +{{eqn | ll= \leadstoandfrom + | o = + | r = \paren {x \in S_1 \land y \in T_1} +}} +{{eqn | o = \lor + | r = \paren {x \in S_2 \land y \in T_1} +}} +{{eqn | o = \lor + | r = \paren {x \in S_1 \land y \in T_2} +}} +{{eqn | o = \lor + | r = \paren {x \in S_2 \land y \in T_2} + | c = [[Rule of Distribution]] +}} +{{eqn | ll= \leadstoandfrom + | o = + | r = \tuple {x, y} \in \paren {S_1 \times T_1} \cup \paren {S_2 \times T_2} \cup \paren {S_1 \times T_2} \cup \paren {S_2 \times T_1} + | c = {{Defof|Cartesian Product}} and {{Defof|Set Union}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Cartesian Product Distributes over Set Difference} +Tags: Cartesian Product, Set Difference, Distributive Operations + +\begin{theorem} +[[Definition:Cartesian Product|Cartesian product]] is [[Definition:Distributive Operation|distributive]] over [[Definition:Set Difference|set difference]]: +:$(1): \quad S \times \paren {T_1 \setminus T_2} = \paren {S \times T_1} \setminus \paren {S \times T_2}$ +:$(2): \quad \paren {T_1 \setminus T_2} \times S = \paren {T_1 \times S} \setminus \paren {T_2 \times S}$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | o = + | r = \tuple {x, y} \in S \times \paren {T_1 \setminus T_2} + | c = +}} +{{eqn | o = \leadstoandfrom + | r = \paren {x \in S} \land \paren {y \in \paren {T_1 \setminus T_2} } + | c = {{Defof|Cartesian Product}} +}} +{{eqn | o = \leadstoandfrom + | r = \paren {x \in S} \land \paren {y \in T_1} \land \paren {y \notin T_2} + | c = {{Defof|Set Difference}} +}} +{{eqn | o = \leadstoandfrom + | r = \paren {\tuple {x, y} \in S \times T_1} \land \paren {\tuple {x, y} \notin S \times T_2} + | c = {{Defof|Cartesian Product}} +}} +{{eqn | o = \leadstoandfrom + | r = \tuple {x, y} \in \paren {S \times T_1} \setminus \paren {S \times T_2} + | c = {{Defof|Set Difference}} +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | o = + | r = \tuple {x, y} \in \paren {T_1 \setminus T_2} \times S + | c = +}} +{{eqn | o = \leadstoandfrom + | r = \paren {x \in \paren {T_1 \setminus T_2} } \land \paren {y \in S} + | c = {{Defof|Cartesian Product}} +}} +{{eqn | o = \leadstoandfrom + | r = \paren {x \in T_1} \land \paren {x \notin T_2} \land \paren {y \in S} + | c = {{Defof|Set Difference}} +}} +{{eqn | o = \leadstoandfrom + | r = \paren {\tuple {x, y} \in T_1 \times S} \land \paren {\tuple {x, y} \notin T_2 \times S} + | c = {{Defof|Cartesian Product}} +}} +{{eqn | o = \leadstoandfrom + | r = \tuple {x, y} \in \paren {T_1 \times S} \setminus \paren {T_2 \times S} + | c = {{Defof|Set Difference}} +}} +{{end-eqn}} +{{Qed}} +\end{proof}<|endoftext|> +\section{Set Difference of Cartesian Products} +Tags: Cartesian Product, Set Difference + +\begin{theorem} +:$\paren {S_1 \times S_2} \setminus \paren {T_1 \times T_2} = \paren {S_1 \times \paren {S_2 \setminus T_2} } \cup \paren {\paren {S_1 \setminus T_1} \times S_2}$ +\end{theorem} + +\begin{proof} +Let $\tuple {x, y} \in \paren {S_1 \times S_2} \setminus \paren {T_1 \times T_2}$. +Then: +{{begin-eqn}} +{{eqn | l = \tuple {x, y} \in S_1 \times S_2 + | o = \land + | r = \tuple {x, y} \notin T_1 \times T_2 + | c = {{Defof|Set Difference}} +}} +{{eqn | ll= \leadstoandfrom + | l = x \in S_1 \land y \in S_2 + | o = \land + | r = \neg \paren {x \in T_1 \land y \in T_2} + | c = {{Defof|Cartesian Product}} +}} +{{eqn | ll= \leadstoandfrom + | l = x \in S_1 \land y \in S_2 + | o = \land + | r = \paren {x \notin T_1 \lor y \notin T_2} + | c = [[De Morgan's Laws (Logic)/Disjunction of Negations|De Morgan's Laws: Disjunction of Negations]] +}} +{{eqn | ll= \leadstoandfrom + | l = \paren {x \in S_1 \land y \in S_2 \land x \notin T_1} + | o = \lor + | r = \paren {x \in S_1 \land y \in S_2 \land y \notin T_2} + | c = [[Rule of Distribution]] +}} +{{eqn | ll= \leadstoandfrom + | l = \paren {x \in S_1 \setminus T_1 \land y \in S_2} + | o = \lor + | r = \paren {x \in S_1 \land y \in S_2 \setminus T_2} + | c = {{Defof|Set Difference}} +}} +{{eqn | ll= \leadstoandfrom + | l = \paren {\tuple {x, y} \in \paren {S_1 \setminus T_1} \times S_2} + | o = \lor + | r = \paren {\tuple {x, y} \in S_1 \times \paren {S_2 \setminus T_2} } + | c = {{Defof|Cartesian Product}} +}} +{{eqn | ll= \leadstoandfrom + | l = \tuple {x, y} + | o = \in + | r = \paren {S_1 \times \paren {S_2 \setminus T_2} } \cup \paren {\paren {S_1 \setminus T_1} \times S_2} + | c = {{Defof|Set Union}} +}} +{{end-eqn}} +The result follows from the definition of [[Definition:Subset|subset]] and [[Definition:Set Equality|set equality]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Inverse of Inverse Relation} +Tags: Inverse Relations + +\begin{theorem} +The inverse of an [[Definition:Inverse Relation|inverse relation]] is the [[Definition:Relation|relation]] itself: +:$\paren {\RR^{-1} }^{-1} = \RR$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \tuple {s, t} + | o = \in + | r = \RR + | c = +}} +{{eqn | ll= \leadstoandfrom + | l = \tuple {t, s} + | o = \in + | r = \RR^{-1} + | c = {{Defof|Inverse Relation}} +}} +{{eqn | ll= \leadstoandfrom + | l = \tuple {s, t} + | o = \in + | r = \paren {\RR^{-1} }^{-1} + | c = {{Defof|Inverse Relation}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Diagonal Relation is Equivalence} +Tags: Diagonal Relation, Examples of Equivalence Relations, Equality + +\begin{theorem} +The [[Definition:Diagonal Relation|diagonal relation]] $\Delta_S$ on a [[Definition:Set|set]] $S$ is always an [[Definition:Equivalence Relation|equivalence]] in $S$. +\end{theorem} + +\begin{proof} +Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: +=== [[Diagonal Relation is Reflexive|Reflexive]] === +{{:Diagonal Relation is Reflexive}} +{{qed|lemma}} +=== [[Diagonal Relation is Symmetric|Symmetric]] === +{{:Diagonal Relation is Symmetric}} +{{qed|lemma}} +=== [[Diagonal Relation is Transitive|Transitive]] === +{{:Diagonal Relation is Transitive}} +{{qed}} +\end{proof}<|endoftext|> +\section{Trivial Relation is Equivalence} +Tags: Equivalence Relations, Trivial Relation + +\begin{theorem} +The [[Definition:Trivial Relation|trivial relation]] on $S$: +:$\RR = S \times S$ +is always an [[Definition:Equivalence Relation|equivalence]] in $S$. +\end{theorem} + +\begin{proof} +Let us verify the conditions for an [[Definition:Equivalence Relation|equivalence]] in turn. +=== Reflexivity === +For $\RR$ to be [[Definition:Reflexive Relation|reflexive]] means: +:$\forall x \in S: \tuple {x, x} \in S \times S$ +which is trivial by definition of the [[Definition:Cartesian Product|Cartesian product]] $S \times S$. +{{qed|lemma}} +=== Symmetry === +For $\RR$ to be [[Definition:Symmetric Relation|symmetric]] means: +:$\forall x, y \in S: \tuple {x, y} \in S \times S \land \tuple {y, x} \in S \times S$ +Since we have by definition of [[Definition:Cartesian Product|Cartesian product]] that: +:$\forall x, y \in S: \tuple {y, x} \in S \times S$ +this follows by [[True Statement is implied by Every Statement]]. +{{qed|lemma}} +=== Transitivity === +For $\RR$ to be [[Definition:Transitive Relation|transitive]] means: +:$\tuple {x, y} \in S \times S \land \tuple {y, z} \in S \times S \implies \tuple {x, z} \in S \times S$ +By definition of [[Definition:Cartesian Product|Cartesian product]], we have that: +:$\forall x, z \in S: \tuple {x, z} \in S \times S$ +hence by [[True Statement is implied by Every Statement]], it follows that $\RR$ is [[Definition:Transitive Relation|transitive]]. +{{qed|lemma}} +Having verified all three conditions, we conclude $\RR$ is an [[Definition:Equivalence Relation|equivalence]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Equality of Relations} +Tags: Relation Theory, Equality + +\begin{theorem} +Let $\RR_1$ and $\RR_2$ be [[Definition:Relation|relations]] on $S_1 \times T_1$ and $S_2 \times T_2$ respectively. +Then $\RR_1$ and $\RR_2$ are '''[[Definition:Equality|equal]]''' {{iff}}: +:$S_1 = S_2$ +:$T_1 = T_2$ +:$\tuple {s, t} \in \RR_1 \iff \tuple {s, t} \in \RR_2$ +It is worth labouring the point that for two relations to be '''equal''', not only must their [[Definition:Domain of Relation|domains]] be equal, but so must their [[Definition:Codomain of Relation|codomains]]. +\end{theorem} + +\begin{proof} +This follows from [[Definition:Set Equality|set equality]] and [[Equality of Ordered Pairs]]. +A '''relation''' on $S \times T$ is an [[Definition:Ordered Triple|ordered triple]], thus: +{{begin-eqn}} +{{eqn | l = \RR_1 + | r = \tuple {S_1, T_1, R_1} + | c = where $R_1 \subseteq S_1 \times T_1$ +}} +{{eqn | l = \RR_2 + | r = \tuple {S_2, T_2, R_2} + | c = where $R_2 \subseteq S_2 \times T_2$ +}} +{{end-eqn}} +From [[Equality of Ordered Tuples]], $\RR_1 = \RR_1$ {{iff}}: +{{begin-eqn}} +{{eqn | l = S_1 + | r = S_2 +}} +{{eqn | l = T_1 + | r = T_2 +}} +{{eqn | l = R_1 + | r = R_2 +}} +{{end-eqn}} +Then we have that $R_1$ and $R_2$ are [[Definition:Set|sets]] of [[Definition:Ordered Pair|ordered pairs]]. +Thus by definition of [[Definition:Set Equality|set equality]]: +:$R_1 = R_2 \iff \paren {\forall \tuple {x, y} \in \S_1 \times T_2: \tuple {x, y} \in R_1 \iff \tuple {x, y} \in R_2}$ +Hence the result. +{{Qed}} +\end{proof}<|endoftext|> +\section{Image of Singleton under Relation} +Tags: Relation Theory, Singletons + +\begin{theorem} +Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]]. +Then the [[Definition:Image of Element under Relation|image]] of an [[Definition:Element|element]] of $S$ is equal to the [[Definition:Image of Subset under Relation|image]] of a [[Definition:Singleton|singleton]] containing that element, the singleton being a [[Definition:Subset|subset]] of $S$: +:$\forall s \in S: \map \RR s = \RR \sqbrk {\set s}$ +\end{theorem} + +\begin{proof} +We have the definitions: +{{begin-eqn}} +{{eqn | l = \map \RR s + | r = \set {t \in T: \tuple {s, t} \in \RR} + | c = {{Defof|Image of Element under Relation}} +}} +{{eqn | l = \RR \sqbrk {\set s} + | r = \set {t \in T: \exists s \in \set s: \tuple {s, t} \in \RR} + | c = {{Defof|Image of Subset under Relation}} +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | l = t + | o = \in + | r = \map \RR s +}} +{{eqn | ll= \leadsto + | l = \tuple {s, t} + | o = \in + | r = \RR +}} +{{eqn | ll= \leadsto + | lo= \exists s \in \set s: + | l = \tuple {s, t} + | o = \in + | r = \RR + | c = {{Defof|Singleton}} +}} +{{eqn | ll= \leadsto + | l = t + | o = \in + | r = \RR \sqbrk {\set s} +}} +{{eqn | ll= \leadsto + | l = \map \RR s + | o = \subseteq + | r = \RR \sqbrk {\set s} + | c = {{Defof|Subset}} +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | l = t + | o = \in + | r = \RR \sqbrk {\set s} +}} +{{eqn | ll= \leadsto + | lo= \exists s \in \set s: + | l = \tuple {s, t} + | o = \in + | r = \RR +}} +{{eqn | ll= \leadsto + | l = \tuple {s, t} + | o = \in + | r = \RR + | c = as $r \in \set s \implies r = s$ by [[Singleton Equality]] +}} +{{eqn | ll= \leadsto + | l = t + | o = \in + | r = \map \RR s +}} +{{eqn | ll= \leadsto + | l = \RR \sqbrk {\set s} + | o = \subseteq + | r = \map \RR s +}} +{{end-eqn}} +Finally: +:$\map \RR s = \RR \sqbrk {\set s}$ +by definition of [[Definition:Set Equality|set equality]]. +{{Qed}} +[[Category:Relation Theory]] +[[Category:Singletons]] +8ype9fcu1jses6a8x537szz5bvbbcnt +\end{proof}<|endoftext|> +\section{Image of Subset under Relation is Subset of Image} +Tags: Subsets, Relation Theory + +\begin{theorem} +Let $S$ and $T$ be [[Definition:Set|sets]]. +Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]] from $S$ to $T$. +Let $A, B \subseteq S$ such that $A \subseteq B$. +Then the [[Definition:Image of Subset under Relation|image]] of $A$ is a [[Definition:Subset|subset]] of the [[Definition:Image of Subset under Relation|image]] of $B$: +:$A \subseteq B \implies \RR \sqbrk A \subseteq \RR \sqbrk B$ +In the notation of [[Definition:Direct Image Mapping of Relation|direct image mappings]], this can be written: +:$A \subseteq B \implies \map {\RR^\to} A \subseteq \map {\RR^\to} B$ +\end{theorem} + +\begin{proof} +Suppose $\RR \sqbrk A \not \subseteq \RR \sqbrk B$. +{{begin-eqn}} +{{eqn | l = \RR \sqbrk A + | o = \nsubseteq + | r = \RR \sqbrk B +}} +{{eqn | ll= \leadsto + | l = \exists t \in \RR \sqbrk A: \exists \tuple {s, t} \in \RR: s + | o = \notin + | r = B + | c = {{Defof|Image of Subset under Relation}} +}} +{{eqn | ll= \leadsto + | l = \exists s \notin B: s + | o = \in + | r = A + | c = {{Defof|Ordered Pair}} +}} +{{eqn | ll= \leadsto + | l = A + | o = \nsubseteq + | r = B + | c = {{Defof|Subset}} +}} +{{end-eqn}} +The result follows by the [[Rule of Transposition]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Image of Element is Subset} +Tags: Relation Theory + +\begin{theorem} +Let $S$ and $T$ be [[Definition:Set|sets]]. +Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]]. +Let $A \subseteq S$. +Then: +:$s \in A \implies \map \RR s \subseteq \RR \sqbrk A$ +\end{theorem} + +\begin{proof} +From [[Image of Singleton under Relation]]: +:$\map \RR s = \RR \sqbrk {\set s}$ +From [[Singleton of Element is Subset]]: +:$s \in A \implies \set s \subseteq A$ +The result follows from [[Image of Subset is Subset of Image]]. +{{qed}} +[[Category:Relation Theory]] +81ns2wlo73dzga5xv99qsw98k988h3w +\end{proof}<|endoftext|> +\section{Image is Subset of Codomain} +Tags: Relation Theory + +\begin{theorem} +Let $\RR = S \times T$ be a [[Definition:Relation|relation]]. +For all [[Definition:Subset|subsets]] $A$ of the [[Definition:Domain of Relation|domain]] of $\RR$, the [[Definition:Image of Subset under Relation|image]] of $A$ is a subset of the [[Definition:Codomain of Relation|codomain]] of $\RR$: +:$\forall A \subseteq \Dom \RR: \RR \sqbrk A \subseteq T$ +In the notation of [[Definition:Direct Image Mapping of Relation|direct image mappings]], this can be written as: +:$\forall A \in \powerset S: \map {\RR^\to} A \in \powerset T$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = A + | o = \subseteq + | r = \Dom \RR + | c = +}} +{{eqn | ll= \leadsto + | l = \RR \sqbrk A + | o = \subseteq + | r = \Img \RR + | c = [[Image of Subset is Subset of Image]] +}} +{{eqn | o = \subseteq + | r = T + | c = +}} +{{end-eqn}} +{{qed}} +[[Category:Relation Theory]] +hy4x25yvdjd9rj1cqotvh3g8hhde5km +\end{proof}<|endoftext|> +\section{Image of Empty Set is Empty Set} +Tags: Relation Theory, Empty Set + +\begin{theorem} +Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]]. +The [[Definition:Image of Subset under Relation|image]] of the [[Definition:Empty Set|empty set]] is the [[Definition:Empty Set|empty set]]: +:$\RR \sqbrk \O = \O$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \RR \sqbrk \O + | r = \set {t \in \Rng \RR: \exists s \in \O: \tuple {s, t} \in \RR} + | c = {{Defof|Image of Subset under Relation}} +}} +{{eqn | l = \neg \exists s + | o = \in + | r = \O + | c = {{Defof|Empty Set}} +}} +{{eqn | ll= \leadsto + | l = \neg \exists t + | o = \in + | r = \set {t \in \Rng \RR: \exists s \in \O: \tuple {s, t} \in \RR} + | c = +}} +{{eqn | ll= \leadsto + | l = \RR \sqbrk \O + | r = \O + | c = {{Defof|Empty Set}} +}} +{{end-eqn}} +{{qed}} +[[Category:Relation Theory]] +[[Category:Empty Set]] +t7v5c6z280lgdvtczg7afcei0864pgv +\end{proof}<|endoftext|> +\section{Domain of Composite Relation} +Tags: Composite Relations + +\begin{theorem} +Let $\RR_2 \circ \RR_1$ be a [[Definition:Composition of Relations|composite relation]]. +Then the [[Definition:Domain of Relation|domain]] of $\RR_2 \circ \RR_1$ is the [[Definition:Domain of Relation|domain]] of $\RR_1$: +:$\Dom {\RR_2 \circ \RR_1} = \Dom {\RR_1}$ +\end{theorem} + +\begin{proof} +Let $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$. +The [[Definition:Domain of Relation|domain]] of $\RR_1$ is $S_1$. +The composite of $\RR_1$ and $\RR_2$ is defined as: +:$\RR_2 \circ \RR_1 = \set {\tuple {x, z}: x \in S_1, z \in S_3: \exists y \in S_2: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2}$ +From this definition: +:$\RR_2 \circ \RR_1 \subseteq S_1 \times S_3$. +Thus the [[Definition:Domain of Relation|domain]] of $\RR_2 \circ \RR_1$ is $S_1$. +Thus: +:$\Dom {\RR_2 \circ \RR_1} = S_1 = \Dom {\RR_1}$ +{{qed}} +[[Category:Composite Relations]] +174sj1pl3roals6ya0vhrygzissq3nb +\end{proof}<|endoftext|> +\section{Codomain of Composite Relation} +Tags: Composite Relations + +\begin{theorem} +Let $\RR_2 \circ \RR_1$ be a [[Definition:Composition of Relations|composite relation]]. +Then the [[Definition:Codomain of Relation|codomain]] of $\RR_2 \circ \RR_1$ is the [[Definition:Codomain of Relation|codomain]] of $\RR_2$: +:$\Cdm {\RR_2 \circ \RR_1} = \Cdm {\RR_2}$ +\end{theorem} + +\begin{proof} +Let $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$. +The [[Definition:Codomain of Relation|codomain]] of $\RR_2$ is $S_3$. +The composite of $\RR_1$ and $\RR_2$ is defined as: +:$\RR_2 \circ \RR_1 = \set {\tuple {x, z}: x \in S_1, z \in S_3: \exists y \in S_2: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2}$ +From this definition: +:$\RR_2 \circ \RR_1 \subseteq S_1 \times S_3$ +Thus the [[Definition:Codomain of Relation|codomain]] of $\RR_2 \circ \RR_1$ is $S_3$. +Thus: +:$\Cdm {\RR_2 \circ \RR_1} = S_3 = \Cdm {\RR_2}$ +{{qed}} +[[Category:Composite Relations]] +1ud451rt4knvb7jiqbis137ljeuaqx9 +\end{proof}<|endoftext|> +\section{Preimage of Mapping equals Domain} +Tags: Preimages under Mappings + +\begin{theorem} +The [[Definition:Preimage of Mapping|preimage of a mapping]] is the same [[Definition:Set|set]] as its [[Definition:Domain of Mapping|domain]]: +:$\Preimg f = \Dom f$ +\end{theorem} + +\begin{proof} +Let $f \subseteq S \times T$ be a [[Definition:Mapping|mapping]]. +Then: +{{begin-eqn}} +{{eqn | o = + | r = \forall x \in S: \exists y \in T: \tuple {x, y} \in f + | c = {{Defof|Mapping}} +}} +{{eqn | o = \implies + | r = \forall x \in S: x \in \Preimg f + | c = {{Defof|Preimage of Mapping}} +}} +{{eqn | o = \implies + | r = S \subseteq \Preimg f + | c = {{Defof|Subset}} +}} +{{end-eqn}} +From [[Preimage of Relation is Subset of Domain]], we have that $\Preimg f \subseteq S$. +The result follows from the definition of [[Definition:Set Equality|set equality]]. +{{qed}} +[[Category:Preimages under Mappings]] +1u6shpayoujr0isjcfwbj0lesidhakp +\end{proof}<|endoftext|> +\section{Preimage of Relation is Subset of Domain} +Tags: Relation Theory + +\begin{theorem} +Let $\mathcal R \subseteq S \times T$ be a [[Definition:Relation|relation]]. +Then the [[Definition:Preimage of Relation|preimage]] of $\mathcal R$ is a [[Definition:Subset|subset]] of its [[Definition:Domain of Relation|domain]]: +:$\Preimg {\mathcal R} \subseteq S$ +\end{theorem} + +\begin{proof} +The [[Definition:Preimage of Relation|preimage]] of $\mathcal R$ is defined as: +:$\Preimg {\mathcal R} = \set {s \in \Dom {\mathcal R}: \exists t \in \Rng {\mathcal R}: \tuple {s, t} \in \mathcal R}$ +Hence: +{{begin-eqn}} +{{eqn | l = s + | o = \in + | r = \Preimg {\mathcal R} + | c = +}} +{{eqn | ll= \leadsto + | l = s + | o = \in + | r = \Dom {\mathcal R} + | c = {{Defof|Preimage of Relation}} +}} +{{eqn | ll= \leadsto + | l = \Preimg {\mathcal R} + | o = \subseteq + | r = \Dom {\mathcal R} + | c = [[Subset of Set with Propositional Function]] +}} +{{end-eqn}} +{{qed}} +[[Category:Relation Theory]] +k59tjbklrx0yccrj8xlj3fexm3y3uju +\end{proof}<|endoftext|> +\section{Preimage of Image under Left-Total Relation is Superset} +Tags: Preimages under Relations, Composite Relations + +\begin{theorem} +Let $\RR \subseteq S \times T$ be a [[Definition:Left-Total Relation|left-total relation]]. +Then: +:$A \subseteq S \implies A \subseteq \paren {\RR^{-1} \circ \RR} \sqbrk A$ +where: +:$\RR \sqbrk A$ denotes the [[Definition:Image of Subset under Relation|image of $A$ under $\RR$]] +:$\RR^{-1} \sqbrk A$ denotes the [[Definition:Preimage of Subset under Relation|preimage of $A$ under $\RR$]] +:$\RR^{-1} \circ \RR$ denotes [[Definition:Composition of Relations|composition]] of $\RR^{-1}$ and $\RR$. +This can be expressed in the language and notation of [[Definition:Direct Image Mapping of Relation|direct image mappings]] and [[Definition:Inverse Image Mapping of Relation|inverse image mappings]] as: +:$\forall A \in \powerset S: A \subseteq \map {\paren {\RR^\gets \circ \RR^\to} } A$ +\end{theorem} + +\begin{proof} +Suppose $A \subseteq S$. +We have: +{{begin-eqn}} +{{eqn | l = x + | o = \in + | r = A +}} +{{eqn | ll= \leadsto + | l = \set x + | o = \subseteq + | r = A + | c = [[Singleton of Element is Subset]] +}} +{{eqn | ll= \leadsto + | l = \RR \sqbrk {\set x} + | o = \subseteq + | r = \RR \sqbrk A + | c = [[Image of Subset under Relation is Subset of Image]] +}} +{{eqn | ll= \leadsto + | l = \RR^{-1} \sqbrk {\RR \sqbrk {\set x} } + | o = \subseteq + | r = \RR^{-1} \sqbrk {\RR \sqbrk A} + | c = [[Image of Subset under Relation is Subset of Image/Corollary 1|Image of Subset under Relation is Subset of Image: Corollary 1]] +}} +{{end-eqn}} +Then we have that $\RR$ is a [[Definition:Left-Total Relation|left-total relation]]. +Thus: +:$\exists t \in T: \tuple {x, t} \in \RR$ +Hence by definition of [[Definition:Inverse Relation|inverse relation]]: +:$\tuple {t, x} \in \RR^{-1}$ +Hence: +{{begin-eqn}} +{{eqn | ll= \leadsto + | l = x + | o = \in + | r = \RR^{-1} \sqbrk {\RR \sqbrk A} + | c = {{Defof|Relation}} +}} +{{eqn | ll= \leadsto + | l = x + | o = \in + | r = \paren {\RR^{-1} \circ \RR} \sqbrk A + | c = {{Defof|Composition of Relations}} +}} +{{end-eqn}} +So by definition of [[Definition:Subset|subset]]: +:$A \subseteq S \implies A \subseteq \paren {\RR^{-1} \circ \RR} \sqbrk A$ +{{qed}} +\end{proof}<|endoftext|> +\section{Inverse of Composite Relation} +Tags: Inverse Relations, Composite Relations + +\begin{theorem} +Let $\RR_2 \circ \RR_1 \subseteq S_1 \times S_3$ be the [[Definition:Composition of Relations|composite]] of the two [[Definition:Relation|relations]] $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$. +Then: +:$\paren {\RR_2 \circ \RR_1}^{-1} = \RR_1^{-1} \circ \RR_2^{-1}$ +\end{theorem} + +\begin{proof} +Let $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$ be [[Definition:Relation|relations]]. +We assume that: +:$\Dom {\RR_2} = \Cdm {\RR_1}$ +where $\Dom \RR$ denotes [[Definition:Domain of Relation|domain]] and $\Cdm \RR$ denotes [[Definition:Codomain of Relation|codomain]] of a [[Definition:Relation|relation]] $\RR$. +This is necessary for $\RR_2 \circ \RR_1$ to exist. +From the [[Definition:Inverse Relation|definition of an inverse relation]], we have: +:$\Dom {\RR_2} = \Cdm {\RR_2^{-1} }$ +:$\Cdm {\RR_1} = \Dom {\RR_1^{-1} }$ +So we confirm that $\RR_1^{-1} \circ \RR_2^{-1}$ is defined. +{{begin-eqn}} +{{eqn | l = \RR_2 \circ \RR_1 + | r = \set {\tuple {x, z}: x \in S_1, z \in S_3: \exists y \in S_2: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2} + | c = {{Defof|Composition of Relations}} +}} +{{eqn | ll= \leadsto + | l = \paren {\RR_2 \circ \RR_1}^{-1} + | r = \set {\tuple {z, x}: \tuple {x, z} \in \RR_2 \circ \RR_1} + | c = {{Defof|Inverse Relation}} +}} +{{eqn | r = \set {\tuple {z, x}: x \in S_1, z \in S_3: \exists y \in S_2: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2} + | c = {{Defof|Composition of Relations}} +}} +{{eqn | r = \set {\tuple {z, x}: z \in S_3, x \in S_1: \exists y \in S_2: \tuple {z, y} \in \RR_2^{-1} \land \tuple {y, x} \in \RR_1^{-1} } + | c = {{Defof|Inverse Relation}} +}} +{{eqn | r = \RR_1^{-1} \circ \RR_2^{-1} + | c = {{Defof|Composition of Relations}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Image of Union under Relation} +Tags: Relation Theory, Set Union, Image of Union under Relation + +\begin{theorem} +Let $S$ and $T$ be [[Definition:Set|sets]]. +Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]]. +Let $S_1$ and $S_2$ be [[Definition:Subset|subsets]] of $S$. +Then: +:$\RR \sqbrk {S_1 \cup S_2} = \RR \sqbrk {S_1} \cup \RR \sqbrk {S_2}$ +That is, the [[Definition:Image of Subset under Relation|image]] of the [[Definition:Set Union|union]] of [[Definition:Subset|subsets]] of $S$ is equal to the [[Definition:Set Union|union]] of their [[Definition:Image of Subset under Relation|images]]. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = t + | o = \in + | r = \RR \sqbrk {S_1 \cup S_2} + | c = +}} +{{eqn | ll= \leadstoandfrom + | l = \exists s \in S_1 \cup S_2: t + | o = \in + | r = \RR \sqbrk s + | c = {{Defof|Image of Subset under Relation}} +}} +{{eqn | ll= \leadstoandfrom + | l = \exists s: s \in S_1 \lor s \in S_2: t + | o = \in + | r = \RR \sqbrk s + | c = {{Defof|Set Union}} +}} +{{eqn | ll= \leadstoandfrom + | l = t + | o = \in + | r = \RR \sqbrk {S_1} \lor t \in \RR \sqbrk {S_2} + | c = {{Defof|Image of Subset under Relation}} +}} +{{eqn | ll= \leadstoandfrom + | l = t + | o = \in + | r = \RR \sqbrk {S_1} \cup \RR \sqbrk {S_2} + | c = {{Defof|Set Union}} +}} +{{end-eqn}} +{{Qed}} +\end{proof}<|endoftext|> +\section{Image of Intersection under Relation} +Tags: Relation Theory, Set Intersection + +\begin{theorem} +Let $S$ and $T$ be [[Definition:Set|sets]]. +Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]]. +Let $S_1$ and $S_2$ be [[Definition:Subset|subsets]] of $S$. +Then: +:$\RR \sqbrk {S_1 \cap S_2} \subseteq \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$ +That is, the [[Definition:Image of Subset under Relation|image]] of the [[Definition:Set Intersection|intersection]] of [[Definition:Subset|subsets]] of $S$ is a [[Definition:Subset|subset]] of the [[Definition:Set Intersection|intersection]] of their [[Definition:Image of Subset under Relation|images]]. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = S_1 \cap S_2 + | o = \subseteq + | r = S_1 + | c = [[Intersection is Subset]] +}} +{{eqn | ll= \leadsto + | l = \RR \sqbrk {S_1 \cap S_2} + | o = \subseteq + | r = \RR \sqbrk {S_1} + | c = [[Image of Subset is Subset of Image]] +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | l = S_1 \cap S_2 + | o = \subseteq + | r = S_2 + | c = [[Intersection is Subset]] +}} +{{eqn | ll= \leadsto + | l = \RR \sqbrk {S_1 \cap S_2} + | o = \subseteq + | r = \RR \sqbrk {S_2} + | c = [[Image of Subset is Subset of Image]] +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | ll= \leadsto + | l = \RR \sqbrk {S_1 \cap S_2} + | o = \subseteq + | r = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2} + | c = [[Intersection is Largest Subset]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Image of Set Difference under Relation} +Tags: Relation Theory, Set Difference, Image of Set Difference under Relation + +\begin{theorem} +Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]]. +Let $A$ and $B$ be [[Definition:Subset|subsets]] of $S$. +Then: +:$\RR \sqbrk A \setminus \RR \sqbrk B \subseteq \RR \sqbrk {A \setminus B}$ +where: +:$\setminus$ denotes [[Definition:Set Difference|set difference]] +:$\RR \sqbrk A$ denotes [[Definition:Image of Subset under Relation|image of $A$ under $\RR$]]. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = y + | o = \in + | r = \RR \sqbrk A \setminus \RR \sqbrk B + | c = +}} +{{eqn | ll= \leadsto + | l = \exists x \in A: x \notin B: \tuple {x, y} + | o = \in + | r = \RR + | c = {{Defof|Image of Subset under Relation}} +}} +{{eqn | ll= \leadsto + | l = \exists x \in A \setminus B: \tuple {x, y} + | o = \in + | r = \RR + | c = {{Defof|Set Difference}} +}} +{{eqn | ll= \leadsto + | l = y + | o = \in + | r = \RR \sqbrk {A \setminus B} + | c = {{Defof|Image of Subset under Relation}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Preimage of Union under Relation} +Tags: Preimages under Relations, Set Union, Preimage of Union under Relation + +\begin{theorem} +Let $S$ and $T$ be [[Definition:Set|sets]]. +Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]]. +Let $T_1$ and $T_2$ be [[Definition:Subset|subsets]] of $T$. +Then: +:$\RR^{-1} \sqbrk {T_1 \cup T_2} = \RR^{-1} \sqbrk {T_1} \cup \RR^{-1} \sqbrk {T_2}$ +\end{theorem} + +\begin{proof} +We have that $\RR^{-1}$ is a [[Definition:Relation|relation]] +The result follows from [[Image of Union under Relation]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Preimage of Intersection under Relation} +Tags: Relation Theory, Set Intersection + +\begin{theorem} +Let $S$ and $T$ be [[Definition:Set|sets]]. +Let $\mathcal R \subseteq S \times T$ be a [[Definition:Relation|relation]]. +Let $C$ and $D$ be [[Definition:Subset|subsets]] of $T$. +Then: +:$\mathcal R^{-1} \left[{C \cap D}\right] \subseteq \mathcal R^{-1} \left[{C}\right] \cap \mathcal R^{-1} \left[{D}\right]$ +\end{theorem} + +\begin{proof} +This follows from [[Image of Intersection under Relation]], and the fact that $\mathcal R^{-1}$ is itself a [[Definition:Relation|relation]], and therefore obeys the same rules. +{{qed}} +[[Category:Relation Theory]] +[[Category:Set Intersection]] +bvvhm3d3k4ssf054yk53qc4wbzl6epq +\end{proof}<|endoftext|> +\section{Preimage of Set Difference under Relation} +Tags: Preimages under Relations, Set Difference + +\begin{theorem} +Let $\mathcal R \subseteq S \times T$ be a [[Definition:Relation|relation]]. +Let $C$ and $D$ be [[Definition:Subset|subsets]] of $T$. +Then: +:$\mathcal R^{-1} \sqbrk C \setminus \mathcal R^{-1} \sqbrk D \subseteq \mathcal R^{-1} \sqbrk {C \setminus D}$ +where: +:$\setminus$ denotes [[Definition:Set Difference|set difference]] +:$\mathcal R^{-1} \sqbrk C$ denotes the [[Definition:Preimage of Subset under Relation|preimage of $C$ under $\mathcal R$]]. +\end{theorem} + +\begin{proof} +We have that $\mathcal R^{-1}$ is itself a [[Definition:Relation|relation]] +The result then follows from [[Image of Set Difference under Relation]]. +{{qed}} +[[Category:Preimages under Relations]] +[[Category:Set Difference]] +09fl9rygc9mkeiqhl5fucr63f9pru6w +\end{proof}<|endoftext|> +\section{Restriction is Subset of Relation} +Tags: Restrictions + +\begin{theorem} +Let $\mathcal R \subseteq S \times T$ be a [[Definition:Relation|relation]]. +Let $X \subseteq S$. +Then the [[Definition:Restriction of Relation|restriction]] of $\mathcal R$ to $X$ is a [[Definition:Subset|subset]] of $\mathcal R$. +\end{theorem} + +\begin{proof} +From the definition of [[Definition:Restriction of Relation|restriction]]: +:$\forall x \in X: \mathcal R \restriction_X \left({x}\right) = \mathcal R \left({x}\right)$ +Thus: +:$\forall x \in X: \exists t \in T: \left({x, t}\right) \in \mathcal R \restriction_X$ +But $\left({x, t}\right)$ is also (by definition) in $\mathcal R$. +It follows that: +:$\mathcal R \restriction_X \subseteq \mathcal R$ +{{qed}} +[[Category:Restrictions]] +kb9s90y6df7l76lo3j7hpdxrnt7sg0e +\end{proof}<|endoftext|> +\section{Null Relation is Antireflexive, Symmetric and Transitive} +Tags: Reflexive Relations, Symmetric Relations, Transitive Relations, Null Relation + +\begin{theorem} +Let $S$ be a [[Definition:Set|set]] which is [[Definition:Non-Empty Set|non-empty]]. +Let $\RR \subseteq S \times S$ be the [[Definition:Null Relation|null relation]]. +Then $\RR$ is [[Definition:Antireflexive Relation|antireflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]]. +If $S = \O$ then [[Relation on Empty Set is Equivalence]] applies. +\end{theorem} + +\begin{proof} +From the definition of [[Definition:Null Relation|null relation]]: +:$\RR = \O$ +=== Antireflexivity === +This follows directly from the definition: +:$\RR = \O \implies \forall x \in S: \tuple {x, x} \notin \RR$ +and so $\RR$ is [[Definition:Antireflexive Relation|antireflexive]]. +{{qed|lemma}} +=== Symmetry === +It follows [[Definition:Vacuous Truth|vacuously]] that: +:$\tuple {x, y} \in \RR \implies \tuple {y, x} \in \RR$ +and so $\RR$ is [[Definition:Symmetric Relation|symmetric]]. +{{qed|lemma}} +=== Transitivity === +It follows [[Definition:Vacuous Truth|vacuously]] that: +:$\tuple {x, y}, \tuple {y, z} \in \RR \implies \tuple {x, z} \in \RR$ +and so $\RR$ is [[Definition:Transitive Relation|transitive]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Symmetry of Relations is Symmetric} +Tags: Symmetric Relations + +\begin{theorem} +Let $\mathcal R$ be a [[Definition:Relation|relation]] on $S$ which is [[Definition:Symmetric Relation|symmetric]]. Then: +:$\left({x, y}\right) \in \mathcal R \iff \left({y, x}\right) \in \mathcal R$. +\end{theorem} + +\begin{proof} +Let $\mathcal R$ be [[Definition:Symmetric Relation|symmetric]]. +{{begin-eqn}} +{{eqn | l=\left({x, y}\right) \in \mathcal R + | o=\implies + | r=\left({y, x}\right) \in \mathcal R + | c=Definition of [[Definition:Symmetric Relation|Symmetric]] +}} +{{eqn | l=\left({y, x}\right) \in \mathcal R + | o=\implies + | r=\left({x, y}\right) \in \mathcal R + | c=Definition of [[Definition:Symmetric Relation|Symmetric]] +}} +{{eqn | ll=\implies + | l=(\left({x, y}\right) \in \mathcal R + | o=\iff + | r=\left({y, x}\right) \in \mathcal R) + | c=Definition of [[Definition:Biconditional|Biconditional]] +}} +{{end-eqn}} +{{qed}} +[[Category:Symmetric Relations]] +1ivm5w9dfywugeqhvcxevx91ds9sauu +\end{proof}<|endoftext|> +\section{Relation is Symmetric and Antisymmetric iff Coreflexive} +Tags: Symmetric Relations, Reflexive Relations + +\begin{theorem} +Let $S$ be a [[Definition:Set|set]]. +Let $\RR \subseteq S \times S$ be a [[Definition:Relation|relation]] in $S$. +Then: +:$\RR$ is both [[Definition:Symmetric Relation|symmetric]] and [[Definition:Antisymmetric Relation|antisymmetric]] +{{iff}}: +:$\RR$ is [[Definition:Coreflexive Relation|coreflexive]]. +\end{theorem} + +\begin{proof} +=== Necessary Condition === +Let $\RR$ be both [[Definition:Symmetric Relation|symmetric]] and [[Definition:Antisymmetric Relation|antisymmetric]] +Suppose $\RR \not \subseteq \Delta_S$. +Then: +:$\exists \tuple {x, y} \in \RR: x \ne y$ +But then as $\RR$ is [[Definition:Symmetric Relation|symmetric]]: +:$\tuple {y, x} \in \RR$ +So we have: +:$\tuple {x, y} \in \RR$ +and: +:$\tuple {y, x} \in \RR$ +where $x \ne y$. +Thus $\RR$ is not [[Definition:Antisymmetric Relation|antisymmetric]], contrary to hypothesis. +So our assumption that $\RR \not \subseteq \Delta_S$ is false. +That is: +:$\RR \subseteq \Delta_S$ +The result follows by definition of [[Definition:Coreflexive Relation/Definition 2|coreflexive]]. +{{qed|lemma}} +=== Sufficient Condition === +Let $\RR$ be [[Definition:Coreflexive Relation|coreflexive]]. +Then: +{{begin-eqn}} +{{eqn | l = \tuple {x, y} + | o = \in + | r = \RR +}} +{{eqn | ll= \leadsto + | l = x + | r = y + | c = Definition of [[Definition:Coreflexive Relation|Coreflexive Relation]] +}} +{{eqn | ll= \leadsto + | l = \tuple {x, y} + | r = \tuple {y, x} + | c = [[Equality of Ordered Pairs]] +}} +{{eqn | ll= \leadsto + | l = \tuple {y, x} + | o = \in + | r = \RR +}} +{{end-eqn}} +and so by definition $\RR$ is [[Definition:Symmetric Relation|symmetric]]. +Let $\tuple {x, y} \in \RR$ and $\tuple {y, x} \in \RR$. +From the above, this can happen only when $x = y$. +That is, by definition, $\RR$ is [[Definition:Antisymmetric Relation|antisymmetric]]. +So $\RR$ is both [[Definition:Symmetric Relation|symmetric]] and [[Definition:Antisymmetric Relation|antisymmetric]]. +Hence the result. +{{qed}} +\end{proof}<|endoftext|> +\section{Relation both Symmetric and Asymmetric is Null} +Tags: Symmetric Relations + +\begin{theorem} +Let $\RR$ be a [[Definition:Relation|relation]] in $S$ which is both [[Definition:Symmetric Relation|symmetric]] and [[Definition:Asymmetric Relation|asymmetric]]. +Then: +:$\RR = \O$ +\end{theorem} + +\begin{proof} +Let $\RR \ne \O$. +Then: +:$\exists \tuple {x, y} \in \RR$ +Now, either $\tuple {y, x} \in \RR$ or $\tuple {y, x} \notin \RR$. +If $\tuple {y, x} \in \RR$, then $\RR$ is not [[Definition:Asymmetric Relation|asymmetric]]. +If $\tuple {y, x} \notin \RR$, then $\RR$ is not [[Definition:Symmetric Relation|symmetric]]. +Therefore, if $\RR \ne \O$, $\RR$ can not be both [[Definition:Symmetric Relation|symmetric]] and [[Definition:Asymmetric Relation|asymmetric]]. +Let $\RR = \O$. +Then: +:$\forall x, y \in S: \tuple {x, y} \notin \RR \land \tuple {y, x} \notin \RR$ +From [[Biconditional as Disjunction of Conjunctions]]: +:$\tuple {p \land q} \lor \tuple {\neg p \land \neg q} \vdash p \iff q$ +Thus: +:$\forall x, y \in S: \tuple {x, y} \in \RR \iff \tuple {y, x} \in \RR$ +So the condition for symmetry is fulfilled, and $\RR = \O$ is [[Definition:Symmetric Relation|symmetric]]. +As $\RR = \O$: +:$\forall x, y \in S: \tuple {x, y} \notin \RR$ +From [[False Statement implies Every Statement]] $\neg p \vdash p \implies q$, we can deduce: +:$\forall x, y \in S: \tuple {x, y} \notin \RR$ +therefore: +:$\forall x, y \in S: \tuple {x, y} \in \RR \implies \forall x, y \in S: \tuple {y, x} \notin \RR$ +This is the definition of [[Definition:Asymmetric Relation|asymmetric]]. +Thus the only [[Definition:Relation|relation]] $\RR$ to be both [[Definition:Symmetric Relation|symmetric]] and [[Definition:Asymmetric Relation|asymmetric]] is $\RR = \O$. +{{qed}} +\end{proof}<|endoftext|> +\section{Equivalent Characterizations of Abelian Group} +Tags: Abelian Groups + +\begin{theorem} +Let $G$ be a [[Definition:Group|group]]. +{{TFAE}} +{{begin-eqn}} +{{eqn | n = 1 + | o = + | c = $G$ is [[Definition:Abelian Group|abelian]] +}} +{{eqn | n = 2 + | o = + | c = $\forall a, b \in G: \paren {a b}^{-1} = a^{-1} b^{-1}$ +}} +{{eqn | n = 3 + | o = + | c = [[Definition:Cross Cancellation Property|Cross cancellation property]]: $\forall a, b, c \in G: a b = c a \implies b = c$ +}} +{{eqn | n = 4 + | o = + | c = [[Definition:Middle Cancellation Property|Middle cancellation property]]: $\forall a, b, c, d, x \in G: a x b = c x d \implies a b = c d$ +}} +{{end-eqn}} +\end{theorem} + +\begin{proof} +=== 1 iff 2 === +See [[Inversion Mapping is Automorphism iff Group is Abelian]]. +{{qed|lemma}} +{{explain|There's a better link that proves this one, but IIRC it's difficult to find. I've been tidying up this recently, I'll see if I can find it in a bit.}} +=== 1 iff 3 === +See [[Group Abelian iff Cross Cancellation Property]]. +{{qed|lemma}} +=== 1 iff 4 === +See [[Group Abelian iff Middle Cancellation Property]]. +{{qed|lemma}} +Hence all four statements are [[Definition:Logical Equivalence|logically equivalent]]. +{{qed}} +[[Category:Abelian Groups]] +lagjvdvlr6t1nx0th4h4fsxoubxkp92 +\end{proof}<|endoftext|> +\section{Relation Symmetry} +Tags: Symmetric Relations + +\begin{theorem} +Every [[Definition:Empty Set|non-null]] [[Definition:Relation|relation]] has exactly one of these properties: it is either: +:[[Definition:Symmetric Relation|symmetric]], +:[[Definition:Asymmetric Relation|asymmetric]] or +:[[Definition:Non-Symmetric Relation|non-symmetric]]. +\end{theorem} + +\begin{proof} +From [[Relation both Symmetric and Asymmetric is Null]], the [[Definition:Empty Set|empty set]] ''is'' both symmetric and asymmetric. +This is why the assumption that $\mathcal R \ne \varnothing$. +Let $\mathcal R$ be [[Definition:Symmetric Relation|symmetric]]. +Then from the definition of [[Definition:Asymmetric Relation|Asymmetric]], $\mathcal R$ is not asymmetric. +Also, from the definition of [[Definition:Non-Symmetric Relation|Non-Symmetric]], it is not non-symmetric. +Let $\mathcal R$ be [[Definition:Asymmetric Relation|asymmetric]]. +Then from the definition of [[Definition:Symmetric Relation|Symmetric]], $\mathcal R$ is not symmetric. +Also, from the definition of [[Definition:Non-Symmetric Relation|Non-Symmetric]], it is not non-symmetric. +If $\mathcal R$ is not symmetric, and $\mathcal R$ is not asymmetric, then from the definition of [[Definition:Non-Symmetric Relation|Non-Symmetric]], it is non-symmetric. +{{Qed}} +\end{proof}<|endoftext|> +\section{Asymmetric Relation is Antisymmetric} +Tags: Symmetric Relations + +\begin{theorem} +Let $\RR$ be an [[Definition:Asymmetric Relation|asymmetric relation]]. +Then $\RR$ is also [[Definition:Antisymmetric Relation|antisymmetric]]. +\end{theorem} + +\begin{proof} +Let $\RR$ be [[Definition:Asymmetric Relation|asymmetric]]. +Then from the definition of [[Definition:Asymmetric Relation|asymmetric]]: +:$\tuple {x, y} \in \RR \implies \tuple {y, x} \notin \RR$ +Thus: +:$\neg \exists \tuple {x, y} \in \RR: \tuple {y, x} \in \RR$ +Thus: +:$\set {\tuple {x, y} \in \RR \land \tuple {y, x} \in \RR} = \O$ +Thus: +:$\tuple {x, y} \in \RR \land \tuple {y, x} \in \RR \implies x = y$ +is [[Definition:Vacuous Truth|vacuously true]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Asymmetric Relation is Antireflexive} +Tags: Symmetric Relations, Reflexive Relations + +\begin{theorem} +Let $S$ be a [[Definition:Set|set]]. +Let $\RR \subseteq S \times S$ be a [[Definition:Endorelation|relation]] on $S$. +Let $\RR$ be [[Definition:Asymmetric Relation|asymmetric]]. +Then $\RR$ is also [[Definition:Antireflexive Relation|antireflexive]]. +\end{theorem} + +\begin{proof} +Let $\RR$ be [[Definition:Asymmetric Relation|asymmetric]]. +Then, by definition: +:$\tuple {x, y} \in \RR \implies \tuple {y, x} \notin \RR$ +{{AimForCont}} $\tuple {x, x} \in \RR$. +Then: +{{begin-eqn}} +{{eqn | l = \tuple {x, x} \in \RR + | o = \implies + | r = \tuple {x, x} \notin \RR + | c = {{Defof|Asymmetric Relation}} +}} +{{eqn | ll= \leadsto + | l = \tuple {x, x} \notin \RR + | o = + | c = [[Proof by Contradiction]] +}} +{{end-eqn}} +Thus $\RR$ is [[Definition:Antireflexive Relation|antireflexive]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Antireflexive and Transitive Relation is Asymmetric} +Tags: Reflexive Relations, Symmetric Relations, Transitive Relations + +\begin{theorem} +Let $\mathcal R \subseteq S \times S$ be a [[Definition:Relation|relation]] which is not [[Definition:Null Relation|null]]. +Let $\mathcal R$ be [[Definition:Antireflexive Relation|antireflexive]] and [[Definition:Transitive Relation|transitive]]. +Then $\mathcal R$ is also [[Definition:Asymmetric Relation|asymmetric]]. +\end{theorem} + +\begin{proof} +Let $\mathcal R \subseteq S \times S$ be [[Definition:Antireflexive Relation|antireflexive]] and [[Definition:Transitive Relation|transitive]]. +That is: +{{begin-eqn}} +{{eqn | l=\forall x \in S: + | o= + | r=\left({x, x}\right) \notin \mathcal R + | c=Definition of [[Definition:Antireflexive Relation|Antireflexive]] +}} +{{eqn | l=\left({x, y}\right) \in \mathcal R \land \left({y, z}\right) \in \mathcal R + | o=\implies + | r=\left({x, z}\right) \in \mathcal R + | c=Definition of [[Definition:Transitive Relation|Transitive]] +}} +{{end-eqn}} +We have that $\mathcal R$ is not [[Definition:Null Relation|null]]. +Suppose $\mathcal R$ is not [[Definition:Asymmetric Relation|asymmetric]]. +So, by definition, $\exists \left({x, y}\right) \in \mathcal R: \left({y, x}\right) \in \mathcal R$. +Then from the transitivity of $\mathcal R$ that would mean $\left({x, x}\right) \in \mathcal R$. +But that would contradict the antireflexivity of $\mathcal R$. +Therefore $\left({x, y}\right) \in \mathcal R \implies \left({y, x}\right) \notin \mathcal R$ and $\mathcal R$ has been shown to be asymmetric. +{{qed}} +\end{proof}<|endoftext|> +\section{Antitransitive Relation is Antireflexive} +Tags: Reflexive Relations, Transitive Relations + +\begin{theorem} +Let $S$ be a [[Definition:Set|set]]. +Let $\mathcal R \subseteq S \times S$ be a [[Definition:Endorelation|relation]] on $S$. +Let $\mathcal R$ be [[Definition:Antitransitive Relation|antitransitive]]. +Then $\mathcal R$ is also [[Definition:Antireflexive Relation|antireflexive]]. +\end{theorem} + +\begin{proof} +Suppose $\mathcal R \subseteq S \times S$ is not [[Definition:Antireflexive Relation|antireflexive]]. +Then $\exists x \in S: \left({x, x}\right) \in \mathcal R$. (In the case of $\mathcal R$ being [[Definition:Reflexive Relation|reflexive]], the property holds for ''all'' $x \in S$.) +Thus $\mathcal R$ is not [[Definition:Antitransitive Relation|antitransitive]]: +{{begin-eqn}} +{{eqn | o = + | r = \left({x, x}\right) \in \mathcal R \land \left({x, x}\right) \in \mathcal R + | c = +}} +{{eqn | o = \implies + | r = \left({x, x}\right) \in \mathcal R + | c = [[Rule of Idempotence]] +}} +{{end-eqn}} +... which means $\exists x \in S$ such that the condition for [[Definition:Antitransitive Relation|antitransitivity]] is broken. +So $\mathcal R \subseteq S \times S$ has to be [[Definition:Antireflexive Relation|antireflexive]]. +{{Qed}} +\end{proof}<|endoftext|> +\section{Symmetric Transitive and Serial Relation is Reflexive} +Tags: Reflexive Relations, Symmetric Relations, Transitive Relations, Serial Relations, Equivalence Relations + +\begin{theorem} +Let $\RR$ be a [[Definition:Relation|relation]] which is: +:[[Definition:Symmetric Relation|symmetric]] +:[[Definition:Transitive Relation|transitive]] +:[[Definition:Serial Relation|serial]]. +Then $\RR$ is [[Definition:Reflexive Relation|reflexive]]. +Thus such a [[Definition:Relation|relation]] is an [[Definition:Equivalence Relation|equivalence]]. +\end{theorem} + +\begin{proof} +Let $S$ be a [[Definition:Set|set]] on which $\RR$ is a [[Definition:Relation|relation]] which is [[Definition:Symmetric Relation|symmetric]], [[Definition:Transitive Relation|transitive]] and [[Definition:Serial Relation|serial]]. +As $\RR$ is [[Definition:Symmetric Relation|symmetric]]: +:$x \mathrel \RR y \implies y \mathrel \RR x$ +As $\RR$ is [[Definition:Transitive Relation|transitive]]: +:$x \mathrel \RR y \land y \mathrel \RR x \implies x \mathrel \RR x$ +As $\RR$ is [[Definition:Serial Relation|serial]]: +:$\forall x \in S: \exists y \in S: x \mathrel \RR y$ +Let $x \in S$. +Then +{{begin-eqn}} +{{eqn | lo= \exists y \in S: + | l = \tuple {x, y} + | o = \in + | r = \RR + | c = as $\RR$ is [[Definition:Serial Relation|serial]] +}} +{{eqn | ll= \leadsto + | l = \tuple {y, x} + | o = \in + | r = \RR + | c = as $\RR$ is [[Definition:Symmetric Relation|symmetric]] +}} +{{eqn | ll= \leadsto + | l = \tuple {x, x} + | o = \in + | r = \RR + | c = as $\RR$ is [[Definition:Transitive Relation|transitive]] +}} +{{end-eqn}} +Thus: +:$\forall x: x \mathrel \RR x$ +and by definition $\RR$ is [[Definition:Reflexive Relation|reflexive]]. +It follows by definition that such a relation is an [[Definition:Equivalence Relation|equivalence relation]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Inverse Relation Properties} +Tags: Inverse Relations + +\begin{theorem} +Let $\mathcal R$ be a [[Definition:Relation|relation]] on a set $S$. +If $\mathcal R$ has any of the properties: +* [[Definition:Reflexive Relation|Reflexive]] +* [[Definition:Antireflexive Relation|Antireflexive]] +* [[Definition:Non-Reflexive Relation|Non-reflexive]] +* [[Definition:Symmetric Relation|Symmetric]] +* [[Definition:Asymmetric Relation|Asymmetric]] +* [[Definition:Antisymmetric Relation|Antisymmetric]] +* [[Definition:Non-Symmetric Relation|Non-symmetric]] +* [[Definition:Transitive Relation|Transitive]] +* [[Definition:Antitransitive Relation|Antitransitive]] +* [[Definition:Non-Transitive Relation|Non-transitive]] +... then its [[Definition:Inverse Relation|inverse]] $\mathcal R^{-1}$ has the same properties. +\end{theorem} + +\begin{proof} +=== [[Inverse of Reflexive Relation is Reflexive|Reflexivity]] === +{{:Inverse of Reflexive Relation is Reflexive}} +=== [[Inverse of Antireflexive Relation is Antireflexive|Antireflexivity]] === +{{:Inverse of Antireflexive Relation is Antireflexive}} +=== [[Inverse of Non-Reflexive Relation is Non-Reflexive|Non-Reflexivity]] === +{{:Inverse of Non-Reflexive Relation is Non-Reflexive}} +=== [[Inverse of Symmetric Relation is Symmetric|Symmetry]] === +{{:Inverse of Symmetric Relation is Symmetric}} +=== [[Inverse of Asymmetric Relation is Asymmetric|Asymmetry]] === +{{:Inverse of Asymmetric Relation is Asymmetric}} +=== [[Inverse of Antisymmetric Relation is Antisymmetric|Antisymmetry]] === +{{:Inverse of Antisymmetric Relation is Antisymmetric}} +=== [[Inverse of Non-Symmetric Relation is Non-Symmetric|Non-Symmetry]] === +{{:Inverse of Non-Symmetric Relation is Non-Symmetric}} +=== [[Inverse of Transitive Relation is Transitive|Transitivity]] === +{{:Inverse of Transitive Relation is Transitive}} +=== [[Inverse of Antitransitive Relation is Antitransitive|Antitransitivity]] === +{{:Inverse of Antitransitive Relation is Antitransitive}} +=== [[Inverse of Non-Transitive Relation is Non-Transitive|Non-Transitivity]] === +{{:Inverse of Non-Transitive Relation is Non-Transitive}} +[[Category:Inverse Relations]] +2ruzcjj4s8bs7p8muxq0jmun8d4duzo +\end{proof}<|endoftext|> +\section{Properties of Restriction of Relation} +Tags: Restrictions + +\begin{theorem} +Let $S$ be a [[Definition:Set|set]]. +Let $\mathcal R \subseteq S \times S$ be a [[Definition:Relation|relation]] on $S$. +Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$. +Let $\mathcal R \restriction_T \ \subseteq T \times T$ be the [[Definition:Restriction of Relation|restriction]] of $\mathcal R$ to $T$. +If $\mathcal R$ on $S$ has any of the properties: +* [[Definition:Reflexive Relation|Reflexive]] +* [[Definition:Antireflexive Relation|Antireflexive]] +* [[Definition:Symmetric Relation|Symmetric]] +* [[Definition:Antisymmetric Relation|Antisymmetric]] +* [[Definition:Asymmetric Relation|Asymmetric]] +* [[Definition:Transitive Relation|Transitive]] +* [[Definition:Antitransitive Relation|Antitransitive]] +* [[Definition:Connected Relation|Connected]] +... then $\mathcal R \restriction_T$ on $T$ has the same properties. +\end{theorem} + +\begin{proof} +=== Reflexivity === +==== [[Restriction of Reflexive Relation is Reflexive]] ==== +{{:Restriction of Reflexive Relation is Reflexive}} +==== [[Restriction of Antireflexive Relation is Antireflexive]] ==== +{{:Restriction of Antireflexive Relation is Antireflexive}} +=== Symmetry === +==== [[Restriction of Symmetric Relation is Symmetric]] ==== +{{:Restriction of Symmetric Relation is Symmetric}} +==== [[Restriction of Asymmetric Relation is Asymmetric]] ==== +{{:Restriction of Asymmetric Relation is Asymmetric}} +==== [[Restriction of Antisymmetric Relation is Antisymmetric]] ==== +{{:Restriction of Antisymmetric Relation is Antisymmetric}} +=== Transitivity === +==== [[Restriction of Transitive Relation is Transitive]] ==== +{{:Restriction of Transitive Relation is Transitive}} +==== [[Restriction of Antitransitive Relation is Antitransitive]] ==== +{{:Restriction of Antitransitive Relation is Antitransitive}} +=== Connectedness === +==== [[Restriction of Connected Relation is Connected]] ==== +{{:Restriction of Connected Relation is Connected}} +\end{proof}<|endoftext|> +\section{Relation is Antireflexive iff Disjoint from Diagonal Relation} +Tags: Reflexive Relations + +\begin{theorem} +Let $\mathcal R \subseteq S \times S$ be a [[Definition:Relation|relation]] on a [[Definition:Set|set]] $S$. +Then: +: $\mathcal R$ is [[Definition:Antireflexive Relation|antireflexive]] +[[Definition:Iff|iff]] +: $\Delta_S \cap \mathcal R = \varnothing$ +where $\Delta_S$ is the [[Definition:Diagonal Relation|diagonal relation]]. +\end{theorem} + +\begin{proof} +=== Necessary Condition === +Let $\mathcal R$ be an [[Definition:Antireflexive Relation|antireflexive relation]]. +Let $\left({x, y}\right) \in \Delta_S \cap \mathcal R$. +By definition: +: $\left({x, y}\right) \in \Delta_S \implies x = y$ +Likewise, by definition: +: $\left({x, y}\right) \in \mathcal R \implies x \ne y$. +Thus: +: $\Delta_S \cap \mathcal R = \left\{{\left({x, y}\right): x = y \land x \ne y}\right\}$ +and so: +: $\Delta_S \cap \mathcal R = \varnothing$ +=== Sufficient Condition === +Let $\Delta_S \cap \mathcal R = \varnothing$. +Then by definition: +: $\forall \left({x, y}\right) \in \mathcal R: \left({x, y}\right) \notin \Delta_S$ +Thus: +: $\not \exists \left({x, y}\right) \in \mathcal R: x = y$ +Thus by definition, $\mathcal R$ is [[Definition:Antireflexive Relation|antireflexive]]. +{{Qed}} +[[Category:Reflexive Relations]] +oylgje196y4qtnwbu8ovx5rjfvuzzyd +\end{proof}<|endoftext|> +\section{Inverse Relation Equal iff Subset} +Tags: Inverse Relations + +\begin{theorem} +If a [[Definition:Relation|relation]] $\RR$ is a [[Definition:Subset|subset]] or [[Definition:Superset|superset]] of its [[Definition:Inverse Relation|inverse]], then it equals its inverse. +That is, the following are [[Definition:Logical Equivalence|equivalent]]: +{{begin-axiom}} +{{axiom | n = 1 + | m = \RR \subseteq \RR^{-1} +}} +{{axiom | n = 2 + | m = \RR^{-1} \subseteq \RR +}} +{{axiom | n = 3 + | m = \RR = \RR^{-1} +}} +{{end-axiom}} +\end{theorem} + +\begin{proof} +=== $(1)$ implies $(2)$ === +Suppose $\RR \subseteq \RR^{-1}$. +Then: +{{begin-eqn}} +{{eqn | l = \tuple {a, b} + | o = \in + | r = \RR^{-1} + | c = +}} +{{eqn | ll= \leadsto + | l = \tuple {b, a} + | o = \in + | r = \RR + | c = {{Defof|Inverse Relation}} +}} +{{eqn | ll= \leadsto + | l = \tuple {b, a} + | o = \in + | r = \RR^{-1} + | c = {{Defof|Subset}} (as $\RR \subseteq \RR^{-1}$) +}} +{{eqn | ll= \leadsto + | l = \tuple {a, b} + | o = \in + | r = \RR + | c = {{Defof|Inverse Relation}} +}} +{{end-eqn}} +Hence $\RR^{-1} \subseteq \RR$. +{{qed|lemma}} +=== $(2)$ implies $(1)$ === +Since $(1)$ implies $(2)$ was already established above, interpreting it on $\RR^{-1}$ yields: +:$\RR^{-1} \subseteq \paren {\RR^{-1} }^{-1}$ +By [[Inverse of Inverse Relation]], $\paren {\RR^{-1} }^{-1} = \RR$. +Hence $(2)$ implies $(1)$. +{{qed|lemma}} +=== $(3)$ {{iff}} $(1)$ and $(2)$ === +This is precisely the definition of [[Definition:Set Equality/Definition 2|set equality]]. +{{qed}} +[[Category:Inverse Relations]] +qrykpsv23jatvbaui9xdj6kmnxr2493 +\end{proof}<|endoftext|> +\section{Relation is Symmetric iff Inverse is Symmetric} +Tags: Symmetric Relations + +\begin{theorem} +A [[Definition:Relation|relation]] $\mathcal R$ is [[Definition:Symmetric Relation|symmetric]] {{iff}} its [[Definition:Inverse Relation|inverse]] $\mathcal R^{-1}$ is also [[Definition:Symmetric Relation|symmetric]]. +\end{theorem} + +\begin{proof} +Let $\mathcal R$ be [[Definition:Symmetric Relation|symmetric]]. +Then from [[Relation equals Inverse iff Symmetric]]: +:$\mathcal R = \mathcal R^{-1}$ +The result follows. +{{qed}} +\end{proof}<|endoftext|> +\section{Composite Relation with Inverse is Symmetric} +Tags: Symmetric Relations + +\begin{theorem} +Let $\mathcal R \subseteq S \times T$ be a [[Definition:Relation|relation]]. +Then the [[Definition:Composition of Relations|composition]] of $\mathcal R$ with its [[Definition:Inverse Relation|inverse]] $\mathcal R^{-1}$ is [[Definition:Symmetric Relation|symmetric]]: +* $\mathcal R^{-1} \circ \mathcal R$ is a symmetric relation on $S$ +* $\mathcal R \circ \mathcal R^{-1}$ is a symmetric relation on $T$. +\end{theorem} + +\begin{proof} +Note that this result holds for any $\mathcal R \subseteq S \times T$, and does not require that $\left({S, \mathcal R}\right)$ necessarily be a [[Definition:Relational Structure|relational structure]]. +{{begin-eqn}} +{{eqn | o= + | r=\left({a, b}\right) \in \mathcal R^{-1} \circ \mathcal R + | c= +}} +{{eqn | o=\implies + | r=\exists c \in T: \left({a, c}\right) \in \mathcal R \land \left({c, b}\right) \in \mathcal R^{-1} + | c=[[Definition:Composition of Relations|Definition of Composite Relation]] +}} +{{eqn | o=\implies + | r=\exists c \in T: \left({b, c}\right) \in \mathcal R \land \left({c, a}\right) \in \mathcal R^{-1} + | c=[[Definition:Inverse Relation|Definition of Inverse Relation]] +}} +{{eqn | o=\implies + | r=\left({b, a}\right) \in \mathcal R^{-1} \circ \mathcal R + | c=[[Definition:Composition of Relations|Definition of Composite Relation]] +}} +{{end-eqn}} +Thus $\left({a, b}\right) \in \mathcal R^{-1} \circ \mathcal R \implies \left({b, a}\right) \in \mathcal R^{-1} \circ \mathcal R$ and thus $\mathcal R^{-1} \circ \mathcal R$ is symmetric. +As $\mathcal R = \left({\mathcal R^{-1}}\right)^{-1}$ from [[Inverse of Inverse Relation]], it follows that $\mathcal R \circ \mathcal R^{-1} = \left({\mathcal R^{-1}}\right)^{-1} \circ \mathcal R^{-1}$ is likewise a symmetric relation. +The domain of $\mathcal R^{-1} \circ \mathcal R$ is $S$ from [[Domain of Composite Relation]], as is its codomain from [[Codomain of Composite Relation]] and the [[Definition:Inverse Relation|definition of Inverse Relation]]. +Similarly, the codomain of $\mathcal R \circ \mathcal R^{-1}$ is $T$, as is its domain. +This completes the proof. +{{Qed}} +[[Category:Symmetric Relations]] +is2tjx47f8rfg6k18o9p64rn6vhq4nw +\end{proof}<|endoftext|> +\section{Inverse of Many-to-One Relation is One-to-Many} +Tags: Inverse Relations + +\begin{theorem} +The [[Definition:Inverse Relation|inverse]] of a [[Definition:Many-to-One Relation|many-to-one relation]] is a [[Definition:One-to-Many Relation|one-to-many relation]], and vice versa. +\end{theorem} + +\begin{proof} +From the definition, a [[Definition:Many-to-One Relation|many-to-one relation]] $\RR \subseteq S \times T$ is such that: +:$\RR \subseteq S \times T: \forall x \in \Dom \RR: \tuple {x, y_1} \in \RR \land \tuple {x, y_2} \in \RR \implies y_1 = y_2$ +Also from the definition, a [[Definition:One-to-Many Relation|one-to-many relation]] $\RR \subseteq S \times T$ is such that: +:$\RR \subseteq S \times T: \forall y \in \Img \RR: \tuple {x_1, y} \in \RR \land \tuple {x_2, y} \in \RR \implies x_1 = x_2$ +The [[Definition:Inverse Relation|inverse]] of a [[Definition:Relation|relation]] $\RR$ is defined as: +:$\RR^{-1} = \set {\tuple {t, s}: \tuple {s, t} \in \RR$ +Also from the definition of [[Definition:Inverse Relation|inverse relation]], we have that: +:$\Dom \RR = \Rng {\RR^{-1} }$ +:$\Rng \RR = \Dom {\RR^{-1} }$ +So, let $\RR$ be a [[Definition:Many-to-One Relation|many-to-one relation]]. +Putting the above all together, we have: +:$\RR^{-1} \subseteq T \times S: \forall x \in \Img {\RR^{-1} }: \tuple {y_1, x} \in \RR^{-1} \land \tuple {y_2, x} \in \RR^{-1} \implies y_1 = y_2$ +and it can be seen that $\RR^{-1}$ is [[Definition:One-to-Many Relation|one-to-many]]. +Similarly, let $\RR$ be a [[Definition:One-to-Many Relation|one-to-many relation]]. +This gives us: +:$\RR^{-1} \subseteq T \times S: \forall y \in \Dom {\RR^{-1} }: \tuple {y, x_1} \in \RR^{-1} \land \tuple {y, x_2} \in \RR^{-1} \implies x_1 = x_2$ +and it can be seen that $\RR^{-1}$ is [[Definition:Many-to-One Relation|many-to-one]]. +{{qed}} +[[Category:Inverse Relations]] +g2tynfvzl6y05zu20qqtctz07oh9pna +\end{proof}<|endoftext|> +\section{Many-to-One Relation Composite with Inverse is Transitive} +Tags: Composite Relations, Transitive Relations + +\begin{theorem} +Let $\mathcal R \subseteq S \times T$ be a [[Definition:Relation|relation]] which is [[Definition:Many-to-One Relation|many-to-one]]. +Then the [[Definition:Composition of Relations|composites]] (both ways) of $\mathcal R$ and its [[Definition:Inverse Relation|inverse]] $\mathcal R^{-1}$, that is, both $\mathcal R^{-1} \circ \mathcal R$ and $\mathcal R \circ \mathcal R^{-1}$, are [[Definition:Transitive Relation|transitive]]. +\end{theorem} + +\begin{proof} +Let $\mathcal R \subseteq S \times T$ be [[Definition:Many-to-One Relation|many-to-one]]. +Then, from the definition of [[Definition:Many-to-One Relation|many-to-one]]: +:$\mathcal R \subseteq S \times T: \forall x \in S: \left({x, y_1}\right) \in \mathcal R \land \left({x, y_2}\right) \in \mathcal R \implies y_1 = y_2$ +Also, note that from [[Inverse of Many-to-One Relation is One-to-Many]], $\mathcal R^{-1}$ is [[Definition:One-to-Many Relation|one-to-many]]. +Let $\left({a, b}\right), \left({b, c}\right) \in \mathcal R^{-1} \circ \mathcal R$. +{{begin-eqn}} +{{eqn | o = + | r = \left({a, b}\right), \left({b, c}\right) \in \mathcal R^{-1} \circ \mathcal R + | c = +}} +{{eqn | ll= \implies + | o = + | r = \exists p \in T: \left({a, p}\right) \in \mathcal R \land \left({p, b}\right) \in \mathcal R^{-1} + | c = +}} +{{eqn | o = \land + | r = \exists q \in T: \left({b, q}\right) \in \mathcal R \land \left({q, c}\right) \in \mathcal R^{-1} + | c = Definition of [[Definition:Composition of Relations|Composite Relation]] +}} +{{eqn | ll= \implies + | o = + | r = \left({b, p}\right) \in \mathcal R \land \left({p, a}\right) \in \mathcal R^{-1} + | c = +}} +{{eqn | o = \land + | r = \left({c, q}\right) \in \mathcal R \land \left({q, b}\right) \in \mathcal R^{-1} + | c = Definition of [[Definition:Inverse Relation|Inverse Relation]] +}} +{{eqn | ll= \implies + | o = + | r = \left({c, p}\right) \in \mathcal R \land \left({p, b}\right) \in \mathcal R^{-1} + | c = as $\mathcal R$ is [[Definition:Many-to-One Relation|many-to-one]] +}} +{{eqn | ll= \implies + | o = + | r = \left({a, p}\right) \in \mathcal R \land \left({p, c}\right) \in \mathcal R^{-1} + | c = +}} +{{eqn | ll= \implies + | o = + | r = \left({a, c}\right) \in \mathcal R + | c = Definition of [[Definition:Composition of Relations|Composite Relation]] +}} +{{end-eqn}} +Thus $\mathcal R^{-1} \circ \mathcal R$ is transitive. +Now let $\left({p, q}\right), \left({q, r}\right) \in \mathcal R \circ \mathcal R^{-1}$. +{{begin-eqn}} +{{eqn | o = + | r = \left({p, q}\right), \left({q, r}\right) \in \mathcal R \circ \mathcal R^{-1} + | c = +}} +{{eqn | ll= \implies + | o = + | r = \exists a \in S: \left({p, a}\right) \in \mathcal R^{-1} \land \left({a, q}\right) \in \mathcal R + | c = +}} +{{eqn | o = \land + | r = \exists b \in S: \left({q, b}\right) \in \mathcal R^{-1} \land \left({b, r}\right) \in \mathcal R + | c = Definition of [[Definition:Composition of Relations|Composite Relation]] +}} +{{eqn | ll= \implies + | o = + | r = \left({a, p}\right) \in \mathcal R \land \left({q, a}\right) \in \mathcal R^{-1} + | c = +}} +{{eqn | o = \land + | r = \left({b, q}\right) \in \mathcal R \land \left({r, b}\right) \in \mathcal R^{-1} + | c = Definition of [[Definition:Inverse Relation|Inverse Relation]] +}} +{{end-eqn}} +But $\mathcal R$ is [[Definition:Many-to-One Relation|many-to-one]]. +This means that +:$\forall x \in S: \left({x, y_1}\right) \in \mathcal R \land \left({x, y_2}\right) \in \mathcal R \implies y_1 = y_2$ +So: +{{begin-eqn}} +{{eqn | l = \left({a, p}\right) \in \mathcal R \land \left({a, q}\right) \in \mathcal R + | o = \implies + | r = p = q + | c = as $\mathcal R$ is [[Definition:Many-to-One Relation|many-to-one]] +}} +{{eqn | l = \left({b, q}\right) \in \mathcal R \land \left({b, r}\right) \in \mathcal R + | o = \implies + | r = q = r + | c = as $\mathcal R$ is [[Definition:Many-to-One Relation|many-to-one]] +}} +{{eqn | o = \implies + | r = p = q = r + | c = +}} +{{eqn | o = \implies + | r = \left({p, r}\right) \in \mathcal R \circ \mathcal R^{-1} + | c = +}} +{{end-eqn}} +Thus (trivially) $\mathcal R \circ \mathcal R^{-1}$ is [[Definition:Transitive Relation|transitive]]. +{{qed}} +[[Category:Composite Relations]] +[[Category:Transitive Relations]] +q2f53o6ksd17yxkbuerw6pcni4r6f4n +\end{proof}<|endoftext|> +\section{One-to-Many Relation Composite with Inverse is Coreflexive} +Tags: Reflexive Relations + +\begin{theorem} +Let $\mathcal R \subseteq S \times S$ be a [[Definition:Relation|relation]] which is [[Definition:One-to-Many Relation|one-to-many]]. +Then the [[Definition:Composition of Relations|composite]] of $\mathcal R$ with its [[Definition:Inverse Relation|inverse]] is a [[Definition:Coreflexive Relation|coreflexive relation]]: +:$\mathcal R^{-1} \circ \mathcal R \subseteq \Delta_X$ +{{refactor|This is a separate result.}} +That is, by [[Relation is Symmetric and Antisymmetric iff Coreflexive]], $\mathcal R^{-1} \circ \mathcal R$ is both [[Definition:Symmetric Relation|symmetric]] and [[Definition:Antisymmetric Relation|antisymmetric]]. +\end{theorem} + +\begin{proof} +As $\mathcal R$ is [[Definition:One-to-Many Relation|one-to-many]], it follows from [[Inverse of Many-to-One Relation is One-to-Many]] that $\mathcal R^{-1}$ is [[Definition:Many-to-One Relation|many-to-one]]. +Let $\left({x, z}\right) \in \mathcal R^{-1} \circ \mathcal R$. +Then: +: $\exists y \in S: \left({x, y}\right) \in \mathcal R, \left({y, z}\right) \in \mathcal R^{-1}$ +As $\left({x, y}\right) \in \mathcal R$, from the definition of [[Definition:Inverse Relation|inverse relation]]: +:$\left({y, x}\right) \in \mathcal R^{-1}$ +As $\mathcal R^{-1}$ is [[Definition:Many-to-One Relation|many-to-one]]: +: $\left({y, z}\right) = \left({y, x}\right)$ +Thus if $\left({x, z}\right) \in \mathcal R^{-1} \circ \mathcal R$, it follows that: +: $\left({x, z}\right) = \left({x, x}\right)$ +Thus: +: $\mathcal R^{-1} \circ \mathcal R \subseteq \Delta_X$ +The result follows by definition of [[Definition:Coreflexive Relation|coreflexive relation]]. +{{Qed}} +[[Category:Reflexive Relations]] +k8wchgaag457a6dv3zi7wjzu5032jhb +\end{proof}<|endoftext|> +\section{Element in its own Equivalence Class} +Tags: Equivalence Classes, Fundamental Theorem on Equivalence Relations + +\begin{theorem} +Let $\mathcal R$ be an [[Definition:Equivalence Relation|equivalence relation]] on a [[Definition:Set|set]] $S$. +Then every [[Definition:Element|element]] of $S$ is in its own [[Definition:Equivalence Class|$\mathcal R$-class]]: +:$\forall x \in S: x \in \eqclass x {\mathcal R}$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \forall x \in S: \tuple {x, x} + | o = \in + | r = \mathcal R + | c = {{Defof|Equivalence Relation}}: $\mathcal R$ is [[Definition:Reflexive Relation|Reflexive]] +}} +{{eqn | ll= \leadsto + | l = x + | o = \in + | r = \eqclass x {\mathcal R} + | c = {{Defof|Equivalence Class}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Equivalence Class of Element is Subset} +Tags: Equivalence Classes + +\begin{theorem} +Let $\mathcal R$ be an [[Definition:Equivalence Relation|equivalence relation]] on a [[Definition:Set|set]] $S$. +The [[Definition:Equivalence Class|$\mathcal R$-class]] of every [[Definition:Element|element]] of $S$ is a [[Definition:Subset|subset]] of the [[Definition:Set|set]] the [[Definition:Element|element]] is in: +:$\forall x \in S: \eqclass x {\mathcal R} \subseteq S$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = y + | o = \in + | r = \eqclass x {\mathcal R} + | c = +}} +{{eqn | ll= \leadsto + | l = \tuple {x, y} + | o = \in + | r = \mathcal R + | c = {{Defof|Equivalence Class}} +}} +{{eqn | ll= \leadsto + | l = x + | o = \in + | r = S \land y \in S + | c = {{Defof|Relation}} +}} +{{eqn | ll= \leadsto + | l = \eqclass x {\mathcal R} + | o = \subseteq + | r = S + | c = {{Defof|Subset}} +}} +{{end-eqn}} +{{Qed}} +[[Category:Equivalence Classes]] +s5lvveejd3j7w5rsiuthw62yckm7grw +\end{proof}<|endoftext|> +\section{Equivalence Class is not Empty} +Tags: Equivalence Classes, Empty Set, Fundamental Theorem on Equivalence Relations + +\begin{theorem} +Let $\mathcal R$ be an [[Definition:Equivalence Relation|equivalence relation]] on a [[Definition:Set|set]] $S$. +Then no [[Definition:Equivalence Class|$\mathcal R$-class]] is [[Definition:Empty Set|empty]]. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | lo= \forall \eqclass x {\mathcal R} \subseteq S: \exists x \in S: + | l = x + | o = \in + | r = \eqclass x {\mathcal R} + | c = {{Defof|Equivalence Class}} +}} +{{eqn | ll= \leadsto + | l = \eqclass x {\mathcal R} + | o = \ne + | r = \O + | c = {{Defof|Empty Set}} +}} +{{end-eqn}} +\end{proof}<|endoftext|> +\section{Handshake Lemma} +Tags: Graph Theory, Vertices of Graphs, Direct Proofs, Named Theorems, Handshake Lemma + +\begin{theorem} +Let $G$ be a $\tuple {p, q}$-[[Definition:Undirected Graph|undirected graph]], which may be a [[Definition:Multigraph|multigraph]] or a [[Definition:Loop-Graph|loop-graph]], or both. +Let $V = \set {v_1, v_2, \ldots, v_p}$ be the [[Definition:Vertex Set|vertex set]] of $G$. +Then: +:$\displaystyle \sum_{i \mathop = 1}^p \map {\deg_G} {v_i} = 2 q$ +where $\map {\deg_G} {v_i}$ is the [[Definition:Degree of Vertex|degree]] of [[Definition:Vertex of Graph|vertex]] $v_i$. +That is, the [[Definition:Integer Addition|sum]] of all the [[Definition:Degree of Vertex|degrees]] of all the [[Definition:Vertex of Graph|vertices]] of an [[Definition:Undirected Graph|graph]] is equal to twice its [[Definition:Size of Graph|size]]. +This result is known as the '''Handshake Lemma''' or '''Handshaking Lemma'''. +\end{theorem} + +\begin{proof} +In the [[Definition:Graph Notation|notation]] $\tuple {p, q}$-[[Definition:Graph (Graph Theory)|graph]], $p$ is its [[Definition:Order of Graph|order]] and $q$ its [[Definition:Size of Graph|size]]. +That is, $p$ is the number of [[Definition:Vertex of Graph|vertices]] in $G$, and $q$ is the number of [[Definition:Edge of Graph|edges]] in $G$. +Each [[Definition:Edge of Graph|edge]] is [[Definition:Incident (Undirected Graph)|incident]] to exactly two [[Definition:Vertex of Graph|vertices]]. +The [[Definition:Degree of Vertex|degree]] of each [[Definition:Vertex of Graph|vertex]] is defined as the number of [[Definition:Edge of Graph|edges]] to which it is [[Definition:Incident (Undirected Graph)|incident]]. +So when we add up the [[Definition:Degree of Vertex|degrees]] of all the [[Definition:Vertex of Graph|vertices]], we are counting all the [[Definition:Edge of Graph|edges]] of the graph twice. +{{qed}} +\end{proof}<|endoftext|> +\section{Equivalence Class holds Equivalent Elements} +Tags: Equivalence Classes + +\begin{theorem} +Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on a [[Definition:Set|set]] $S$. +Then: +:$\tuple {x, y} \in \RR \iff \eqclass x \RR = \eqclass y \RR$ +\end{theorem} + +\begin{proof} +=== Necessary Condition === +First we prove that $\tuple {x, y} \in \RR \implies \eqclass x \RR = \eqclass y \RR$. +Suppose: +: $\tuple {x, y} \in \RR: x, y \in S$ +Then: +{{begin-eqn}} +{{eqn | l = z + | o = \in + | r = \eqclass x \RR + | c = +}} +{{eqn | ll= \leadsto + | l = \tuple {x, z} + | o = \in + | r = \RR + | c = {{Defof|Equivalence Class}} +}} +{{eqn | ll= \leadsto + | l = \tuple {z, x} + | o = \in + | r = \RR + | c = {{Defof|Equivalence Relation}}: $\RR$ is [[Definition:Symmetric Relation|symmetric]] +}} +{{eqn | ll= \leadsto + | l = \tuple {z, y} + | o = \in + | r = \RR + | c = {{Defof|Equivalence Relation}}: $\RR$ is [[Definition:Transitive Relation|transitive]] +}} +{{eqn | ll= \leadsto + | l = \tuple {y, z} + | o = \in + | r = \RR + | c = {{Defof|Equivalence Relation}}: $\RR$ is [[Definition:Symmetric Relation|symmetric]] +}} +{{eqn | ll= \leadsto + | l = z + | o = \in + | r = \eqclass y \RR + | c = {{Defof|Equivalence Class}} +}} +{{end-eqn}} +So: +:$\eqclass x \RR \subseteq \eqclass y \RR$ +Now: +{{begin-eqn}} +{{eqn | l = \tuple {x, y} \in \RR + | o = \implies + | r = \eqclass x \RR \subseteq \eqclass y \RR + | c = (see above) +}} +{{eqn | l = \tuple {x, y} \in \RR + | o = \implies + | r = \tuple {y, x} \in \RR + | c = {{Defof|Equivalence Relation}}: $\RR$ is [[Definition:Symmetric Relation|symmetric]] +}} +{{eqn | o = \leadsto + | r = \eqclass y \RR \subseteq \eqclass x \RR + | c = from above +}} +{{eqn | o = \leadsto + | r = \eqclass y \RR = \eqclass x \RR + | c = {{Defof|Set Equality}} +}} +{{end-eqn}} +... so we have shown that: +:$\tuple {x, y} \in \RR \implies \eqclass x \RR = \eqclass y \RR$. +{{qed|lemma}} +=== Sufficient Condition === +Next we prove that $\eqclass x \RR = \eqclass y \RR \implies \tuple {x, y} \in \RR$. +By definition of [[Definition:Set Equality|set equality]]: +:$\eqclass x \RR = \eqclass y \RR$ +means: +:$\paren {x \in \eqclass x \RR \iff x \in \eqclass y \RR}$ +So by definition of [[Definition:Equivalence Class|equivalence class]]: +:$\tuple {y, x} \in \RR$ +Hence by definition of [[Definition:Equivalence Relation|equivalence relation]]: $\RR$ is [[Definition:Symmetric Relation|symmetric]] +:$\tuple {x, y} \in \RR$ +So we have shown that +:$\eqclass x \RR = \eqclass y \RR \implies \tuple {x, y} \in \RR$ +Thus, we have: +{{begin-eqn}} +{{eqn | l = \tuple {x, y} \in \RR + | o = \implies + | r = \eqclass x \RR = \eqclass y \RR + | c = +}} +{{eqn | l = \eqclass x \RR = \eqclass y \RR + | o = \implies + | r = \tuple {x, y} \in \RR + | c = +}} +{{end-eqn}} +{{qed|lemma}} +So by [[Definition:Biconditional|equivalence]]: +:$\tuple {x, y} \in \RR \iff \eqclass x \RR = \eqclass y\RR$ +{{qed}} +\end{proof}<|endoftext|> +\section{Equivalence Classes are Disjoint} +Tags: Equivalence Classes, Equivalence Classes are Disjoint, Fundamental Theorem on Equivalence Relations + +\begin{theorem} +Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on a [[Definition:Set|set]] $S$. +Then all [[Definition:Equivalence Class|$\RR$-classes]] are [[Definition:Pairwise Disjoint|pairwise disjoint]]: +:$\tuple {x, y} \notin \RR \iff \eqclass x \RR \cap \eqclass y \RR = \O$ +\end{theorem} + +\begin{proof} +First we show that: +:$\tuple {x, y} \notin \RR \implies \eqclass x \RR \cap \eqclass y \RR = \O$ +Suppose two $\RR$-classes are ''not'' [[Definition:Disjoint Sets|disjoint]]: +{{begin-eqn}} +{{eqn | l = \eqclass x \RR \cap \eqclass y \RR \ne \O + | o = \leadsto + | r = \exists z: z \in \eqclass x \RR \cap \eqclass y \RR + | c = {{Defof|Empty Set}} +}} +{{eqn | o = \leadsto + | r = \exists z: z \in \eqclass x \RR \land z \in \eqclass y \RR + | c = {{Defof|Set Intersection}} +}} +{{eqn | o = \leadsto + | r = \exists z: \paren {\tuple {x, z} \in \RR} \land \paren {\tuple {y, z} \in \RR} + | c = {{Defof|Equivalence Class}} +}} +{{eqn | o = \leadsto + | r = \exists z: \paren {\tuple {x, z} \in \RR} \land \paren {\tuple {x, y} \in \RR} + | c = {{Defof|Symmetric Relation}}: $\RR$ is symmetric +}} +{{eqn | o = \leadsto + | r = \tuple {x, y} \in \RR + | c = {{Defof|Transitive Relation}}: $\RR$ is transitive +}} +{{end-eqn}} +Thus we have shown that $\eqclass x \RR \cap \eqclass y \RR \ne \O \implies \tuple {x, y} \in \RR$. +Therefore, by the [[Rule of Transposition]]: +:$\tuple {x, y} \notin \RR \implies \eqclass x \RR \cap \eqclass y \RR = \O$ +Now we show that: +:$\eqclass x \RR \cap \eqclass y \RR = \O \implies \tuple {x, y} \notin \RR$ +Suppose $\tuple {x, y} \in \RR$. +{{begin-eqn}} +{{eqn | o = + | r = y \in \eqclass y \RR + | c = {{Defof|Equivalence Class}} +}} +{{eqn | o = + | r = \tuple {x, y} \in \RR + | c = [[Definition:By Hypothesis|by hypothesis]] +}} +{{eqn | o = \leadsto + | r = y \in \eqclass x \RR + | c = {{Defof|Equivalence Class}} +}} +{{eqn | o = \leadsto + | r = y \in \eqclass x \RR \land y \in \eqclass y \RR + | c = [[Rule of Conjunction]] +}} +{{eqn | o = \leadsto + | r = y \in \eqclass x \RR \cap \eqclass y \RR + | c = {{Defof|Set Intersection}} +}} +{{eqn | o = \leadsto + | r = \eqclass x \RR \cap \eqclass y \RR \ne \O + | c = {{Defof|Empty Set}} +}} +{{end-eqn}} +Thus we have shown that: +:$\tuple {x, y} \in \RR \implies \eqclass x \RR \cap \eqclass y \RR \ne \O$ +Therefore, by the [[Rule of Transposition]]: +:$\eqclass x \RR \cap \eqclass y \RR = \O \implies \paren {x, y} \notin \RR$ +Using the rule of [[Biconditional Introduction]] on these results: +:$\eqclass x \RR \cap \eqclass y \RR = \O \iff \paren {x, y} \notin \RR$ +and the proof is complete. +\end{proof} + +\begin{proof} +Suppose that for $x, y \in S$: +:$\eqclass x {\RR} \cap \eqclass y \RR \ne \O$ +Let: +:$z \in \eqclass x \RR$ +:$z \in \eqclass y \RR$ +Then by definition of [[Definition:Equivalence Class|equivalence class]]: +:$\tuple {x, z} \in \RR$ +:$\tuple {y, z} \in \RR$ +Let $c \in \eqclass x \RR$. +That is: +:$\tuple {x, c} \in \RR$ +By definition of [[Definition:Equivalence Relation|equivalence relation]], $\RR$ is [[Definition:Symmetric Relation|symmetric]] so: +:$\tuple {z, x} \in \RR$ +By definition of [[Definition:Equivalence Relation|equivalence relation]], $\RR$ is [[Definition:Transitive Relation|transitive]] so: +:$\tuple {z, x} \in \RR \land \tuple {x, c} \in \RR \implies \tuple {z, c} \in \RR$ +and +:$\tuple {y, z} \in \RR \land \tuple {z, c} \in \RR \implies \tuple {y, c} \in \RR$ +So we have $c \in \eqclass y \RR$. +By definition of [[Definition:Subset|subset]]: +:$\eqclass x \RR \subseteq \eqclass y \RR$ +Similarly, let $c \in \eqclass y \RR$. +That is: +:$\tuple {y, c} \in \RR$ +By definition of [[Definition:Equivalence Relation|equivalence relation]], $\RR$ is [[Definition:Symmetric Relation|symmetric]] so: +:$\tuple {z, y} \in \RR$ +By definition of [[Definition:Equivalence Relation|equivalence relation]], $\RR$ is [[Definition:Transitive Relation|transitive]] so: +:$\tuple {z, y} \in \RR \land \tuple {y, c} \in \RR \implies \tuple {z, c} \in \RR$ +and +:$\tuple {x, z} \in \RR \land \tuple {z, c} \in \RR \implies \tuple {x, c} \in \RR$ +So we have $c \in \eqclass x \RR$. +By definition of [[Definition:Subset|subset]]: +:$\eqclass y \RR \subseteq \eqclass x \RR$ +That is: +:$\eqclass x \RR \subseteq \eqclass y \RR$ +and +:$\eqclass y \RR \subseteq \eqclass x \RR$ +By definition of [[Definition:Set Equality/Definition 2|set equality]]: +:$\eqclass x \RR = \eqclass y \RR$ +Thus: +:$\eqclass x \RR \cap \eqclass y \RR \ne \O \implies \eqclass x \RR = \eqclass y \RR$ +and the result follows. +{{qed}} +\end{proof}<|endoftext|> +\section{Fundamental Theorem on Equivalence Relations} +Tags: Equivalence Relations, Quotient Sets, Fundamental Theorems, Fundamental Theorem on Equivalence Relations + +\begin{theorem} +Let $\RR \subseteq S \times S$ be an [[Definition:Equivalence Relation|equivalence]] on a [[Definition:Set|set]] $S$. +Then the [[Definition:Quotient Set|quotient]] $S / \RR$ of $S$ by $\RR$ forms a [[Definition:Set Partition|partition]] of $S$. +\end{theorem} + +\begin{proof} +To prove that $S / \RR$ is a [[Definition:Set Partition|partition]] of $S$, we have to prove: +:$(1): \quad \displaystyle \bigcup {S / \RR} = S$ +:$(2): \quad \eqclass x \RR \ne \eqclass y \RR \iff \eqclass x \RR \cap \eqclass y \RR = \O$ +:$(3): \quad \forall \eqclass x \RR \in S / \RR: \eqclass x \RR \ne \O$ +Taking each proposition in turn: +=== [[Union of Equivalence Classes is Whole Set]] === +The set of [[Definition:Equivalence Class|$\RR$-classes]] constitutes the whole of $S$: +{{:Union of Equivalence Classes is Whole Set}}{{qed|lemma}} +=== [[Equivalence Classes are Disjoint]] === +{{:Equivalence Classes are Disjoint/Proof 1}}{{qed|lemma}} +=== [[Equivalence Class is not Empty]] === +{{:Equivalence Class is not Empty}}{{qed|lemma}} +Thus all conditions for $S / \RR$ to be a [[Definition:Set Partition|partition]] are fulfilled. +{{qed}} +\end{proof}<|endoftext|> +\section{Equivalence Class is Unique} +Tags: Equivalence Classes + +\begin{theorem} +Let $\mathcal R$ be an [[Definition:Equivalence Relation|equivalence relation]] on $S$. +For each $x \in S$, the [[Definition:Exactly One|one and only one]] [[Definition:Equivalence Class|$\mathcal R$-class]] to which $x$ belongs is $\eqclass x {\mathcal R}$. +\end{theorem} + +\begin{proof} +This follows directly from the [[Fundamental Theorem on Equivalence Relations]]: the [[Definition:Quotient Set|set]] of $\mathcal R$-classes forms a [[Definition:Set Partition|partition]] of $S$. +{{qed}} +\end{proof}<|endoftext|> +\section{Equivalence Class Equivalent Statements} +Tags: Equivalence Classes + +\begin{theorem} +Let $\mathcal R$ be an [[Definition:Equivalence Relation|equivalence]] on $S$. +{{refactor|Use TFAE template}} +Then $\forall x, y \in S$, the following statements are all equivalent: +:$(1): \quad x$ and $y$ are in the same [[Definition:Equivalence Class|$\mathcal R$-class]] +:$(2): \quad \eqclass x {\mathcal R} = \eqclass y {\mathcal R}$ +:$(3): \quad x \mathrel {\mathcal R} y$ +:$(4): \quad x \in \eqclass y {\mathcal R}$ +:$(5): \quad y \in \eqclass x {\mathcal R}$ +:$(6): \quad \eqclass x {\mathcal R} \cap \eqclass y {\mathcal R} \ne \O$ +\end{theorem} + +\begin{proof} +: $(1)$ and $(2)$ are equivalent because, by [[Equivalence Class is Unique]], $\eqclass x {\mathcal R}$ is the unique [[Definition:Equivalence Class|$\mathcal R$-class]] to which $x$ belongs, and $\eqclass y {\mathcal R}$ is the unique [[Definition:Equivalence Class|$\mathcal R$-class]] to which $y$ belongs. As these are unique for each, they must therefore be the same set. +: $(2)$ is equivalent to $(3)$ by [[Equivalence Class holds Equivalent Elements]]. +: $(3)$ is equivalent to $(4)$ by the definition of [[Definition:Equivalence Class|Equivalence Class]]. +: $(3)$ is equivalent to $(5)$ through dint of the [[Definition:Symmetric Relation|symmetry]] of $\mathcal R$ and the definition of [[Definition:Equivalence Class|Equivalence Class]]. +: $(3)$ is equivalent to $(6)$ from [[Equivalence Classes are Disjoint]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Relation Induced by Partition is Equivalence} +Tags: Quotient Sets, Equivalence Relations + +\begin{theorem} +Let $\mathbb S$ be a [[Definition:Set Partition|partition]] of a [[Definition:Set|set]] $S$. +Let $\RR$ be the [[Definition:Relation Induced by Partition|relation induced by $\mathbb S$]]. +Then: +:$(1): \quad \RR$ is [[Definition:Unique|unique]] +:$(2): \quad \RR$ is an [[Definition:Equivalence Relation|equivalence relation]] on $S$. +Hence $\mathbb S$ is the [[Definition:Quotient Set|quotient set]] of $S$ by $\RR$, that is: +:$\mathbb S = S / \RR$ +\end{theorem} + +\begin{proof} +Let $\mathbb S$ be a [[Definition:Set Partition|partition]] of $S$. +From [[Definition:Relation Induced by Partition|Relation Induced by Partition]], we define the [[Definition:Relation|relation]] $\RR$ on $S$ by: +:$\RR = \set {\tuple {x, y}: \paren {\exists T \in \mathbb S: x \in T \land y \in T} }$ +=== Test for Equivalence === +We are to show that $\RR$ is an [[Definition:Equivalence Relation|equivalence relation]]. +Checking in turn each of the critera for [[Definition:Equivalence Relation|equivalence]]: +==== Reflexive ==== +$\RR$ is [[Definition:Reflexive Relation|reflexive]]: +{{begin-eqn}} +{{eqn | o = + | r = \paren {x \in T \land x \in T} + | c = [[Rule of Idempotence]] +}} +{{eqn | o = \leadstoandfrom + | r = \tuple {x, x} \in \RR + | c = Definition of $\RR$ +}} +{{end-eqn}} +{{qed|lemma}} +==== Symmetric ==== +$\RR$ is [[Definition:Symmetric Relation|symmetric]]: +{{begin-eqn}} +{{eqn | o = + | r = \tuple {x, y} \in \RR + | c = +}} +{{eqn | o = \leadsto + | r = x \in T \land y \in T + | c = Definition of $\RR$ +}} +{{eqn | o = \leadsto + | r = y \in T \land x \in T + | c = [[Rule of Commutation]] +}} +{{eqn | o = \leadsto + | r = \tuple {y, x} \in \RR + | c = Definition of $\RR$ +}} +{{end-eqn}} +{{qed|lemma}} +==== Transitive ==== +$\RR$ is [[Definition:Transitive Relation|transitive]]: +{{begin-eqn}} +{{eqn | l = x \in T \land y \in T + | o = \leadstoandfrom + | r = \tuple {x, y} \in \RR + | c = Definition of $\RR$ +}} +{{eqn | l = y \in T \land z \in T + | o = \leadstoandfrom + | r = \tuple {y, z} \in \RR + | c = Definition of $\RR$ +}} +{{eqn | l = \paren {x \in T \land y \in T} \land \paren {y \in T \land z \in T} + | o = \leadstoandfrom + | r = \tuple {x, z} \in \RR + | c = Definition of $\RR$ +}} +{{end-eqn}} +{{qed|lemma}} +So $\RR$ is [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]]. +Hence, by definition, $\RR$ is an [[Definition:Equivalence Relation|equivalence relation]]. +{{qed|lemma}} +=== Test for Uniqueness === +Now by definition of a [[Definition:Set Partition|partition]], we have that: +:$\mathbb S$ [[Definition:Set Partition|partition]]s $S \implies \forall x \in S: \exists T \in \mathbb S: x \in T$ +Also: +{{begin-eqn}} +{{eqn | l = x, y \in T + | o = \leadsto + | r = \tuple {x, y} \in \RR + | c = Definition of $\RR$ +}} +{{eqn | l = x \in T_1, y \in T_2, T_1 \ne T_2 + | o = \leadsto + | r = \tuple{x, y} \notin \RR + | c = {{Defof|Set Partition}} +}} +{{end-eqn}} +Thus $\mathbb S$ is the [[Definition:Set of Sets|set]] of [[Definition:Equivalence Class|$\RR$-classes]] constructed above, and no other [[Definition:Relation|relation]] can be constructed in this way. +{{qed}} +\end{proof}<|endoftext|> +\section{Relation Partitions Set iff Equivalence} +Tags: Equivalence Relations, Quotient Sets, Set Partitions, Relation Partitions Set iff Equivalence + +\begin{theorem} +Let $\RR$ be a [[Definition:Relation|relation]] on a [[Definition:Set|set]] $S$. +Then $S$ can be [[Definition:Set Partition|partitioned]] into [[Definition:Subset|subsets]] by $\RR$ {{iff}} $\RR$ is an [[Definition:Equivalence Relation|equivalence relation]] on $S$. +The [[Definition:Set Partition|partition]] of $S$ defined by $\RR$ is the [[Definition:Quotient Set|quotient set]] $S / \RR$. +\end{theorem} + +\begin{proof} +Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on $S$. +From the [[Fundamental Theorem on Equivalence Relations]], we have that the [[Definition:Equivalence Class|equivalence classes]] of $\RR$ form a [[Definition:Set Partition|partition]]. +{{qed|lemma}} +Let $S$ be [[Definition:Set Partition|partitioned]] into [[Definition:Subset|subsets]] by a [[Definition:Relation|relation]] $\RR$. +From [[Relation Induced by Partition is Equivalence]], $\RR$ is an [[Definition:Equivalence Relation|equivalence relation]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Intersection of Equivalences} +Tags: Equivalence Relations, Set Intersection + +\begin{theorem} +The [[Definition:Set Intersection|intersection]] of two [[Definition:Equivalence Relation|equivalence relations]] is itself an [[Definition:Equivalence Relation|equivalence relation]]. +\end{theorem} + +\begin{proof} +Let $\mathcal R_1$ and $\mathcal R_2$ be [[Definition:Equivalence Relation|equivalence relations]] on $S$. +Let $\mathcal R_3 = \mathcal R_1 \cap \mathcal R_2$. +Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: +=== Reflexive === +[[Definition:Equivalence Relation|Equivalence relations]] are by definition [[Definition:Reflexive Relation|reflexive]]. +So, by [[Intersection of Reflexive Relations is Reflexive]], so is $\mathcal R_3$. +=== Symmetric === +[[Definition:Equivalence Relation|Equivalence relations]] are by definition [[Definition:Symmetric Relation|symmetric]]. +So, by [[Intersection of Symmetric Relations is Symmetric]], so is $\mathcal R_3$. +=== Transitive === +[[Definition:Equivalence Relation|Equivalence relations]] are by definition [[Definition:Transitive Relation|transitive]]. +So, by [[Intersection of Transitive Relations is Transitive]], so is $\mathcal R_3$. +Thus $\mathcal R_3$ is an [[Definition:Equivalence Relation|equivalence]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Union of Equivalences} +Tags: Union, Set Union, Equivalence Relations, Set Union, Union of Equivalences + +\begin{theorem} +The [[Definition:Set Union|union]] of two [[Definition:Equivalence Relation|equivalence relations]] is '''not''' necessarily an [[Definition:Equivalence Relation|equivalence relation]] itself. +\end{theorem} + +\begin{proof} +This can be shown by giving an example. +Let $S = \left\{{a, b, c}\right\}$, and let $\mathcal R_1$ and $\mathcal R_2$ be [[Definition:Equivalence Relation|equivalences]] on $S$ such that: +: $\left[\!\left[{a}\right]\!\right]_{\mathcal R_1} = \left[\!\left[{b}\right]\!\right]_{\mathcal R_1} = \left\{{a, b}\right\}$ +: $\left[\!\left[{c}\right]\!\right]_{\mathcal R_1} = \left\{{c}\right\}$ +: $\left[\!\left[{a}\right]\!\right]_{\mathcal R_2} = \left\{{a}\right\}$ +: $\left[\!\left[{b}\right]\!\right]_{\mathcal R_2} = \left[\!\left[{c}\right]\!\right]_{\mathcal R_2} = \left\{{b, c}\right\}$ +Let $\mathcal R_3 = \mathcal R_1 \cup \mathcal R_2$. +Then: +{{begin-eqn}} +{{eqn | o = + | r = \left({a, b}\right) \in \mathcal R_1 + | c = +}} +{{eqn | o = + | r = \left({b, c}\right) \in \mathcal R_2 + | c = +}} +{{eqn | o = \implies + | r = \left({a, b}\right) \in \mathcal R_1 \cup \mathcal R_2 \land \left({b, c}\right) \in \mathcal R_1 \cup \mathcal R_2 + | c = +}} +{{eqn | o = \implies + | r = \left({a, b}\right) \in \mathcal R_3 \land \left({b, c}\right) \in \mathcal R_3 + | c = +}} +{{end-eqn}} +However: +{{begin-eqn}} +{{eqn | o = + | r = \left({a, c}\right) \notin \mathcal R_1 + | c = +}} +{{eqn | o = + | r = \left({a, c}\right) \notin \mathcal R_2 + | c = +}} +{{eqn | o = \implies + | r = \left({a, c}\right) \notin \mathcal R_1 \cup \mathcal R_2 + | c = +}} +{{eqn | o = \implies + | r = \left({a, c}\right) \notin \mathcal R_3 + | c = +}} +{{end-eqn}} +So $\mathcal R_3$ is not [[Definition:Transitive Relation|transitive]], and therefore $\mathcal R_3 = \mathcal R_1 \cup \mathcal R_2$ is not an [[Definition:Equivalence Relation|equivalence relation]]. +{{qed}} +\end{proof} + +\begin{proof} +We have that the [[Union of Reflexive Relations is Reflexive]]. +We also have that the [[Union of Symmetric Relations is Symmetric]]. +However, we also have that the [[Union of Transitive Relations Not Always Transitive]]. +Hence the union of two [[Definition:Equivalence Relation|equivalence relations]] is not always an [[Definition:Equivalence Relation|equivalence relation]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Equivalence of Definitions of Equivalence Relation} +Tags: Equivalence Relations + +\begin{theorem} +Let $\RR$ be a [[Definition:Endorelation|relation]] on a [[Definition:Set|set]] $S$. +{{TFAE|def = Equivalence Relation}} +\end{theorem} + +\begin{proof} +=== Definition 1 implies Definition 2 === +Let $\RR$ be an [[Definition:Equivalence Relation/Definition 1|equivalence relation by definition 1]]. +By definition, $\RR$ is [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]]. +From [[Relation Contains Diagonal Relation iff Reflexive]]: +:$\Delta_S \subseteq \RR$ +From [[Relation equals Inverse iff Symmetric]]: +:$\RR^{-1} = \RR$ +and so by definition of [[Definition:Set Equality/Definition 2|set equality]]: +:$\RR^{-1} \subseteq \RR$ +From [[Relation contains Composite with Self iff Transitive]]: +:$\RR \circ \RR \subseteq \RR$ +From [[Union is Smallest Superset]]: +:$\Delta_S \cup \RR^{-1} \cup \RR \circ \RR \subseteq \RR$ +Thus $\RR$ is an [[Definition:Equivalence Relation/Definition 2|equivalence relation by definition 2]]. +{{qed|lemma}} +=== Definition 2 implies Definition 1 === +Let $\RR$ be an [[Definition:Equivalence Relation/Definition 2|equivalence relation by definition 2]]. +That is: +:$\Delta_S \cup \RR^{-1} \cup \RR \circ \RR \subseteq \RR$ +By [[Union is Smallest Superset]]: +:$(1): \quad \Delta_S \subseteq \RR$ +:$(2): \quad \RR^{-1} \subseteq \RR$ +:$(3): \quad \RR \circ \RR \subseteq \RR$ +From [[Relation Contains Diagonal Relation iff Reflexive]], $(1)$ gives directly that $\RR$ is [[Definition:Reflexive Relation|reflexive]]. +From [[Inverse Relation Equal iff Subset]], $(2)$ gives that $\RR^{-1} = \RR$. +So from [[Relation equals Inverse iff Symmetric]] $\RR$ is [[Definition:Symmetric Relation|symmetric]]. +From [[Relation contains Composite with Self iff Transitive]], $(3)$ gives that $\RR$ is [[Definition:Transitive Relation|transitive]]. +So $\RR$ has been shown to be [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]]. +Thus $\RR$ is an [[Definition:Equivalence Relation/Definition 1|equivalence relation by definition 1]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Equivalence iff Diagonal and Inverse Composite} +Tags: Equivalence Relations + +\begin{theorem} +Let $\RR$ be a [[Definition:Relation|relation]] on $S$. +Then $\RR$ is an [[Definition:Equivalence Relation|equivalence relation]] on $S$ {{iff}}: +:$\Delta_S \subseteq \RR$ +and: +:$\RR = \RR \circ \RR^{-1}$ +\end{theorem} + +\begin{proof} +=== Necessary Condition === +Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]]. +Then by definition, it is [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]]. +As $\RR$ is [[Definition:Reflexive Relation|reflexive]], we have $\Delta_S \subseteq \RR$ from [[Relation Contains Diagonal Relation iff Reflexive]]. +As $\RR$ is [[Definition:Transitive Relation|transitive]]: +:$\RR \circ \RR \subseteq \RR$ +from [[Relation contains Composite with Self iff Transitive]]. +But as $\RR$ is [[Definition:Symmetric Relation|symmetric]]: +:$\RR = \RR^{-1}$ +from [[Relation equals Inverse iff Symmetric]]. +Thus: +:$\RR \circ \RR^{-1} \subseteq \RR$ +Now we need to show that $\RR \subseteq \RR \circ \RR^{-1}$: +Let $\tuple {x, y} \in \RR$. +Then as $\RR$ is [[Definition:Reflexive Relation|reflexive]]: +:$\tuple {x, y} \in \RR \land \tuple {y, y} \in \RR$ +and so: +:$\tuple {x, y} \in \RR \circ \RR$ +As $\RR = \RR^{-1}$, it follows that: +:$\RR \subseteq \RR \circ \RR^{-1}$ +So it has been shown: +:$\Delta_S \subseteq \RR$ and $\RR = \RR \circ \RR^{-1}$ +{{qed|lemma}} +=== Sufficient Condition === +Now, let $\Delta_S \subseteq \RR$ and $\RR = \RR \circ \RR^{-1}$. +From [[Relation Contains Diagonal Relation iff Reflexive]], $\RR$ is [[Definition:Reflexive Relation|reflexive]]. +Let $\tuple {x, y} \in \RR$. +Then: +:$\tuple {x, y} \in \RR \circ \RR^{-1}$ +So: +:$\exists z \in S: \tuple {x, z} \in \RR, \tuple {z, y} \in \RR^{-1}$ +and so: +:$\tuple {y, z} \in \RR$ +That is, by definition, $\RR$ is [[Definition:Transitive Relation|transitive]]. +As $\tuple {x, z} \in \RR$: +:$\tuple {z, x} \in \RR^{-1}$ +and so: +:$\tuple {y, x} \in \RR$ +So it follows that $\RR$ is [[Definition:Symmetric Relation|symmetric]]. +Thus $\RR$ is [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]]. +Therefore, by definition, $\RR$ is an [[Definition:Equivalence Relation|equivalence relation]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Equality of Elements in Range of Mapping} +Tags: Mapping Theory + +\begin{theorem} +Let $f: S \to T$ be a [[Definition:Mapping|mapping]]. +Then: +:$\exists y \in \Rng f: \tuple {x_1, y} \in f \land \tuple {x_2, y} \in f \iff \map f {x_1} = \map f {x_2}$ +\end{theorem} + +\begin{proof} +=== Necessary Condition === +Let: +:$\exists y \in \Rng f: \tuple {x_1, y} \in f \land \tuple {x_2, y} \in f$ +Then: +{{begin-eqn}} +{{eqn | o = + | r = \tuple {x_1, y} \in f \land \tuple {x_2, y} \in f + | c = +}} +{{eqn | o = \leadsto + | r = \map f {x_1} = y \land \map f {x_2} = y + | c = {{Defof|Mapping}} +}} +{{eqn | o = \leadsto + | r = \map f {x_1} = \map f {x_2} + | c = [[Equality is Equivalence Relation]] +}} +{{end-eqn}} +{{qed|lemma}} +=== Sufficient Condition === +Let: +:$\map f {x_1} = \map f {x_2}$ +Then: +{{begin-eqn}} +{{eqn | o = + | r = \exists y \in \Rng f: \map f {x_1} = y = \map f {x_2} + | c = +}} +{{eqn | o = \leadsto + | r = \exists y \in \Rng f: \tuple {x_1, y} \in f \land \tuple {x_2, y} \in f + | c = {{Defof|Mapping}} +}} +{{end-eqn}} +{{qed|lemma}} +The result follows from the definition of [[Definition:Logical Equivalence|logical equivalence]]. +{{qed}} +[[Category:Mapping Theory]] +r2h6pslsz6g6qeely8104vu5cthaxa1 +\end{proof}<|endoftext|> +\section{Inverse of Mapping is One-to-Many Relation} +Tags: Inverse Relations, Inverse Mappings + +\begin{theorem} +Let $f$ be a [[Definition:Mapping|mapping]]. +Then its [[Definition:Inverse of Mapping|inverse]] $f^{-1}$ is a [[Definition:One-to-Many Relation|one-to-many relation]]. +\end{theorem} + +\begin{proof} +We have that $f$ is a [[Definition:Mapping|mapping]]. +Hence $f$ is [[Definition:A Fortiori|a fortiori]] a [[Definition:Many-to-One Relation|many-to-one relation]]. +Then from [[Inverse of Many-to-One Relation is One-to-Many]], $f^{-1}$ is [[Definition:One-to-Many Relation|one-to-many]]. +{{Qed}} +[[Category:Inverse Relations]] +[[Category:Inverse Mappings]] +ayh6r7f3eifj9ml49b63dzgczo2bazq +\end{proof}<|endoftext|> +\section{Equality of Mappings} +Tags: Mapping Theory + +\begin{theorem} +Two [[Definition:Mapping|mappings]] $f_1: S_1 \to T_1, f_2: S_2 \to T_2$ are equal {{iff}}: +:$(1): \quad S_1 = S_2$ +:$(2): \quad T_1 = T_2$ +:$(3): \quad \forall x \in S_1: \map {f_1} x = \map {f_2} x$ +\end{theorem} + +\begin{proof} +This follows directly from [[Equality of Relations]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Mapping is Constant iff Image is Singleton} +Tags: Constant Mappings, Singletons + +\begin{theorem} +A [[Definition:Mapping|mapping]] is a [[Definition:Constant Mapping|constant mapping]] {{iff}} its [[Definition:Image of Mapping|image]] is a [[Definition:Singleton|singleton]]. +\end{theorem} + +\begin{proof} +=== Necessary Condition === +Let $f_c: S \to T$ be a [[Definition:Constant Mapping|constant mapping]]. Then from the definition of the [[Definition:Image of Element under Mapping|image of a element]]: +:$\forall x \in S: \map {f_c} x = c \implies \Img S = \set c$ +Thus the [[Definition:Image of Mapping|image]] of $f_c: S \to T$ is a [[Definition:Singleton|singleton]]. +=== Sufficient Condition === +Let $f: S \to T$ such that its image is a singleton $\set c \subseteq T$. +{{begin-eqn}} +{{eqn | l = \forall x \in S: \map f x + | o = \in + | r = \Img f + | c = {{Defof|Image of Mapping}} +}} +{{eqn | ll= \leadsto + | l = \forall x \in S: \map f x + | o = \in + | r = \set c + | c = [[Definition:By Hypothesis|By Hypothesis]] +}} +{{eqn | ll= \leadsto + | l = \forall x \in S: \map f x + | r = c + | c = {{Defof|Singleton}} +}} +{{end-eqn}} +Thus $f: S \to T$ is a [[Definition:Constant Mapping|constant mapping]] (and we can write it $f_c: S \to T$). +{{qed}} +\end{proof}<|endoftext|> +\section{Diagonal Relation is Right Identity} +Tags: Relation Theory, Identity Mappings + +\begin{theorem} +Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]] on $S \times T$. +Then: +:$\RR \circ \Delta_S = \RR$ +where $\Delta_S$ is the [[Definition:Diagonal Relation|diagonal relation]] on $S$, and $\circ$ signifies [[Definition:Composition of Relations|composition of relations]]. +\end{theorem} + +\begin{proof} +We use the [[Equality of Relations|definition of relation equality]], as follows: +=== Equality of Codomains === +The [[Definition:Codomain of Relation|codomains]] of $\RR$ and $\RR \circ \Delta_S$ are both equal to $T$ from [[Codomain of Composite Relation]]. +=== Equality of Domains === +The [[Definition:Domain of Relation|domains]] of $\RR$ and $\RR \circ \Delta_S$ are also easily shown to be equal. +From [[Domain of Composite Relation]]: +:$\Dom {\RR \circ \Delta_S} = \Dom {\Delta_S}$ +But from the definition of the [[Definition:Diagonal Relation|diagonal relation]]: +:$\Dom {\Delta_S} = \Img {\Delta_S} = S$ +=== Equality of Relations === +The [[Definition:Composition of Relations|composite of $\Delta_S$ and $\RR$]] is defined as: +:$\RR \circ \Delta_S = \set {\tuple {x, z} \in S \times T: \exists y \in S: \tuple {x, y} \in \Delta_S \land \tuple {y, z} \in \RR}$ +But by definition of the [[Definition:Diagonal Relation|diagonal relation]] on $S$, we have that: +:$\tuple {x, y} \in \Delta_S \implies x = y$ +Hence: +:$\RR \circ \Delta_S = \set {\tuple {y, z} \in S \times T: \exists y \in S: \tuple {y, y} \in \Delta_S \land \tuple {y, z} \in \RR}$ +But as $\forall y \in S: \tuple {y, y} \in \Delta_S$, this means: +:$\RR \circ \Delta_S = \set {\tuple {y, z} \in S \times T: \tuple {y, z} \in \RR}$ +That is: +:$\RR \circ \Delta_S = \RR$ +Hence the result. +{{qed}} +\end{proof}<|endoftext|> +\section{Identity Mapping is Left Identity} +Tags: Identity Mappings + +\begin{theorem} +Let $S$ and $T$ be [[Definition:Set|sets]]. +Let $f: S \to T$ be a [[Definition:Mapping|mapping]]. +Then: +:$I_T \circ f = f$ +where $I_T$ is the [[Definition:Identity Mapping|identity mapping]] on $T$, and $\circ$ signifies [[Definition:Composition of Mappings|composition of mappings]]. +\end{theorem} + +\begin{proof} +=== Equality of Domains === +The [[Definition:Domain of Mapping|domains]] of $f$ and $I_T \circ f$ are equal from [[Domain of Composite Relation]]: +:$\Dom {I_T \circ f} = \Dom f$ +{{qed|lemma}} +=== Equality of Codomains === +The [[Definition:Codomain of Mapping|codomains]] of $f$ and $f \circ I_S$ are also easily shown to be equal. +From [[Codomain of Composite Relation]], the [[Definition:Codomain of Mapping|codomains]] of $I_T \circ f$ and $I_T$ are both equal to $T$. +But from the definition of the [[Definition:Identity Mapping|identity mapping]], the [[Definition:Codomain of Mapping|codomain]] of $I_T$ is $\Dom {I_T} = T$ +{{qed|lemma}} +=== Equality of Relations === +The [[Definition:Composition of Mappings|composite of $f$ and $I_T$]] is defined as: +:$I_T \circ f = \set {\tuple {x, z} \in S \times T: \exists y \in T: \tuple {x, y} \in f \land \tuple {y, z} \in I_T}$ +But by definition of the [[Definition:Identity Mapping|identity mapping]] on $T$, we have that: +:$\tuple {y, z} \in I_T \implies y = z$ +Hence: +:$I_T \circ f = \set {\tuple {x, y} \in S \times T: \exists y \in T: \tuple {x, y} \in f \land \tuple {y, y} \in I_T}$ +But as $\forall y \in T: \tuple {y, y} \in I_T$, this means: +:$I_T \circ f = \set {\tuple {x, y} \in S \times T: \tuple {x, y} \in f}$ +That is: +:$I_T \circ f = f$ +{{qed|lemma}} +Hence the result, by [[Equality of Mappings]]. +{{qed}} +\end{proof} + +\begin{proof} +By definition, a [[Definition:Mapping|mapping]] is also a [[Definition:Relation|relation]]. +Also by definition, the [[Definition:Identity Mapping|identity mapping]] is the same as the [[Definition:Diagonal Relation|diagonal relation]]. +Thus [[Diagonal Relation is Left Identity]] can be applied directly. +{{qed}} +\end{proof}<|endoftext|> +\section{Image of Element under Inverse Mapping} +Tags: Inverse Mappings + +\begin{theorem} +Let $S$ and $T$ be [[Definition:Set|sets]]. +Let $f: S \to T$ be a [[Definition:Mapping|mapping]] such that its [[Definition:Inverse of Mapping|inverse]] $f^{-1}: T \to S$ is also a [[Definition:Mapping|mapping]]. +Then: +:$\forall x \in S, y \in T: \map f x = y \iff \map {f^{-1} } y = x$ +\end{theorem} + +\begin{proof} +=== Sufficient Condition === +Let $f: S \to T$ be a [[Definition:Mapping|mapping]]. +From the definition of [[Definition:Inverse of Mapping|inverse mapping]]: +:$f^{-1} = \set {\tuple {y, x}: \tuple {x, y} \in f}$ +Let $y = f \paren x$. +From the definition of the [[Definition:Preimage of Element under Mapping|preimage of an element]]: +:$\map {f^{-1} } y = \set {s \in S: \tuple {y, x} \in f}$ +Thus: +:$x \in \map {f^{-1} } y$ +However, $f^{-1}$ is a [[Definition:Mapping|mapping]]. +Therefore, by definition: +:$\forall y \in T: \tuple {y, x_1} \in f^{-1} \land \tuple {y, x_2} \in f^{-1} \implies x_1 = x_2$ +Thus: +:$\forall s \in f^{-1} \paren y: s = x$ +Thus: +:$\map {f^{-1} } y = \set x$ +That is: +:$x = \map {f^{-1} } y$ +{{qed|lemma}} +=== Necessary Condition === +Let $\map {f^{-1} } y = x$. +By definition of [[Definition:Inverse Mapping|inverse mapping]]: +:$\map f x = y$ +{{qed}} +\end{proof}<|endoftext|> +\section{Injection iff Left Cancellable} +Tags: Injections, Cancellability, Injection iff Left Cancellable + +\begin{theorem} +A [[Definition:Mapping|mapping]] $f$ is an [[Definition:Injection|injection]] {{iff}} $f$ is [[Definition:Left Cancellable Mapping|left cancellable]]. +\end{theorem} + +\begin{proof} +From the definition: a [[Definition:Mapping|mapping]] $f: Y \to Z$ is [[Definition:Left Cancellable Mapping|left cancellable]] {{iff}}: +:$\forall X: \forall g_1: X \to Y, g_2: X \to Y: f \circ g_1 = f \circ g_2 \implies g_1 = g_2$ +=== [[Injection iff Left Cancellable/Necessary Condition|Necessary Condition]] === +{{:Injection iff Left Cancellable/Necessary Condition}}{{qed|lemma}} +=== [[Injection iff Left Cancellable/Sufficient Condition|Sufficient Condition]] === +{{:Injection iff Left Cancellable/Sufficient Condition}}{{Qed}} +\end{proof}<|endoftext|> +\section{Identity Mapping is Injection} +Tags: Injections, Identity Mappings + +\begin{theorem} +On any [[Definition:Set|set]] $S$, the [[Definition:Identity Mapping|identity mapping]] $I_S: S \to S$ is an [[Definition:Injection|injection]]. +\end{theorem} + +\begin{proof} +From the definition of the [[Definition:Identity Mapping|identity mapping]]: +:$\forall x \in S: \map {I_S} x = x$ +So: +:$\map {I_S} x = \map {I_S} y \implies x = y$ +So by definition $I_S$ is an [[Definition:Injection|injection]]. +{{Qed}} +\end{proof}<|endoftext|> +\section{Composite of Injections is Injection} +Tags: Injections, Composite Mappings + +\begin{theorem} +A [[Definition:Composition of Mappings|composite]] of [[Definition:Injection|injections]] is an [[Definition:Injection|injection]]. +That is: +:If $f$ and $g$ are [[Definition:Injection|injections]], then so is $f \circ g$. +\end{theorem} + +\begin{proof} +Let $f$ and $g$ be [[Definition:Injection|injections]]. +Then: +{{begin-eqn}} +{{eqn | l = \map {f \circ g} x + | r = \map {f \circ g} y + | c = +}} +{{eqn | ll= \leadsto + | l = \map f {\map g x} + | r = \map f {\map g y} + | c = {{Defof|Composition of Mappings}} +}} +{{eqn | ll= \leadsto + | l = \map g x + | r = \map g y + | c = as $f$ is [[Definition:Injection|injective]] +}} +{{eqn | ll= \leadsto + | l = x + | r = y + | c = as $g$ is [[Definition:Injection|injective]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Injection if Composite is Injection} +Tags: Injections, Composite Mappings + +\begin{theorem} +Let $f$ and $g$ be [[Definition:Mapping|mappings]] such that their [[Definition:Composition of Mappings|composite]] $g \circ f$ is an [[Definition:Injection|injection]]. +Then $f$ is an [[Definition:Injection|injection]]. +\end{theorem} + +\begin{proof} +Let $g \circ f$ be [[Definition:Injection|injective]]. +We need to show that $\map f a = \map f b \implies a = b$. +So suppose $\map f a = \map f b$. +Then: +{{begin-eqn}} +{{eqn | l = \map {g \circ f} a + | r = \map g {\map f a} + | c = {{Defof|Composition of Mappings}} +}} +{{eqn | r = \map g {\map f b} + | c = [[Definition:By Hypothesis|by hypothesis]] +}} +{{eqn | r = \map {g \circ f} b + | c = {{Defof|Composition of Mappings}} +}} +{{end-eqn}} +and so $a = b$ as $g \circ f$ is an [[Definition:Injection|injection]]. +So we have shown that: +:$\map f a = \map f b \implies a = b$ +Hence the result from the definition of [[Definition:Injection|injection]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Injection iff Left Inverse} +Tags: Injections, Left Inverse Mappings, Injection iff Left Inverse + +\begin{theorem} +A [[Definition:Mapping|mapping]] $f: S \to T, S \ne \O$ is an [[Definition:Injection|injection]] {{iff}}: +:$\exists g: T \to S: g \circ f = I_S$ +where $g$ is a [[Definition:Mapping|mapping]]. +That is, {{iff}} $f$ has a [[Definition:Left Inverse Mapping|left inverse]]. +\end{theorem} + +\begin{proof} +Let: +:$\exists g: T \to S: g \circ f = I_S$ +From [[Identity Mapping is Injection]], $I_S$ is [[Definition:Injection|injective]], so $g \circ f$ is [[Definition:Injection|injective]]. +So from [[Injection if Composite is Injection]], $f$ is an [[Definition:Injection|injection]]. +Note that the existence of such a $g$ ''requires'' that $S \ne \O$. +{{qed|lemma}} +Now, assume $f$ is an [[Definition:Injection|injection]]. +We now define a [[Definition:Mapping|mapping]] $g: T \to S$ as follows. +As $S \ne \O$, we choose $x_0 \in S$. +By definition of [[Definition:Injection/Definition 3|injection]]: +:$f^{-1} {\restriction_{\Img f} } \to S$ is a [[Definition:Mapping|mapping]] +so it is possible to define: +:$\map g y = \begin{cases} +x_0: & y \in T \setminus \Img f \\ +\map {f^{-1} } y: & y \in \Img f \end{cases}$ +:[[File:InjectionIffLeftInverse.png|750px]] +It does not matter what the [[Definition:Element|elements]] of $T \setminus \Img f$ are. +Using the construction given, the equation $g \circ f = I_S$ holds whatever value (or values) we choose for $g \sqbrk {T \setminus \Img f}$. +The remaining [[Definition:Element|elements]] of $T$ can be mapped arbitrarily, and will not affect the [[Definition:Image of Element under Mapping|image]] of any $x \in S$ under the [[Definition:Mapping|mapping]] $g \circ f$. +So, for all $x \in S$: +:$\map {g \circ f} x = \map g {\map f x}$ +is the [[Definition:Unique|unique]] [[Definition:Element|element]] of $S$ which $f$ maps to $\map f x$. +This unique element is $x$. +Thus $g \circ f = I_S$. +{{Qed}} +\end{proof} + +\begin{proof} +Take the result [[Condition for Composite Mapping on Left]]: +Let $A, B, C$ be [[Definition:Set|sets]]. +Let $f: A \to B$ and $g: A \to C$ be [[Definition:Mapping|mappings]]. +Then: +:$\forall x, y \in A: \map f x = \map f y \implies \map g x = \map g y$ +{{iff}}: +:$\exists h: B \to C$ such that $h$ is a [[Definition:Mapping|mapping]] and $h \circ f = g$. +Let $C = A = S$, let $B = T$ and let $g = I_S$. +Then the above translates into: +:$\exists h: T \to S$ such that $h$ is a [[Definition:Mapping|mapping]] and $h \circ f = g$ +{{iff}}: +:$\forall x, y \in S: \map f x = \map f y \implies \map {I_S} x = \map {I_S} y$ +But as $\map {I_S} x = x$ and $\map {I_S} y = y$ by definition of [[Definition:Identity Mapping|identity mapping]], it follows that: +:$\exists h: T \to S$ such that $h$ is a [[Definition:Mapping|mapping]] and $h \circ f = g$ +{{iff}}: +:$\forall x, y \in S: \map f x = \map f y \implies x = y$ +which is our result. +{{qed}} +\end{proof} + +\begin{proof} +Let $f: S \to T$ be an [[Definition:Injection|injection]]. +Then $f$ is a [[Definition:One-to-Many Relation|one-to-many relation]]. +By [[Inverse of Many-to-One Relation is One-to-Many]], $f^{-1}: T \to S$ is [[Definition:Many-to-One Relation|many-to-one]]. +By [[Many-to-One Relation Extends to Mapping]], there is a [[Definition:Mapping|Mapping]] $g: T \to S$ such that $f^{-1} \subseteq g$. +Let $\tuple {x, y} \in g \circ f$. +Then: +:$\exists z \in T: \tuple {x, z} \in f, \tuple {z, y} \in g$ +Since $\tuple {x, z} \in f$: +:$\tuple {z, x} \in f^{-1} \subseteq g$ +Since $\tuple {z, y} \in g$, $\tuple {z, x} \in g$ and $g$ is a [[Definition:Mapping|mapping]]: +:$x = y$ +so: +:$\tuple {x, y} \in I_S$ +So we see that: +:$g \circ f \subseteq I_S$ +Let $x \in S$. +Then: +:$\tuple {x, \map f x} \in f$ +and: +:$\tuple {\map f x, x} \in f^{-1} \subseteq g$ +So: +:$\tuple {x, x} \in g \circ f$ +Thus: +:$I_S \subseteq g \circ f$ +By definition of [[Definition:Set Equality/Definition 2|set equality]]: +:$I_S = g \circ f$ +{{qed}} +\end{proof}<|endoftext|> +\section{Inclusion Mapping is Injection} +Tags: Injections, Inclusion Mappings + +\begin{theorem} +Let $S, T$ be [[Definition:Set|sets]] such that $S$ is a [[Definition:Subset|subset]] of $T$. +Then the [[Definition:Inclusion Mapping|inclusion mapping]] $i_S: S \to T$ defined as: +:$\forall x \in S: \map {i_S} x = x$ +is an [[Definition:Injection|injection]]. +For this reason the [[Definition:Inclusion Mapping|inclusion mapping]] can be known as the '''canonical injection of $S$ to $T$'''. +\end{theorem} + +\begin{proof} +Suppose $\map {i_S} {s_1} = \map {i_S} {s_2}$. +{{begin-eqn}} +{{eqn | l = \map {i_S} {s_1} + | r = s_1 + | c = {{Defof|Inclusion Mapping}} +}} +{{eqn | l = \map {i_S} {s_2} + | r = s_2 + | c = {{Defof|Inclusion Mapping}} +}} +{{eqn | l = \map {i_S} {s_1} + | r = \map {i_S} {s_2} + | c = by definition +}} +{{eqn | ll= \leadsto + | l = s_1 + | r = s_2 + | c = from above +}} +{{end-eqn}} +Thus $i_S$ is an injection by [[Definition:Injection|definition]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Surjection iff Right Cancellable} +Tags: Surjections, Cancellability, Surjection iff Right Cancellable + +\begin{theorem} +Let $f$ be a [[Definition:Mapping|mapping]]. +Then $f$ is a [[Definition:Surjection|surjection]] {{iff}} $f$ is [[Definition:Right Cancellable Mapping|right cancellable]]. +\end{theorem}<|endoftext|> +\section{Identity Mapping is Surjection} +Tags: Surjections, Identity Mappings + +\begin{theorem} +On any set $S$, the [[Definition:Identity Mapping|identity mapping]] $I_S: S \to S$ is a [[Definition:Surjection|surjection]]. +\end{theorem} + +\begin{proof} +The [[Definition:Identity Mapping|identity mapping]] is defined as: +:$\forall y \in S: \map {I_S} y = y$ +Then we have: +{{begin-eqn}} +{{eqn | l = \forall y \in S: \exists x \in S: x + | r = y + | c = that is, $y$ itself +}} +{{eqn | ll= \leadsto + | l = \forall y \in S: \exists x \in S: \map {I_S} x + | r = y + | c = Definition of $I_S$ +}} +{{end-eqn}} +Hence the result. +{{qed}} +\end{proof}<|endoftext|> +\section{Restriction of Mapping to Image is Surjection} +Tags: Restrictions, Surjections + +\begin{theorem} +Let $f: S \to T$ be a [[Definition:Mapping|mapping]]. +Let $g: S \to \Img f$ be the [[Definition:Restriction of Mapping|restriction]] of $f$ to $S \times \Img f$. +Then $g$ is a [[Definition:Surjective Restriction|surjective restriction]] of $f$. +\end{theorem} + +\begin{proof} +From [[Image is Subset of Codomain]]: +:$\Img g \subseteq T$ +Furthermore, by definition of [[Definition:Image of Mapping|image]], we have: +:$\forall s \in S: g \paren s \in \Img g$ +Therefore, $g$ may be viewed as a [[Definition:Mapping|mapping]] $g: S \to \Img g$. +Thus, by definition, $g$ is a [[Definition:Surjection/Definition 2|surjection]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Composite of Surjections is Surjection} +Tags: Surjections, Composite Mappings + +\begin{theorem} +A [[Definition:Composition of Mappings|composite]] of [[Definition:Surjection|surjections]] is a [[Definition:Surjection|surjection]]. +That is: +:If $g$ and $f$ are [[Definition:Surjection|surjections]], then so is $g \circ f$. +\end{theorem} + +\begin{proof} +Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be [[Definition:Surjection|surjections]]. +Then: +{{begin-eqn}} +{{eqn | l = \forall z \in S_3: \exists y \in S_2: \map g y + | r = z + | c = {{Defof|Surjection}} +}} +{{eqn | ll= \leadsto + | l = \exists x \in S_1: \map f x + | r = y + | c = {{Defof|Surjection}} +}} +{{end-eqn}} +By definition of a [[Definition:Composition of Mappings|composite mapping]]: +:$\map {g \circ f} x = \map g {\map f x} = \map g y = z$ +Hence $g \circ f$ is [[Definition:Surjection|surjective]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Surjection if Composite is Surjection} +Tags: Surjections, Composite Mappings + +\begin{theorem} +Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be [[Definition:Mapping|mappings]] such that $g \circ f$ is a [[Definition:Surjection|surjection]]. +Then $g$ is a [[Definition:Surjection|surjection]]. +\end{theorem} + +\begin{proof} +Let $g \circ f$ be [[Definition:Surjection|surjective]]. +Fix $z \in S_3$. +Now find an $x \in S_1: \map {g \circ f} x = z$. +The surjectivity of $g \circ f$ guarantees this can be done. +Now find an $y \in S_2: f \paren x = y$. +$f$ is a [[Definition:Mapping|mapping]] and therefore a [[Definition:Left-Total Relation|left-total relation]]; which guarantees this too can be done. +It follows that: +{{begin-eqn}} +{{eqn | l = \map g y + | r = \map g {\map f x} +}} +{{eqn | r = \map {g \circ f} x + | c = {{Defof|Composition of Mappings}} +}} +{{eqn | r = z + | c = Choice of $x$ +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Surjection iff Right Inverse} +Tags: Surjections, Right Inverse Mappings, Surjection iff Right Inverse + +\begin{theorem} +A [[Definition:Mapping|mapping]] $f: S \to T, S \ne \O$ is a [[Definition:Surjection|surjection]] {{iff}}: +:$\exists g: T \to S: f \circ g = I_T$ +where: +:$g$ is a [[Definition:Mapping|mapping]] +:$I_T$ is the [[Definition:Identity Mapping|identity mapping]] on $T$. +That is, {{iff}} $f$ has a [[Definition:Right Inverse Mapping|right inverse]]. +\end{theorem} + +\begin{proof} +Assume $\exists g: T \to S: f \circ g = I_T$. +From [[Identity Mapping is Surjection]], $I_T$ is [[Definition:Surjection|surjective]], so $f \circ g$ is [[Definition:Surjection|surjective]]. +So from [[Surjection if Composite is Surjection]], $f$ is a [[Definition:Surjection|surjection]]. +Note that the existence of such a $g$ ''requires'' that $S$ is [[Definition:Non-Empty Set|non-empty]]. +Now, assume $f$ is a [[Definition:Surjection|surjection]]. +Consider the [[Definition:Indexed Family of Sets|indexed family]] of [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|sets]] $\set {\map {f^{-1} } y}_{y \mathop \in T}$ where $\map {f^{-1} } y$ denotes the [[Definition:Preimage of Element under Mapping|preimage]] of $y$ under $f$. +Using the [[Axiom:Axiom of Choice|axiom of choice]], there exists a [[Definition:Mapping|mapping]] $g: T \to S$ such that $\map g y \in \set {\map {f^{-1} } y}$ for all $y \in T$. +:[[File:SurjectionIffRightInverse.png|550px]] +That is, $\map {\paren {f \circ g} } y = y$, as desired. +{{qed}} +\end{proof} + +\begin{proof} +Take the result [[Condition for Composite Mapping on Right]]: +Let $A, B, C$ be [[Definition:Set|sets]]. +Let $f: B \to A$ and $g: C \to A$ be [[Definition:Mapping|mappings]]. +Then: +:$\Img g \subseteq \Img f$ +{{iff}}: +:$\exists h: C \to B$ such that $h$ is a [[Definition:Mapping|mapping]] and $f \circ h = g$. +Let $C = A = T$, let $B = S$ and let $g = I_T$. +Then the above translates into: +:$\Img {I_T} \subseteq \Img f$ +{{iff}}: +:$\exists g: T \to S$ such that $g$ is a [[Definition:Mapping|mapping]] and $f \circ g = I_T$. +But we know that: +:$\Img f \subseteq T = \Img {I_T}$ +So by [[Definition:Set Equality|definition of set equality]], the result follows. +{{qed}} +\end{proof}<|endoftext|> +\section{Identity Mapping is Bijection} +Tags: Bijections, Identity Mappings + +\begin{theorem} +The [[Definition:Identity Mapping|identity mapping]] $I_S: S \to S$ on the [[Definition:Set|set]] $S$ is a [[Definition:Bijection|bijection]]. +\end{theorem} + +\begin{proof} +The [[Definition:Identity Mapping|identity mapping]] is: +:an [[Definition:Injection|injection]], from [[Identity Mapping is Injection]] +:a [[Definition:Surjection|surjection]], from [[Identity Mapping is Surjection]] +and therefore a [[Definition:Bijection|bijection]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Identity Mapping is Permutation} +Tags: Identity Mappings + +\begin{theorem} +The [[Definition:Identity Mapping|identity mapping]] $I_S: S \to S$ on the [[Definition:Set|set]] $S$ is a [[Definition:Permutation|permutation]]. +\end{theorem} + +\begin{proof} +The identity mapping $I_S$ is a [[Definition:Bijection|bijection]] from $S$ to itself, by [[Identity Mapping is Bijection]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Bijection iff Left and Right Inverse} +Tags: Bijections, Inverse Mappings + +\begin{theorem} +Let $f: S \to T$ be a [[Definition:Mapping|mapping]]. +$f$ is a [[Definition:Bijection|bijection]] {{iff}}: +:$(1): \quad \exists g_1: T \to S: g_1 \circ f = I_S$ +:$(2): \quad \exists g_2: T \to S: f \circ g_2 = I_T$ +where both $g_1$ and $g_2$ are mappings. +It also follows that it is necessarily the case that $g_1 = g_2$ for such to be possible. +\end{theorem} + +\begin{proof} +=== Necessary Condition === +Let $f: S \to T$ be a [[Definition:Mapping|mapping]]. +Let $f$ be such that: +:$\exists g_1: T \to S: g_1 \circ f = I_S$ +:$\exists g_2: T \to S: f \circ g_2 = I_T$ +where both $g_1$ and $g_2$ are [[Definition:Mapping|mappings]]. +Then from [[Left and Right Inverse Mappings Implies Bijection]] it follows that $f$ is a [[Definition:Bijection|bijection]]. +From [[Left and Right Inverses of Mapping are Inverse Mapping]] it follows that: +:$g_1 = g_2 = f^{-1}$ +where $f^{-1}$ is the [[Definition:Inverse Mapping|inverse]] of $f$. +=== Sufficient Condition === +Let $f: S \to T$ be a [[Definition:Bijection|bijection]]. +Then from [[Bijection has Left and Right Inverse]] it follows that: +:$f^{-1} \circ f = I_S$ and +:$f \circ f^{-1} = I_T$ +where $f^{-1}$ is the [[Definition:Inverse Mapping|inverse]] of $f$. +\end{proof}<|endoftext|> +\section{Inverse of Bijection is Bijection} +Tags: Bijections, Inverse Mappings + +\begin{theorem} +Let $f: S \to T$ be a [[Definition:Bijection/Definition 1|bijection]] in the sense that: +:$(1): \quad f$ is an [[Definition:Injection|injection]] +:$(2): \quad f$ is a [[Definition:Surjection|surjection]]. +Then the [[Definition:Inverse of Mapping|inverse]] $f^{-1}$ of $f$ is itself a [[Definition:Bijection/Definition 1|bijection]] by the same definition. +\end{theorem} + +\begin{proof} +Let $f$ be both an [[Definition:Injection|injection]] and a [[Definition:Surjection|surjection]]. +From [[Mapping is Injection and Surjection iff Inverse is Mapping]] it follows that its [[Definition:Inverse of Mapping|inverse $f^{-1}$]] is a [[Definition:Mapping|mapping]]. +From [[Inverse of Inverse Relation]]: +: $\paren {f^{-1} }^{-1} = f$ +Thus the [[Definition:Inverse of Mapping|inverse of $f^{-1}$]] is $f$. +But then $f$, being a [[Definition:Bijection|bijection]], is by definition a [[Definition:Mapping|mapping]]. +So from [[Mapping is Injection and Surjection iff Inverse is Mapping]] it follows that $f^{-1}$ is a [[Definition:Bijection|bijection]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Inverse Element of Bijection} +Tags: Bijections, Inverse Mappings + +\begin{theorem} +Let $S$ and $T$ be [[Definition:Set|sets]]. +Let $f: S \to T$ be a [[Definition:Bijection|bijection]]. +Then: +:$\map {f^{-1} } y = x \iff \map f x = y$ +where $f^{-1}$ is the [[Definition:Inverse Mapping|inverse mapping]] of $f$. +\end{theorem} + +\begin{proof} +Suppose $f$ is a [[Definition:Bijection|bijection]]. +Because $f^{-1}$ is a [[Definition:Bijection|bijection]] from [[Bijection iff Inverse is Bijection]], it is by definition a [[Definition:Mapping|mapping]]. +The result follows directly from [[Image of Element under Inverse Mapping]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Composite of Bijection with Inverse is Identity Mapping} +Tags: Bijections, Inverse Mappings + +\begin{theorem} +Let $f: S \to T$ be a [[Definition:Bijection|bijection]]. +Then: +: $f^{-1} \circ f = I_S$ +: $f \circ f^{-1} = I_T$ +where $I_S$ and $I_T$ are the [[Definition:Identity Mapping|identity mappings]] on $S$ and $T$ respectively. +\end{theorem} + +\begin{proof} +Let $f: S \to T$ be a [[Definition:Bijection|bijection]]. +From [[Inverse of Bijection is Bijection]], $f^{-1}$ is also a [[Definition:Bijection|bijection]]. +Let $x \in S$. +From [[Inverse Element of Bijection]]: +:$\exists y \in T: y = \map f x \implies x = \map {f^{-1} } y$ +Then: +{{begin-eqn}} +{{eqn | l = \map {f^{-1} \circ f} x + | r = \map {f^{-1} } {\map f x} + | c = {{Defof|Composition of Mappings}} +}} +{{eqn | r = \map {f^{-1} } y + | c = [[Definition:By Hypothesis|By Hypothesis]] +}} +{{eqn | r = x + | c = [[Definition:By Hypothesis|By Hypothesis]] +}} +{{eqn | r = \map {I_S} x + | c = {{Defof|Identity Mapping}} +}} +{{end-eqn}} +From [[Domain of Composite Relation]] and [[Codomain of Composite Relation]], the [[Definition:Domain of Mapping|domain]] and [[Definition:Codomain of Mapping|codomain]] of $f^{-1} \circ f$ are both $S$, and so are those of $I_S$ by [[Definition:Identity Mapping|definition]]. +So all the criteria for [[Equality of Mappings]] are met and thus $f^{-1} \circ f = I_S$. +Let $y \in T$. +From [[Inverse Element of Bijection]]: +:$\exists x \in S: x = \map {f^{-1} } y \implies y = \map f x$ +Then: +{{begin-eqn}} +{{eqn | l = \map {f \circ f^{-1} } y + | r = \map f {\map {f^{-1} } y} + | c = {{Defof|Composition of Mappings}} +}} +{{eqn | r = \map f x + | c = [[Definition:By Hypothesis|By Hypothesis]] +}} +{{eqn | r = y + | c = [[Definition:By Hypothesis|By Hypothesis]] +}} +{{eqn | r = \map {I_T} y + | c = {{Defof|Identity Mapping}} +}} +{{end-eqn}} +From [[Domain of Composite Relation]] and [[Codomain of Composite Relation]], the [[Definition:Domain of Mapping|domain]] and [[Definition:Codomain of Mapping|codomain]] of $f \circ f^{-1}$ are both $T$, and so are those of $I_T$ by [[Definition:Identity Mapping|definition]]. +So all the criteria for [[Equality of Mappings]] are met and thus $f \circ f^{-1} = I_T$. +{{Qed}} +\end{proof}<|endoftext|> +\section{Inverse of Inverse of Bijection} +Tags: Bijections, Inverse Mappings, Inverse of Inverse of Bijection + +\begin{theorem} +Let $f: S \to T$ be a [[Definition:Bijection|bijection]]. +Then: +:$\paren {f^{-1} }^{-1} = f$ +where $f^{-1}$ is the [[Definition:Inverse of Mapping|inverse of $f$]]. +\end{theorem} + +\begin{proof} +A [[Definition:Mapping|mapping]] is a [[Definition:Relation|relation]]. +Thus it follows that [[Inverse of Inverse Relation]] can be applied directly. +{{qed}} +\end{proof} + +\begin{proof} +Let $f: S \to T$ be a [[Definition:Bijection|bijection]]. +From [[Composite of Bijection with Inverse is Identity Mapping]] we have: +:$f^{-1} \circ f = I_S$ +:$f \circ f^{-1} = I_T$ +where $I_S$ and $I_T$ are the [[Definition:Identity Mapping|identity mappings]] on $S$ and $T$ respectively. +The result follows from [[Left and Right Inverses of Mapping are Inverse Mapping]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Composite of Bijections is Bijection} +Tags: Bijections, Composite Mappings, Composite of Bijections is Bijection + +\begin{theorem} +Let $f$ and $g$ be [[Definition:Mapping|mappings]] such that $\Dom f = \Cdm g$. +Then: +:If $f$ and $g$ are both [[Definition:Bijection|bijections]], then so is $f \circ g$ +where $f \circ g$ is the [[Definition:Composition of Mappings|composite mapping]] of $f$ with $g$. +\end{theorem} + +\begin{proof} +As every [[Definition:Bijection|bijection]] is also by definition an [[Definition:Injection|injection]], a [[Definition:Composition of Mappings|composite]] of [[Definition:Bijection|bijections]] is also a [[Definition:Composition of Mappings|composite]] of [[Definition:Injection|injections]]. +Every [[Definition:Composition of Mappings|composite]] of [[Definition:Injection|injections]] is also an [[Definition:Injection|injection]] by [[Composite of Injections is Injection]]. +As every [[Definition:Bijection|bijection]] is also by definition a [[Definition:Surjection|surjection]], a [[Definition:Composition of Mappings|composite]] of [[Definition:Bijection|bijections]] is also a [[Definition:Composition of Mappings|composite]] of [[Definition:Surjection|surjections]]. +Every [[Definition:Composition of Mappings|composite]] of [[Definition:Surjection|surjections]] is also a [[Definition:Surjection|surjection]] by [[Composite of Surjections is Surjection]]. +As a [[Definition:Composition of Mappings|composite]] of [[Definition:Bijection|bijections]] is therefore both an [[Definition:Injection|injection]] and a [[Definition:Surjection|surjection]], it is also a [[Definition:Bijection|bijection]]. +{{Qed}} +\end{proof} + +\begin{proof} +Let $g: X \to Y$ and $f: Y \to Z$ be [[Definition:Bijection|bijections]]. +Then from [[Bijection iff Inverse is Bijection]], both $f^{-1}$ and $g^{-1}$ are [[Definition:Bijection|bijections]]. +From [[Inverse of Composite Relation]] we have that $g^{-1} \circ f^{-1}$ is the [[Definition:Inverse Mapping|inverse]] of $f \circ g$. +Then: +:$\paren {f \circ g} \circ \paren {g^{-1} \circ f^{-1} } = I_Z$ +:$\paren {g^{-1} \circ f^{-1} } \circ \paren {f \circ g} = I_X$ +Hence the result. +\end{proof}<|endoftext|> +\section{Inverse of Composite Bijection} +Tags: Bijections, Inverse Mappings, Composite Mappings, Inverse of Composite Bijection + +\begin{theorem} +Let $f$ and $g$ be [[Definition:Bijection|bijections]] such that $\Dom g = \Cdm f$. +Then: +:$\paren {g \circ f}^{-1} = f^{-1} \circ g^{-1}$ +and $f^{-1} \circ g^{-1}$ is itself a [[Definition:Bijection|bijection]]. +\end{theorem} + +\begin{proof} +$\left({g \circ f}\right)^{-1} = f^{-1} \circ g^{-1}$ is a specific example of [[Inverse of Composite Relation]]. +As $f$ and $g$ are [[Definition:Bijection|bijections]] then by [[Bijection iff Inverse is Bijection]], so are both $f^{-1}$ and $g^{-1}$. +By [[Composite of Bijections is Bijection]], it follows that $f^{-1} \circ g^{-1}$ is a [[Definition:Bijection|bijection]]. +{{qed}} +\end{proof} + +\begin{proof} +Let $f: X \to Y$ and $g: Y \to Z$ be [[Definition:Bijection|bijections]]. +Then: +{{begin-eqn}} +{{eqn | l = \paren {g \circ f} \circ \paren {f^{-1} \circ g^{-1} } + | r = g \circ \paren {\paren {f \circ f^{-1} } \circ g^{-1} } + | c = [[Composition of Mappings is Associative]] +}} +{{eqn | r = g \circ \paren {I_Y \circ g^{-1} } + | c = [[Composite of Bijection with Inverse is Identity Mapping]] +}} +{{eqn | r = g \circ g^{-1} + | c = [[Identity Mapping is Left Identity]] +}} +{{eqn | r = I_Z + | c = [[Composite of Bijection with Inverse is Identity Mapping]] +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | l = \paren {f^{-1} \circ g^{-1} } \circ \paren {g \circ f} + | r = \paren {f^{-1} \circ \paren {g^{-1} \circ g} } \circ f + | c = [[Composition of Mappings is Associative]] +}} +{{eqn | r = \paren {f^{-1} \circ I_Y} \circ f + | c = [[Composite of Bijection with Inverse is Identity Mapping]] +}} +{{eqn | r = f^{-1} \circ f + | c = [[Identity Mapping is Right Identity]] +}} +{{eqn | r = I_X + | c = [[Composite of Bijection with Inverse is Identity Mapping]] +}} +{{end-eqn}} +Hence the result. +\end{proof}<|endoftext|> +\section{Composite of Permutations is Permutation} +Tags: Permutation Theory + +\begin{theorem} +Let $f, g$ are [[Definition:Permutation|permutations]] of a [[Definition:Set|set]] $S$. +Then their [[Definition:Composition of Mappings|composite]] $g \circ f$ is also a [[Definition:Permutation|permutation]] of $S$. +\end{theorem} + +\begin{proof} +This follows from the fact that a [[Definition:Permutation|permutation]] is a [[Definition:Bijection|bijection]]. +The [[Definition:Domain of Mapping|domain]] and [[Definition:Codomain of Mapping|codomain]] are coincident. +The result follows from [[Composite of Bijections is Bijection]]. +{{Qed}} +\end{proof}<|endoftext|> +\section{Inverse of Permutation is Permutation} +Tags: Permutation Theory + +\begin{theorem} +If $f$ is a [[Definition:Permutation|permutation]] of $S$, then so is its [[Definition:Inverse of Mapping|inverse]] $f^{-1}$. +\end{theorem} + +\begin{proof} +Let $f: S \to S$ is a [[Definition:Permutation|permutation]] of $S$. +By definition, a [[Definition:Permutation|permutation]] is a [[Definition:Bijection|bijection]] such that the [[Definition:Domain of Mapping|domain]] and [[Definition:Codomain of Mapping|codomain]] are the same [[Definition:Set|set]]. +From [[Bijection iff Inverse is Bijection]], it follows $f^{-1}$ is a [[Definition:Bijection|bijection]]. +From the definition of [[Definition:Inverse Relation|inverse relation]], the [[Definition:Domain of Mapping|domain]] of a [[Definition:Relation|relation]] is the [[Definition:Codomain of Mapping|codomain]] of its [[Definition:Inverse Relation|inverse]] and vice versa. +Thus the [[Definition:Domain of Mapping|domain]] and [[Definition:Codomain of Mapping|codomain]] of $f^{-1}$ are both $S$ and it follows that $f^{-1}$ is a [[Definition:Permutation|permutation]]. +{{Qed}} +\end{proof}<|endoftext|> +\section{Left Inverse Mapping is Surjection} +Tags: Left Inverse Mappings, Surjections + +\begin{theorem} +Let $f: S \to T$ be an [[Definition:Injection|injection]]. +Let $g: T \to S$ be a [[Definition:Left Inverse Mapping|left inverse]] of $f$. +Then $g$ is a [[Definition:Surjection|surjection]]. +\end{theorem} + +\begin{proof} +Let $f: S \to T$ be an [[Definition:Injection|injection]]. +Then from [[Injection iff Left Inverse]] there exists at least one [[Definition:Left Inverse Mapping|left inverse]] $g: T \to S$ of $f$ such that $g \circ f = I_S$. +$I_S$ is a [[Identity Mapping is Surjection|surjection]]. +Thus $g \circ f$ is a [[Definition:Surjection|surjection]]. +By [[Surjection if Composite is Surjection]] it follows that $g$ is also a [[Definition:Surjection|surjection]]. +{{Qed}} +\end{proof}<|endoftext|> +\section{Right Inverse Mapping is Injection} +Tags: Injections, Right Inverse Mappings + +\begin{theorem} +Let $f: S \to T$ be a [[Definition:Mapping|mapping]]. +Let $g: T \to S$ be a [[Definition:Right Inverse Mapping|right inverse]] of $f$. +Then $g$ is an [[Definition:Injection|injection]]. +\end{theorem} + +\begin{proof} +By the definition of [[Definition:Right Inverse Mapping|right inverse]]: +:$f \circ g = I_T$ +where $I_T$ is the [[Definition:Identity Mapping|identity mapping]] on $T$. +By [[Identity Mapping is Injection]], $I_T$ is an [[Definition:Injection|injection]]. +By [[Injection if Composite is Injection]], it follows that $g$ is an [[Definition:Injection|injection]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Set Equivalence is Equivalence Relation} +Tags: Equivalence Relations, Set Equivalence + +\begin{theorem} +[[Definition:Set Equivalence|Set equivalence]] is an [[Definition:Equivalence Relation|equivalence relation]]. +\end{theorem} + +\begin{proof} +For two [[Definition:Set|sets]] to be [[Definition:Set Equivalence|equivalent]], there needs to exist a [[Definition:Bijection|bijection]] between them. +In the following, let $\phi$, $\phi_1$, $\phi_2$ etc. be understood to be [[Definition:Bijection|bijections]]. +=== Reflexive === +From [[Identity Mapping is Bijection]], the [[Definition:Identity Mapping|identity mapping]] $I_S: S \to S$ is the required [[Definition:Bijection|bijection]]. +Thus there exists a [[Definition:Bijection|bijection]] from $S$ to itself and $S$ is therefore equivalent to itself. +Therefore [[Definition:Set Equivalence|set equivalence]] is [[Definition:Reflexive Relation|reflexive]]. +{{qed|lemma}} +=== Symmetric === +{{begin-eqn}} +{{eqn | o = + | r = S \sim T + | c = +}} +{{eqn | o = \leadsto + | r = \exists \phi: S \to T + | c = {{Defof|Set Equivalence}} +}} +{{eqn | o = \leadsto + | r = \exists \phi^{-1}: T \to S + | c = [[Bijection iff Inverse is Bijection]] +}} +{{eqn | o = \leadsto + | r = T \sim S + | c = {{Defof|Set Equivalence}}: $\phi^{-1}$ is also a [[Definition:Bijection|bijection]] +}} +{{end-eqn}} +Therefore [[Definition:Set Equivalence|set equivalence]] is [[Definition:Symmetric Relation|symmetric]]. +{{qed|lemma}} +=== Transitive === +{{begin-eqn}} +{{eqn | o = + | r = S_1 \sim S_2 \land S_2 \sim S_3 + | c = +}} +{{eqn | o = \leadsto + | r = \exists \phi_1: S_1 \to S_2 \land \exists \phi_2: S_2 \to S_3 + | c = {{Defof|Set Equivalence}} +}} +{{eqn | o = \leadsto + | r = \exists \phi_2 \circ \phi_1: S_1 \to S_3 + | c = [[Composite of Bijections is Bijection]]: $\phi_2 \circ \phi_1$ is a [[Definition:Bijection|bijection]] +}} +{{eqn | o = \leadsto + | r = S_1 \sim S_3 + | c = {{Defof|Set Equivalence}} +}} +{{end-eqn}} +Therefore [[Definition:Set Equivalence|set equivalence]] is [[Definition:Transitive Relation|transitive]]. +{{qed}} +{{expand|See talk}} +\end{proof}<|endoftext|> +\section{Set Equivalence Less One Element} +Tags: Equivalence Relations + +\begin{theorem} +Let $S$ and $T$ be [[Definition:Set|sets]] which are [[Definition:Set Equivalence|equivalent]]: +:$S \sim T$ +Let $a \in S$ and $b \in T$. +Then: +: $S \setminus \left\{{a}\right\} \sim T \setminus \left\{{b}\right\}$ +where $\setminus$ denotes [[Definition:Set Difference|set difference]]. +\end{theorem} + +\begin{proof} +As $S \sim T$, there exists a [[Definition:Bijection|bijection]] $f: S \to T$. +Let $g: \left({S \setminus \left\{{a}\right\}}\right) \to \left({T \setminus \left\{{b}\right\}}\right)$ be the [[Definition:Mapping|mapping]] defined as follows: +$\forall x \in S \setminus \left\{{a}\right\}: g \left({x}\right) = +\begin{cases} +f \left({x}\right): f \left({x}\right) \ne b \\ +f \left({a}\right): f \left({x}\right) = b +\end{cases}$ +First it is shown that $g$ is an [[Definition:Injection|injection]]. +Let $x, y \in S \setminus \left\{{a}\right\}$ and $x \ne y$. +The following cases apply: +$(1): \quad f \left({x}\right) \ne b$ and $f \left({y}\right) \ne b$ +By definition of $g$: +:$f \left({x}\right) = g \left({x}\right)$ +and: +:$f \left({y}\right) = g \left({y}\right)$ +It follows from the [[Definition:Injection|injectivity]] of $f$ that: +:$g \left({x}\right) \ne g \left({y}\right)$ +$(2): \quad f \left({x}\right) = b$ +By definition of $g$: +:$g \left({x}\right) = f \left({a}\right)$ +We have that: +:$y \in S \setminus \left\{{a}\right\}$ +Therefore by definition of [[Definition:Set Difference|set difference]]: +:$y \ne a$ +So by the [[Definition:Injection|injectivity]] of $f$: +:$f \left({y}\right) \ne f \left({a}\right) = g \left({x}\right)$ +Also by the [[Definition:Injection|injectivity]] of $f$: +:$f \left({y}\right) \ne b$ +Thus, by the definition of $g$: +:$g \left({y}\right) = f \left({y}\right) \ne f \left({a}\right) = g \left({x}\right)$ +$(2): \quad f \left({y}\right) = b$ +The proof is the same as that of case $2$, with the roles of $x$ and $y$ reversed. +{{qed|lemma}} +Next it is shown that $g$ is a [[Definition:Surjection|surjection]]. +Let $y \in T \setminus \left\{{b}\right\}$. +By definition of [[Definition:Surjection|surjectivity]] of $f$: +:$\exists x \in S: f \left({x}\right) = y$ +It is to be shown that: +:$\exists x' \in S, x \ne a: f \left({x'}\right) = y$ +$(1): \quad f \left({a}\right) = b$ +[[Definition:By Hypothesis|By hypothesis]]: +:$f \left({x}\right) = y \ne b$ +Then by the [[Rule of Transposition|rule of transposition]]: +:$x \ne a$ +so let: +:$x' = x$ +$(2): \quad f \left({a}\right) \ne b$ +By the [[Rule of Transposition|rule of transposition]]: +:$f^{-1} \left({b}\right) \ne a$ +where $f^{-1} \left({b}\right)$ denotes the [[Definition:Preimage of Element under Mapping|preimage]] of $b$ under $f$. +:Case $(2.1): \quad f \left({a}\right) = y$ +By the definition of $g$: +:$g \left({f^{-1} \left({b}\right)}\right) = y$ +so let: +:$x' = f^{-1} \left({b}\right)$ +:Case $(2.2): \quad f \left({a}\right) \ne y$ +[[Definition:By Hypothesis|By hypothesis]]: +:$f \left({x}\right) = y \ne f \left({a}\right)$ +Then by the [[Rule of Transposition|rule of transposition]]: +:$x \ne a$ +so let: +:$x' = x$ +Thus: +:$\forall y \in T \setminus \left\{{b}\right\}: \exists x \in S \setminus \left\{{a}\right\}: g \left({x}\right) = y$ +and so $g$ is a [[Definition:Surjection|surjection]]. +{{qed|lemma}} +Thus $g$ is both an [[Definition:Injection|injection]] and a [[Definition:Surjection|surjection]], and is by definition a [[Definition:Bijection|bijection]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Image of Union under Mapping} +Tags: Mapping Theory, Set Union, Image of Union under Mapping + +\begin{theorem} +Let $S$ and $T$ be [[Definition:Set|sets]]. +Let $f: S \to T$ be a [[Definition:Mapping|mapping]]. +Let $A$ and $B$ be [[Definition:Subset|subsets]] of $S$. +Then: +:$f \sqbrk {A \cup B} = f \sqbrk A \cup f \sqbrk B$ +This can be expressed in the language and notation of [[Definition:Direct Image Mapping|direct image mappings]] as: +:$\forall A, B \in \powerset S: \map {f^\to} {A \cup B} = \map {f^\to} A \cup \map {f^\to} B$ +\end{theorem} + +\begin{proof} +As $f$, being a [[Definition:Mapping|mapping]], is also a [[Definition:Relation|relation]], we can apply [[Image of Union under Relation]]: +:$\RR \sqbrk {A \cup B} = \RR \sqbrk A \cup \RR \sqbrk B$ +{{qed}} +\end{proof} + +\begin{proof} +First we have: +{{begin-eqn}} +{{eqn | l = A + | o = \subseteq + | r = A \cup B + | c = [[Set is Subset of Union]] +}} +{{eqn | ll= \leadsto + | l = f \sqbrk A + | o = \subseteq + | r = f \sqbrk {A \cup B} + | c = [[Image of Subset under Mapping is Subset of Image]] +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | l = B + | o = \subseteq + | r = A \cup B + | c = [[Set is Subset of Union]] +}} +{{eqn | ll= \leadsto + | l = f \sqbrk B + | o = \subseteq + | r = f \sqbrk {A \cup B} + | c = [[Image of Subset under Mapping is Subset of Image]] +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | ll= \leadsto + | l = f \sqbrk A \cup f \sqbrk B + | o = \subseteq + | r = f \sqbrk {A \cup B} + | c = [[Union is Smallest Superset]] +}} +{{end-eqn}} +{{Qed|lemma}} +Then: +{{begin-eqn}} +{{eqn | l = y + | o = \in + | r = f \sqbrk {A \cup B} + | c = +}} +{{eqn | ll= \leadsto + | l = \exists x \in A \cup B: y + | r = \map f x + | c = {{Defof|Image of Subset under Relation}} +}} +{{eqn | ll= \leadsto + | l = \exists x: x \in A \lor x \in B: y + | r = \map f x + | c = {{Defof|Set Union}} +}} +{{eqn | ll= \leadsto + | l = y + | o = \in + | r = f \sqbrk A \lor y \in f \sqbrk B + | c = {{Defof|Image of Subset under Relation}} +}} +{{eqn | ll= \leadsto + | l = y + | o = \in + | r = f \sqbrk A \cup f \sqbrk B + | c = {{Defof|Set Union}} +}} +{{eqn | ll= \leadsto + | l = f \sqbrk {A \cup B} + | o = \subseteq + | r = f \sqbrk A \cup f \sqbrk B + | c = {{Defof|Subset}} +}} +{{end-eqn}} +{{Qed|lemma}} +Thus we have: +{{begin-eqn}} +{{eqn | l = f \sqbrk A \cup f \sqbrk B + | o = \subseteq + | r = f \sqbrk {A \cup B} + | c = +}} +{{eqn | l = f \sqbrk {A \cup B} + | o = \subseteq + | r = f \sqbrk A \cup f \sqbrk B + | c = +}} +{{eqn | ll= \leadsto + | l = f \sqbrk A \cup f \sqbrk B + | r = f \sqbrk {A \cup B} + | c = {{Defof|Set Equality}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Image of Intersection under Mapping} +Tags: Mapping Theory, Set Intersection, Image of Intersection under Mapping + +\begin{theorem} +Let $S$ and $T$ be [[Definition:Set|sets]]. +Let $f: S \to T$ be a [[Definition:Mapping|mapping]]. +Let $S_1$ and $S_2$ be [[Definition:Subset|subsets]] of $S$. +Then: +:$f \sqbrk {S_1 \cap S_2} \subseteq f \sqbrk {S_1} \cap f \sqbrk {S_2}$ +This can be expressed in the language and notation of [[Definition:Direct Image Mapping|direct image mappings]] as: +:$\forall S_1, S_2 \in \powerset S: \map {f^\to} {S_1 \cap S_2} \subseteq \map {f^\to} {S_1} \cap \map {f^\to} {S_2}$ +\end{theorem} + +\begin{proof} +As $f$, being a [[Definition:Mapping|mapping]], is also a [[Definition:Relation|relation]], we can apply [[Image of Intersection under Relation]]: +:$\RR \sqbrk {S_1 \cap S_2} \subseteq \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$ +{{qed}} +\end{proof} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = S_1 \cap S_2 + | o = \subseteq + | r = S_1 + | c = [[Intersection is Subset]] +}} +{{eqn | ll= \leadsto + | l = f \sqbrk {S_1 \cap S_2} + | o = \subseteq + | r = f \sqbrk {S_1} + | c = [[Image of Subset under Mapping is Subset of Image]] +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | l = S_1 \cap S_2 + | o = \subseteq + | r = S_2 + | c = [[Intersection is Subset]] +}} +{{eqn | ll= \leadsto + | l = f \sqbrk {S_1 \cap S_2} + | o = \subseteq + | r = f \sqbrk {S_2} + | c = [[Image of Subset under Mapping is Subset of Image]] +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | ll= \leadsto + | l = f \sqbrk {S_1 \cap S_2} + | o = \subseteq + | r = f \sqbrk {S_1} \cap f \sqbrk {S_2} + | c = [[Intersection is Largest Subset]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Image of Intersection under Injection} +Tags: Injections, Set Intersection, Image of Intersection under Injection + +\begin{theorem} +Let $S$ and $T$ be [[Definition:Set|sets]]. +Let $f: S \to T$ be a [[Definition:Mapping|mapping]]. +Then: +:$\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$ +{{iff}} $f$ is an [[Definition:Injection|injection]]. +\end{theorem} + +\begin{proof} +An [[Definition:Injection|injection]] is a type of [[Definition:One-to-One Relation|one-to-one relation]], and therefore also a [[Definition:One-to-Many Relation|one-to-many relation]]. +Therefore [[Image of Intersection under One-to-Many Relation]] applies: +:$\forall A, B \subseteq S: \mathcal R \left[{A \cap B}\right] = \mathcal R \left[{A}\right] \cap \mathcal R \left[{B}\right]$ +{{iff}} $\mathcal R$ is a [[Definition:One-to-Many Relation|one-to-many relation]]. +We have that $f$ is a [[Definition:Mapping|mapping]] and therefore a [[Definition:Many-to-One Relation|many-to-one relation]]. +So $f$ is a [[Definition:One-to-Many Relation|one-to-many relation]] {{iff}} $f$ is also an [[Definition:Injection|injection]]. +It follows that: +:$\forall A, B \subseteq S: f \left[{A \cap B}\right] = f \left[{A}\right] \cap f \left[{B}\right]$ +{{iff}} $f$ is an [[Definition:Injection|injection]]. +{{qed}} +\end{proof} + +\begin{proof} +From [[Image of Intersection under Mapping]]: +:$f \sqbrk {A \cap B} \subseteq f \sqbrk A \cap f \sqbrk B$ +which holds for all [[Definition:Mapping|mappings]]. +It remains to be shown that: +:$f \sqbrk A \cap f \sqbrk B \subseteq f \sqbrk {A \cap B}$ +{{iff}} $f$ is an [[Definition:Injection|injection]]. +=== Sufficient Condition === +Suppose that: +:$\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$ +If $S$ is [[Definition:Singleton|singleton]], then $f$ is [[Definition:A Fortiori|a fortiori]] an [[Definition:Injection|injection]] from [[Mapping from Singleton is Injection]]. +So, assume $S$ is not [[Definition:Singleton|singleton]]. +{{AimForCont}} $f$ is specifically ''not'' an [[Definition:Injection|injection]]. +Then: +:$\exists x, y \in S: \exists z \in T: \tuple {x, z} \in T, \tuple {y, z} \in T, x \ne y$ +and of course $\set x \subseteq S, \set y \subseteq S$. +So: +:$z \in f \sqbrk {\set x}$ +:$z \in f \sqbrk {\set y}$ +and so by definition of [[Definition:Set Intersection|intersection]]: +:$z \in f \sqbrk {\set x} \cap f \sqbrk {\set y}$ +But: +:$\set x \cap \set y = \O$ +Thus from the [[Image of Empty Set is Empty Set/Corollary|corollary to Image of Empty Set is Empty Set]]: +:$f \sqbrk {\set x \cap \set y} = \O$ +and so: +:$f \sqbrk {\set x \cap \set y} \ne f \sqbrk {\set x} \cap f \sqbrk {\set y}$ +But [[Definition:By Hypothesis|by hypothesis]]: +:$\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$ +From this [[Definition:Contradiction|contradiction]] it follows that $f$ is an [[Definition:Injection|injection]]. +{{qed|lemma}} +=== Necessary Condition === +Let $f$ be an [[Definition:Injection|injection]]. +It is necessary to show: +:$f \sqbrk {S_1} \cap f \sqbrk {S_2} \subseteq f \sqbrk {S_1 \cap S_2}$ +Let $t \in f \sqbrk {S_1} \cap f \sqbrk {S_2}$. +Then: +{{begin-eqn}} +{{eqn | l = t + | o = \in + | r = f \sqbrk {S_1} \cap f \sqbrk {S_2} + | c = +}} +{{eqn | lll=\leadsto + | l = t + | o = \in + | r = f \sqbrk {S_1} + | c = {{Defof|Set Intersection}} +}} +{{eqn | lo= \land + | l = t + | o = \in + | r = f \sqbrk {S_2} + | c = +}} +{{eqn | lll=\leadsto + | ll= \exists s_1 \in S_1, s_2 \in S_2: + | l = t + | r = \map f {s_1} + | c = {{Defof|Image of Subset under Mapping}} +}} +{{eqn | lo= \land + | l = t + | r = \map f {s_2} + | c = +}} +{{eqn | ll= \leadsto + | l = \map f {s_1} + | r = \map f {s_2} + | c = +}} +{{eqn | ll= \leadsto + | l = s_1 + | r = s_2 + | c = {{Defof|Injection}} +}} +{{eqn | ll= \leadsto + | l = s_1 = s_2 + | o = \in + | r = S_1 + | c = +}} +{{eqn | lo= \land + | l = s_1 = s_2 + | o = \in + | r = S_2 + | c = +}} +{{eqn | ll=\leadsto + | l = s_1 = s_2 + | o = \in + | r = S_1 \cap S_2 + | c = {{Defof|Set Intersection}} +}} +{{eqn | ll= \leadsto + | l = \map f {s_1} = \map f {s_2} + | o = \in + | r = f \sqbrk {S_1 \cap S_2} + | c = [[Image of Element is Subset]] +}} +{{eqn | ll= \leadsto + | l = f \sqbrk {S_1} \cap f \sqbrk {S_2} + | o = \subseteq + | r = f \sqbrk {S_1 \cap S_2} + | c = {{Defof|Subset}} +}} +{{end-eqn}} +So if $f$ is an [[Definition:Injection|injection]], it follows that: +:$f \sqbrk {S_1 \cap S_2} = f \sqbrk {S_1} \cap f \sqbrk {S_2}$ +{{qed|lemma}} +Putting the results together, we see that: +:$f \sqbrk {S_1 \cap S_2} = f \sqbrk {S_1} \cap f \sqbrk {S_2}$ {{iff}} $f$ is an [[Definition:Injection|injection]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Image of Intersection under One-to-Many Relation} +Tags: Relation Theory, Set Intersection + +\begin{theorem} +Let $S$ and $T$ be [[Definition:Set|sets]]. +Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]]. +Then: +:$\forall S_1, S_2 \subseteq S: \RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$ +{{iff}} $\RR$ is [[Definition:One-to-Many Relation|one-to-many]]. +\end{theorem} + +\begin{proof} +=== Sufficient Condition === +Suppose that: +:$\forall S_1, S_2 \subseteq S: \RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$ +If $S$ is [[Definition:Singleton|singleton]], the result follows immediately as $\RR$ would ''have'' to be [[Definition:One-to-Many Relation|one-to-many]]. +So, assume $S$ is not [[Definition:Singleton|singleton]]. +Suppose $\RR$ is specifically ''not'' [[Definition:One-to-Many Relation|one-to-many]]. +So: +:$\exists x, y \in S: \exists z \in T: \tuple {x, z} \in T, \tuple {y, z} \in T, x \ne y$ +and of course $\set x \subseteq S, \set y \subseteq S$. +So: +:$z \in \RR \sqbrk {\set x}$ +:$z \in \RR \sqbrk {\set y}$ +and so by definition of [[Definition:Set Intersection|intersection]]: +:$z \in \RR \sqbrk {\set x} \cap \RR \sqbrk {\set y}$ +But: +:$\set x \cap \set y = \O$ +Thus from [[Image of Empty Set is Empty Set]]: +:$\RR \sqbrk {\set x \cap \set y} = \O$ +and so: +:$\RR \sqbrk {\set x \cap \set y} \ne \RR \sqbrk {\set x} \cap \RR \sqbrk {\set y}$ +{{qed|lemma}} +=== Necessary Condition === +Let $\RR$ be [[Definition:One-to-Many Relation|one-to-many]]. +From [[Image of Intersection under Relation]], we already have: +:$\RR \sqbrk {S_1 \cap S_2} \subseteq \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$ +So we just need to show: +:$\RR \sqbrk {S_1} \cap \RR \sqbrk {S_2} \subseteq \RR \sqbrk {S_1 \cap S_2}$ +Let $t \in \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$. +Then: +{{begin-eqn}} +{{eqn | l = t + | o = \in + | r = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2} + | c = +}} +{{eqn | ll= \leadsto + | l = t + | o = \in + | r = \RR \sqbrk {S_1} \land t \in \RR \sqbrk {S_2} + | c = {{Defof|Set Intersection}} +}} +{{eqn | ll= \leadsto + | l = \exists s_1 \in S_1: \tuple {s_1, t} + | o = \in + | r = \RR \land \exists s_2 \in S_2: \tuple {s_2, t} \in \RR + | c = {{Defof|Relation}} +}} +{{eqn | ll= \leadsto + | l = s_1 + | r = s_2 + | c = {{Defof|One-to-Many Relation}} +}} +{{eqn | ll= \leadsto + | l = s_1 = s_2 + | o = \in + | r = S_1 \land s_1 = s_2 \in S_2 + | c = +}} +{{eqn | ll= \leadsto + | l = s_1 = s_2 + | o = \in + | r = S_1 \cap S_2 + | c = {{Defof|Set Intersection}} +}} +{{eqn | ll= \leadsto + | l = \map \RR {s_1} = \map \RR {s_2} + | o = \in + | r = \RR \sqbrk {S_1 \cap S_2} + | c = [[Image of Element is Subset]] +}} +{{eqn | ll= \leadsto + | l = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2} + | o = \subseteq + | r = \RR \sqbrk {S_1 \cap S_2} + | c = {{Defof|Subset}} +}} +{{end-eqn}} +So if $\RR$ is [[Definition:One-to-Many Relation|one-to-many]], it follows that: +:$\RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$ +{{qed|lemma}} +Putting the results together, we see that: +:$\RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$ {{iff}} $\RR$ is [[Definition:One-to-Many Relation|one-to-many]]. +{{qed}} +[[Category:Relation Theory]] +[[Category:Set Intersection]] +sotqrohpes80rjpomanui4ipx1n4v3s +\end{proof}<|endoftext|> +\section{One-to-Many Image of Set Difference} +Tags: Set Difference, Relation Theory + +\begin{theorem} +Let $\mathcal R \subseteq S \times T$ be a [[Definition:Relation|relation]]. +Let $A$ and $B$ be [[Definition:Subset|subsets]] of $S$. +Then: +:$(1): \quad \mathcal R \left[{A}\right] \setminus \mathcal R \left[{B}\right] = \mathcal R \left[{A \setminus B}\right]$ +{{iff}} $\mathcal R$ is [[Definition:One-to-Many Relation|one-to-many]]. +\end{theorem} + +\begin{proof} +=== Sufficient Condition === +First, to show that $(1)$ holds if $\mathcal R$ is [[Definition:One-to-Many Relation|one-to-many]]. +From [[Image of Set Difference under Relation]], we already have: +:$\mathcal R \left[{A}\right] \setminus \mathcal R \left[{B}\right] \subseteq \mathcal R \left[{A \setminus B}\right]$ +So we just need to show: +:$\mathcal R \left[{A \setminus B}\right] \subseteq \mathcal R \left[{A}\right] \setminus \mathcal R \left[{B}\right]$ +Let $t \notin \mathcal R \left[{A}\right] \setminus \mathcal R \left[{B}\right]$. +Then $t \notin \mathcal R \left[{A}\right] \lor t \in \mathcal R \left[{B}\right]$ by [[De Morgan's Laws (Logic)|De Morgan's Laws]]. +{{explain|Go into more detail}} +Suppose $t \notin \mathcal R \left[{A}\right]$. +Then by definition of a [[Definition:Relation|relation]]: +: $\neg \exists s \in A: \left({s, t}\right) \in \mathcal R$ +By [[Image of Subset is Subset of Image]]: +: $\mathcal R \left[{A \setminus B}\right] \subseteq \mathcal R \left[{A}\right]$ +Thus, by [[Definition:Subset|definition of subset]] and [[Rule of Transposition]]: +: $t \notin \mathcal R \left[{A}\right] \implies t \notin \mathcal R \left[{A \setminus B}\right]$ +Now suppose $t \in \mathcal R \left[{B}\right]$. +Then: +: $\exists s \in B: \left({s, t}\right) \in \mathcal R$ +Because $\mathcal R$ is [[Definition:One-to-Many Relation|one-to-many]]: +: $\forall x \in S: \left({x, t}\right) \in \mathcal R \implies x = s$ +and thus: +: $x \in B$ +Thus: +: $x \notin A \setminus B$ +and hence: +: $t \notin \mathcal R \left[{A \setminus B}\right]$ +So by [[Proof by Cases]]: +: $t \notin \mathcal R \left[{A}\right] \setminus \mathcal R \left[{B}\right] \implies t \notin \mathcal R \left[{A \setminus B}\right]$ +The result follows from [[Set Complement inverts Subsets]]: +:$S \subseteq T \iff \complement \left({T}\right) \subseteq \complement \left({S}\right)$ +{{qed|lemma}} +=== Necessary Condition === +Now for the converse: If $(1)$ holds, it is to be shown that $\mathcal R$ is [[Definition:One-to-Many Relation|one-to-many]]. +Let $s, t \in S$ be [[Definition:Distinct Elements|distinct]]. +That is, $s \ne t$. +Then in particular: +: $\left\{{s}\right\} \setminus \left\{{t}\right\} = \left\{{s}\right\}$ +Applying $(1)$ to these two sets, it follows that: +:$\mathcal R \left[{\left\{{s}\right\}}\right] \setminus \mathcal R \left[{\left\{{t}\right\}}\right] = \mathcal R \left[{\left\{{s}\right\}}\right]$ +By [[Set Difference with Disjoint Set]], this implies that: +:$\mathcal R \left[{\left\{{s}\right\}}\right] \cap \mathcal R \left[{\left\{{t}\right\}}\right] = \varnothing$ +It follows that every element of $T$ can be related to at most one element of $S$. +That is, $\mathcal R$ is [[Definition:One-to-Many Relation|one-to-many]]. +{{qed}} +[[Category:Set Difference]] +[[Category:Relation Theory]] +i0devwmi7g85kpinon19sptljw9f9jj +\end{proof}<|endoftext|> +\section{Preimage of Union under Mapping} +Tags: Preimages under Mappings, Set Union, Preimage of Union under Mapping + +\begin{theorem} +Let $S$ and $T$ be [[Definition:Set|sets]]. +Let $f: S \to T$ be a [[Definition:Mapping|mapping]]. +Let $T_1$ and $T_2$ be [[Definition:Subset|subsets]] of $T$. +Then: +:$f^{-1} \sqbrk {T_1 \cup T_2} = f^{-1} \sqbrk {T_1} \cup f^{-1} \sqbrk {T_2}$ +This can be expressed in the language and notation of [[Definition:Inverse Image Mapping|inverse image mappings]] as: +:$\forall T_1, T_2 \in \powerset T: \map {f^\gets} {T_1 \cup T_2} = \map {f^\gets} {T_1} \cup \map {f^\gets} {T_2}$ +\end{theorem} + +\begin{proof} +As $f$, being a [[Definition:Mapping|mapping]], is also a [[Definition:Relation|relation]], we can apply [[Preimage of Union under Relation]]: +:$\RR^{-1} \sqbrk {T_1 \cup T_2} = \RR^{-1} \sqbrk {T_1} \cup \RR^{-1} \sqbrk {T_2}$ +{{qed}} +\end{proof}<|endoftext|> +\section{Preimage of Intersection under Mapping} +Tags: Preimages under Mappings, Set Intersection, Preimage of Intersection under Mapping + +\begin{theorem} +Let $f: S \to T$ be a [[Definition:Mapping|mapping]]. +Let $T_1$ and $T_2$ be [[Definition:Subset|subsets]] of $T$. +Then: +:$f^{-1} \sqbrk {T_1 \cap T_2} = f^{-1} \sqbrk {T_1} \cap f^{-1} \sqbrk {T_2}$ +This can be expressed in the language and notation of [[Definition:Inverse Image Mapping|inverse image mappings]] as: +:$\forall T_1, T_2 \in \powerset T: \map {f^\gets} {T_1 \cap T_2} = \map {f^\gets} {T_1} \cap \map {f^\gets} {T_2}$ +\end{theorem} + +\begin{proof} +As $f$, being a [[Definition:Mapping|mapping]], is also a [[Definition:Many-to-One Relation|many-to-one relation]], it follows from [[Inverse of Many-to-One Relation is One-to-Many]] that its [[Definition:Inverse of Mapping|inverse]] $f^{-1}$ is a [[Definition:One-to-Many Relation|one-to-many relation]]. +Thus [[Image of Intersection under One-to-Many Relation]] applies: +:$\mathcal R \sqbrk {T_1 \cap T_2} = \mathcal R \sqbrk {T_1} \cap \mathcal R \sqbrk {T_2}$ +where here $\mathcal R = f^{-1}$. +{{Qed}} +\end{proof}<|endoftext|> +\section{Preimage of Set Difference under Mapping} +Tags: Preimages under Mappings, Set Difference + +\begin{theorem} +Let $f: S \to T$ be a [[Definition:Mapping|mapping]]. +Let $T_1$ and $T_2$ be [[Definition:Subset|subsets]] of $T$. +Then: +:$f^{-1} \sqbrk {T_1 \setminus T_2} = f^{-1} \sqbrk {T_1} \setminus f^{-1} \sqbrk {T_2}$ +where: +:$\setminus$ denotes [[Definition:Set Difference|set difference]] +:$f^{-1} \sqbrk {T_1}$ denotes [[Definition:Preimage of Subset under Mapping|preimage]]. +\end{theorem} + +\begin{proof} +From [[Inverse of Mapping is One-to-Many Relation]], we have that $f^{-1}: T \to S$ is [[Definition:One-to-Many Relation|one-to-many]]. +Thus we can apply [[One-to-Many Image of Set Difference]]: +:$\mathcal R \sqbrk {T_1 \setminus T_2} = \mathcal R \sqbrk {T_1} \setminus \mathcal R \sqbrk {T_2}$ +where in this context $\mathcal R = f^{-1}$. +{{qed}} +\end{proof}<|endoftext|> +\section{Projection is Surjection} +Tags: Surjections, Projections + +\begin{theorem} +Let $S$ and $T$ be [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|sets]]. +Let $S \times T$ be the [[Definition:Cartesian Product|Cartesian product]] of $S$ and $T$. +Let $\pr_1: S \times T \to T$ and $\pr_2: S \times T \to T$ be the [[Definition:First Projection|first projection]] and [[Definition:Second Projection|second projection]] respectively on $S \times T$. +Then $\pr_1$ and $\pr_2$ are both [[Definition:Surjection|surjections]]. +\end{theorem} + +\begin{proof} +Let $S$ and $T$ be [[Definition:Set|sets]] such that neither is [[Definition:Empty Set|empty]]. +Let $\pr_1: S \times T \to S$ be the [[Definition:First Projection|first projection]] on $S \times T$. +Then by definition of [[Definition:First Projection|first projection]]: +:$\forall x \in S: \exists \tuple {x, t} \in S \times T: \map {\pr_1} {x, t} = x$ +Similarly, let $\pr_2: S \times T \to T$ be the [[Definition:Second Projection|second projection]] on $S \times T$. +Then by definition of [[Definition:Second Projection|second projection]]: +:$\forall x \in T: \exists \tuple {s, x} \in S \times T: \map {\pr_2} {s, x} = x$ +Hence the result. +{{qed}} +\end{proof}<|endoftext|> +\section{Sum Rule for Derivatives} +Tags: Differential Calculus, Sum Rule for Derivatives + +\begin{theorem} +Let $\map f x, \map j x, \map k x$ be [[Definition:Real Function|real functions]] defined on the [[Definition:Open Real Interval|open interval]] $I$. +Let $\xi \in I$ be a point in $I$ at which both $j$ and $k$ are [[Definition:Differentiable Real Function at Point|differentiable]]. +Let $\map f x = \map j x + \map k x$. +Then $f$ is [[Definition:Differentiable Real Function at Point|differentiable]] at $\xi$ and: +:$\map {f'} \xi = \map {j'} \xi + \map {k'} \xi$ +It follows from the definition of [[Definition:Derivative on Interval|derivative]] that if $j$ and $k$ are both [[Definition:Differentiable on Interval|differentiable]] on the interval $I$, then: +:$\forall x \in I: \map {f'} x = \map {j'} x + \map {k'} x$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \map {f'} \xi + | r = \lim_{h \mathop \to 0} \frac {\map f {\xi + h} - \map f \xi} h + | c = {{Defof|Derivative}} +}} +{{eqn | r = \lim_{h \mathop \to 0} \frac {\paren {\map j {\xi + h} + \map k {\xi + h} } - \paren {\map j \xi + \map k \xi } } h + | c = +}} +{{eqn | r = \lim_{h \mathop \to 0} \frac {\map j {\xi + h} + \map k {\xi + h} - \map j \xi - \map k \xi} h + | c = +}} +{{eqn | r = \lim_{h \mathop \to 0} \frac {\paren {\map j {\xi + h} - \map j \xi} + \paren {\map k {\xi + h} - \map k \xi} } h + | c = +}} +{{eqn | r = \lim_{h \mathop \to 0} \paren {\frac {\map j {\xi + h} - \map j \xi} h + \frac {\map k {\xi + h} - \map k \xi} h} + | c = +}} +{{eqn | r = \lim_{h \mathop \to 0} \frac {\map j {\xi + h} - \map j \xi} h + \lim_{h \mathop \to 0} \frac {\map k {\xi + h} - \map k \xi} h + | c = +}} +{{eqn | r = \map {j'} \xi + \map {k'} \xi + | c = {{Defof|Derivative}} +}} +{{end-eqn}} +{{qed}} +\end{proof} + +\begin{proof} +It can be observed that this is an example of a [[Linear Combination of Derivatives]] with $\lambda = \mu = 1$. +{{qed}} +\end{proof}<|endoftext|> +\section{Quotient Mapping is Surjection} +Tags: Quotient Mappings, Surjections + +\begin{theorem} +Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on $S$. +Then the [[Definition:Quotient Mapping|quotient mapping]] $q_\RR: S \to S / \RR$ is a [[Definition:Surjection|surjection]]. +\end{theorem} + +\begin{proof} +From [[Equivalence Class is not Empty]], we have that: +:$\forall \eqclass x \RR \in S / \RR: \exists x \in S: x \in \eqclass x \RR$ +and the result follows. +{{qed}} +\end{proof}<|endoftext|> +\section{Trivial Quotient is a Bijection} +Tags: Quotient Sets, Bijections + +\begin{theorem} +Let $\Delta_S$ be the [[Definition:Diagonal Relation|diagonal relation]] on a [[Definition:Set|set]] $S$. +Let $q_{\Delta_S}: S \to S / \Delta_S$ be the [[Definition:Trivial Quotient|trivial quotient of $S$]]. +Then $q_{\Delta_S}: S \to S / \Delta_S$ is a [[Definition:Bijection|bijection]]. +\end{theorem} + +\begin{proof} +The [[Definition:Diagonal Relation|diagonal relation]] is defined as: +:$\Delta_S = \set {\tuple {x, x}: x \in S} \subseteq S \times S$ +From the fact that $q_{\Delta_S}$ is a [[Definition:Quotient Mapping|quotient mapping]], we know that it is a [[Quotient Mapping is Surjection|surjection]]. +It relates each $x \in S$ to the [[Definition:Singleton|singleton]] $\set x$. +Thus: +:$\set x = \eqclass x {\Delta_S} \subseteq S$ +So $\map {q_{\Delta_S} } x = \set x$, and it follows that: +:$\eqclass x {\Delta_S} = \eqclass y {\Delta_S} \implies x = y$ +Thus $q_{\Delta_S}$ is [[Definition:Injection|injective]], and therefore by definition a [[Definition:Bijection|bijection]]. +{{Qed}} +\end{proof}<|endoftext|> +\section{Composite of Quotient Mappings} +Tags: Quotient Sets, Equivalence Relations + +\begin{theorem} +Let $S$ be a [[Definition:Set|set]]. +Let $\mathcal R_1$ be an [[Definition:Equivalence Relation|equivalence]] on $S$, and $\mathcal R_2$ be an [[Definition:Equivalence Relation|equivalence]] on the [[Definition:Quotient Set|quotient set]] $S / \mathcal R_1$. +We can find an [[Definition:Equivalence Relation|equivalence]] $\mathcal R_3$ on $S$ such that $\paren {S / \mathcal R_1} / \mathcal R_2$ is in [[Definition:Bijection|one-to-one correspondence]] with $S / \mathcal R_3$ under the [[Definition:Mapping|mapping]]: +:$\phi: \paren {S / \mathcal R_1} / \mathcal R_2 \to S / \mathcal R_3: \eqclass {\eqclass x {\mathcal R_1} } {\mathcal R_2} \mapsto \eqclass x {\mathcal R_3}$. +\end{theorem} + +\begin{proof} +Define $\mathcal R_3$ to be the [[Definition:Equivalence Relation Induced by Mapping|equivalence induced by]]: +:$x \mapsto \eqclass {\eqclass x {\mathcal R_1} } {\mathcal R_2}$ +By definition of $\mathcal R_3$: +:$\eqclass {\eqclass x {\mathcal R_1} } {\mathcal R_2} = \eqclass {\eqclass y {\mathcal R_1} } {\mathcal R_2} \implies \eqclass x {\mathcal R_3} = \eqclass y {\mathcal R_3}$ +Therefore, $\phi$ is [[Definition:Well-Defined Mapping|well-defined]]. +Again by definition of $\mathcal R_3$: +:$\eqclass x {\mathcal R_3} = \eqclass y {\mathcal R_3} \implies \eqclass {\eqclass x {\mathcal R_1} } {\mathcal R_2} = \eqclass {\eqclass y {\mathcal R_1} } {\mathcal R_2}$ +which means $\phi$ is an [[Definition:Injection|injection]]. +Lastly, note that every element of $S / \mathcal R_3$ is of the form $\eqclass x {\mathcal R_3}$ for some $x \in S$. +It is now immediate from the definition of $\phi$ that it is [[Definition:Surjective|surjective]]. +Hence $\phi$ is a [[Definition:Bijection|bijection]], as desired. +{{qed}} +\end{proof}<|endoftext|> +\section{Renaming Mapping is Well-Defined} +Tags: Mapping Theory, Quotient Sets, Equivalence Relations, Renaming Mapping is Well-Defined + +\begin{theorem} +Let $f: S \to T$ be a [[Definition:Mapping|mapping]]. +Let $r: S / \RR_f \to \Img f$ be the [[Definition:Renaming Mapping|renaming mapping]], defined as: +:$r: S / \RR_f \to \Img f: \map r {\eqclass x {\RR_f} } = \map f x$ +where: +:$\RR_f$ is the [[Definition:Equivalence Relation Induced by Mapping|equivalence induced]] by the [[Definition:Mapping|mapping]] $f$ +:$S / \RR_f$ is the [[Definition:Quotient Set|quotient set]] of $S$ determined by $\RR_f$ +:$\eqclass x {\RR_f}$ is the [[Definition:Equivalence Class|equivalence class]] of $x$ under $\RR_f$. +The renaming mapping is always [[Definition:Well-Defined Mapping|well-defined]]. +\end{theorem} + +\begin{proof} +By [[Relation Induced by Mapping is Equivalence Relation]], we have that $\RR_f$ is an [[Definition:Equivalence Relation|equivalence relation]]. +To determine whether $r$ is [[Definition:Well-Defined Mapping|well-defined]], we have to determine whether $r: S / \RR_f \to \Img f$ actually defines a [[Definition:Mapping|mapping]] at all. +Consider a typical element $\eqclass x {\RR_f}$ of $S / \RR_f$. +Suppose we were to choose another name for the class $\eqclass x {\RR_f}$. +Assume that $\eqclass x {\RR_f}$ is not a [[Definition:Singleton|singleton]]. +For example, let us choose $y \in \eqclass x {\RR_f}, y \ne x$ such that: +:$\eqclass x {\RR_f} = \eqclass y {\RR_f}$ +then: +:$\map r {\eqclass x {\RR_f} } = \map r {\eqclass y {\RR_f} }$ +Hence from the definition, we have: +{{begin-eqn}} +{{eqn | l = y + | o = \in + | r = \eqclass x {\RR_f} + | c = +}} +{{eqn | ll= \leadsto + | l = \map f y + | r = \map f x + | c = +}} +{{eqn | r = \map r {\eqclass x {\RR_f} } + | c = +}} +{{eqn | r = \map r {\eqclass y {\RR_f} } + | c = +}} +{{end-eqn}} +Thus $r$ is well-defined. +{{qed}} +\end{proof} + +\begin{proof} +From [[Condition for Mapping from Quotient Set to be Well-Defined]]: +:there exists a [[Definition:Mapping|mapping]] $\phi: S / \RR \to T$ such that $\phi \circ q_\RR = f$ +{{iff}}: +:$\forall x, y \in S: \tuple {x, y} \in \RR \implies \map f x = \map f y$ +But by definition of the [[Definition:Equivalence Relation Induced by Mapping|equivalence induced]] by the [[Definition:Mapping|mapping]] $f$: +:$\forall x, y \in S: \tuple {x, y} \in \RR_f \implies \map f x = \map f y$ +The result follows directly. +{{qed}} +\end{proof}<|endoftext|> +\section{Powers of Group Elements} +Tags: Group Theory, Powers (Abstract Algebra), Index Laws + +\begin{theorem} +Let $\left({G, \circ}\right)$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$. +Let $a \in G$. +Then the following results hold: +\end{theorem}<|endoftext|> +\section{Renaming Mapping is Bijection} +Tags: Equivalence Relations, Bijections, Quotient Sets, Renaming Mapping is Bijection + +\begin{theorem} +Let $f: S \to T$ be a [[Definition:Mapping|mapping]]. +Let $r: S / \RR_f \to \Img f$ be the [[Definition:Renaming Mapping|renaming mapping]], defined as: +:$r: S / \RR_f \to \Img f: \map r {\eqclass x {\RR_f} } = \map f x$ +where: +:$\RR_f$ is the [[Definition:Equivalence Relation Induced by Mapping|equivalence induced]] by the [[Definition:Mapping|mapping]] $f$ +:$S / \RR_f$ is the [[Definition:Quotient Set|quotient set]] of $S$ determined by $\RR_f$ +:$\eqclass x {\RR_f}$ is the [[Definition:Equivalence Class|equivalence class]] of $x$ under $\RR_f$. +The [[Definition:Renaming Mapping|renaming mapping]] is a [[Definition:Bijection|bijection]]. +\end{theorem} + +\begin{proof} +From [[Renaming Mapping is Well-Defined]], $r: S / \RR_f \to \Img f$ is a [[Definition:Well-Defined Mapping|well-defined mapping]]. +By definition, $\RR_f$ is the [[Definition:Equivalence Relation Induced by Mapping|equivalence relation induced]] by the [[Definition:Mapping|mapping]] $f$. +Hence by definition: +:$\tuple {s_1, s_2} \in \RR_f \iff \map f {s_1} = \map f {s_2}$ +From [[Condition for Mapping from Quotient Set to be Injection]], this is precisely the condition required for $r$ to be an [[Definition:Injection|injection]]. +Next it is noted that the [[Definition:Codomain of Mapping|codomain]] of $r$ is $\Img f$. +Then from [[Restriction of Mapping to Image is Surjection]], we have that $f_{\restriction \Img f}$ is a [[Definition:Surjection|surjection]]. +Hence from [[Condition for Mapping from Quotient Set to be Surjection]] it follows that $r$ is also a [[Definition:Surjection|surjection]]. +Thus $r$ is shown to be both an [[Definition:Injection|injection]] and a [[Definition:Surjection|surjection]], and so by definition is a [[Definition:Bijection|bijection]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Composite of Mapping with Inverse} +Tags: Equivalence Relations, Inverse Mappings, Composite Mappings + +\begin{theorem} +Let $f: S \to T$ be a [[Definition:Mapping|mapping]]. +Then: +:$\forall x \in S: \map {f^{-1} \circ f} x = \eqclass x {\RR_f}$ +where: +:$\RR_f$ is the [[Definition:Equivalence Relation Induced by Mapping|equivalence induced by $f$]] +:$\eqclass x {\RR_f}$ is the [[Definition:Equivalence Class|$\RR_f$-equivalence class of $x$]]. +\end{theorem} + +\begin{proof} +Let $y = \map f x$. +Then by the definition of [[Definition:Equivalence Relation Induced by Mapping|induced equivalence]]: +:$x \in \eqclass x {\RR_f}$ +By the definition of the [[Definition:Inverse of Mapping|inverse of a mapping]]: +:$f^{-1} = \set {\tuple {y, x}: \tuple {x, y} \in f}$ +Thus: +:$\eqclass x {\RR_f} = \set {s \in \Dom f: \map f s = \map f x}$ +By definition: +:$\map {f^{-1} } y = \eqclass x {\RR_f}$ +Hence the result. +{{qed}} +\end{proof}<|endoftext|> +\section{Factoring Mapping into Surjection and Inclusion} +Tags: Surjections, Inclusion Mappings + +\begin{theorem} +Every [[Definition:Mapping|mapping]] $f:S \to T$ can be uniquely '''factored into''' a [[Definition:Surjection|surjection]] $g$ followed by the [[Definition:Inclusion Mapping|inclusion mapping]] $i_T$. +That is, $f = i_T \circ g$ where: +:$g: S \to \Img f: \map g x = \map f x$ +:$i_T: \Img f \to T: \map {i_T} x = x$ +This can be illustrated using a [[Definition:Commutative Diagram|commutative diagram]] as follows: +::$\begin{xy}\xymatrix@L+2mu@+1em { + S \ar@{-->}[r]^*{g} + \ar[rd]_*{f = i_T \circ g} +& + \Img f \ar@{-->}[d]^*{i_T} +\\ +& + T +}\end{xy}$ +\end{theorem} + +\begin{proof} +From [[Surjection by Restriction of Codomain]], any $f: S \to \Img f$ is a [[Definition:Surjection|surjection]]. +The [[Definition:Mapping|mapping]] $g: S \to \Img f$ where $\map g x = \map f x$ is therefore also clearly a [[Definition:Surjection|surjection]]. +The [[Definition:Mapping|mapping]] $g: S \to \Img f: \map g x = \map f x$ is clearly unique, by [[Equality of Mappings]]. +From [[Inclusion Mapping is Injection]], $i_T: \Img f \to T$ is an [[Definition:Injection|injection]]. +Likewise, the mapping $i_T: \Img f \to T : \map {i_T} x = x$ is also unique, by its own definition. +{{qed}} +\end{proof}<|endoftext|> +\section{Quotient Theorem for Surjections} +Tags: Surjections, Quotient Sets, Quotient Mappings, Named Theorems + +\begin{theorem} +Let $f: S \to T$ be a [[Definition:Surjection|surjection]]. +Then there is one and only one [[Definition:Bijection|bijection]] $r: S / \RR_f \to T$ such that: +:$r \circ q_{\RR_f} = f$ +where: +:$\RR_f$ is the [[Definition:Equivalence Relation Induced by Mapping|equivalence induced by $f$]] +:$r: S / \RR_f \to T$ is the [[Definition:Renaming Mapping|renaming mapping]] +:$q_{\RR_f}: S \to S / \RR_f$ is the [[Definition:Quotient Mapping|quotient mapping]] induced by $\RR_f$. +This can be illustrated using a [[Definition:Commutative Diagram|commutative diagram]] as follows: +::$\begin {xy} \xymatrix@L + 2mu@ + 1em { + S \ar@{-->}[rr]^*{f = r \circ q_{\RR_f} } + \ar[dd]_*{q_{\RR_f} } +&& T \\ +\\ +S / \RR_f \ar[urur]_*{r} +} \end {xy}$ +\end{theorem} + +\begin{proof} +From the definition of [[Definition:Equivalence Relation Induced by Mapping|Induced Equivalence]], the [[Definition:Mapping|mapping]] $f: S \to T$ induces an [[Definition:Equivalence Relation|equivalence]] $\RR_f$ on its [[Definition:Domain of Mapping|domain]]. +As $f: S \to T$ is a [[Definition:Surjection|surjection]], $T = \Img f$ by definition. +From [[Renaming Mapping is Bijection]], the [[Definition:Renaming Mapping|renaming mapping]] $r: S / \RR_f \to T$ is a [[Definition:Bijection|bijection]], where $S / \RR_f$ is the [[Definition:Quotient Set|quotient set]] of $S$ by $\RR_f$. +Hence: +:$r \circ q_{\RR_f} = f$ +:$r$ is the only mapping $r: S / \RR_f \to T$ that satisfies this equality. +{{qed}} +\end{proof}<|endoftext|> +\section{Quotient Theorem for Sets} +Tags: Mapping Theory, Quotient Mappings, Quotient Sets, Named Theorems, Quotient Theorem for Sets + +\begin{theorem} +A [[Definition:Mapping|mapping]] $f: S \to T$ can be [[Definition:Unique|uniquely]] '''factored into''' a [[Definition:Surjection|surjection]], followed by a [[Definition:Bijection|bijection]], followed by an [[Definition:Injection|injection]]. +Thus: +:$f = i \circ r \circ q_{\RR_f}$ +where: +{{begin-eqn}} +{{eqn | ll= q_{\RR_f}: + | lo= S \to S / \RR_f: + | l = \map {q_{\RR_f} } s + | r = \eqclass s {\RR_f} + | c = [[Definition:Quotient Mapping|Quotient Mapping]] +}} +{{eqn | ll= r: + | lo= S / \RR_f \to \Img f: + | l = \map r {\eqclass s {\RR_f} } + | r = \map f s + | c = [[Definition:Renaming Mapping|Renaming Mapping]] +}} +{{eqn | ll= i: + | lo= \Img f \to T: + | l = \map i t + | r = t + | c = [[Definition:Inclusion Mapping|Inclusion Mapping]] +}} +{{end-eqn}} +where: +:$\RR_f$ is the [[Definition:Equivalence Relation Induced by Mapping|equivalence induced by $f$]] +:$S / \RR_f$ is the [[Definition:Quotient Set|quotient set of $S$ induced by $\RR_f$]] +This can be illustrated using a [[Definition:Commutative Diagram|commutative diagram]] as follows: +::$\begin {xy} \xymatrix@L + 2mu@ + 1em { + S \ar@{-->}[rrr]^*{f = i_T \circ r \circ q_{\RR_f} } + \ar[d]_*{q_{\RR_f} } +& & & T \\ +S / \RR_f \ar[rrr]_*{r} & & & \Img f \ar[u]_*{i_T} } \end {xy}$ +\end{theorem} + +\begin{proof} +From [[Factoring Mapping into Surjection and Inclusion]], $f$ can be factored uniquely into: +:A [[Definition:Surjection|surjection]] $g: S \to \Img f$, followed by: +:The [[Definition:Inclusion Mapping|inclusion mapping]] $i: \Img f \to T$ (an [[Definition:Injection|injection]]). +::$\begin{xy}\xymatrix@L+2mu@+1em { + S \ar[drdr]_*{g} + \ar@{-->}[rr]^*{f = i_T \circ g} +& & T +\\ \\ +& & \Img f \ar[uu]_*{i_T} +}\end{xy}$ +From the [[Quotient Theorem for Surjections]], the [[Definition:Surjection|surjection]] $g$ can be factored uniquely into: +:The [[Definition:Quotient Mapping|quotient mapping]] $q_{\RR_f}: S \to S / \RR_f$ (a [[Definition:Surjection|surjection]]), followed by: +:The [[Definition:Renaming Mapping|renaming mapping]] $r: S / \RR_f \to \Img f$ (a [[Definition:Bijection|bijection]]). +Thus: +:$f = i_T \circ \paren {r \circ q_{\RR_f} }$ +As [[Composition of Mappings is Associative]] it can be seen that $f = i_T \circ r \circ q_{\RR_f}$. +The [[Definition:Commutative Diagram|commutative diagram]] is as follows: +::$\begin {xy} \xymatrix@L + 2mu@ + 1em { + S \ar@{-->}[rrr]^*{f = i_T \circ r \circ q_{\RR_f} } + \ar[ddd]_*{q_{\RR_f} } + \ar@{..>}[drdrdr]_*{g = r \circ q_{\RR_f} } +& & & T \\ \\ \\ +S / \RR_f \ar[rrr]_*{r} & & & \Img f \ar[uuu]_*{i_T} } \end {xy}$ +{{qed}} +\end{proof}<|endoftext|> +\section{Empty Set is Element of Power Set} +Tags: Empty Set, Power Set + +\begin{theorem} +The [[Definition:Empty Set|empty set]] is an [[Definition:Element|element]] of all [[Definition:Power Set|power sets]]: +:$\forall S: \O \in \powerset S$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \forall S: \O + | o = \subseteq + | r = S + | c = [[Empty Set is Subset of All Sets]] +}} +{{eqn | ll= \leadsto + | l = \forall S: \O + | o = \in + | r = \powerset S + | c = {{Defof|Power Set}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Direct Image Mapping of Relation is Mapping} +Tags: Direct Image Mappings + +\begin{theorem} +Let $S$ and $T$ be [[Definition:Set|sets]]. +Let $\mathcal R \subseteq S \times T$ be a [[Definition:Relation|relation]] on $S \times T$. +Let $\mathcal R^\to: \powerset S \to \powerset T$ be the [[Definition:Direct Image Mapping of Relation|direct image mapping]] of $\mathcal R$: +:$\forall X \in \powerset S: \map {\mathcal R^\to} X = \set {t \in T: \exists s \in X: \tuple {s, t} \in \mathcal R}$ +Then $\mathcal R^\to$ is indeed a [[Definition:Mapping|mapping]]. +\end{theorem} + +\begin{proof} +Take the general relation $\mathcal R \subseteq S \times T$. +Let $X \subseteq S$, that is $X \in \powerset S$. +Suppose $X = \O$. +Then from [[Image of Empty Set is Empty Set]]: +:$\map {\mathcal R^\to} X = \O \subseteq T$ +Suppose $X = S$. +Then from [[Image is Subset of Codomain/Corollary 1|Image is Subset of Codomain: Corollary 1]]: +:$\map {\mathcal R^\to} X = \Img {\mathcal R} \subseteq T$ +Finally, suppose $\O \subset X \subset S$. +From [[Image is Subset of Codomain]], again: +:$\map {\mathcal R^\to} X \subseteq T$ +Now, from the definition of the [[Definition:Power Set|power set]], we have that: +:$Y \subseteq T \iff Y \in \powerset T$ +We defined $R^\to \subseteq \powerset S \times \powerset T$ such that: +:$R^\to: \powerset S \to \powerset T: \map {\mathcal R^\to} X = \set {t \in T: \exists s \in X: \tuple {s, t} \in \mathcal R}$ +By definition, there is only one $\map {\mathcal R^\to} X$ for any given $X$, and so $R^\to$ is [[Definition:Many-to-One Relation|many-to-one]]. +We have shown that: +:$\forall X \subseteq S: \map {\mathcal R^\to} X \in \powerset T$ +So: +:$\forall X \in \powerset S: \exists_1 Y \in \powerset T: \map {\mathcal R^\to} X = Y$ +So: +:$R^\to$ is defined for all $X \in \powerset S$ +and therefore $R^\to$ is a [[Definition:Mapping|mapping]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Inverse Image Mapping of Relation is Mapping} +Tags: Inverse Image Mappings + +\begin{theorem} +Let $S$ and $T$ be [[Definition:Set|sets]]. +Let $\mathcal R \subseteq S \times T$ be a [[Definition:Relation|relation]] on $S \times T$. +Let $\mathcal R^\gets$ be the [[Definition:Inverse Image Mapping of Relation|inverse image mapping]] of $\mathcal R$: +:$\mathcal R^\gets: \powerset T \to \powerset S: \map {\mathcal R^\gets} Y = \mathcal R^{-1} \sqbrk Y$ +Then $\mathcal R^\gets$ is indeed a [[Definition:Mapping|mapping]]. +\end{theorem} + +\begin{proof} +$\mathcal R^{-1}$, being a [[Definition:Relation|relation]], obeys the same laws as $\mathcal R$. +So [[Direct Image Mapping of Relation is Mapping]] applies directly. +{{qed}} +\end{proof}<|endoftext|> +\section{Direct Image Mapping of Injection is Injection} +Tags: Injections, Direct Image Mappings + +\begin{theorem} +Let $f: S \to T$ be an [[Definition:Injection|injection]]. +Then the [[Definition:Direct Image Mapping of Mapping|direct image mapping]] of $f$: +:$f^\to: \powerset S \to \powerset T$ +is an [[Definition:Injection|injection]]. +\end{theorem} + +\begin{proof} +Suppose $f^\to: \powerset S \to \powerset T$ is not an [[Definition:Injection|injection]]. +Then: +:$\exists Y \in \powerset T: \exists X_1, X_2 \in \powerset S: X_1 \ne X_2 \land \map {f^\to} {X_1} = \map {f^\to} {X_2} = Y$ +There are two cases to consider: +:$(1): \quad$ Either $X_1$ or $X_2$ is the [[Definition:Empty Set|empty set]] +:$(2): \quad$ Neither $X_1$ nor $X_2$ is the [[Definition:Empty Set|empty set]]. +=== Either subset is the empty set === +{{WLOG}} assume that $X_1 = \O$. +Then, from [[Image of Empty Set is Empty Set]]: +:$Y = \map {f^\to} {X_1} = \O$ +But that means: +:$\map {f^\to} {X_2} = \O$ +Now $X_1 \ne X_2$, so $X_2 \ne \O$. +Thus: +:$\exists x_2 \in X_2: \neg \exists y \in T: \map f {x_2} = y$ +which means that $f$ is not even a [[Definition:Mapping|mapping]], let alone an [[Definition:Injection|injection]]. +The same argument applies if $X_2$ is [[Definition:Empty Set|empty]]. +{{qed|lemma}} +=== Neither subset is the empty set === +Now, assume that neither $X_1$ nor $X_2$ is the [[Definition:Empty Set|empty set]]. +As $X_1 \ne X_2$, there must be at least one [[Definition:Element|element]] in either one which is not in the other. +{{WLOG}} assume that $\exists x_1 \in X_1: x_1 \notin X_2$. +Now suppose $\map f {x_1} = y \in Y$. +However, as $\map {f^\to} {X_2} = Y$, there must be some $x_2 \in X_2$ such that $\map f {x_2} = y \in Y$. +So: +:$\exists x_1 \ne x_2 \in S: \map f {x_1} = \map f {x_2}$ +which means, by definition, that $f$ is not [[Definition:Injection|injective]]. +Thus, if $f^\to: \powerset S \to \powerset T$ is not an [[Definition:Injection|injection]] then neither is $f: S \to T$. +Thus, by the [[Rule of Transposition]], the result follows. +{{qed}} +[[Category:Injections]] +[[Category:Direct Image Mappings]] +1wwgkycxtfjzr2g8406zchd8gyumae6 +\end{proof}<|endoftext|> +\section{Direct Image Mapping of Surjection is Surjection} +Tags: Surjections, Direct Image Mappings, Power Set, Direct Image Mapping of Surjection is Surjection + +\begin{theorem} +Let $f: S \to T$ be a [[Definition:Surjection|surjection]]. +Then the [[Definition:Direct Image Mapping of Mapping|direct image mapping]] of $f$: +:$f^\to: \powerset S \to \powerset T$ +is a [[Definition:Surjection|surjection]]. +\end{theorem} + +\begin{proof} +Let $f^\gets$ be the [[Definition:Inverse Image Mapping of Mapping|inverse image mapping]] of $f$. +Let $Y \in \powerset T$. +Let $X = \map {f^\gets} Y$. +By [[Subset equals Image of Preimage iff Mapping is Surjection]]: +:$\map {f^\to} X = Y$ +As such an $X$ exists for each $Y \in \powerset S$, $f^\to$ is [[Definition:Surjection|surjective]]. +{{qed}} +\end{proof} + +\begin{proof} +Let $f: S \to T$ be a [[Definition:Surjection|surjection]]. +Then: +:$\forall y \in T: \exists x \in S: f \paren x = y$ +From the [[Quotient Theorem for Surjections]], there is [[Definition:Unique|one and only one]] [[Definition:Bijection|bijection]] $r: S / \mathcal R_f \to T$ such that $r \circ q_{\mathcal R_f} = f$. +Each [[Definition:Element|element]] of $S / \mathcal R_f$ is a [[Definition:Subset|subset]] of $S$ and therefore an [[Definition:Element|element]] of $\powerset S$. +Thus as $r: S / \mathcal R_f \to T$ is a [[Definition:Bijection|bijection]], and hence [[Definition:A Fortiori|a fortiori]] also an [[Definition:Injection|injection]]: +:$\forall X_1, X_2 \in \powerset S: \map r {X_1} = \map r {X_2} \implies X_1 = X_2$ +Because $f$ is a [[Definition:Surjection|surjection]], every $y \in T$ is mapped to by [[Definition:Unique|exactly one]] [[Definition:Element|element]] of the [[Definition:Set Partition|partition]] of $S$ defined by $\mathcal R_f$. +Let $T = \set {y_1, y_2, \ldots}$. +Let the [[Definition:Set Partition|partition]] defined by $\mathcal R_f$ be $\bigcup \paren {X_1, X_2, \ldots}$ where $\map r {X_n} = y_n$. +Let $Y_r \in \powerset T$, such that $Y_r = \set {y_{r_1}, y_{r_2}, \ldots}$. +Then: +:$\map {f^\to} {X_r} = Y_r$ +where $\displaystyle X_r = \bigcup \paren {X_{r_1}, X_{r_2}, \ldots}$. +As $\set {X_1, X_2, \ldots}$ is a [[Definition:Set Partition|partition]] of $S$, $\forall Y_r \in \powerset T: X_r$ is [[Definition:Unique|unique]]. +Thus $f^\to: \powerset S \to \powerset T$ is a [[Definition:Surjection|surjection]]. +{{Qed}} +\end{proof} + +\begin{proof} +Let $f: S \to T$ be a [[Definition:Surjection|surjection]]. +By definition, $f^\to$ is defined by sending [[Definition:Subset|subsets]] of $S$ to their [[Definition:Image of Subset under Mapping|image]] under $f$. +That is: +:$\forall X \subseteq S: \map {f^\to} X = \set{\map f x: x \in X} \subseteq T$ +To prove that $f^\to$ is a [[Definition:Surjection|surjection]], we need to show that every [[Definition:Subset|subset]] of $T$ is the [[Definition:Image of Element under Mapping|image]] under $f^\to$ of some [[Definition:Subset|subset]] of $S$. +Let $Y \subseteq T$. +Since $f$ is a [[Definition:Surjection|surjection]], by definition we have that: +:$\forall t \in T: \exists s \in S: \map f s = t$ +Hence (possibly requiring the [[Axiom:Axiom of Choice|Axiom of Choice]]), we can select for each $y \in Y$ some $s_y \in S$ such that $\map f {s_y} = y$. +Define $X_Y$ to be: +:$X_Y := \set {s_y: y \in Y}$ +Then: +{{begin-eqn}} +{{eqn | l = \map {f^\to} {X_Y} + | r = \set {\map f {s_y}: s_y \in X_Y} + | c = +}} +{{eqn | r = \set {\map f {s_y}: y \in Y} + | c = +}} +{{eqn | r = \set {y : y \in Y} + | c = +}} +{{eqn | r = Y + | c = +}} +{{end-eqn}} +Thus $f^\to: \powerset S \to \powerset T$ is a [[Definition:Surjection|surjection]]. +{{qed}} +{{AoC||4}} +\end{proof}<|endoftext|> +\section{Cantor's Theorem} +Tags: Surjections, Power Set, Cantor's Theorem + +\begin{theorem} +There is no [[Definition:Surjection|surjection]] from a [[Definition:Set|set]] $S$ to its [[Definition:Power Set|power set]] for any set $S$. +That is, $S$ is [[Definition:Strictly Smaller Set|strictly smaller]] than its [[Definition:Power Set|power set]]. +\end{theorem} + +\begin{proof} +Let $S$ be a [[Definition:Set|set]]. +Let $\mathcal P \left({S}\right)$ be the [[Definition:Power Set|power set]] of $S$. +Let $f: S \to \mathcal P \left({S}\right)$ be a [[Definition:Mapping|mapping]]. +Let $T = \left\{{x \in S: \neg \left({x \in f \left({x}\right)}\right)}\right\}$. +Then $T \subseteq S$, so $T \in \mathcal P \left({S}\right)$ by the definition of [[Definition:Power Set|power set]]. +We will show that $T$ is not in the [[Definition:Image of Mapping|image]] of $f$ and therefore $f$ is not [[Definition:Surjection|surjective]]. +Aiming for a [[Proof by Contradiction|contradiction]], suppose that: +: $\exists a \in S: T = f \left({a}\right)$ +Suppose that: +: $a \in f \left({a}\right)$ +Then by the definition of $T$: +: $\neg \left({a \in T}\right)$ +Thus since $T = f \left({a}\right)$: +: $\neg \left({a \in f \left({a}\right) }\right)$ +By [[Rule of Implication]]: +: $(1) \quad a \in f \left({a}\right) \implies \neg \left({ a \in f \left({a}\right) }\right)$ +Suppose instead that: +: $\neg \left({a \in f \left({a}\right)}\right)$ +Then by the definition of $T$: +: $a \in T$ +Thus since $T = f \left({a}\right)$: +: $a \in f \left({a}\right)$ +By [[Rule of Implication]]: +: $(2) \quad \neg \left({ a \in f \left({a}\right) }\right) \implies a \in f \left({a}\right)$ +By [[Non-Equivalence of Proposition and Negation/Formulation 1|Non-Equivalence of Proposition and Negation]], applied to $(1)$ and $(2)$, this is a [[Definition:Contradiction|contradiction]]. +As the specific choice of $a$ did not matter, we derive a contradiction by [[Existential Instantiation]]. +Thus by [[Proof by Contradiction]], the supposition that $\exists a \in S: T = f \left({a}\right)$ must be false. +It follows that $f$ is not a [[Definition:Surjection|surjection]]. +{{qed}} +\end{proof} + +\begin{proof} +{{AimForCont}} $S$ is a [[Definition:Set|set]] with a [[Definition:Surjection|surjection]] $f: S \to \powerset S$. +Then: +{{begin-eqn}} +{{eqn | l = \forall x \in S: \map f x + | o = \in + | r = \powerset S + | c = [[Definition:By Hypothesis|By Hypothesis]] +}} +{{eqn | ll= \leadsto + | l = \forall x \in S: \map f x + | o = \subseteq + | r = S + | c = {{Defof|Power Set}} +}} +{{end-eqn}} +Now by [[Law of Excluded Middle]], there are two choices for every $x \in S$: +: $x \in \map f x$ +: $x \notin \map f x$ +Let $T = \set {x \in S: x \notin \map f x}$. +As $f$ is supposed to be a [[Definition:Surjection|surjection]], $\exists a \in S: T = \map f a$. +Thus: +: $a \in \map f a \implies a \notin \map f a$ +: $a \notin \map f a \implies a \in \map f a$ +This is a [[Proof by Contradiction|contradiction]], so the initial supposition that there is such a [[Definition:Surjection|surjection]] must be false. +{{qed}} +{{LEM||3}} +\end{proof} + +\begin{proof} +The proof proceeds by [[Principle of Mathematical Induction|induction]]. +For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]: +:There is no [[Definition:Surjection|surjection]] from $S$ onto $\map {\PP^n} S$. +=== Basis for the Induction === +$\map P 1$ is [[Cantor's Theorem]]. +This is our [[Definition:Basis for the Induction|basis for the induction]]. +=== Induction Hypothesis === +Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. +So this is our [[Definition:Induction Hypothesis|induction hypothesis]]: +:There is no [[Definition:Surjection|surjection]] from $S$ onto $\map {\PP^k} S$. +Then we need to show: +:There is no [[Definition:Surjection|surjection]] from $S$ onto $\map {\PP^{k + 1} } S$. +=== Induction Step === +This is our [[Definition:Induction Step|induction step]]: +Suppose that $\map P k$ is true. +{{AimForCont}} that $f: S \to \map {\PP^{k + 1} } S$ is a [[Definition:Surjection|surjection]]. +Define the [[Definition:Mapping|mapping]] $g: S \to \map {\PP^k} S$ as: +:$\displaystyle \map g x = \bigcup \map f x$ +This is actually a mapping into $\map {\PP^k} S$, as follows: +:$\map f x \in \map {\PP^{k + 1} } S = \powerset {\map {\mathcal P^k} S}$ +By the definition of [[Definition:Power Set|power set]]: +:$\map f x \subseteq \map {\PP^k} S$ +Thus each element of $\map f x$ is a [[Definition:Subset|subset]] of $\map {\PP^{k - 1} } S$. +Thus by [[Union of Subsets is Subset/Set of Sets|Union of Subsets is Subset]]: +:$\displaystyle \bigcup \map f x \subseteq \map {\PP^{k - 1} } S$ +Therefore: +:$\displaystyle \bigcup \map f x \in \map {\PP^k} S$ +That is, $\displaystyle \map g x$ is a mapping into $\map {\PP^k} S$. +Next we have that: +:$\forall y \in \map {\PP^k} S: \set y \in \map {\PP^{n + 1} } S$ +But [[Definition:By Hypothesis|by hypothesis]] $f$ is [[Definition:Surjection|surjective]], and so: +:$\exists x \in S: \map f x = \set y$ +Then: +:$\displaystyle \map g x = \bigcup \set y = y$ +As this holds for all such $y$, $g$ is [[Definition:Surjection|surjective]]. +But this contradicts the [[Cantor's Theorem (Strong Version)/Proof 2#Induction Hypothesis|induction hypothesis]]. +Thus we conclude that the theorem holds for all $n$. +{{qed}} +\end{proof} + +\begin{proof} +We temporarily introduce the notation: +:$a^n = \begin{cases} +a : & n = 0 \\ +\set {a^{n - 1} } : & n > 0 +\end{cases}$ +where $a \in S$. +Thus $a^n$ consists of a single element $a^{n - 1} \in \map {\PP^{n - 1} } S$. +Let there be a [[Definition:Bijection|bijection]] $f: S \to \QQ^n$ where $\QQ^n \subseteq \map {\PP^n} S$. +Then define: +:$\AA^{n - 1} = \set {s^{n - 1} \in \map {\PP^{n - 1} } S: s^{n - 1} \ne \map f s}$ +Each $\QQ^{n - 1} \in \QQ^n$ is the [[Definition:Image of Element under Mapping|image]] of some [[Definition:Element|element]] of $S$ under $f$. +Let $\QQ^{n - 1} = \map f x$ for some $x \in S$. +Let $x^{n - 1} \in \QQ^{n - 1} = \map f x$. +Then $x^{n - 1} \notin \AA^{n - 1}$, and so $\AA^{n - 1} \notin \QQ^{n - 1}$. +On the other hand, let $x^{n - 1} \notin \QQ^{n - 1} = \map f x$. +Then $x^{n - 1} \in \AA^{n - 1}$, and so again $ A^{n - 1} \notin \QQ^{n - 1}$. +Thus $\AA^{n - 1} \notin Q^n$. +Therefore $Q^n$ is a [[Definition:Proper Subset|proper subset]] of $\map {\PP^n} S$. +Hence the result. +{{qed}} +\end{proof}<|endoftext|> +\section{Surjection Induced by Powerset is Induced by Surjection} +Tags: Surjections, Direct Image Mappings + +\begin{theorem} +Let $\mathcal R \subseteq S \times T$ be a [[Definition:Relation|relation]]. +Let $\mathcal R^\to: \powerset S \to \powerset T$ be the [[Definition:Direct Image Mapping of Relation|direct image mapping]] $\mathcal R$. +Let $\mathcal R^\to$ be a [[Definition:Surjection|surjection]]. +Let $X = \Preimg {\mathcal R}$, that is, the [[Definition:Preimage of Relation|preimage]] of $\mathcal R$. +Then $\mathcal R {\restriction_X} \subseteq X \times T$, that is, the [[Definition:Restriction of Relation|restriction of $\mathcal R$ to $X$]], is a [[Definition:Surjection|surjection]]. +\end{theorem} + +\begin{proof} +Let $X$ be the [[Definition:Preimage of Relation|preimage]] of $\mathcal R$. +Suppose $\mathcal R {\restriction_X} \subseteq X \times T$ is a [[Definition:Mapping|mapping]], but not a [[Definition:Surjection|surjection]]. +Then: +:$\exists y \in T: \neg \exists x \in S: \tuple {x, y} \in \mathcal R$ +Because no [[Definition:Element|element]] of $S$ relates to $y$, no [[Definition:Subset|subset]] of $S$ contains any [[Definition:Element|element]] of $S$ that relates to the [[Definition:Subset|subset]] $\left\{{y}\right\} \subseteq T$. +Thus: +:$\exists \set y \in \powerset T: \neg \exists X \in \powerset S: \map {\mathcal R^\to} X = \set y$ +So we see that $\mathcal R^\to: \powerset S \to \powerset T$ is not a [[Definition:Surjection|surjection]]. +Now, suppose $\mathcal R {\restriction_X} \subseteq X \times T$ is not even a [[Definition:Mapping|mapping]]. +This could happen, according to the definition of a [[Definition:Mapping|mapping]], in one of two ways: +:$(1): \quad \exists x \in X: \neg \exists y \in T: \tuple {x, y} \in \mathcal R$ +:$(2): \quad \exists x \in S: \tuple {x, y_1} \in \mathcal R \land \tuple {x, y_2} \in \mathcal R: y_1 \ne y_2$ +Because $X$ is already the [[Definition:Preimage of Relation|preimage of $\mathcal R$]], the first of these cannot happen here. +For the second, it can be seen that neither $\set {y_1}$ nor $\set {y_2}$ can be in $\Rng {\map {\mathcal R^\to} {\powerset S} }$. +Therefore $\mathcal R^\to: \powerset S \to \powerset T$ can not be a [[Definition:Surjection|surjection]]. +{{questionable|See [[Talk:Surjection Induced by Powerset is Induced by Surjection|talk page]]}} +Thus, by the [[Rule of Transposition]], the result follows. +{{qed}} +[[Category:Surjections]] +[[Category:Direct Image Mappings]] +3go48kp22mh6x6rmyb0svhk154d6im9 +\end{proof}<|endoftext|> +\section{Mapping is Injection if its Direct Image Mapping is Injection} +Tags: Injections, Direct Image Mappings + +\begin{theorem} +Let $f: S \to T$ be a [[Definition:Mapping|mapping]]. +Let $f^\to: \powerset S \to \powerset T$ be the [[Definition:Direct Image Mapping of Mapping|direct image mapping]] of $f$. +Let $f^\to$ be an [[Definition:Injection|injection]]. +Then $f: S \to T$ is also an [[Definition:Injection|injection]]. +\end{theorem} + +\begin{proof} +Suppose $f: S \to T$ is a [[Definition:Mapping|mapping]], but not [[Definition:Injection|injective]]. +Then: +: $\exists x_1 \ne x_2 \in S: \map f {x_1} = \map f {x_2} = y$ +Let: +:$X_1 = \set {x_1}$ +:$X_2 = \set {x_2}$ +:$Y = \set y$ +Then it follows that: +:$\map {f^\to} {X_1} = \map {f^\to} {X_2} = Y$ +Thus $f^\to: \powerset S \to \powerset T$ is not [[Definition:Injection|injective]]. +So by the [[Rule of Transposition]], the result follows. +{{qed}} +[[Category:Injections]] +[[Category:Direct Image Mappings]] +fh8xao0jy03y7omejf9wlya8ginzydw +\end{proof}<|endoftext|> +\section{Direct Image Mapping is Bijection iff Mapping is Bijection} +Tags: Bijections, Direct Image Mappings + +\begin{theorem} +Let $\mathcal R \subseteq S \times T$ be a [[Definition:Relation|relation]]. +Let $\mathcal R^\to: \powerset S \to \powerset T$ be the [[Definition:Direct Image Mapping of Relation|direct image mapping]] of $\mathcal R$. +Then $\mathcal R \subseteq S \times T$ is a [[Definition:Bijection|bijection]] {{iff}} $\mathcal R^\to: \powerset S \to \powerset T$ is a [[Definition:Bijection|bijection]]. +\end{theorem} + +\begin{proof} +{{ProofWanted}} +[[Category:Bijections]] +[[Category:Direct Image Mappings]] +8yjejqsu9mr2iys6uy4s9omkw7o88na +\end{proof}<|endoftext|> +\section{No Bijection from Set to its Power Set} +Tags: Power Set, Bijections + +\begin{theorem} +Let $S$ be a [[Definition:Set|set]]. +Let $\powerset S$ denote the [[Definition:Power Set|power set]] of $S$. +There is no [[Definition:Bijection|bijection]] $f: S \to \powerset S$. +\end{theorem} + +\begin{proof} +A [[Definition:Bijection|bijection]] is by its definition also a [[Definition:Surjection|surjection]]. +By [[Cantor's Theorem]] there is no [[Definition:Surjection|surjection]] from $S$ to $\powerset S$. +Hence the result. +{{Qed}} +\end{proof}<|endoftext|> +\section{Set is Equivalent to Proper Subset of Power Set} +Tags: Power Set, Subsets + +\begin{theorem} +Every [[Definition:Set|set]] is [[Definition:Set Equivalence|equivalent]] to a [[Definition:Proper Subset|proper subset]] of its [[Definition:Power Set|power set]]: +: $\forall S: \exists T \subset \mathcal P \left({S}\right): S \sim T$ +\end{theorem} + +\begin{proof} +To show equivalence between two sets, we need to demonstrate that a [[Definition:Bijection|bijection]] exists between them. +We will now define such a bijection. +Let $T = \left\{{\left\{{x}\right\}: x \in S}\right\}$. +{{begin-eqn}} +{{eqn | ll= \forall x \in S: + | l = \left\{ {x}\right\} + | o = \subseteq + | r = S + | c = {{Defof|Subset}} +}} +{{eqn | ll= \implies + | l = \left\{ {x}\right\} + | o = \in + | r = \mathcal P \left({S}\right) + | c = {{Defof|Power Set}} +}} +{{eqn | ll= \implies + | l = T + | o = \subseteq + | r = \mathcal P \left({S}\right) + | c = {{Defof|Subset}} +}} +{{end-eqn}} +Now we define the mapping $\phi: S \to T$: +: $\phi: S \to T: \forall x \in S: \phi \left({x}\right) = \left\{{x}\right\}$ +$\phi$ is an [[Definition:Injection|injection]]: +: $\forall x, y \in S: \left\{{x}\right\} = \left\{{y}\right\} \implies x = y$ by definition of [[Definition:Set Equality|set equality]] +$\phi$ is a [[Definition:Surjection|surjection]]: +:$\forall \left\{{x}\right\} \in T: \exists x \in S: \phi \left({x}\right) = \left\{{x}\right\}$ +So $\phi$, being both an [[Definition:Injection|injection]] and a [[Definition:Surjection|surjection]], is a [[Definition:Bijection|bijection]] by definition. +To show that $T \subset \mathcal P \left({S}\right)$, that is, is a [[Definition:Proper Subset|proper subset]] of $\mathcal P \left({S}\right)$, we merely note that $\varnothing \in \mathcal P \left({S}\right)$ by [[Empty Set is Element of Power Set]], but $\varnothing \notin T$. +Thus $T \subseteq \mathcal P \left({S}\right)$ but $\mathcal P \left({S}\right) \not \subseteq T$. +Hence the result. +{{qed}} +[[Category:Power Set]] +[[Category:Subsets]] +phjkamlgupbj8jjllksmk1viqo6xv8z +\end{proof}<|endoftext|> +\section{Union of Power Sets} +Tags: Power Set, Set Union, Subsets + +\begin{theorem} +The [[Definition:Set Union|union]] of the [[Definition:Power Set|power sets]] of two [[Definition:Set|sets]] $S$ and $T$ is a [[Definition:Subset|subset]] of the [[Definition:Power Set|power set]] of their [[Definition:Set Union|union]]: +:$\powerset S \cup \powerset T \subseteq \powerset {S \cup T}$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = X + | o = \in + | r = \powerset S \cup \powerset T + | c = +}} +{{eqn | ll= \leadsto + | l = X + | o = \subseteq + | r = S \lor X \subseteq T + | c = {{Defof|Set Union}} and {{Defof|Power Set}} +}} +{{eqn | ll= \leadsto + | l = X + | o = \subseteq + | r = S \cup T + | c = {{Defof|Set Union}} +}} +{{eqn | ll= \leadsto + | l = X + | o = \in + | r = \powerset {S \cup T} + | c = {{Defof|Power Set}} +}} +{{end-eqn}} +{{Qed}} +\end{proof}<|endoftext|> +\section{Intersection of Power Sets} +Tags: Power Set, Set Intersection + +\begin{theorem} +The [[Definition:Set Intersection|intersection]] of the [[Definition:Power Set|power sets]] of two [[Definition:Set|sets]] $S$ and $T$ is equal to the [[Definition:Power Set|power set]] of their [[Definition:Set Intersection|intersection]]: +:$\powerset S \cap \powerset T = \powerset {S \cap T}$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = X + | o = \in + | r = \powerset {S \cap T} + | c = +}} +{{eqn | ll= \leadstoandfrom + | l = X + | o = \subseteq + | r = S \cap T + | c = {{Defof|Power Set}} +}} +{{eqn | ll= \leadstoandfrom + | l = X + | o = \subseteq + | r = S \land X \subseteq T + | c = {{Defof|Set Intersection}} +}} +{{eqn | ll= \leadstoandfrom + | l = X + | o = \in + | r = \powerset S \land X \in \powerset T + | c = {{Defof|Power Set}} +}} +{{eqn | ll= \leadstoandfrom + | l = X + | o = \in + | r = \powerset S \cap \powerset T + | c = {{Defof|Set Intersection}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Cantor-Bernstein-Schröder Theorem} +Tags: Set Theory, Injections, Bijections, Cardinals, Cantor-Bernstein-Schröder Theorem + +\begin{theorem} +If a [[Definition:Subset|subset]] of one [[Definition:Set|set]] is [[Definition:Set Equivalence|equivalent]] to the other, and a [[Definition:Subset|subset]] of the other is [[Definition:Set Equivalence|equivalent]] to the first, then the two [[Definition:Set|sets]] are themselves [[Definition:Set Equivalence|equivalent]]: +:$\forall S, T: T \sim S_1 \subseteq S \land S \sim T_1 \subseteq T \implies S \sim T$ +\end{theorem} + +\begin{proof} +Let $S, T$ be [[Definition:Set|sets]]. +Let $\powerset S, \powerset T$ denote their [[Definition:Power Set|power sets]]. +Let $f: S \to T$ and $g: T \to S$ be [[Definition:Injection|injections]] that we know to exist between $S$ and $T$. +Consider the [[Definition:Relative Complement|relative complements]] of elements of $\powerset S$ and $\powerset T$ as [[Definition:Mapping|mappings]]: +:$\complement_S: \powerset S \to \powerset S: \forall X \in \powerset S: \relcomp S X = S \setminus X$ +:$\complement_T: \powerset T \to \powerset T: \forall Y \in \powerset T: \relcomp T Y = T \setminus Y$ +which follow directly from the definition of [[Definition:Relative Complement|relative complement]]. +Let $\alpha$ and $\beta$ denote the [[Definition:Direct Image Mapping of Mapping|direct image mappings]] of $f$ and $g$, respectively. +Consider the [[Definition:Mapping|mapping]] $z: \powerset S \to \powerset S$ defined by the [[Definition:Composite Mapping|composition]]: +:$z = \complement_S \circ \beta \circ \complement_T \circ \alpha$ +=== $z$ is an increasing mapping === +Let $A, B \in \powerset S$ with $A \subseteq B$. +We have: +{{begin-eqn}} +{{eqn | l = \map \alpha A + | o = \subseteq + | r = \map \alpha B + | c = [[Image of Subset under Relation is Subset of Image/Corollary 2|Image of Subset under Relation is Subset of Image: Corollary 2]] +}} +{{eqn | ll= \leadsto + | l = \map {\paren {\complement_T \circ \alpha} } A + | o = \supseteq + | r = \map {\paren {\complement_T \circ \alpha} } B + | c = [[Set Complement inverts Subsets]] +}} +{{eqn | ll= \leadsto + | l = \map {\paren {\beta \circ \complement_T \circ \alpha} } A + | o = \supseteq + | r = \map {\paren {\beta \circ \complement_T \circ \alpha} } B + | c = [[Image of Subset under Relation is Subset of Image/Corollary 2|Image of Subset under Relation is Subset of Image: Corollary 2]] +}} +{{eqn | ll= \leadsto + | l = \map {\paren {\complement_S \circ \beta \circ \complement_T \circ \alpha} } A + | o = \subseteq + | r = \map {\paren {\complement_S \circ \beta \circ \complement_T \circ \alpha} } B + | c = [[Set Complement inverts Subsets]] +}} +{{eqn | ll= \leadsto + | l = \map z A + | o = \subseteq + | r = \map z B + | c = Definition of $z$ +}} +{{end-eqn}} +{{qed|lemma}} +By the [[Knaster-Tarski Lemma/Power Set|Knaster-Tarski Lemma: Power Set]], there is a $\mathbb G \in \powerset S$ such that: +:$\map z {\mathbb G} = \mathbb G$ +From [[Relative Complement of Relative Complement]] we have that $\complement_S \circ \complement_S$ is the [[Definition:Identity Mapping|identity mapping]] on $\powerset S$. +Thus we obtain: +{{begin-eqn}} +{{eqn | l = \relcomp S {\mathbb G} + | r = \map {\paren {\complement_S \circ z} } {\mathbb G} + | c = +}} +{{eqn | r = \map {\paren {\complement_S \circ \complement_S \circ \beta \circ \complement_T \circ \alpha} } {\mathbb G} + | c = Definition of $z$ +}} +{{eqn | r = g \sqbrk {\map {\paren {\complement_T \circ \alpha} } {\mathbb G} } + | c = as $\complement_S \circ \complement_S = I_{\powerset S}$ +}} +{{end-eqn}} +At this stage, a diagram can be helpful: +:[[File:Cantor-Bernstein-Schroeder.png|500px]] +Let $h: S \to T$ be the [[Definition:Mapping|mapping]] defined as: +:$\forall x \in S: \map h x = \begin {cases} +\map f x & : x \in \mathbb G \\ +\map {g^{-1} } x & : x \in \relcomp S {\mathbb G} +\end {cases}$ +From the above, we have that: +:$\relcomp S {\mathbb G} \subseteq g \sqbrk T$ +Therefore, as $g$ is an [[Definition:Injection|injection]], it follows that the [[Definition:Preimage of Element under Mapping|preimage]] $\map {g^{-1} } x$ is a [[Definition:Singleton|singleton]]. +So $h$ is a [[Definition:Bijection|bijection]] by dint of the [[Definition:Injection|injective]] nature of both $f$ and $g^{-1}$. +{{qed}} +\end{proof} + +\begin{proof} +From the facts that $T \sim S_1$ and $S \sim T_1$, we can set up the two [[Definition:Bijection|bijections]]: +:$f: S \to T_1$ +:$g: T \to S_1$ +Let: +:$S_2 = g \sqbrk {f \sqbrk S} = g \sqbrk {T_1} \subseteq S_1$ +and: +:$T_2 = f \sqbrk {g \sqbrk T} = f \sqbrk {S_1} \subseteq T_1$ +So $S_2 \subseteq S_1$ and $S_2 \sim S$, while $T_2 \subseteq T_1$, and $T_2 \sim T$. +For each [[Definition:Natural Number|natural number]] $k$, let $S_{k + 2} \subseteq S$ be the image of $S_k$ under the mapping $g \circ f$. +{{explain|Why do these superset relations hold?}} +Then: +:$S \supseteq S_1 \supseteq S_2 \supseteq \ldots \supseteq S_k \supseteq S_{k + 1} \ldots$ +Let $\displaystyle D = \bigcap_{k \mathop = 1}^\infty S_k$. +{{explain|Why can we represent it so?}} +Now we can represent $S$ as: +{{begin-eqn}} +{{eqn | n = 1 + | l = S + | r = \paren {S \setminus S_1} \cup \paren {S_1 \setminus S_2} \cup \ldots \cup \paren {S_k \setminus S_{k + 1} } \cup \ldots \cup D +}} +{{end-eqn}} +where $S \setminus S_1$ denotes [[Definition:Set Difference|set difference]]. +{{explain}} +Similarly, we can represent $S_1$ as: +{{begin-eqn}} +{{eqn | n = 2 + | l = S_1 + | r = \paren {S_1 \setminus S_2} \cup \paren {S_2 \setminus S_3} \cup \ldots \cup \paren {S_k \setminus S_{k + 1} } \cup \ldots \cup D +}} +{{end-eqn}} +Now let: +{{begin-eqn}} +{{eqn | l = M + | r = \paren {S_1 \setminus S_2} \cup \paren {S_3 \setminus S_4} \cup \paren {S_5 \setminus S_6} \cup \ldots + | c = +}} +{{eqn | l = N + | r = \paren {S \setminus S_1} \cup \paren {S_2 \setminus S_3} \cup \paren {S_4 \setminus S_5} \cup \ldots + | c = +}} +{{eqn | l = N_1 + | r = \paren {S_2 \setminus S_3} \cup \paren {S_4 \setminus S_5} \cup \paren {S_6 \setminus S_7} \cup \ldots + | c = +}} +{{end-eqn}} +and rewrite $(1)$ and $(2)$ as: +{{begin-eqn}} +{{eqn | n = 3 + | l = S + | r = D \cup M \cup N +}} +{{eqn | n = 4 + | l = S_1 + | r = D \cup M \cup N_1 +}} +{{end-eqn}} +Now: +{{begin-eqn}} +{{eqn | l = g \circ f \sqbrk {S \setminus S_1} = \paren {S_2 \setminus S_3} + | o = \leadsto + | r = \paren {S \setminus S_1} \sim \paren {S_2 \setminus S_3} + | c = +}} +{{eqn | l = g \circ f \sqbrk {S_2 \setminus S_3} = \paren {S_4 \setminus S_5} + | o = \leadsto + | r = \paren {S_2 \setminus S_3} \sim \paren {S_4 \setminus S_5} + | c = +}} +{{end-eqn}} +{{handwaving}} +and so on. +So $N \sim N_1$. +It follows from $(3)$ and $(4)$ that a [[Definition:Bijection|bijection]] can be set up between $S$ and $S_1$. +But $S_1 \sim T$. +Therefore $S \sim T$. +{{qed}} +\end{proof} + +\begin{proof} +Suppose $S \preccurlyeq T$ and $T \preccurlyeq S$. +By [[Definition:Dominate (Set Theory)|definition]], we have that there exist [[Definition:Injection|injections]] $f: S \to T$ and $g: T \to S$. +We are going to try to build a [[Definition:Sequence|sequence]] $t_1, s_1, t_2, s_2, t_3 \ldots$ as follows. +Consider any $t_1 \in T$. +By [[Law of Excluded Middle]] there are two choices: +{{begin-eqn}} +{{eqn | l = \exists s_1 \in S: \map f {s_1} + | r = t_1 + | c = +}} +{{eqn | l = \neg \exists s_1 \in S: \map f {s_1} + | r = t_1 + | c = +}} +{{end-eqn}} +Suppose $\exists s_1 \in S: \map f {s_1} = t_1$. +Because $f$ is [[Definition:Injection|injective]], such an $s_1$ is unique. +So we can choose $s_1 = \map{f^{-1} } {t_1}$. +Again, by [[Law of Excluded Middle]] there are two further choices: +{{begin-eqn}} +{{eqn | l = \exists t_2 \in T: \map g {t_2} + | r = s_1 + | c = +}} +{{eqn | l = \neg \exists t_2 \in T: \map g {t_2} + | r = s_1 + | c = +}} +{{end-eqn}} +Suppose $\exists t_2 \in T: \map g {t_2} = s_1$. +Because $f$ is [[Definition:Injection|injective]], such an $t_2$ is unique. +Similarly, we choose $s_2 = \map {f^{-1} } {t_2}$, if it exists. +This process goes on until one of the following happens: +:We reach some $s_n \in S$ such that $\neg \exists t \in T: \map g t = s_n$. This may be possible because $g$ may not be a [[Definition:Surjection|surjection]]. +:We reach some $t_n \in T$ such that $\neg \exists s \in S: \map f s = t_n$.This may be possible because $f$ may not be a [[Definition:Surjection|surjection]]. +:The process goes on for ever. +For each $t \in T$, then, there is a well-defined process which turns out in one of the above three ways. +We [[Definition:Partition (Set Theory)|partition]] $T$ up into three [[Definition:Subset|subsets]] that are [[Definition:Disjoint Sets|mutually disjoint]]: +: Let $T_A = \{$ all $t \in T$ such that the process ends with some $s_n\}$ +: Let $T_B = \{$ all $t \in T$ such that the process ends with some $t_n\}$ +: Let $T_C = \{$ all $t \in T$ such that the process goes on for ever$\}$. +We can do exactly the same thing with the elements of $S$: +: Let $S_A = \{$ all $s \in S$ such that the process ends with some $s_n\}$ +: Let $S_B = \{$ all $s \in S$ such that the process ends with some $t_n\}$ +: Let $S_C = \{$ all $s \in S$ such that the process goes on for ever$\}$. +What we need to do is show that $S \sim T$. +We do this by showing that $S_A \sim T_A$, $S_B \sim T_B$ and $S_C \sim T_C$. +* The [[Definition:Restriction of Mapping|restriction]] of $f$ to $S_A$ is a [[Definition:Bijection|bijection]] from $S_A$ to $T_A$. +To do this we need to show that: +# $s \in S_A \implies \map f s \in T_A$; +# $\forall t \in T_A: \exists s \in S_A: \map f s = t$. +Let $s \in S_A$. Then the process applied to $s$ ends in $S$. +Now consider the process applied to $\map f s$. Its first step leads us back to $s$. Then it continues the process, applied to $s$, and ends up in $S$. Thus $\map f s \in T_A$. +Thus $s \in S_A \implies \map f s \in T_A$. +Now suppose $t \in T_A$. Then the process applied to $t$ ends in $S$. +In particular, it must have a first stage, otherwise it would end in $T$ with $t$ itself. +Hence $t = \map f s$ for some $s$. +But the process applied to this $s$ is the same as the continuation of the process applied to $t$, and therefore it ends in $S$. +Thus $s \in S_A$ as required. +Hence the [[Definition:Restriction of Mapping|restriction]] of $f$ to $S_A$ is a [[Definition:Bijection|bijection]] from $S_A$ to $T_A$. +We can use the same argument to show that $g: T_B \to S_B$ is also a [[Definition:Bijection|bijection]]. Hence $g^{-1}: S_B \to T_B$ is a bijection. +Finally, suppose $t \in T_C$. +Because $f$ is an [[Definition:Injection|injection]], $t = \map f s$ for some $s$, and the process applied to $t$ must start. +And this $s$ must belong to $S_C$, because the process starting from $s$ is the same as the process starting from $t$ after the first step. This never ends, as $t \in T_C$. +Now we can define a [[Definition:Bijection|bijection]] $h: S \to T$ as follows: +:$\map h x = \begin{cases} +\map f s: x \in S_A \\ +\map f s: x \in S_C \\ +\map {g^{-1} } x: x \in S_B \end{cases}$ +The fact that $h$ is a bijection follows from the facts that: +:$(1): \quad S_A$, $S_B$ and $S_C$ are [[Definition:Disjoint Sets|mutually disjoint]] +:$(2): \quad T_A$, $T_B$ and $T_C$ are [[Definition:Disjoint Sets|mutually disjoint]] +:$(3): \quad f$, $f$ and $g^{-1}$ are the [[Definition:Bijection|bijections]] which respectively do the mappings between them. +{{Qed}} +\end{proof} + +\begin{proof} +From the facts that $T \sim S_1$ and $S \sim T_1$, we can set up the two [[Definition:Injection|injections]]: +:$f \sqbrk S = T_1 \subseteq T$ +:$g \sqbrk T = S_1 \subseteq S$ +Using $f$ and $g$, $S$ and $T$ are divided into [[Definition:Disjoint Sets|disjoint subsets]] such that there exists a [[Definition:Bijection|bijection]] between the [[Definition:Subset|subsets]] of $S$ and those of $T$, as follows. +Let $a \in S$. +Then we define the [[Definition:Element|elements]] of $S$: +:$\ldots, a_{-2 n}, \ldots, a_{-2}, a_0, a_2, \ldots, a_{2 n}, \ldots$ +and the [[Definition:Element|elements]] of $T$: +:$\ldots, a_{-2 n + 1}, \ldots, a_{-1}, a_1, \ldots, a_{2 n - 1}, \ldots$ +[[Principle of Recursive Definition|recursively]] as follows: +Let: +{{begin-eqn}} +{{eqn | l = a_0 + | r = a +}} +{{eqn | l = a_1 + | r = \map f {a_0} +}} +{{eqn | l = a_2 + | r = \map g {a_1} +}} +{{eqn | o = \vdots +}} +{{eqn | l = a_{2 n - 1} + | r = \map f {a_{2 n - 2} } +}} +{{eqn | l = a_{2 n} + | r = \map g {a_{2 n - 1} } +}} +{{eqn | o = \vdots +}} +{{end-eqn}} +This construction is valid for all $n \ge 1$, but note that some of the $a_n$'s may coincide with others. +We set up a similar construction for [[Definition:Negative Integer|negative integers]]: +{{begin-eqn}} +{{eqn | l = a_{-1} + | r = \map {g^{-1} } {a_0} + | c = if such an [[Definition:Element|element]] exists in $T$ +}} +{{eqn | l = a_{-2} + | r = \map {f^{-1} } {a_{-1} } + | c = if such an [[Definition:Element|element]] exists in $S$ +}} +{{eqn | o = \vdots +}} +{{eqn | l = a_{-2 n + 1} + | r = \map {g^{-1} } {a_{-2 n + 2} } + | c = if such an [[Definition:Element|element]] exists in $T$ +}} +{{eqn | l = a_{2 n} + | r = \map {f^{-1} } {a_{-2 n + 1} } + | c = if such an [[Definition:Element|element]] exists in $S$ +}} +{{eqn | o = \vdots +}} +{{end-eqn}} +As $f$ and $g$ are [[Definition:Injection|injections]], it follows that if $\map {f^{-1} } x$ and $\map {g^{-1} } y$ exist for any $x \in T$, $y \in S$, then those [[Definition:Element|elements]] are [[Definition:Unique|unique]]. +Let: +:the [[Definition:Element|elements]] of $S$ with an [[Definition:Even Integer|even]] index be denoted $\sqbrk a_S$ +and +:the [[Definition:Element|elements]] of $T$ with an [[Definition:Odd Integer|odd]] index be denoted $\sqbrk a_T$ +that is: +{{begin-eqn}} +{{eqn | l = \sqbrk a_S + | r = \set {\ldots, a_{-2 n}, \ldots, a_{-2}, a_0, a_2, \ldots, a_{2 n}, \ldots} \subseteq S +}} +{{eqn | l = \sqbrk a_T + | r = \set {\ldots, a_{-2 n + 1}, \ldots, a_{-1}, a_1, \ldots, a_{2 n - 1}, \ldots} \subseteq T +}} +{{end-eqn}} +We are given that $f$ and $g$ are [[Definition:Injection|injections]]. +So by the definition of $\sqbrk a_S$, for any two $a, b \in S$, $\sqbrk a_S$ and $\sqbrk b_S$ are either [[Definition:Disjoint Sets|disjoint]] or [[Definition:Set Equality|equal]]. +The same applies to $\sqbrk a_T$ and $\sqbrk b_T$ for any $a, b \in T$. +It follows that: +:$\AA_S = \set {\sqbrk a_S: a \in S}$ is a [[Definition:Partition (Set Theory)|partition]] of $S$ +and +:$\AA_T = \set {\sqbrk a_T: a \in S}$ is a [[Definition:Partition (Set Theory)|partition]] of $T$. +So: +:every [[Definition:Element|element]] of $S$ belongs to exactly one [[Definition:Element|element]] of $\AA_S$ +and: +:every [[Definition:Element|element]] of $T$ belongs to exactly one [[Definition:Element|element]] of $\AA_T$. +So let $a \in S$ and $b \in T$ such that $\map f a = b$. +It follows that a [[Definition:Bijection|bijection]] can be constructed from $\AA_S$ to $\AA_T$. +Now there are two different kinds of $\sqbrk a_S$ sets: +:$(1):$ It is possible that no repetition occurs in the [[Definition:Sequence|sequence]] $\sequence {a_{2 n} }$. +As a consequence, no repetition occurs in the [[Definition:Sequence|sequence]] $\sequence {a_{2 n - 1} }$ either. +By the method of construction it can be seen that $\sqbrk a_S$ and $\sqbrk a_T$ are both [[Definition:Countable|countably infinite]]. +So a [[Definition:Bijection|bijection]] can be constructed between $\sqbrk a_S$ and $\sqbrk a_T$. +:$(2):$ There may be a repetition in $\sqbrk a_S$. +Suppose such a repetition is $a_{2 m} = a_{2 n}$ for some $m \ne n$. +Then $a_{2 m + 1} = a_{2 n + 1}$ and $a_{2 m + 2} = a_{2 n + 2}$ and so on. +In general $a_{2 m + k} = a_{2 n + k}$ for all $k \in \N$. +But because $f$ and $g$ are [[Definition:Injection|injections]] it follows that $a_{2 m - 1} = a_{2 n - 1}$ and $a_{2 m - 2} = a_{2 n - 2}$ and so on, where they exist. +So in general $a_{2 m - k} = a_{2 n - k}$ for all $k \in \N$, where they exist. +Given $m$, we can choose $n$ so that the [[Definition:Element|elements]] $a_{2 m}, a_{2 m + 2}, \ldots, a_{2 n - 2}$ are [[Definition:Distinct Elements|distinct]]. +Then $a_{2 m + 1}, a_{2 m + 3}, \ldots, a_{2 n - 1}$ are likewise distinct [[Definition:Element|elements]] of $T$. +Thus we can set up a [[Definition:Bijection|bijection]]: +:$a_{2 m} \mapsto a_{2 m + 1}, a_{2 m + 2} \mapsto a_{2 m + 3}, \ldots, a_{2 n - 2} \mapsto a_{2 n - 1}$ +between the [[Definition:Element|elements]] of $\sqbrk a_S$ and $\sqbrk a_T$. +It follows that a [[Definition:Bijection|bijection]] can be constructed between any two [[Definition:Element|elements]] of the partitions of $S$ and $T$. +These maps then automatically yield a [[Definition:Bijection|bijection]] from $S$ to $T$. +Hence the result. +{{qed}} +\end{proof} + +\begin{proof} +By [[Injection to Image is Bijection]]: +:$g: T \to g \sqbrk T$ is a [[Definition:Bijection|bijection]]. +Thus $T$ is [[Definition:Set Equivalence|equivalent]] to $g \sqbrk T$. +By [[Composite of Injections is Injection]] $g \circ f: S \to g \sqbrk T \subset S$ is also an [[Definition:Injection|injection]] (to a [[Definition:Subset|subset]] of the [[Definition:Domain of Mapping|domain]] of $g \circ f$). +Then by [[Cantor-Bernstein-Schröder Theorem/Lemma|Cantor-Bernstein-Schröder Theorem: Lemma]]: +:There exists a [[Definition:Bijection|bijection]] $h: S \to g \sqbrk T$. +Thus $S$ is [[Definition:Set Equivalence|equivalent]] to $g \sqbrk T$. +We already know that $T$ is equivalent to $g \sqbrk T$. +Thus by [[Set Equivalence is Equivalence Relation]], $S$ is [[Definition:Set Equivalence|equivalent]] to $T$. +By the definition of [[Definition:Set Equivalence|set equivalence]]: +:There is a [[Definition:Bijection|bijection]] $\phi: S \to T$. +{{qed}} +\end{proof} + +\begin{proof} +Let $\mathcal P \left({A}\right)$ be the [[Definition:Power Set|power set]] of $A$. +Define a mapping $E: \mathcal P \left({A}\right) \to \mathcal P \left({A}\right)$ thus: +: $E \left({S}\right) = A \setminus g \left({B \setminus f \left({S}\right)}\right)$ +=== $E$ is increasing === +Let $S, T \in \mathcal P \left({A}\right)$ such that $S \subseteq T$. +Then: +{{begin-eqn}} +{{eqn | l = f \left({S}\right) + | o = \subseteq + | r = f \left({T}\right) + | c = [[Image of Subset is Subset of Image]] +}} +{{eqn | ll= \implies + | l = B \setminus f \left({T}\right) + | o = \subseteq + | r = B \setminus f \left({S}\right) + | c = [[Set Difference with Subset is Superset of Set Difference]] +}} +{{eqn | ll= \implies + | l = g \left({B \setminus f \left({T}\right)}\right) + | o = \subseteq + | r = g \left({B \setminus f \left({S}\right)}\right) + | c = [[Image of Subset is Subset of Image]] +}} +{{eqn | ll= \implies + | l = A \setminus g \left({B \setminus f \left({S}\right)}\right) + | o = \subseteq + | r = A \setminus g \left({B \setminus f \left({T}\right)}\right) + | c = [[Set Difference with Subset is Superset of Set Difference]] +}} +{{end-eqn}} +That is, $E \left({S}\right) \subseteq E \left({T}\right)$. +{{qed|lemma}} +By the [[Knaster-Tarski Lemma/Power Set|Knaster-Tarski Lemma]], $E$ has a [[Definition:Fixed Point|fixed point]] $X$. +By the definition of [[Definition:Fixed Point|fixed point]]: +: $E \left({X}\right) = X$ +Thus by the definition of $E$: +: $A \setminus g \left({B \setminus f \left({X}\right)}\right) = X$ +Therefore: +: $(1): \quad A \setminus \left({A \setminus g \left({B \setminus f \left({X}\right)}\right)}\right) = A \setminus X$ +Since $g$ is a [[Definition:Mapping|mapping]] into $A$: +: $g \left({B \setminus f \left({X}\right)}\right) \subseteq A$ +Thus by [[Relative Complement of Relative Complement]]: +: $A \setminus \left({A \setminus g \left({B \setminus f \left({X}\right)}\right)}\right) = g \left({B \setminus f \left({X}\right)}\right)$ +Thus by $(1)$: +:$g \left({B \setminus f \left({X}\right)}\right) = A \setminus X$ +Let $f' = f \restriction_{X \times f \left({X}\right)}$ be the [[Definition:Restriction of Mapping|restriction]] of $f$ to $X \times f \left({X}\right)$. +Similarly, let $g' = g \restriction_{\left({B \setminus f \left({X}\right)}\right) \times \left({A \setminus X}\right)} = g \restriction_{\left({B \setminus f \left({X}\right)}\right) \times g \left({B \setminus f \left({X}\right)}\right)}$. +By [[Injection to Image is Bijection]], $f'$ and $g'$ are both [[Definition:Bijection|bijections]]. +Define a [[Definition:Relation|relation]] $h: A \to B$ by $h = f' \cup {g'}^{-1}$. +We will show that $h$ is a [[Definition:Bijection|bijection]] from $A$ onto $B$. +The [[Definition:Domain of Mapping|domain]] of $f'$ is $X$, which is [[Definition:Disjoint Sets|disjoint]] from the [[Definition:Codomain of Mapping|codomain]], $A \setminus X$, of $g'$. +The [[Definition:Domain of Mapping|domain]] of $g'$ is $B \setminus f \left({X}\right)$, which is [[Definition:Disjoint Sets|disjoint]] from the [[Definition:Codomain of Mapping|codomain]], $f \left({X}\right)$, of $f'$. +Let $h = f' \cup {g'}^{-1}$. +By the [[Union of Bijections with Disjoint Domains and Codomains is Bijection/Corollary|corollary to Union of Bijections with Disjoint Domains and Codomains is Bijection]]: +: $h$ is a [[Definition:Bijection|bijection]] from $X \cup \left({A \setminus X}\right)$ onto $f \left({X}\right) \cup \left({B \setminus f \left({X}\right)}\right)$. +By [[Union with Relative Complement]], $h$ is a [[Definition:Bijection|bijection]] from $A$ onto $B$. +Since $f' \subseteq f$ and $g' \subseteq g$, each element of $h$ is an element of $f$ or of $g^{-1}$. +That is, if $y = h \left({x}\right)$ then either $y = f \left({x}\right)$ or $x = g \left({y}\right)$. +{{Qed}} +\end{proof}<|endoftext|> +\section{Trivial Ordering is Universally Compatible} +Tags: Order Theory + +\begin{theorem} +Let $S$ be a [[Definition:Set|set]]. +Let $\RR$ be the [[Definition:Trivial Ordering|trivial ordering]] on $S$. +Then $\RR$ is [[Definition:Universally Compatible Relation|universally compatible]]. +\end{theorem} + +\begin{proof} +To prove that the trivial ordering is in fact an ordering, we need to checking each of the criteria for an [[Definition:Ordering|ordering]]: +=== Reflexivity === +:$\forall a \in S: a \mathrel \RR a$: +From its definition, we have $\forall a, b \in S: a = b \implies a \mathrel \RR b$. +Thus, as $a = a$, we have $\forall a \in S: a \mathrel \RR a$. +So [[Definition:Reflexive Relation|reflexivity]] is proved. +=== Transitivity === +:$\forall a, b, c \in S: a \mathrel \RR b \land b \mathrel \RR c \implies a \mathrel \RR c$: +From the definition: +:$a \mathrel \RR b \iff a = b$ +:$b \mathrel \RR c \iff b = c$ +So as $a = b \land b = c \implies a = c$ from [[Equality is Equivalence Relation|transitivity of equals]], we have that $a \mathrel \RR c$ and thus [[Definition:Transitive Relation|transitivity]] is proved. +=== Antisymmetry === +:$\forall a, b \in S: a \mathrel \RR b \land b \mathrel \RR a \implies a = b$: +From the definition: +:$a \mathrel \RR b \iff a = b$ +:$b \mathrel \RR a \iff b = a$ +[[Definition:Antisymmetric Relation|Antisymmetry]] follows from [[Equality is Equivalence Relation|symmetry of equals]]. +The [[Definition:Trivial Ordering|trivial ordering]] is by definition the same as the [[Definition:Diagonal Relation|diagonal relation]], and is therefore [[Diagonal Relation is Universally Compatible|universally compatible]]. +{{qed}} +[[Category:Order Theory]] +fabrlm275a3wq81prjzwuimdfgrewj4 +\end{proof}<|endoftext|> +\section{Identity Mapping is Order Isomorphism} +Tags: Order Isomorphisms, Identity Mappings, Identity Mapping is Order Isomorphism + +\begin{theorem} +Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]]. +The [[Definition:Identity Mapping|identity mapping]] $I_S$ is an [[Definition:Order Isomorphism|order isomorphism]] from $\left({S, \preceq}\right)$ to itself. +\end{theorem} + +\begin{proof} +By definition: +:$\forall x \in S: I_S \left({x}\right) = x$ +So: +:$x \preceq y \implies I_S \left({x}\right) \preceq I_S \left({y}\right)$ +As $I_S$ is a [[Identity Mapping is Bijection|bijection]], we also have: +:$I_S^{-1} \left({x}\right) = x$ +So: +:$x \preceq y \implies I_S^{-1} \left({x}\right) \preceq I_S^{-1} \left({y}\right)$ +{{qed}} +\end{proof} + +\begin{proof} +An [[Definition:Ordered Set|ordered set]] is a [[Definition:Relational Structure|relational structure]] where [[Definition:Order Isomorphism|order isomorphism]] is a special case of [[Definition:Relation Isomorphism|relation isomorphism]]. +The result follows directly from [[Identity Mapping is Relation Isomorphism]]. +{{Qed}} +\end{proof}<|endoftext|> +\section{Inverse of Order Isomorphism is Order Isomorphism} +Tags: Order Isomorphisms + +\begin{theorem} +Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be [[Definition:Ordered Set|ordered sets]]. +Let $\phi$ be a [[Definition:Bijection|bijection]] from $\left({S, \preceq_1}\right)$ to $\left({T, \preceq_2}\right)$. +Then: +: $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ +is an [[Definition:Order Isomorphism|order isomorphism]] {{iff}}: +: $\phi^{-1}: \left({T, \preceq_2}\right) \to \left({S, \preceq_1}\right)$ +is also an [[Definition:Order Isomorphism|order isomorphism]]. +\end{theorem} + +\begin{proof} +Follows directly from the definition of [[Definition:Order Isomorphism|order isomorphism]]. +Let $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ be an [[Definition:Order Isomorphism|order isomorphism]]. +Then: +:$\forall x, y, \in S: x \preceq_1 y \implies \phi \left({x}\right) \preceq_2 \phi \left({y}\right)$ +and: +:$\forall \phi \left({x}\right), \phi \left({y}\right) \in T: \phi \left({x}\right) \preceq_2 \phi \left({y}\right) \implies \phi^{-1} \left({\phi \left({x}\right)}\right) \preceq_1 \phi^{-1} \left({\phi \left({y}\right)}\right)$ +That is: +:$\forall \phi \left({x}\right), \phi \left({y}\right) \in T: \phi \left({x}\right) \preceq_2 \phi \left({y}\right) \implies x \preceq_1 y$ +Hence the result. +{{Qed}} +\end{proof}<|endoftext|> +\section{Composite of Order Isomorphisms is Order Isomorphism} +Tags: Order Isomorphisms + +\begin{theorem} +Let $\left({S_1, \preceq_1}\right)$, $\left({S_2, \preceq_2}\right)$ and $\left({S_3, \preceq_3}\right)$ be [[Definition:Ordered Set|ordered sets]]. +Let: +: $\phi: \left({S_1, \preceq_1}\right) \to \left({S_2, \preceq_2}\right)$ +and: +: $\psi: \left({S_2, \preceq_2}\right) \to \left({S_3, \preceq_3}\right)$ +be [[Definition:Order Isomorphism|order isomorphisms]]. +Then $\psi \circ \phi: \left({S_1, \preceq_1}\right) \to \left({S_3, \preceq_3}\right)$ is also an [[Definition:Order Isomorphism|order isomorphism]]. +\end{theorem} + +\begin{proof} +From [[Composite of Bijections is Bijection]], $\psi \circ \phi$ is a [[Definition:Bijection|bijection]], as, by definition, an [[Definition:Order Isomorphism|order isomorphism]] is also a [[Definition:Bijection|bijection]]. +By definition of [[Definition:Composition of Mappings|composition of mappings]], $\psi \circ \phi \left({x}\right) = \psi \left({\phi \left({x}\right)}\right)$. +As $\phi$ is an [[Definition:Order Isomorphism|order isomorphism]], we have: +:$\forall x_1, y_1 \in S_1: x_1 \preceq_1 y_1 \implies \phi \left({x_1}\right) \preceq_2 \phi \left({y_1}\right)$ +As $\psi$ is an [[Definition:Order Isomorphism|order isomorphism]], we have: +:$\forall x_2, y_2 \in S_2: x_2 \preceq_2 y_2 \implies \psi \left({x_2}\right) \preceq_3 \psi \left({y_2}\right)$ +By setting $x_2 = \phi \left({x_1}\right), y_2 = \phi \left({y_1}\right)$, it follows that: +:$\forall x_1, y_1 \in S_1: x_1 \preceq_1 y_1 \implies \psi \left({\phi \left({x_1}\right)}\right) \preceq_3 \psi \left({\phi \left({y_1}\right)}\right)$ +Similarly we can show that: +:$\forall x_3, y_3 \in S_3: x_3 \preceq_3 y_3 \implies \phi^{-1} \left({\psi^{-1} \left({x_3}\right)}\right) \preceq_1 \phi^{-1} \left({\psi^{-1} \left({y_3}\right)}\right)$ +{{qed}} +\end{proof}<|endoftext|> +\section{Order Isomorphism is Equivalence Relation} +Tags: Order Isomorphisms, Equivalence Relations, Order Isomorphism is Equivalence Relation + +\begin{theorem} +[[Definition:Order Isomorphism|Order isomorphism]] between [[Definition:Ordered Set|ordered sets]] is an [[Definition:Equivalence Relation|equivalence relation]]. +So any given family of [[Definition:Ordered Set|ordered sets]] can be [[Definition:Partition (Set Theory)|partitioned]] into [[Definition:Disjoint Sets|disjoint]] [[Definition:Equivalence Class|classes]] of isomorphic [[Definition:Set|sets]]. +\end{theorem} + +\begin{proof} +{{Finish|Needs filling out.}} +* [[Definition:Reflexive Relation|Reflexive]]: +Follows directly from [[Identity Mapping is Order Isomorphism]]. +* [[Definition:Symmetric Relation|Symmetric]]: +Follows directly from [[Inverse of Order Isomorphism is Order Isomorphism]]. +* [[Definition:Transitive Relation|Transitive]]: +Follows directly from [[Composite of Order Isomorphisms is Order Isomorphism]]. +{{qed}} +\end{proof} + +\begin{proof} +An [[Definition:Ordered Set|ordered set]] is a [[Definition:Relational Structure|relational structure]] where [[Definition:Order Isomorphism|order isomorphism]] is a special case of [[Definition:Relation Isomorphism|relation isomorphism]]. +The result follows directly from [[Relation Isomorphism is Equivalence Relation]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Dual Ordering is Ordering} +Tags: Order Theory + +\begin{theorem} +Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]]. +Let $\succeq$ denote the [[Definition:Dual Ordering|dual ordering]] of $\preceq$. +Then $\succeq$ is an [[Definition:Ordering|ordering]] on $S$. +\end{theorem} + +\begin{proof} +By definition, $\succeq$ is the [[Definition:Inverse Relation|inverse relation]] to $\preceq$. +By [[Inverse Relation Properties]], if a [[Definition:Relation|relation]] is [[Definition:Reflexive Relation|reflexive]], [[Definition:Transitive Relation|transitive]] and/or [[Definition:Antisymmetric Relation|antisymmetric]], then so is its [[Definition:Inverse Relation|inverse]]. +The result follows. +{{qed}} +\end{proof}<|endoftext|> +\section{Power Set is Complete Lattice} +Tags: Lattice Theory, Power Set + +\begin{theorem} +Let $S$ be a [[Definition:Set|set]]. +Let $\left({\mathcal P \left({S}\right), \subseteq}\right)$ be the [[Definition:Relational Structure|relational structure]] defined on $\mathcal P \left({S}\right)$ by the [[Definition:Relation|relation]] $\subseteq$. +Then $\left({\mathcal P \left({S}\right), \subseteq}\right)$ is a [[Definition:Complete Lattice|complete lattice]]. +\end{theorem} + +\begin{proof} +From [[Subset Relation on Power Set is Partial Ordering]], we have that $\subseteq$ is a [[Definition:Partial Ordering|partial ordering]]. +We note in passing that for any set $S$: +: From [[Supremum of Power Set]], $\mathcal P \left({S}\right)$ has a [[Definition:Supremum of Set|supremum]], that is, $S$ itself +: From [[Infimum of Power Set]], $\mathcal P \left({S}\right)$ has an [[Definition:Infimum of Set|infimum]], that is, $\varnothing$. +These are also the [[Definition:Maximal Element|maximal]] and [[Definition:Minimal Element|minimal]] elements of $\mathcal P \left({S}\right)$. +Let $\mathbb S$ be a [[Definition:Subset|subset]] of $\mathcal P \left({S}\right)$. +Then from [[Union is Smallest Superset]]: +:$\left({\forall X \in \mathbb S: X \subseteq T}\right) \iff \bigcup \mathbb S \subseteq T$ +and from [[Intersection is Largest Subset]]: +:$\left({\forall X \in \mathbb S: T \subseteq X}\right) \iff T \subseteq \bigcap \mathbb S$ +So $\bigcap \mathbb S$ is the [[Definition:Infimum of Set|infimum]] and $\bigcup \mathbb S$ is the [[Definition:Supremum of Set|supremum]] of $\left({\mathbb S, \subseteq}\right)$. +Hence by definition $\mathcal P \left({S}\right)$ is a [[Definition:Complete Lattice|complete lattice]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Totally Ordered Set is Lattice} +Tags: Total Orderings, Lattice Theory + +\begin{theorem} +Every [[Definition:Totally Ordered Set|totally ordered set]] is a [[Definition:Lattice|lattice]]. +\end{theorem} + +\begin{proof} +Let $\left({S, \preceq}\right)$ be a [[Definition:Totally Ordered Set|totally ordered set]]. +Then we have: +: $\forall x, y \in S: x \preceq y \lor y \preceq x$ +: $\forall x, y \in S: x \preceq y \implies \sup \left\{{x, y}\right\} = y \land \inf \left\{{x, y}\right\} = x$ +: $\forall x, y \in S: y \preceq x \implies \sup \left\{{x, y}\right\} = x \land \inf \left\{{x, y}\right\} = y$ +Thus the conditions for $\left({S, \preceq}\right)$ to be a [[Definition:Lattice|lattice]] are fulfilled. +{{qed}} +\end{proof}<|endoftext|> +\section{Trichotomy Law (Ordering)} +Tags: Trichotomies, Trichotomy Law, Total Orderings, Named Theorems + +\begin{theorem} +Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]]. +Then $\preceq$ is a [[Definition:Total Ordering|total ordering]] {{iff}}: +:$\forall a, b \in S: \paren {a \prec b} \lor \paren {a = b} \lor \paren {a \succ b}$ +That is, every element either [[Definition:Strictly Precede|strictly precedes]], is [[Definition:Equals|the same as]], or [[Definition:Strictly Succeed|strictly succeeds]], every other element. +In other words, {{iff}} $\prec$ is a [[Definition:Trichotomy|trichotomy]]. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | o = + | r = \forall a, b \in S: a \preceq b \lor b \preceq a + | c = {{Defof|Total Ordering}} +}} +{{eqn | o = \leadstoandfrom + | r = \forall a, b \in S: a \preceq b \lor a \succeq b + | c = {{Defof|Dual Ordering}} +}} +{{eqn | o = \leadstoandfrom + | r = \forall a, b \in S: \paren {a = b \lor a \prec b} \lor \paren {a = b \lor a \succ b} + | c = [[Strictly Precedes is Strict Ordering]] +}} +{{eqn | o = \leadstoandfrom + | r = \forall a, b \in S: a \prec b \lor a = b \lor a \succ b + | c = Rules of [[Rule of Commutation|Commutation]], [[Rule of Association|Association]] and [[Rule of Idempotence|Idempotence]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Complement of Reflexive Relation} +Tags: Reflexive Relations, Relative Complement + +\begin{theorem} +Let $\RR \subseteq S \times S$ be a [[Definition:Relation|relation]]. +Then $\RR$ is [[Definition:Reflexive Relation|reflexive]] {{iff}} its [[Definition:Complement of Relation|complement]] $\relcomp {S \times S} \RR \subseteq S \times S$ is [[Definition:Antireflexive Relation|antireflexive]]. +Likewise, $\RR$ is [[Definition:Antireflexive Relation|antireflexive]] {{iff}} its [[Definition:Complement of Relation|complement]] $\relcomp {S \times S} \RR \subseteq S \times S$ is [[Definition:Reflexive Relation|reflexive]]. +\end{theorem} + +\begin{proof} +Let $\RR \subseteq S \times T$ be [[Definition:Reflexive Relation|reflexive]]. +Then: +:$\forall x \in S: \tuple {x, x} \in \RR$ +By the definition of [[Definition:Complement of Relation|complement]] of $\RR$: +:$\tuple {x, y} \in \RR \implies \tuple {x, y} \notin \relcomp {S \times S} \RR$ +The same applies to $\tuple {x, x}$, and thus: +:$\forall x \in S: \tuple {x, x} \notin \relcomp {S \times S} \RR$ +Thus $\relcomp {S \times S} \RR$ is [[Definition:Antireflexive Relation|antireflexive]]. +Similarly, by definition: +:$\forall x \in S: \tuple {x, x} \notin \RR \implies \neg \tuple {x, x} \notin \relcomp {S \times S} \RR$ +By [[Double Negation]] it follows that: +:$\tuple {x, x} \in \relcomp {S \times S} \RR$ +The [[Definition:Converse Statement|converses]] of these results follow from the fact that: +:$\relcomp {S \times S} {\relcomp {S \times S} \RR} = \RR$ +by [[Relative Complement of Relative Complement]]. +{{Qed}} +[[Category:Reflexive Relations]] +[[Category:Relative Complement]] +iqnuxo8bs5jeattlg5gkrvi05m9rgsj +\end{proof}<|endoftext|> +\section{Complement of Symmetric Relation} +Tags: Symmetric Relations, Relative Complement + +\begin{theorem} +Let $\mathcal R \subseteq S \times S$ be a [[Definition:Relation|relation]]. +Then $\mathcal R$ is [[Definition:Symmetric Relation|symmetric]] {{iff}} its [[Definition:Complement of Relation|complement]] $\relcomp {S \times S} {\mathcal R} \subseteq S \times S$ is also [[Definition:Symmetric Relation|symmetric]]. +\end{theorem} + +\begin{proof} +Let $\mathcal R \subseteq S \times S$ be [[Definition:Symmetric Relation|symmetric]]. +Then from [[Symmetry of Relations is Symmetric]]: +:$\tuple {x, y} \in \mathcal R \iff \tuple {y, x} \in \mathcal R$ +{{AimForCont}} $\relcomp {S \times S} {\mathcal R} \subseteq S \times S$ is not [[Definition:Symmetric Relation|symmetric]]. +Then: +:$\exists \tuple {x, y} \in \relcomp {S \times S} {\mathcal R}: \tuple {y, x} \notin \relcomp {S \times S} {\mathcal R}$ +But then by definition of [[Definition:Complement of Relation|complement]] of $\mathcal R$: +:$\tuple {y, x} \in \mathcal R$ +As $\mathcal R$ is [[Definition:Symmetric Relation|symmetric]] it follows that: +:$\tuple {x, y} \in \mathcal R$ +This [[Definition:Contradiction|contradicts]] the premise that $\tuple {x, y} \in \relcomp {S \times S} {\mathcal R}$. +Hence by [[Proof by Contradiction]] it follows that $\relcomp {S \times S} {\mathcal R}$ is [[Definition:Symmetric Relation|symmetric]]. +{{qed}} +The converse follows similarly. +[[Category:Symmetric Relations]] +[[Category:Relative Complement]] +og8lpnl3stud43136bscyl1muu27udf +\end{proof}<|endoftext|> +\section{Strictly Increasing Mapping is Increasing} +Tags: Increasing Mappings + +\begin{theorem} +A [[Definition:Mapping|mapping]] that is [[Definition:Strictly Increasing Mapping|strictly increasing]] is an [[Definition:Increasing Mapping|increasing]] mapping. +\end{theorem} + +\begin{proof} +Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be [[Definition:Ordered Set|ordered sets]]. +Let $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ be [[Definition:Strictly Increasing Mapping|strictly increasing]]. +From [[Strictly Precedes is Strict Ordering]]: +:$x \preceq_1 y \implies x = y \lor x \prec_1 y$ +So: +{{begin-eqn}} +{{eqn | l = x + | r = y + | c = +}} +{{eqn | ll= \implies + | l = \phi \left({x}\right) + | r = \phi \left({y}\right) + | c = Definition of [[Definition:Mapping|Mapping]] +}} +{{eqn | ll= \implies + | l = \phi \left({x}\right) + | o = \preceq_2 + | r = \phi \left({y}\right) + | c = as $\preceq_2$, being an [[Definition:Ordering|ordering]], is [[Definition:Reflexive Relation|reflexive]] +}} +{{end-eqn}} +This leaves us with: +{{begin-eqn}} +{{eqn | l = x + | o = \prec_1 + | r = y + | c = +}} +{{eqn | ll= \implies + | l = \phi \left({x}\right) + | o = \prec_2 + | r = \phi \left({y}\right) + | c = Definition of [[Definition:Strictly Increasing Mapping|Strictly Increasing]] +}} +{{eqn | ll= \implies + | l = \phi \left({x}\right) + | o = \preceq_2 + | r = \phi \left({y}\right) + | c = [[Strictly Precedes is Strict Ordering]] +}} +{{end-eqn}} +{{Qed}} +\end{proof}<|endoftext|> +\section{Strictly Decreasing Mapping is Decreasing} +Tags: Decreasing Mappings + +\begin{theorem} +A [[Definition:Mapping|mapping]] that is [[Definition:Strictly Decreasing Mapping|strictly decreasing]] is a [[Definition:Decreasing Mapping|decreasing]] mapping. +\end{theorem} + +\begin{proof} +Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be [[Definition:Ordered Set|ordered sets]]. +Let $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ be [[Definition:Strictly Decreasing Mapping|strictly decreasing]]. +From [[Strictly Precedes is Strict Ordering]]: +:$x \preceq_1 y \iff x = y \lor x \prec_1 y$ +So: +{{begin-eqn}} +{{eqn | l = x + | r = y + | c = +}} +{{eqn | ll= \implies + | l = \phi \left({y}\right) + | r = \phi \left({x}\right) + | c = Definition of [[Definition:Mapping|mapping]] +}} +{{eqn | ll= \implies + | l = \phi \left({y}\right) + | o = \preceq_2 + | r = \phi \left({x}\right) + | c = as $\preceq_2$, being an [[Definition:Ordering|ordering]], is [[Definition:Reflexive Relation|reflexive]] +}} +{{end-eqn}} +This leaves us with: +{{begin-eqn}} +{{eqn | l = x + | o = \prec_1 + | r = y + | c = +}} +{{eqn | ll= \implies + | l = \phi \left({y}\right) + | o = \prec_2 + | r = \phi \left({x}\right) + | c = Definition of [[Definition:Strictly Decreasing Mapping|Strictly Decreasing]] +}} +{{eqn | ll= \implies + | l = \phi \left({y}\right) + | o = \preceq_2 + | r = \phi \left({x}\right) + | c = [[Strictly Precedes is Strict Ordering]] +}} +{{end-eqn}} +{{Qed}} +\end{proof}<|endoftext|> +\section{Strictly Monotone Mapping is Monotone} +Tags: Monotone Mappings, Order Theory, Mapping Theory + +\begin{theorem} +A [[Definition:Mapping|mapping]] that is [[Definition:Strictly Monotone Mapping|strictly monotone]] is a [[Definition:Monotone Mapping|monotone]] mapping. +\end{theorem} + +\begin{proof} +If $\phi$ is strictly monotone, then it is either [[Definition:Strictly Increasing Mapping|strictly increasing]] or [[Definition:Strictly Decreasing Mapping|strictly decreasing]]. +If $\phi$ is strictly increasing, then by [[Strictly Increasing Mapping is Increasing]], $\phi$ is [[Definition:Increasing Mapping|increasing]]. +If $\phi$ is strictly decreasing, then by [[Strictly Decreasing Mapping is Decreasing]], $\phi$ is [[Definition:Decreasing Mapping|decreasing]]. +Thus $\phi$ is [[Definition:Monotone Mapping|monotone]] by definition. +{{qed}} +\end{proof}<|endoftext|> +\section{Order Embedding into Image is Isomorphism} +Tags: Order Embeddings, Order Isomorphisms + +\begin{theorem} +Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be [[Definition:Ordered Set|ordered sets]]. +Let $S'$ be the [[Definition:Image of Mapping|image]] of a [[Definition:Mapping|mapping]] $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$. +Then: +:$\phi$ is an [[Definition:Order Embedding|order embedding]] from $\struct {S, \preceq_1}$ into $\struct {T, \preceq_2}$ +{{iff}}: +:$\phi$ is an [[Definition:Order Isomorphism|order isomorphism]] from $\struct {S, \preceq_1}$ into $\struct {S', \preceq_2 \restriction_{S'} }$. +\end{theorem} + +\begin{proof} +Let $\phi$ be an [[Definition:Order Embedding|order embedding]] from $\struct {S, \preceq_1}$ into $\struct {T, \preceq_2}$. +Then $\phi$ is an [[Definition:Injection|injection]] into $\struct {T, \preceq_2}$ by definition. +From [[Restriction of Mapping to Image is Surjection]], a [[Definition:Mapping|mapping]] from a [[Definition:Set|set]] to the [[Definition:Image of Mapping|image]] of that [[Definition:Mapping|mapping]] is a [[Definition:Surjection|surjection]]. +Thus the [[Definition:Surjective Restriction|surjective restriction]] of $\phi$ onto $S'$ is an [[Definition:Order Embedding|order embedding]] which is also a [[Definition:Surjection|surjection]]. +Hence the result from [[Order Isomorphism is Surjective Order Embedding]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Strictly Monotone Mapping with Totally Ordered Domain is Injective} +Tags: Order Theory, Injections, Monotone Mappings + +\begin{theorem} +Let $\struct {S, \preceq_1}$ be a [[Definition:Totally Ordered Set|totally ordered set]]. +Let $\struct {T, \preceq_2}$ be an [[Definition:Ordered Set|ordered set]]. +Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be a [[Definition:Strictly Monotone Mapping|strictly monotone mapping]]. +Then $\phi$ is [[Definition:Injection|injective]]. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = x, y + | o = \in + | r = S + | c = +}} +{{eqn | lo= \land + | l = x + | o = \ne + | r = y + | c = +}} +{{eqn | ll= \leadsto + | l = x + | o = \prec_1 + | r = y +}} +{{eqn | lo= \lor + | l = y + | o = \prec_1 + | r = x + | c = [[Trichotomy Law (Ordering)|Trichotomy Law]] +}} +{{eqn | ll= \leadsto + | l = \map \phi x + | o = \prec_2 + | r = \map \phi y +}} +{{eqn | lo= \lor + | l = \map \phi y + | o = \prec_2 + | r = \map \phi x + | c = $\phi$ is [[Definition:Strictly Monotone Mapping|strictly monotone]] +}} +{{eqn | ll= \leadsto + | l = \map \phi x + | o = \ne + | r = \map \phi y + | c = Definition of $\prec_2$ +}} +{{end-eqn}} +Hence the result. +{{qed}} +\end{proof}<|endoftext|> +\section{Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing} +Tags: Order Embeddings, Total Orderings + +\begin{theorem} +Let $\struct {S, \preceq_1}$ be a [[Definition:Totally Ordered Set|totally ordered set]]. +Let $\struct {T, \preceq_2}$ be an [[Definition:Ordered Set|ordered set]]. +Let $\phi: S \to T$ be a [[Definition:Mapping|mapping]]. +Then $\phi$ is an [[Definition:Order Embedding|order embedding]] {{iff}} $\phi$ is [[Definition:Strictly Increasing Mapping|strictly increasing]]. +That is: +:$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$ +{{iff}}: +:$\forall x, y \in S: x \prec_1 y \implies \map \phi x \prec_2 \map \phi y$ +\end{theorem} + +\begin{proof} +=== Forward Implication === +Let $\phi$ be an [[Definition:Order Embedding|order embedding]]. +Let $x, y \in S$ with $x \prec_1 y$, where $\prec_1$ is the [[Definition:Strictly Precedes/Definition 2|strict predecessor]] relation. +Then: +{{begin-eqn}} +{{eqn | o = + | r = x \prec_1 y \land x \ne y + | c = Definition of [[Definition:Strictly Precedes/Definition 2|Strictly Precedes]] +}} +{{eqn | o = \leadsto + | r = \map \phi x \preceq_2 \map \phi y + | c = {{Defof|Order Embedding}} +}} +{{eqn | o = \leadsto + | r = \map \phi x \prec_2 \map \phi y + | c = $\phi$ is [[Definition:Injection|injective]] +}} +{{end-eqn}} +So by definition, $\phi$ is [[Definition:Strictly Increasing Mapping|strictly increasing]]. +{{qed|lemma}} +{{transclude:Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication + |section = proofs + |title = Reverse Implication + |header = 3 + |increase = 2 + |link = true +}} +\end{proof}<|endoftext|> +\section{Finite Totally Ordered Set is Well-Ordered} +Tags: Total Orderings, Well-Orderings + +\begin{theorem} +Every [[Definition:Finite Set|finite]] [[Definition:Totally Ordered Set|totally ordered set]] is [[Definition:Well-Ordered Set|well-ordered]]. +\end{theorem} + +\begin{proof} +Let $\struct {S, \preceq}$ be a [[Definition:Finite Set|finite]] [[Definition:Totally Ordered Set|totally ordered set]]. +From [[Condition for Well-Foundedness]], $\struct {S, \preceq}$ is [[Definition:Well-Founded|well-founded]] {{iff}} there is no [[Definition:Infinite Sequence|infinite sequence]] $\sequence {a_n}$ of [[Definition:Element|elements]] of $S$ such that $\forall n \in \N: a_{n + 1} \prec a_n$. +If it were the case that $\struct {S, \preceq}$ had such an [[Definition:Infinite Sequence|infinite sequence]], then at least some of the [[Definition:Term of Sequence|terms]] would be repeated in that sequence. +So there would be, for example: +:$s_i \preceq s_j \preceq s_k \preceq s_i$ +and $\preceq$ would therefore not be [[Definition:Transitive Relation|transitive]] and so not a [[Definition:Totally Ordered Set|totally ordered set]]. +So $\struct {S, \preceq}$ is [[Definition:Well-Founded|well-founded]]. +The result follows from the definition of [[Definition:Well-Ordered Set|well-ordered set]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Subset of Well-Ordered Set is Well-Ordered} +Tags: Well-Orderings + +\begin{theorem} +Every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of a [[Definition:Well-Ordered Set|well-ordered set]] is itself [[Definition:Well-Ordered Set|well-ordered]]. +\end{theorem} + +\begin{proof} +Let $\struct {S, \preceq}$ be a [[Definition:Well-Ordered Set|well-ordered set]]. +Let $T \subseteq S$. +Let $X \subseteq T$. +By [[Subset Relation is Transitive]], $X \subseteq S$. +By the definition of a [[Definition:Well-Ordered Set|well-ordered set]], $X$ has a [[Definition:Smallest Element|smallest element]]. +It follows by definition that $T$ is [[Definition:Well-Ordered Set|well-ordered]]. +Hence the result. +{{Qed}} +\end{proof}<|endoftext|> +\section{Element Commutes with Product of Commuting Elements} +Tags: Commutativity, Associativity + +\begin{theorem} +Let $(S, \circ)$ be a [[Definition:Semigroup|semigroup]]. +Let $x, y, z \in S$. +If $x$ [[Definition:Commute|commutes]] with both $y$ and $z$, then $x$ [[Definition:Commute|commutes]] with $y \circ z$. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = x \circ \left({y \circ z}\right) + | r = \left({x \circ y}\right) \circ z + | c = $\circ$ is [[Definition:Associative Operation|associative]] +}} +{{eqn | r = \left({y \circ x}\right) \circ z + | c = $x$ [[Definition:Commute|commutes]] with $y$ +}} +{{eqn | r = y \circ \left({x \circ z}\right) + | c = $\circ$ is [[Definition:Associative Operation|associative]] +}} +{{eqn | r = y \circ \left({z \circ x}\right) + | c = $x$ [[Definition:Commute|commutes]] with $z$ +}} +{{eqn | r = \left({y \circ z}\right) \circ x + | c = $\circ$ is [[Definition:Associative Operation|associative]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Associative Idempotent Anticommutative} +Tags: Abstract Algebra, Commutativity, Associativity, Idempotence + +\begin{theorem} +Let $\circ$ be a [[Definition:Binary Operation|binary operation]] on a [[Definition:Set|set]] $S$. +Let $\circ$ be [[Definition:Associative Operation|associative]]. +Then $\circ$ is [[Definition:Anticommutative|anticommutative]] {{iff}}: +: $(1): \quad \circ$ is [[Definition:Idempotent Operation|idempotent]] +and: +: $(2): \quad \forall a, b \in S: a \circ b \circ a = a$. +\end{theorem} + +\begin{proof} +Let $\circ$ be an [[Definition:Associative Operation|associative operation]] on $S$. +=== Necessary Condition === +Suppose $\circ$ is [[Definition:Anticommutative|anticommutative]]. +Let $a \circ a = x$ for some $a \in S$. Then: +{{begin-eqn}} +{{eqn | l = \left({a \circ a}\right) \circ a + | r = a \circ \left({a \circ a}\right) + | c = $\circ$ is [[Definition:Associative Operation|Associative]] +}} +{{eqn | ll= \implies + | l = a \circ x + | r = x \circ a + | c = +}} +{{eqn | ll= \implies + | l = x + | r = a + | c = $\circ$ is [[Definition:Anticommutative|Anticommutative]] +}} +{{eqn | ll= \implies + | l = a \circ a + | r = a + | c = +}} +{{end-eqn}} +So $\circ$ being [[Definition:Associative Operation|associative]] and [[Definition:Anticommutative|anticommutative]] implies that $\circ$ is [[Definition:Idempotent Operation|idempotent]]. +Now, let $a \circ b \circ a = x$ for some $a, b$. +{{begin-eqn}} +{{eqn | l = \left({a \circ a}\right) \circ b \circ a + | r = a \circ b \circ a + | c = [[Definition:Idempotent Operation|Idempotence]] of $\circ$, proved above +}} +{{eqn | r = a \circ b \circ \left({a \circ a}\right) + | c = [[Definition:Idempotent Operation|Idempotence]] of $\circ$ +}} +{{eqn | ll= \implies + | l = a \circ \left({a \circ b \circ a}\right) + | r = \left({a \circ b \circ a}\right) \circ a + | c = $\circ$ is [[Definition:Associative Operation|Associative]] +}} +{{eqn | ll= \implies + | l = a \circ x + | r = x \circ a + | c = +}} +{{eqn | ll= \implies + | l = x + | r = a + | c = $\circ$ is [[Definition:Anticommutative|Anticommutative]] +}} +{{eqn | ll= \implies + | l = a \circ b \circ a + | r = a + | c = +}} +{{end-eqn}} +So $\circ$ being [[Definition:Associative Operation|associative]] and [[Definition:Anticommutative|anticommutative]] implies, via the fact (also proved) that $\circ$ is [[Definition:Idempotent Operation|idempotent]], that $a \circ b \circ a = a$. +{{qed|lemma}} +=== Sufficient Condition === +Now, suppose $\circ$ (which we take to be [[Definition:Associative Operation|associative]]) is [[Definition:Idempotent Operation|idempotent]] and: +:$\forall a, b \in S: a \circ b \circ a = a$ +It remains to be shown that $\circ$ is [[Definition:Anticommutative|anticommutative]]. +Suppose $a \circ b = b \circ a$. +Then: +{{begin-eqn}} +{{eqn | l = a \circ b + | r = b \circ a + | c = +}} +{{eqn | ll= \implies + | l = a \circ b \circ a + | r = a \circ a \circ b = a \circ b = a + | c = +}} +{{eqn | l = a \circ b \circ a + | r = b \circ a \circ a = b \circ a = a + | c = +}} +{{eqn | l = b \circ a \circ b + | r = b \circ b \circ a = b \circ a = b + | c = +}} +{{eqn | l = b \circ a \circ b + | r = a \circ b \circ b = a \circ b = b + | c = +}} +{{eqn | ll= \implies + | l = a + | r = a \circ b = b + | c = +}} +{{eqn | l = a + | r = b \circ a = b + | c = +}} +{{eqn | ll= \implies + | l = a + | r = b + | c = +}} +{{end-eqn}} +So, $\circ$ being [[Definition:Idempotent Operation|idempotent]] and $a \circ b \circ a = a$ implies that $\circ$ is [[Definition:Anticommutative|anticommutative]]. +{{Qed}} +\end{proof}<|endoftext|> +\section{Associative and Anticommutative} +Tags: Commutativity, Associativity + +\begin{theorem} +Let $\circ$ be a [[Definition:Binary Operation|binary operation]] on a set $S$. +Let $\circ$ be both [[Definition:Associative Operation|associative]] and [[Definition:Anticommutative|anticommutative]]. +Then: +:$\forall x, y, z \in S: x \circ y \circ z = x \circ z$ +\end{theorem} + +\begin{proof} +Let $\circ$ be both [[Definition:Associative Operation|associative]] and [[Definition:Anticommutative|anticommutative]]. +Then from [[Associative Idempotent Anticommutative]]: +: $\forall x, z \in S: x \circ z \circ x = x$ +and $\circ$ is [[Definition:Idempotent Operation|idempotent]]. +Consider $x \circ y \circ z \circ x \circ z$. +We have: +{{begin-eqn}} +{{eqn | l = x \circ y \circ z \circ x \circ z + | r = x \circ \left({y \circ z}\right) \circ x \circ z + | c = +}} +{{eqn | r = x \circ z + | c = +}} +{{end-eqn}} +Also: +{{begin-eqn}} +{{eqn | l = x \circ y \circ z \circ x \circ z + | r = x \circ y \circ \left({z \circ x \circ z}\right) + | c = +}} +{{eqn | r = x \circ y \circ z + | c = +}} +{{end-eqn}} +Hence the result. +{{qed}} +\end{proof}<|endoftext|> +\section{Constant Operation is Commutative} +Tags: Constant Operation, Commutativity + +\begin{theorem} +Let $S$ be a [[Definition:Set|set]]. +Let $x \sqbrk c y = c$ be a [[Definition:Constant Operation|constant operation]] on $S$. +Then $\sqbrk c$ is a [[Definition:Commutative Operation|commutative operation]]: +:$\forall x, y \in S: x \sqbrk c y = y \sqbrk c x$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = x \sqbrk c y + | r = c + | c = {{Defof|Constant Operation}} +}} +{{eqn | l = y \sqbrk c x + | r = c + | c = {{Defof|Constant Operation}} +}} +{{eqn | ll= \leadsto + | l = x \sqbrk c y + | r = y \sqbrk c x + | c = +}} +{{end-eqn}} +Hence the result by definition of [[Definition:Commutative Operation|commutativity]]. +{{Qed}} +[[Category:Constant Operation]] +[[Category:Commutativity]] +rgvaxlfl8v7zwybh4pmlxf52y5rnnxb +\end{proof}<|endoftext|> +\section{Constant Operation is Associative} +Tags: Constant Operation + +\begin{theorem} +Let $S$ be a [[Definition:Set|set]]. +Let $x \left[{c}\right] y = c$ be a [[Definition:Constant Operation|constant operation]] on $S$. +Then $\left[{c}\right]$ is an [[Definition:Associative Operation|associative operation]]: +:$\forall x, y, z \in S: \left({x \left[{c}\right] y}\right) \left[{c}\right] z = x \left[{c}\right] \left({y \left[{c}\right] z}\right)$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \left({x \left[{c}\right] y}\right) \left[{c}\right] z + | r = c \left[{c}\right] z + | c = Definition of [[Definition:Constant Operation|Constant Operation]] +}} +{{eqn | r = c + | c = Definition of [[Definition:Constant Operation|Constant Operation]] +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | l = x \left[{c}\right] \left({y \left[{c}\right] z}\right) + | r = x \left[{c}\right] c + | c = Definition of [[Definition:Constant Operation|Constant Operation]] +}} +{{eqn | r = c + | c = Definition of [[Definition:Constant Operation|Constant Operation]] +}} +{{end-eqn}} +{{qed}} +[[Category:Constant Operation]] +jph7k2jvg81vrd0hlsz5994dno3s6nk +\end{proof}<|endoftext|> +\section{Left Operation is Idempotent} +Tags: Left and Right Operations + +\begin{theorem} +The [[Definition:Left Operation|left operation]] is [[Definition:Idempotent Operation|idempotent]]: +:$\forall x: x \leftarrow x = x$ +\end{theorem} + +\begin{proof} +Immediate from the definition of the [[Definition:Left Operation|left operation]]. +{{Qed}} +\end{proof}<|endoftext|> +\section{Right Operation is Idempotent} +Tags: Left and Right Operations + +\begin{theorem} +The [[Definition:Right Operation|right operation]] is [[Definition:Idempotent Operation|idempotent]]: +:$\forall x: x \rightarrow x = x$ +\end{theorem} + +\begin{proof} +Immediate from the definition of the [[Definition:Right Operation|right operation]]. +\end{proof}<|endoftext|> +\section{Left Operation is Anticommutative} +Tags: Left and Right Operations + +\begin{theorem} +The [[Definition:Left Operation|left operation]] is [[Definition:Anticommutative|anticommutative]]: +:$\forall x, y: x \leftarrow y = y \leftarrow x \iff x = y$ +\end{theorem} + +\begin{proof} +Immediate from the definition of the [[Definition:Left Operation|left operation]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Right Operation is Anticommutative} +Tags: Left and Right Operations + +\begin{theorem} +The [[Definition:Right Operation|right operation]] is [[Definition:Anticommutative|anticommutative]]: +:$\forall x, y: x \rightarrow y = y \rightarrow x \iff x = y$ +\end{theorem} + +\begin{proof} +Immediate from the definition of the [[Definition:Right Operation|right operation]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Left Operation is Associative} +Tags: Left and Right Operations + +\begin{theorem} +The [[Definition:Left Operation|left operation]] is [[Definition:Associative Operation|associative]]: +:$\forall x, y, z: \paren {x \leftarrow y} \leftarrow z = x \leftarrow \paren {y \leftarrow z}$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = x \leftarrow \paren {y \leftarrow z} + | r = x \leftarrow y + | c = {{Defof|Left Operation}} +}} +{{eqn | r = x + | c = {{Defof|Left Operation}} +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | l = \paren {x \leftarrow y} \leftarrow z + | r = x \leftarrow z + | c = {{Defof|Left Operation}} +}} +{{eqn | r = x + | c = {{Defof|Left Operation}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Right Operation is Associative} +Tags: Left and Right Operations + +\begin{theorem} +The [[Definition:Right Operation|right operation]] is [[Definition:Associative Operation|associative]]: +:$\forall x, y, z: \paren {x \rightarrow y} \rightarrow z = x \rightarrow \paren {y \rightarrow z}$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \paren {x \rightarrow y} \rightarrow z + | r = y \rightarrow z + | c = {{Defof|Right Operation}} +}} +{{eqn | r = z + | c = {{Defof|Right Operation}} +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | l = x \rightarrow \paren {y \rightarrow z} + | r = x \rightarrow z + | c = {{Defof|Right Operation}} +}} +{{eqn | r = z + | c = {{Defof|Right Operation}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Max and Min are Commutative} +Tags: Max and Min Operations + +\begin{theorem} +The [[Definition:Max Operation|Max]] and [[Definition:Min Operation|Min]] operations are [[Definition:Commutative Operation|commutative]]: +: $\max \left({x, y}\right) = \max \left({y, x}\right)$ +: $\min \left({x, y}\right) = \min \left({y, x}\right)$ +\end{theorem} + +\begin{proof} +To simplify our notation: +: Let $\max \left({x, y}\right)$ be (temporarily) denoted $x \overline \wedge y$ +: Let $\min \left({x, y}\right)$ be (temporarily) denoted $x \underline \vee y$. +There are three cases to consider: +:$(1): \quad x \le y$ +:$(2): \quad y \le x$ +:$(3): \quad x = y$ +$(1): \quad$ Let $x \le y$. +Then: +{{begin-eqn}} +{{eqn | l=x \overline \wedge y + | r=y + | rr==y \overline \wedge x +}} +{{eqn | l=x \underline \vee y + | r=x + | rr==y \underline \vee x +}} +{{end-eqn}} +$(2): \quad$ Let $y \le x$. +Then: +{{begin-eqn}} +{{eqn | l=x \overline \wedge y + | r=x + | rr==y \overline \wedge x +}} +{{eqn | l=x \underline \vee y + | r=y + | rr==y \underline \vee x +}} +{{end-eqn}} +$(3): \quad$ Let $x = y$. +Then: +{{begin-eqn}} +{{eqn | l=x \overline \wedge y + | r=x = y + | rr==y \overline \wedge x +}} +{{eqn | l=x \underline \vee y + | r=x = y + | rr==y \underline \vee x +}} +{{end-eqn}} +Thus $\overline \wedge$, i.e. $\max$, and $\underline \vee$, i.e. $\min$, have been shown to be [[Definition:Commutative Operation|commutative]] in all cases. +{{qed}} +[[Category:Max and Min Operations]] +49tfykvm4ap31r0om7omi80kwm1vsjx +\end{proof}<|endoftext|> +\section{Max and Min are Associative} +Tags: Max and Min Operations + +\begin{theorem} +The [[Definition:Max Operation|Max]] and [[Definition:Min Operation|Min]] operations are [[Definition:Associative Operation|associative]]: +: $\max \left({\max \left({x, y}\right), z}\right) = \max \left({x, \max \left({y, z}\right)}\right)$ +: $\min \left({\min \left({x, y}\right), z}\right) = \min \left({x, \min \left({y, z}\right)}\right)$ +Thus we are justified in writing $\max \left({x, y, z}\right)$ and $\min \left({x, y, z}\right)$. +\end{theorem} + +\begin{proof} +To simplify our notation: +: Let $\max \left({x, y}\right)$ be (temporarily) denoted $x \overline \wedge y$ +: Let $\min \left({x, y}\right)$ be (temporarily) denoted $x \underline \vee y$. +There are the following cases to consider: +:$(1): \quad x \le y \le z$ +:$(2): \quad x \le z \le y$ +:$(3): \quad y \le x \le z$ +:$(4): \quad y \le z \le x$ +:$(5): \quad z \le x \le y$ +:$(6): \quad z \le y \le x$ +Taking each one in turn: +:$(1): \quad$ Let $x \le y \le z$. Then: +{{begin-eqn}} +{{eqn | l = x \overline \wedge \left({y \overline \wedge z}\right) + | r = x \overline \wedge z + | rr= = z +}} +{{eqn | l = \left({x \overline \wedge y}\right) \overline \wedge z + | r = y \overline \wedge z + | rr= = z +}} +{{eqn | l = x \underline \vee \left({y \underline \vee z}\right) + | r = x \underline \vee y + | rr= = x +}} +{{eqn | l = \left({x \underline \vee y}\right) \underline \vee z + | r = x \underline \vee z + | rr= = x +}} +{{end-eqn}} +:$(2): \quad$ Let $x \le z \le y$. Then: +{{begin-eqn}} +{{eqn | l = x \overline \wedge \left({y \overline \wedge z}\right) + | r = x \overline \wedge y + | rr= = y +}} +{{eqn | l = \left({x \overline \wedge y}\right) \overline \wedge z + | r = y \overline \wedge z + | rr= = y +}} +{{eqn | l = x \underline \vee \left({y \underline \vee z}\right) + | r = x \underline \vee z + | rr= = x +}} +{{eqn | l = \left({x \underline \vee y}\right) \underline \vee z + | r = x \underline \vee z + | rr= = x +}} +{{end-eqn}} +:$(3): \quad$ Let $y \le x \le z$. Then: +{{begin-eqn}} +{{eqn | l = x \overline \wedge \left({y \overline \wedge z}\right) + | r = x \overline \wedge z + | rr= = z +}} +{{eqn | l = \left({x \overline \wedge y}\right) \overline \wedge z + | r = x \overline \wedge z + | rr= = z +}} +{{eqn | l = x \underline \vee \left({y \underline \vee z}\right) + | r = x \underline \vee y + | rr= = y +}} +{{eqn | l = \left({x \underline \vee y}\right) \underline \vee z + | r = y \underline \vee z + | rr= = y +}} +{{end-eqn}} +:$(4): \quad$ Let $y \le z \le x$. Then: +{{begin-eqn}} +{{eqn | l = x \overline \wedge \left({y \overline \wedge z}\right) + | r = x \overline \wedge z + | rr= = x +}} +{{eqn | l = \left({x \overline \wedge y}\right) \overline \wedge z + | r = x \overline \wedge z + | rr= = x +}} +{{eqn | l = x \underline \vee \left({y \underline \vee z}\right) + | r = x \underline \vee y + | rr= = y +}} +{{eqn | l = \left({x \underline \vee y}\right) \underline \vee z + | r = y \underline \vee z + | rr= = y +}} +{{end-eqn}} +:$(5): \quad$ Let $z \le x \le y$. Then: +{{begin-eqn}} +{{eqn | l = x \overline \wedge \left({y \overline \wedge z}\right) + | r = x \overline \wedge y + | rr= = y +}} +{{eqn | l = \left({x \overline \wedge y}\right) \overline \wedge z + | r = y \overline \wedge z + | rr= = y +}} +{{eqn | l = x \underline \vee \left({y \underline \vee z}\right) + | r = x \underline \vee z + | rr= = z +}} +{{eqn | l = \left({x \underline \vee y}\right) \underline \vee z + | r = x \underline \vee z + | rr= = z +}} +{{end-eqn}} +:$(6): \quad$ Let $z \le y \le x$. Then: +{{begin-eqn}} +{{eqn | l = x \overline \wedge \left({y \overline \wedge z}\right) + | r = x \overline \wedge y + | rr= = x +}} +{{eqn | l = \left({x \overline \wedge y}\right) \overline \wedge z + | r = x \overline \wedge z + | rr= = x +}} +{{eqn | l = x \underline \vee \left({y \underline \vee z}\right) + | r = y \underline \vee z + | rr= = z +}} +{{eqn | l = \left({x \underline \vee y}\right) \underline \vee z + | r = y \underline \vee z + | rr= = z +}} +{{end-eqn}} +Thus in all cases it can be seen that the result holds. +{{qed}} +\end{proof}<|endoftext|> +\section{Max and Min are Idempotent} +Tags: Max and Min Operations, Idempotence + +\begin{theorem} +The [[Definition:Max Operation|Max]] and [[Definition:Min Operation|Min]] operations are [[Definition:Idempotent Operation|idempotent]]: +: $\max \left({x, x}\right) = x$ +: $\min \left({x, x}\right) = x$ +\end{theorem} + +\begin{proof} +Follows immediately from the definitions of [[Definition:Max Operation|Max]] and [[Definition:Min Operation|Min]]. +{{Qed}} +[[Category:Max and Min Operations]] +[[Category:Idempotence]] +98gysw9vcuahm70qrfopw6qzd1kffdn +\end{proof}<|endoftext|> +\section{Max and Min Operations are Distributive over Each Other} +Tags: Max and Min Operations, Distributive Operations + +\begin{theorem} +The [[Definition:Max Operation|Max]] and [[Definition:Min Operation|Min]] operations are [[Definition:Distributive Operation|distributive]] over each other: +:$\max \set {x, \min \set {y, z} } = \min \set {\max \set {x, y}, \max \set {x, z} }$ +:$\max \set {\min \set {x, y}, z} = \min \set {\max \set {x, z}, \max \set {y, z} }$ +:$\min \set {x, \max \set {y, z} } = \max \set {\min \set {x, y}, \min \set {x, z} }$ +:$\min \set {\max \set {x, y}, z} = \max \set {\min \set {x, z}, \min \set {y, z} }$ +\end{theorem} + +\begin{proof} +To simplify our notation, let $\max \set {x, y}$ be (temporarily) denoted $x \overline \wedge y$, and let $\min \set {x, y}$ be (temporarily) denoted $x \underline \vee y$. +Note that, once we have proved: +:$x \overline \wedge \paren {y \underline \vee z} = \paren {x \overline \wedge y} \underline \vee \paren {x \overline \wedge z}$ +:$x \underline \vee \paren {y \overline \wedge z} = \paren {x \underline \vee y} \overline \wedge \paren {x \underline \vee z}$ +then the other results follow immediately by the fact that [[Definition:Min Operation|Min]] and [[Definition:Max Operation|Max]] are [[Max and Min are Commutative|commutative]]. +There are the following cases to consider: +:$(1): \quad x \le y \le z$ +:$(2): \quad x \le z \le y$ +:$(3): \quad y \le x \le z$ +:$(4): \quad y \le z \le x$ +:$(5): \quad z \le x \le y$ +:$(6): \quad z \le y \le x$ +$(1): \quad $ Let $x \le y \le z$. +Then: +{{begin-eqn}} +{{eqn | l = x \overline \wedge \paren {y \underline \vee z} + | r = x \overline \wedge y = y + | c = +}} +{{eqn | l = \paren {x \overline \wedge y} \underline \vee \paren {x \overline \wedge z} + | r = y \underline \vee z = y + | c = +}} +{{eqn | o = + | c = +}} +{{eqn | l = x \underline \vee \paren {y \overline \wedge z} + | r = x \underline \vee z = x + | c = +}} +{{eqn | l = \paren {x \underline \vee y} \overline \wedge \paren {x \underline \vee z} + | r = x \overline \wedge x = x + | c = +}} +{{end-eqn}} +$(2): \quad $ Let $x \le z \le y$. +Then: +{{begin-eqn}} +{{eqn | l = x \overline \wedge \paren {y \underline \vee z} + | r = x \overline \wedge z = z + | c = +}} +{{eqn | l = \paren {x \overline \wedge y} \underline \vee \paren {x \overline \wedge z} + | r = y \underline \vee z = z + | c = +}} +{{eqn | o = + | c = +}} +{{eqn | l = x \underline \vee \paren {y \overline \wedge z} + | r = x \underline \vee y = x + | c = +}} +{{eqn | l = \paren {x \underline \vee y} \overline \wedge \paren {x \underline \vee z} + | r = x \overline \wedge x = x + | c = +}} +{{end-eqn}} +$(3): \quad $ Let $y \le x \le z$. +Then: +{{begin-eqn}} +{{eqn | l = x \overline \wedge \paren {y \underline \vee z} + | r = x \overline \wedge y = x + | c = +}} +{{eqn | l = \paren {x \overline \wedge y} \underline \vee \paren {x \overline \wedge z} + | r = x \underline \vee z = x + | c = +}} +{{eqn | o = + | c = +}} +{{eqn | l = x \underline \vee \paren {y \overline \wedge z} + | r = x \underline \vee z = x + | c = +}} +{{eqn | l = \paren {x \underline \vee y} \overline \wedge \paren {x \underline \vee z} + | r = y \overline \wedge x = x + | c = +}} +{{end-eqn}} +$(4): \quad $ Let $y \le z \le x$. +Then: +{{begin-eqn}} +{{eqn | l = x \overline \wedge \paren {y \underline \vee z} + | r = x \overline \wedge y = x + | c = +}} +{{eqn | l = \paren {x \overline \wedge y} \underline \vee \paren {x \overline \wedge z} + | r = x \underline \vee x = x + | c = +}} +{{eqn | o = + | c = +}} +{{eqn | l = x \underline \vee \paren {y \overline \wedge z} + | r = x \underline \vee z = z + | c = +}} +{{eqn | l = \paren {x \underline \vee y} \overline \wedge \paren {x \underline \vee z} + | r = y \overline \wedge z = z + | c = +}} +{{end-eqn}} +$(5): \quad $ Let $z \le x \le y$. +Then: +{{begin-eqn}} +{{eqn | l = x \overline \wedge \paren {y \underline \vee z} + | r = x \overline \wedge z = x + | c = +}} +{{eqn | l = \paren {x \overline \wedge y} \underline \vee \paren {x \overline \wedge z} + | r = y \underline \vee x = x + | c = +}} +{{eqn | o = + | c = +}} +{{eqn | l = x \underline \vee \paren {y \overline \wedge z} + | r = x \underline \vee y = x + | c = +}} +{{eqn | l = \paren {x \underline \vee y} \overline \wedge \paren {x \underline \vee z} + | r = x \overline \wedge z = x + | c = +}} +{{end-eqn}} +$(6): \quad $ Let $z \le y \le x$. +Then: +{{begin-eqn}} +{{eqn | l = x \overline \wedge \paren {y \underline \vee z} + | r = x \overline \wedge z = x + | c = +}} +{{eqn | l = \paren {x \overline \wedge y} \underline \vee \paren {x \overline \wedge z} + | r = x \underline \vee x = x + | c = +}} +{{eqn | o = + | c = +}} +{{eqn | l = x \underline \vee \paren {y \overline \wedge z} + | r = x \underline \vee y = y + | c = +}} +{{eqn | l = \paren {x \underline \vee y} \overline \wedge \paren {x \underline \vee z} + | r = y \overline \wedge z = y + | c = +}} +{{end-eqn}} +Thus in all cases it can be seen that the result holds. +{{qed}} +[[Category:Max and Min Operations]] +[[Category:Distributive Operations]] +8aablkhh5eun94ro113f3tfjjhqdm1t +\end{proof}<|endoftext|> +\section{Equality of Algebraic Structures} +Tags: Abstract Algebra, Equality + +\begin{theorem} +Two [[Definition:Algebraic Structure|algebraic structures]] $\struct {S, \circ}$ and $\struct {T, *}$ are [[Definition:Equality|equal]] {{iff}}: +:$S = T$ +:$\forall a, b \in S: a \circ b = a * b$ +\end{theorem} + +\begin{proof} +This follows from [[Definition:Set Equality|set equality]] and [[Equality of Ordered Pairs]]. +[[Category:Abstract Algebra]] +[[Category:Equality]] +lx4ikvvqj4hwbu3b5xw5bd8y3kckaqx +\end{proof}<|endoftext|> +\section{Restriction of Associative Operation is Associative} +Tags: Associativity + +\begin{theorem} +Let $\struct {S, \circ}$ be an [[Definition:Semigroup|semigroup]]. +Let $T \subseteq S$. +Let $T$ be [[Definition:Closed Algebraic Structure|closed]] under $\circ$. +Then $\struct {T, \circ {\restriction_T} }$ is also a [[Definition:Semigroup|semigroup]], where $\circ {\restriction_T}$ is the [[Definition:Restriction of Operation|restriction]] of $\circ$ to $T$. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | o = + | r = T \subseteq S + | c = +}} +{{eqn | ll= \leadsto + | o = + | r = \forall a, b, c \in T: a, b, c \in S + | c = {{Defof|Subset}} +}} +{{eqn | ll= \leadsto + | o = + | r = a \mathop {\circ {\restriction_T} } \paren {b \mathop {\circ {\restriction_T} } c} + | c = +}} +{{eqn | r = a \circ \paren {b \circ c} + | c = +}} +{{eqn | r = \paren {a \circ b} \circ c + | c = as $\circ$ is [[Definition:Associative Operation|associative]] +}} +{{eqn | r = \paren {a \mathop {\circ {\restriction_T} } b} \mathop {\circ {\restriction_T} } c + | c = +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Restriction of Commutative Operation is Commutative} +Tags: Commutativity + +\begin{theorem} +Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]]. +Let $T \subseteq S$. +Let the [[Definition:Binary Operation|operation]] $\circ$ be [[Definition:Commutative Operation|commutative]] on $\struct {S, \circ}$. +Then the [[Definition:Restriction of Operation|restriction]] $\circ {\restriction_T}$ of $\circ$ to $T$ is also [[Definition:Commutative Operation|commutative]]. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = T + | o = \subseteq + | r = S + | c = +}} +{{eqn | ll= \leadsto + | l = \forall a, b \in T + | o = : + | r = a, b \in S + | c = {{Defof|Subset}} +}} +{{eqn | ll= \leadsto + | l = a \mathop {\circ {\restriction_T} } b + | r = a \circ b + | c = +}} +{{eqn | r = b \circ a + | c = $\circ$ is [[Definition:Commutative Operation|commutative]] +}} +{{eqn | r = b \mathop {\circ {\restriction_T} } a + | c = +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Idempotent Magma Element forms Singleton Submagma} +Tags: Magmas + +\begin{theorem} +Let $\struct {S, \circ}$ be a [[Definition:Magma|magma]]. +Let $x \in S$ be an [[Definition:Idempotent Element|idempotent]] element of $\struct {S, \circ}$. +Then $\struct {\set x, \circ}$ is a [[Definition:Submagma|submagma]] of $\struct {S, \circ}$. +\end{theorem} + +\begin{proof} +By [[Singleton of Element is Subset]]: +: $x \in S \iff \set x \subseteq S$ +By the definition of [[Definition:Idempotent Element|idempotence]]: +: $x \circ x = x \in \set x$ +Thus $\set x$ is a [[Definition:Subset|subset]] of $S$ which is [[Definition:Closed Algebraic Structure|closed]] under $\circ$. +By the definition of [[Definition:Submagma|submagma]], the result follows. +{{qed}} +\end{proof}<|endoftext|> +\section{Magma Subset Product with Self} +Tags: Subset Products + +\begin{theorem} +Let $\left({S, \circ}\right)$ be a [[Definition:Magma|magma]]. +Let $T \subseteq S$. +Then $\left({T, \circ}\right)$ is a [[Definition:Magma|magma]] {{iff}} $T \circ T \subseteq T$, where $T \circ T$ is the [[Definition:Subset Product|subset product]] of $T$ with itself. +\end{theorem} + +\begin{proof} +By definition: +: $T \circ T = \left\{{x = a \circ b: a, b \in T}\right\}$ +=== Necessary Condition === +Let $\left({T, \circ}\right)$ be a [[Definition:Magma|magma]]. +Then $T$ is [[Definition:Closed Algebraic Structure|closed]]. +That is: +:$\forall x, y \in T: x \circ y \in T$ +Thus: +: $x \circ y \in T \circ T \implies x \circ y \in T$ +{{qed|lemma}} +=== Sufficient Condition === +Let $T \circ T \subseteq T$. +Then: +: $x \circ y \in T \circ T \implies x \circ y \in T$ +That is, $T$ is [[Definition:Closed Algebraic Structure|closed]]. +Therefore $\left({T, \circ}\right)$ is a [[Definition:Magma|magma]] by definition. +{{qed}} +\end{proof}<|endoftext|> +\section{Subset Product within Semigroup is Associative} +Tags: Subset Products + +\begin{theorem} +Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]]. +Then the [[Definition:Operation Induced on Power Set|operation $\circ_\PP$ induced]] on the [[Definition:Power Set|power set]] of $S$ is also [[Definition:Associative Operation|associative]]. +\end{theorem} + +\begin{proof} +Let $X, Y, Z \in \powerset S$. +Then: +{{begin-eqn}} +{{eqn | l = X \circ_\PP \paren {Y \circ_\PP Z} + | r = \set {x \circ \paren {y \circ z}: x \in X, y \in Y, z \in Z} + | c = +}} +{{eqn | r = \set {\paren {x \circ y} \circ z: x \in X, y \in Y, z \in Z} + | c = +}} +{{eqn | r = \paren {X \circ_\PP Y} \circ_\PP Z + | c = +}} +{{end-eqn}} +demonstrating that $\circ_\PP$ is [[Definition:Associative Operation|associative]] on $\powerset S$. +{{qed}} +\end{proof}<|endoftext|> +\section{Subset Product within Commutative Structure is Commutative} +Tags: Subset Products + +\begin{theorem} +Let $\struct {S, \circ}$ be a [[Definition:Magma|magma]]. +If $\circ$ is [[Definition:Commutative Operation|commutative]], then the [[Definition:Operation Induced on Power Set|operation $\circ_\mathcal P$ induced]] on the [[Definition:Power Set|power set]] of $S$ is also [[Definition:Commutative Operation|commutative]]. +\end{theorem} + +\begin{proof} +Let $\struct {S, \circ}$ be a [[Definition:Magma|magma]] in which $\circ$ is commutative. +Let $X, Y \in \powerset S$. +Then: +{{begin-eqn}} +{{eqn | l = X \circ_\mathcal P Y + | r = \set {x \circ y: x \in X, y \in Y} +}} +{{eqn | l = Y \circ_\mathcal P X + | r = \set {y \circ x: x \in X, y \in Y} +}} +{{end-eqn}} +from which it follows that $\circ_\mathcal P$ is [[Definition:Commutative Operation|commutative]] on $\powerset S$. +{{qed}} +\end{proof}<|endoftext|> +\section{Subset of Subset Product} +Tags: Subset Products + +\begin{theorem} +Let $\struct {S, \circ}$ be a [[Definition:Magma|magma]]. +Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$. +Let $X, Y, Z \in \powerset S$. +Then: +:$X \subseteq Y \implies \paren {X \circ Z} \subseteq \paren {Y \circ Z}$ +:$X \subseteq Y \implies \paren {Z \circ X} \subseteq \paren {Z \circ Y}$ +where $X \circ Z$ etc. denotes [[Definition:Subset Product|subset product]]. +\end{theorem} + +\begin{proof} +Let $x \in X, z \in Z$. +Then: +:$x \circ z \in X \circ Z$ and $z \circ x \in Z \circ X$ +Now: +:$Y \circ Z = \set {y \circ z: y \in Y, z \in Z}$ +:$Z \circ Y = \set {z \circ y: y \in Y, z \in Z}$ +But by the definition of a [[Definition:Subset|subset]]: +:$x \in X \implies x \in Y$ +Thus: +:$x \circ z \in Y \circ Z$ and $z \circ x \in Z \circ Y$ +and the result follows. +{{Qed}} +\end{proof}<|endoftext|> +\section{Cancellable Element is Cancellable in Subset} +Tags: Cancellability + +\begin{theorem} +Let $\left ({S, \circ}\right)$ be an [[Definition:Algebraic Structure|algebraic structure]]. +Let $\left ({T, \circ}\right) \subseteq \left ({S, \circ}\right)$. +Let $x \in T$ be [[Definition:Cancellable Element|cancellable]] in $S$. +Then $x$ is also [[Definition:Cancellable Element|cancellable]] in $T$. +\end{theorem} + +\begin{proof} +Let $x \in T$ be [[Definition:Cancellable Element|cancellable]] in $S$. +Then by definition, $x \in T$ is [[Definition:Left Cancellable Element|left cancellable]] in $S$. +It follows from [[Left Cancellable Element is Left Cancellable in Subset]] that $x \in T$ is [[Definition:Left Cancellable Element|left cancellable]] in $T$. +Again by definition, $x \in T$ is [[Definition:Right Cancellable Element|right cancellable]] in $S$. +It follows from [[Right Cancellable Element is Right Cancellable in Subset]] that $x \in T$ is [[Definition:Right Cancellable Element|right cancellable]] in $T$. +Thus $x$ is also both [[Definition:Left Cancellable Element|left cancellable]] and [[Definition:Right Cancellable Element|right cancellable]] in $T$. +So by definition, $x$ is [[Definition:Cancellable Element|cancellable]] in $T$. +{{qed}} +\end{proof}<|endoftext|> +\section{Subsemigroup Closure Test} +Tags: Subsemigroups + +\begin{theorem} +To show that an [[Definition:Algebraic Structure|algebraic structure]] $\left({T, \circ}\right)$ is a [[Definition:Subsemigroup|subsemigroup]] of a [[Definition:Semigroup|semigroup]] $\struct {S, \circ}$, we need to show only that: +: $(1): \quad T \subseteq S$ +: $(2): \quad \circ$ is a [[Definition:Closed Operation|closed operation]] in $T$. +\end{theorem} + +\begin{proof} +From [[Restriction of Associative Operation is Associative]], if $\circ$ is [[Definition:Associative Operation|associative]] on $\struct {S, \circ}$, then it will also be [[Definition:Associative Operation|associative]] on $\struct {T, \circ}$. +Thus we do not need to check for [[Definition:Associative Operation|associativity]] in $\struct {T, \circ}$, as that has been inherited from its [[Definition:Extension of Operation|extension]] $\struct {S, \circ}$. +So, once we have established that $T \subseteq S$, all we need to do is to check for $\circ$ to be a [[Definition:Closed Operation|closed operation]]. +{{Qed}} +\end{proof}<|endoftext|> +\section{Cancellable iff Regular Representations Injective} +Tags: Regular Representations + +\begin{theorem} +Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]]. +Then $a \in S$ is [[Definition:Cancellable Element|cancellable]] {{iff}}: +::the [[Definition:Left Regular Representation|left regular representation]] $\map {\lambda_a} x$ is [[Definition:Injection|injective]] +and +::the [[Definition:Right Regular Representation|right regular representation]] $\map {\rho_a} x$ is [[Definition:Injection|injective]]. +\end{theorem} + +\begin{proof} +=== [[Left Cancellable iff Left Regular Representation Injective|Left Cancellable]] === +{{:Left Cancellable iff Left Regular Representation Injective}} +=== [[Right Cancellable iff Right Regular Representation Injective|Right Cancellable]] === +{{:Right Cancellable iff Right Regular Representation Injective}} +\end{proof}<|endoftext|> +\section{More than one Left Identity then no Right Identity} +Tags: Identity Elements + +\begin{theorem} +Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]]. +If $\struct {S, \circ}$ has more than one [[Definition:Left Identity|left identity]], then it has no [[Definition:Right Identity|right identity]]. +\end{theorem} + +\begin{proof} +Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]] with more than one [[Definition:Left Identity|left identity]]. +Take any two of these, and call them $e_{L_1}$ and $e_{L_2}$, where $e_{L_1} \ne e_{L_2}$. +Suppose $\struct {S, \circ}$ has a [[Definition:Right Identity|right identity]]. +Call this [[Definition:Right Identity|right identity]] $e_R$. +Then, by the behaviour of $e_R$, $e_{L_1}$ and $e_{L_2}$: +: $e_{L_1} = e_{L_1} \circ e_R = e_R$ +: $e_{L_2} = e_{L_2} \circ e_R = e_R$ +So $e_{L_1} = e_R = e_{L_2}$, which contradicts the supposition that $e_{L_1}$ and $e_{L_2}$ are different. +Therefore, in an [[Definition:Algebraic Structure|algebraic structure]] with more than one [[Definition:Left Identity|left identity]], there can be no [[Definition:Right Identity|right identity]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Element under Left Operation is Right Identity} +Tags: Left and Right Operations, Identity Elements + +\begin{theorem} +Let $\struct {S, \leftarrow}$ be an [[Definition:Algebraic Structure|algebraic structure]] in which the [[Definition:Binary Operation|operation]] $\leftarrow$ is the [[Definition:Left Operation|left operation]]. +Then no matter what $S$ is, $\struct {S, \leftarrow}$ is a [[Definition:Semigroup|semigroup]] all of whose [[Definition:Element|elements]] are [[Definition:Right Identity|right identities]]. +Thus it can be seen that any [[Definition:Right Identity|right identity]] in a [[Definition:Semigroup|semigroup]] is not necessarily [[Definition:Unique|unique]]. +\end{theorem} + +\begin{proof} +From [[Left Operation is Associative]], $\leftarrow$ is [[Definition:Associative Operation|associative]]. +By the nature of the [[Definition:Left Operation|left operation]], $\struct {S, \leftarrow}$ is [[Definition:Closed Algebraic Structure|closed]]: +:$\forall x, y \in S: x \leftarrow y = x \in S$ +whatever $S$ may be. +So $\struct {S, \leftarrow}$ is a [[Definition:Semigroup|semigroup]]. +From the definition of [[Definition:Left Operation|left operation]]: +:$\forall x, y \in S: x \leftarrow y = x$ +from which it is apparent that all [[Definition:Element|elements]] of $S$ are [[Definition:Right Identity|right identities]]. +From [[More than one Right Identity then no Left Identity]], it also follows that there is no [[Definition:Left Identity|left identity]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Element under Right Operation is Left Identity} +Tags: Left and Right Operations, Identity Elements + +\begin{theorem} +Let $\struct {S, \rightarrow}$ be an [[Definition:Algebraic Structure|algebraic structure]] in which the [[Definition:Binary Operation|operation]] $\rightarrow$ is the [[Definition:Right Operation|right operation]]. +Then $\struct {S, \rightarrow}$ is a [[Definition:Semigroup|semigroup]] all of whose [[Definition:Element|elements]] are [[Definition:Left Identity|left identities]]. +Thus it can be seen that any [[Definition:Left Identity|left identity]] in a [[Definition:Semigroup|semigroup]] is not necessarily [[Definition:Unique|unique]]. +\end{theorem} + +\begin{proof} +From [[Right Operation is Associative]], $\rightarrow$ is [[Definition:Associative Operation|associative]]. +By the nature of the [[Definition:Right Operation|right operation]], $\struct {S, \rightarrow}$ is [[Definition:Closed Algebraic Structure|closed]]: +:$\forall x, y \in S: x \rightarrow y = y \in S$ +whatever $S$ may be. +So $\struct {S, \rightarrow}$ is a [[Definition:Semigroup|semigroup]]. +From the definition of [[Definition:Right Operation|right operation]]: +:$\forall x, y \in S: x \rightarrow y = y$ +from which it is apparent that all [[Definition:Element|elements]] of $S$ are [[Definition:Left Identity|left identities]]. +{{qed}} +From [[More than one Left Identity then no Right Identity]], it also follows that there is no [[Definition:Right Identity|right identity]]. +\end{proof}<|endoftext|> +\section{Left and Right Identity are the Same} +Tags: Identity Elements + +\begin{theorem} +Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]]. +Let $e_L \in S$ be a [[Definition:Left Identity|left identity]], and $e_R \in S$ be a [[Definition:Right Identity|right identity]]. +Then: +:$e_L = e_R$ +that is, both the [[Definition:Left Identity|left identity]] and [[Definition:Right Identity|right identity]] are the same, and are therefore an [[Definition:Identity Element|identity]] $e$. +Furthermore, $e$ is the only [[Definition:Left Identity|left identity]] and [[Definition:Right Identity|right identity]] for $\circ$. +\end{theorem} + +\begin{proof} +Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]] such that: +:$\exists e_L \in S: \forall x \in S: e_L \circ x = x$ +:$\exists e_R \in S: \forall x \in S: x \circ e_R = x$ +Then $e_L = e_L \circ e_R = e_R$ by both the above, hence the result. +The uniqueness of the left and right identity is a direct result of [[Identity is Unique]]. +{{Qed}} +\end{proof}<|endoftext|> +\section{Identity Property in Semigroup} +Tags: Semigroups, Identity Elements + +\begin{theorem} +Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]]. +Let $s \in S$ be such that: +:$\forall a \in S: \exists x, y \in S: s \circ x = a = y \circ s$ +Then $\struct {S, \circ}$ has an [[Definition:Identity Element|identity]]. +\end{theorem} + +\begin{proof} +Suppose that: +:$\forall a \in S: \exists x, y \in S: s \circ x = a = y \circ s$. +Since $s \in S$, it follows that: +:$\exists v, w \in S: s \circ v = s = w \circ s$. +Let $a \in S$. +Then: +:$\exists x, y \in S: s \circ x = a = y \circ s$. +Thus: +{{begin-eqn}} +{{eqn | l = a + | r = s \circ x + | c = +}} +{{eqn | ll= \leadsto + | l = w \circ a + | r = w \circ s \circ x + | c = +}} +{{eqn | r = s \circ x + | c = +}} +{{eqn | r = a + | c = +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | l = a + | r = y \circ s + | c = +}} +{{eqn | ll= \leadsto + | l = a \circ v + | r = y \circ s \circ v + | c = +}} +{{eqn | r = y \circ s + | c = +}} +{{eqn | r = a + | c = +}} +{{end-eqn}} +Hence: +:$w \circ a = a$ and $a \circ v = a$ +for any $a \in S$. +In particular: +:Letting $a = v$ in the first of these gives $w \circ v = v$ +:Letting $a = w$ in the second gives $w \circ v = w$. +Thus $v = w \circ v = w$ is the [[Definition:Identity Element|identity element]] in $S$. +{{qed}} +\end{proof}<|endoftext|> +\section{Identity of Monoid is Cancellable} +Tags: Monoids + +\begin{theorem} +The [[Definition:Identity Element|identity]] of a [[Definition:Monoid|monoid]] is [[Definition:Cancellable Element|cancellable]]. +\end{theorem} + +\begin{proof} +Let $\left({S, \circ}\right)$ be a [[Definition:Monoid|monoid]] whose [[Definition:Identity Element|identity]] is $e$. +Let $x, y \in S$ such that $x \circ e = y \circ e$ +Then, by the definition of the [[Definition:Identity Element|identity]]: +:$x = x \circ e = y \circ e = y$ +... thus $x = y$ and the result is proved. +{{qed}} +[[Category:Monoids]] +eatawhcnkyf0usqdi6ji6dxzxuyafou +\end{proof}<|endoftext|> +\section{Identity is only Idempotent Cancellable Element} +Tags: Cancellability, Idempotence, Identity Elements + +\begin{theorem} +Let $e_S$ is the [[Definition:Identity Element|identity]] of an [[Definition:Algebraic Structure|algebraic structure]] $\left({S, \circ}\right)$. +Then $e_S$ is the only [[Definition:Cancellable Element|cancellable element]] of $\left({S, \circ}\right)$ that is [[Definition:Idempotent Element|idempotent]]. +\end{theorem} + +\begin{proof} +By [[Identity Element is Idempotent]], $e_S$ is [[Definition:Idempotent Element|idempotent]]. +Let $x$ be a [[Definition:Cancellable Element|cancellable]] [[Definition:Idempotent Element|idempotent element]] of $\left({S, \circ}\right)$. +{{begin-eqn}} +{{eqn | l = x \circ x + | r = x + | c = [[Definition:Idempotent Element|$x$ is idempotent]] +}} +{{eqn | r = x \circ e_S + | c = Definition of [[Definition:Identity Element|Identity]] +}} +{{end-eqn}} +So $x \circ x = x \circ e_S$. +But because $x$ is also by hypothesis [[Definition:Cancellable Element|cancellable]], it follows that $x = e_S$. +{{Qed}} +\end{proof}<|endoftext|> +\section{Set of all Self-Maps is Monoid} +Tags: Mapping Theory, Monoids + +\begin{theorem} +Let $S$ be a [[Definition:Set|set]]. +Let $S^S$ be the [[Definition:Set of All Mappings|set of all mappings]] from $S$ to itself. +Let the [[Definition:Binary Operation|operation]] $\circ$ represent [[Definition:Composition of Mappings|composition of mappings]]. +Then the [[Definition:Algebraic Structure|algebraic structure]] $\struct {S^S, \circ}$ is a [[Definition:Monoid|monoid]] whose [[Definition:Identity Element|identity element]] is the [[Definition:Identity Mapping|identity mapping]] on $S$. +\end{theorem} + +\begin{proof} +By [[Set of All Self-Maps is Semigroup]], $\struct {S^S, \circ}$ is a [[Definition:Semigroup|semigroup]]. +By [[Identity Mapping is Left Identity]] and [[Identity Mapping is Right Identity]] the [[Definition:Identity Mapping|identity mapping]] on $S$ is the [[Definition:Identity Element|identity element]] of $\struct {S^S, \circ}$. +Since $\struct {S^S, \circ}$ is a semigroup with an identity element, it is a [[Definition:Monoid|monoid]] by definition. +{{qed}} +\end{proof}<|endoftext|> +\section{Test for Submonoid} +Tags: Monoids + +\begin{theorem} +To show that $\left({T, \circ}\right)$ is a [[Definition:Submonoid|submonoid]] of a [[Definition:Monoid|monoid]] $\left({S, \circ}\right)$, we need to show that: +:$(1): \quad T \subseteq S$ +:$(2): \quad \left({T, \circ}\right)$ is a [[Definition:Magma|magma]] (i.e. that it is [[Definition:Closed Algebraic Structure|closed]]) +:$(3): \quad \left({T, \circ}\right)$ has an [[Definition:Identity Element|identity]]. +\end{theorem} + +\begin{proof} +From [[Subsemigroup Closure Test]], $(1)$ and $(2)$ are sufficient to show that $\left({T, \circ}\right)$ is a [[Definition:Subsemigroup|subsemigroup]] of $\left({S, \circ}\right)$. +Demonstrating the presence of an [[Definition:Identity Element|identity]] is then sufficient to show that it is a [[Definition:Monoid|monoid]]. +{{qed}} +[[Category:Monoids]] +7f3509pyo2ofnhy6uphzcd05t64iw1c +\end{proof}<|endoftext|> +\section{Identity of Cancellable Monoid is Identity of Submonoid} +Tags: Monoids, Cancellability, Identity Elements + +\begin{theorem} +Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]], all of whose [[Definition:Element|elements]] are [[Definition:Cancellable Element|cancellable]]. +Let $\struct {T, \circ}$ be a [[Definition:Submonoid|submonoid]] of $\struct {S, \circ}$. +Then the [[Definition:Identity Element|identity]] of $\struct {T, \circ}$ is the same [[Definition:Element|element]] as the [[Definition:Identity Element|identity]] of $\struct {S, \circ}$. +\end{theorem} + +\begin{proof} +By [[Identity of Monoid is Unique]]: +: there is only one [[Definition:Identity Element|identity element]] of $\struct {S, \circ}$ +and: +: there is only one [[Definition:Identity Element|identity element]] of $\struct {T, \circ}$. +Let $e_S$ be the [[Definition:Identity Element|identity]] of $\struct {S, \circ}$, and $e_T$ the identity of $\struct {T, \circ}$. +From [[Identity is only Idempotent Cancellable Element]], $e_S$ is the only [[Definition:Cancellable Element|cancellable element]] of $\struct {S, \circ}$ which is [[Definition:Idempotent Element|idempotent]]. +But all [[Definition:Element|elements]] of $\struct {S, \circ}$ are [[Definition:Cancellable Element|cancellable]]. +Thus $e_S$ is the only [[Definition:Idempotent Element|idempotent element]] of $\struct {S, \circ}$. +Again, all [[Definition:Element|elements]] of $\struct {S, \circ}$ are [[Definition:Cancellable Element|cancellable]]. +Thus from [[Cancellable Element is Cancellable in Subset]], all [[Definition:Element|elements]] of $\struct {T, \circ}$ are [[Definition:Cancellable Element|cancellable]]. +Thus, $e_T$ is the only [[Definition:Element|element]] of $\struct {T, \circ}$ which is [[Definition:Idempotent Element|idempotent]]. +Thus, as $e_T \in S$, we have $e_S \circ e_T = e_T = e_T \circ e_T$ and thus $e_S = e_T$. +{{qed}} +\end{proof}<|endoftext|> +\section{Cancellable Elements of Monoid form Submonoid} +Tags: Monoids + +\begin{theorem} +The [[Definition:Cancellable Element|cancellable elements]] of a [[Definition:Monoid|monoid]] $\struct {S, \circ}$ form a [[Definition:Submonoid|submonoid]] of $\struct {S, \circ}$. +\end{theorem} + +\begin{proof} +Let $C$ be the [[Definition:Set|set]] of [[Definition:Cancellable Element|cancellable elements]] of $\struct {S, \circ}$. +Obviously $C \subseteq S$. +From [[Cancellable Elements of Semigroup form Subsemigroup]], $\struct {C, \circ}$ is a [[Definition:Subsemigroup|subsemigroup]] of $S$. +Let $e_S$ be the [[Definition:Identity Element|identity]] of $\struct {S, \circ}$. +From [[Identity of Monoid is Cancellable]], $e_S$ is cancellable, therefore $e_S \in C$. +As $e_S$ is the identity of $S$, we have that: +:$\forall x \in C: x \circ e_S = x = e_S \circ x$. +Thus $e_S$ is the identity of $\struct {C, \circ}$, which is therefore a [[Definition:Monoid|monoid]]. +As $C \subseteq S$, the result follows. +{{qed}} +[[Category:Monoids]] +kbtbh08ex31xlu4bbjua3p7ib14hisw +\end{proof}<|endoftext|> +\section{Product of Semigroup Element with Left Inverse is Idempotent} +Tags: Semigroups, Idempotence + +\begin{theorem} +Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]] with a [[Definition:Left Identity|left identity]] $e_L$. +Let $x \in S$ such that $\exists x_L: x_L \circ x = e_L$, that is $x$ has a [[Definition:Left Inverse Element|left inverse]] with respect to the [[Definition:Left Identity|left identity]]. +Then: +:$\paren {x \circ x_L} \circ \paren {x \circ x_L} = x \circ x_L$ +That is, $x \circ x_L$ is [[Definition:Idempotent Element|idempotent]]. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \paren {x \circ x_L} \circ \paren {x \circ x_L} + | r = x \circ \paren {x_L \circ x} \circ x_L + | c = as $\circ$ is [[Definition:Associative Operation|associative]] +}} +{{eqn | r = x \circ e_L \circ x_L + | c = {{Defof|Left Inverse Element}} +}} +{{eqn | r = x \circ x_L + | c = {{Defof|Left Identity}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Product of Semigroup Element with Right Inverse is Idempotent} +Tags: Semigroups, Idempotence, Product of Semigroup Element with Right Inverse is Idempotent + +\begin{theorem} +Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]] with a [[Definition:Right Identity|right identity]] $e_R$. +Let $x \in S$ such that $\exists x_R: x \circ x_R = e_R$, i.e. $x$ has a [[Definition:Right Inverse Element|right inverse]] with respect to the [[Definition:Right Identity|right identity]]. +Then: +:$\paren {x_R \circ x} \circ \paren {x_R \circ x} = x_R \circ x$ +That is, $x_R \circ x$ is [[Definition:Idempotent Element|idempotent]]. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \paren {x_R \circ x} \circ \paren {x_R \circ x} + | r = x_R \circ \paren {x \circ x_R} \circ x + | c = as $\circ$ is [[Definition:Associative Operation|associative]] +}} +{{eqn | r = x_R \circ e_R \circ x + | c = {{Defof|Right Inverse Element}} +}} +{{eqn | r = x_R \circ x + | c = {{Defof|Right Identity}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Left Inverse for All is Right Inverse} +Tags: Semigroups, Inverse Elements + +\begin{theorem} +Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]] with a [[Definition:Left Identity|left identity]] $e_L$ such that: +:$\forall x \in S: \exists x_L: x_L \circ x = e_L$ +That is, every [[Definition:Element|element]] of $S$ has a [[Definition:Left Inverse Element|left inverse]] with respect to the [[Definition:Left Identity|left identity]]. +Then $x \circ x_L = e_L$, that is, $x_L$ is also a [[Definition:Right Inverse Element|right inverse]] with respect to the [[Definition:Left Identity|left identity]]. +\end{theorem} + +\begin{proof} +Let $y = x \circ x_L$. Then: +{{begin-eqn}} +{{eqn | l = e_L \circ y + | r = \paren {y_L \circ y} \circ y + | c = {{Defof|Left Inverse Element}} +}} +{{eqn | r = y_L \circ \paren {y \circ y} + | c = as $\circ$ is [[Definition:Associative Operation|associative]] +}} +{{eqn | r = y_L \circ y + | c = [[Product of Semigroup Element with Left Inverse is Idempotent]]: $y = x \circ x_L$ +}} +{{eqn | r = e_L + | c = {{Defof|Left Inverse Element}} +}} +{{end-eqn}} +So $x \circ x_L = e_L$, and $x_L$ behaves as a [[Definition:Right Inverse Element|right inverse]] as well as a [[Definition:Left Inverse Element|left inverse]] with respect to the [[Definition:Left Identity|left identity]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Left Inverse and Right Inverse is Inverse} +Tags: Monoids, Inverse Elements + +\begin{theorem} +Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]] with [[Definition:Identity Element|identity element]] $e_S$. +Let $x \in S$ such that $x$ has both a [[Definition:Left Inverse Element|left inverse]] and a [[Definition:Right Inverse Element|right inverse]]. That is: +:$\exists x_L \in S: x_L \circ x = e_S$ +:$\exists x_R \in S: x \circ x_R = e_S$ +Then $x_L = x_R$, that is, $x$ has an [[Definition:Inverse Element|inverse]]. +Furthermore, that element is the ''only'' [[Definition:Inverse Element|inverse]] (both [[Definition:Right Inverse Element|right]] and [[Definition:Left Inverse Element|left]]) for $x$ +\end{theorem} + +\begin{proof} +We note that as $\struct {S, \circ}$ is a [[Definition:Monoid|monoid]], $\circ$ is [[Definition:Associative Operation|associative]] by definition. +{{begin-eqn}} +{{eqn | l = x_L + | r = x_L \circ e_S + | c = {{Defof|Identity Element}} +}} +{{eqn | r = x_L \circ \paren {x \circ x_R} + | c = {{Defof|Right Inverse Element}} +}} +{{eqn | r = \paren {x_L \circ x} \circ x_R + | c = {{Defof|Associative Operation}} +}} +{{eqn | r = e_S \circ x_R + | c = {{Defof|Left Inverse Element}} +}} +{{eqn | r = x_R + | c = {{Defof|Identity Element}} +}} +{{end-eqn}} +Its uniqueness comes from [[Inverse in Monoid is Unique]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Left and Right Inverses of Product} +Tags: Monoids + +\begin{theorem} +Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]] whose [[Definition:Identity Element|identity]] is $e_S$. +Let $x, y \in S$. +Let: +:$(1): \quad x \circ y$ have a [[Definition:Left Inverse Element|left inverse]] for $\circ$ +:$(2): \quad y \circ x$ have a [[Definition:Right Inverse Element|right inverse]] for $\circ$. +Then both $x$ and $y$ are [[Definition:Invertible Element|invertible]] for $\circ$. +\end{theorem} + +\begin{proof} +Let $z_L$ be the [[Definition:Left Inverse Element|left inverse]] of $x \circ y$ and $z_R$ be the [[Definition:Right Inverse Element|right inverse]] of $y \circ x$. Then: +{{begin-eqn}} +{{eqn | l = z_L \circ \paren {x \circ y} + | r = e_S + | c = +}} +{{eqn | ll= \leadsto + | l = \paren {z_L \circ x} \circ y + | r = e_S + | c = +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | l = \paren {y \circ x} \circ z_R + | r = e_S + | c = +}} +{{eqn | ll= \leadsto + | l = y \circ \paren {x \circ z_R} + | r = e_S + | c = +}} +{{end-eqn}} +Thus $y$ has both a [[Definition:Left Inverse Element|left inverse]] $z_L \circ x$ and a [[Definition:Right Inverse Element|right inverse]] $x \circ z_R$. +From [[Left Inverse and Right Inverse is Inverse]]: +:$z_L \circ x = x \circ z_R$ +and $y$ has an [[Definition:Inverse Element|inverse]], that is, is [[Definition:Invertible Element|invertible]]. +{{qed|lemma}} +From the above, we have: +{{begin-eqn}} +{{eqn | l = \paren {z_L \circ x} \circ y + | r = e_S + | c = +}} +{{eqn | ll= \leadsto + | l = \paren {x \circ z_R} \circ y + | r = e_S + | c = +}} +{{eqn | ll= \leadsto + | l = x \circ \paren {z_R \circ y} + | r = e_S + | c = +}} +{{end-eqn}} +and: +{{begin-eqn}} +{{eqn | l = y \circ \paren {x \circ z_R} + | r = e_S + | c = +}} +{{eqn | ll= \leadsto + | l = y \circ \paren {z_L \circ x} + | r = e_S + | c = +}} +{{eqn | ll= \leadsto + | l = \paren {y \circ z_L} \circ x + | r = e_S + | c = +}} +{{end-eqn}} +Thus $x$ has both a [[Definition:Left Inverse Element|left inverse]] $y \circ z_L$ and a [[Definition:Right Inverse Element|right inverse]] $z_R \circ y$. +From [[Left Inverse and Right Inverse is Inverse]]: +:$y \circ z_L = z_R \circ y$ +and $x$ has an [[Definition:Inverse Element|inverse]], that is, is [[Definition:Invertible Element|invertible]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Equivalence of Definitions of Self-Inverse} +Tags: Monoids, Inverse Elements + +\begin{theorem} +Let $\left({S, \circ}\right)$ be a [[Definition:Monoid|monoid]] whose [[Definition:Identity Element|identity]] is $e_S$. +Let $x \in S$. +{{TFAE|def = Self-Inverse Element|context = Abstract Algebra}} +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = x \circ x + | r = e_S + | c = +}} +{{eqn | ll= \iff + | l = x \circ x \circ x^{-1} + | r = e_S \circ x^{-1} + | c = +}} +{{eqn | ll= \iff + | l = x \circ e_S + | r = x^{-1} + | c = {{Defof|Inverse Element}} +}} +{{eqn | ll= \iff + | l = x + | r = x^{-1} + | c = {{Defof|Identity Element}} +}} +{{end-eqn}} +{{qed}} +[[Category:Monoids]] +[[Category:Inverse Elements]] +l27o6mhkoj3jr3hhe5126ucaxxzovvr +\end{proof}<|endoftext|> +\section{Inverse of Identity Element is Itself} +Tags: Inverse Elements, Identity Elements + +\begin{theorem} +Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]] with an [[Definition:Identity Element|identity element]] $e$. +Let the [[Definition:Inverse Element|inverse]] of $e$ be $e^{-1}$. +Then: +: $e^{-1} = e$ +That is, $e$ is [[Definition:Self-Inverse Element|self-inverse]]. +\end{theorem} + +\begin{proof} +From [[Identity Element is Idempotent]]: +:$e \circ e = e$ +Hence the result by definition of [[Definition:Identity Element|identity element]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Invertible Element of Associative Structure is Cancellable} +Tags: Associativity, Inverse Elements, Cancellability + +\begin{theorem} +Let $\left({S, \circ}\right)$ be an [[Definition:Algebraic Structure|algebraic structure]] where $\circ$ is [[Definition:Associative Operation|associative]]. +Let $\left({S, \circ}\right)$ have an [[Definition:Identity Element|identity element]] $e_S$. +An [[Definition:Element|element]] of $\left({S, \circ}\right)$ which is [[Definition:Invertible Element|invertible]] is also [[Definition:Cancellable Element|cancellable]]. +\end{theorem} + +\begin{proof} +Let $a \in S$ be [[Definition:Invertible Element|invertible]]. +Suppose $a \circ x = a \circ y$. +Then: +{{begin-eqn}} +{{eqn | l = x + | r = e_S \circ x + | c = Behaviour of [[Definition:Identity Element|Identity]] +}} +{{eqn | r = \left({a^{-1} \circ a}\right) \circ x + | c = Behaviour of [[Definition:Inverse Element|Inverse]] +}} +{{eqn | r = a^{-1} \circ \left({a \circ x}\right) + | c = [[Definition:Associative Operation|Associativity]] of $\circ$ +}} +{{eqn | r = a^{-1} \circ \left({a \circ y}\right) + | c = [[Definition:By Hypothesis|By Hypothesis]] +}} +{{eqn | r = \left({a^{-1} \circ a}\right) \circ y + | c = [Definition:Associative Operation|Associativity]] of $\circ$ +}} +{{eqn | r = e_S \circ y + | c = Behaviour of [[Definition:Inverse Element|Inverse]] +}} +{{eqn | r = y + | c = Behaviour of [[Definition:Identity Element|Identity]] +}} +{{end-eqn}} +A similar argument shows that $x \circ a = y \circ a \implies x = y$. +{{Qed}} +\end{proof}<|endoftext|> +\section{Regular Representation of Invertible Element is Permutation} +Tags: Monoids, Regular Representations + +\begin{theorem} +Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]]. +Let $a \in S$ be [[Definition:Invertible Element|invertible]]. +Then the [[Definition:Left Regular Representation|left regular representation]] $\lambda_a$ and the [[Definition:Right Regular Representation|right regular representation]] $\rho_a$ are [[Definition:Permutation|permutations]] of $S$. +\end{theorem} + +\begin{proof} +Suppose $a \in \struct {S, \circ}$ is [[Definition:Invertible Element|invertible]]. +A [[Definition:Permutation|permutations]] is a [[Definition:Bijection|bijection]] from a [[Definition:Set|set]] to itself. +As $\lambda_a: S \to S$ and $\rho_a: S \to S$ are defined from $S$ to $S$, all we need to do is show that they are [[Definition:Bijection|bijections]]. +To do that we can show that they are both [[Definition:Injection|injective]] and [[Definition:Surjection|surjective]]. +=== Injectivity === +From [[Invertible Element of Monoid is Cancellable]], as $a$ is [[Definition:Invertible Element|invertible]], it is also [[Definition:Cancellable Element|cancellable]]. +From [[Cancellable iff Regular Representations Injective]], it follows that both $\lambda_a$ and $\rho_a$ are [[Definition:Injection|injective]]. +{{qed|lemma}} +=== Surjectivity === +Let $y \in S$. +Then: +{{begin-eqn}} +{{eqn | n = 1 + | l = y + | r = \paren {a \circ a^{-1} } \circ y + | c = (as $a$ is [[Definition:Invertible Element|invertible]]) +}} +{{eqn | r = a \circ \paren {a^{-1} \circ y} + | c = [[Definition:Associative Operation|Associativity of $\circ$]] +}} +{{eqn | r = \lambda_a \paren {a^{-1} \circ y} + | c = [[Definition:Left Regular Representation|Definition of $\lambda_a$]] +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | n = 2 + | l = y + | r = y \circ \paren {a^{-1} \circ a} \circ y + | c = (as $a$ is [[Definition:Invertible Element|invertible]]) +}} +{{eqn | r = \paren {y \circ a^{-1} } \circ a + | c = [[Definition:Associative Operation|Associativity of $\circ$]] +}} +{{eqn | r = \rho_a \paren {y \circ a^{-1} } + | c = [[Definition:Right Regular Representation|Definition of $\rho_a$]] +}} +{{end-eqn}} +Thus both $\lambda_a$ and $\rho_a$ are [[Definition:Surjection|surjective]]. +{{qed|lemma}} +So $\lambda_a$ and $\rho_a$ are [[Definition:Injection|injective]] and [[Definition:Surjection|surjective]], and therefore [[Definition:Bijection|bijections]], and thus [[Definition:Permutation|permutationss]] of $S$. +{{qed}} +\end{proof}<|endoftext|> +\section{Group is Inverse Semigroup with Identity} +Tags: Group Theory + +\begin{theorem} +A [[Definition:Group|group]] is an [[Definition:Inverse Semigroup|inverse semigroup]] with an [[Definition:Identity (Abstract Algebra)|identity]]. +\end{theorem} + +\begin{proof} +Let $\left({S, \circ}\right)$ be a [[Definition:Group|group]]. Then for all $a \in S$: +{{begin-eqn}} +{{eqn | l=e + | r=a \circ a^{-1} + | c=Every $a \in G$ is [[Definition:Invertible Element|Invertible]] +}} +{{eqn | ll=\implies + | l=e \circ a + | r=a \circ a^{-1} \circ a + | c= +}} +{{eqn | ll=\implies + | l=a + | r=a \circ a^{-1} \circ a + | c=[[Definition:Identity (Abstract Algebra)|Definition of Identity]] +}} +{{end-eqn}} +and +{{begin-eqn}} +{{eqn | l=e + | r=a \circ a^{-1} + | c=Every $a \in G$ is [[Definition:Invertible Element|Invertible]] +}} +{{eqn | ll=\implies + | l=a^{-1} \circ e + | r=a^{-1} \circ a \circ a^{-1} + | c= +}} +{{eqn | ll=\implies + | l=a^{-1} + | r=a^{-1} \circ a \circ a^{-1} + | c=[[Definition:Identity (Abstract Algebra)|Definition of Identity]] +}} +{{end-eqn}} +Thus the criteria of an [[Definition:Inverse Semigroup|inverse semigroup]] are fulfilled. +{{qed}} +[[Category:Group Theory]] +gkhnvfry9jz2o0x0xbb4iy7aq2uz4nt +\end{proof}<|endoftext|> +\section{Invertible Elements of Monoid form Subgroup of Cancellable Elements} +Tags: Group Theory, Monoids, Cancellability + +\begin{theorem} +Let $\struct {S, \circ}$ be an [[Definition:Monoid|monoid]] whose [[Definition:Identity Element|identity]] is $e_S$. +Let $C$ be the [[Definition:Set|set]] of all [[Definition:Cancellable Element|cancellable elements]] of $S$. +Let $T$ be the [[Definition:Set|set]] of all [[Definition:Invertible Element|invertible elements]] of $S$. +Then $\struct {T, \circ}$ is a [[Definition:Subgroup|subgroup]] of $\struct {C, \circ}$. +\end{theorem} + +\begin{proof} +From [[Cancellable Elements of Monoid form Submonoid]], $\struct {C, \circ}$ is a [[Definition:Submonoid|submonoid]] of $\struct {S, \circ}$. +Let its [[Definition:Identity Element|identity]] be $e_C$ (which may or may not be the same as $e_S$). +Let $T$ be the [[Definition:Set|set]] of all [[Definition:Invertible Element|invertible elements]] of $S$. +From [[Invertible Element of Monoid is Cancellable]], all the [[Definition:Invertible Element|invertible elements]] of $S$ are also all [[Definition:Cancellable Element|cancellable]], so $T \subseteq C$. +Let $x, y \in T$. +Clearly $x^{-1}, y^{-1} \in T$, as if $x, y$ are [[Definition:Invertible Element|invertible]], then so are their [[Definition:Inverse Element|inverses]]. +Taking the [[Definition:Group Axioms|group axioms]] in turn: +=== G0: Closure === +By [[Inverse of Product]], $x^{-1} \circ y^{-1} \in T$. +Thus $\struct {T, \circ}$ is [[Definition:Closed Algebraic Structure|closed]]. +{{qed|lemma}} +=== G1: Associativity === +This is inherited from $S$, by [[Subsemigroup Closure Test]]. +Thus $\struct {T, \circ}$ is [[Definition:Associative Operation|associative]]. +{{qed|lemma}} +=== G2: Identity === +All the elements of $\struct {C, \circ}$ are by definition [[Definition:Cancellable Element|cancellable]] +So, by [[Identity of Cancellable Monoid is Identity of Submonoid]]: +: $e_C \in T$ +Thus $\struct {T, \circ}$ has an [[Definition:Identity Element|identity element]]. +{{qed|lemma}} +=== G3: Inverses === +By [[Inverse of Inverse in Monoid]]: +:$\struct {x^{-1} \circ y^{-1} }^{-1} \in T$ +Thus every element of $\struct {T, \circ}$ has an [[Definition:Inverse Element|inverse]]. +{{qed|lemma}} +All the [[Definition:Group Axioms|group axioms]] are thus seen to be fulfilled, and so $\struct {T, \circ}$ is a [[Definition:Group|group]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Structure Induced by Associative Operation is Associative} +Tags: Associativity, Pointwise Operations + +\begin{theorem} +Let $\struct {T, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]], and let $S$ be a [[Definition:Set|set]]. +Let $\struct {T^S, \oplus}$ be the [[Definition:Induced Structure|structure on $T^S$ induced]] by $\circ$. +Let $\circ$ be [[Definition:Associative Operation|associative]]. +Then the [[Definition:Pointwise Operation|pointwise operation $\oplus$ induced]] on $T^S$ by $\circ$ is also [[Definition:Associative Operation|associative]]. +\end{theorem} + +\begin{proof} +Let $f, g, h \in T^S$. +Let $\struct {T, \circ}$ be an [[Definition:Associative Algebraic Structure|associative algebraic structure]]. +Then: +{{begin-eqn}} +{{eqn | l = \map {\paren {\paren {f \oplus g} \oplus h} } x + | r = \map {\paren {f \oplus g} } x \circ \map h x + | c = {{Defof|Pointwise Operation}} +}} +{{eqn | r = \paren {\map f x \circ \map g x} \circ \map h x + | c = {{Defof|Pointwise Operation}} +}} +{{eqn | r = \map f x \circ \paren {\map g x \circ \map h x} + | c = $\circ$ is [[Definition:Associative Operation|associative]] +}} +{{eqn | r = \map f x \circ \map {\paren {g \oplus h} } x + | c = {{Defof|Pointwise Operation}} +}} +{{eqn | r = \map {\paren {f \oplus \paren {g \oplus h} } } x + | c = {{Defof|Pointwise Operation}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Structure Induced by Commutative Operation is Commutative} +Tags: Pointwise Operations, Commutativity + +\begin{theorem} +Let $\struct {T, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]], and let $S$ be a [[Definition:Set|set]]. +Let $\struct {T^S, \oplus}$ be the [[Definition:Induced Structure|structure on $T^S$ induced]] by $\circ$. +Let $\circ$ be a [[Definition:Commutative Operation|commutative operation]]. +Then the [[Definition:Pointwise Operation|pointwise operation $\oplus$ induced]] on $T^S$ by $\circ$ is also [[Definition:Commutative Operation|commutative]]. +\end{theorem} + +\begin{proof} +Let $\struct {T, \circ}$ be a [[Definition:Commutative Algebraic Structure|commutative algebraic structure]]. +Let $f, g \in T^S$. +Then: +{{begin-eqn}} +{{eqn | l = \map {\paren {f \oplus g} } x + | r = \map f x \circ \map g x + | c = {{Defof|Pointwise Operation}} +}} +{{eqn | r = \map g x \circ \map f x + | c = $\circ$ is [[Definition:Commutative Operation|commutative]] +}} +{{eqn | r = \map {\paren {g \oplus f} } x + | c = {{Defof|Pointwise Operation}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Induced Structure Identity} +Tags: Identity Elements, Pointwise Operations + +\begin{theorem} +Let $\struct {T, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]], and let $S$ be a [[Definition:Set|set]]. +Let $\struct {T^S, \oplus}$ be the [[Definition:Induced Structure|structure on $T^S$ induced]] by $\circ$. +Let $e$ be an [[Definition:Identity Element|identity]] for $\circ$. +Then the [[Definition:Constant Mapping|constant mapping]] $f_e: S \to T$ defined as: +:$\forall x \in S: \map {f_e} x = e$ +is the [[Definition:Identity Element|identity]] for the [[Definition:Pointwise Operation|pointwise operation $\oplus$ induced]] on $T^S$ by $\circ$. +\end{theorem} + +\begin{proof} +Let $\struct {T, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]] with an [[Definition:Identity Element|identity]] $e$. +Let $f \in T^S$. +Then: +{{begin-eqn}} +{{eqn | l = \map {\paren {f \oplus f_e} } x + | r = \map f x \circ \map {f_e} x + | c = {{Defof|Pointwise Operation}} +}} +{{eqn | r = \map f x \circ e + | c = {{Defof|Constant Mapping}} $f_e$ +}} +{{eqn | r = \map f x + | c = $e$ is the [[Definition:Identity Element|identity]] of $\struct {T, \circ}$ +}} +{{eqn | r = e \circ \map f x + | c = $e$ is the [[Definition:Identity Element|identity]] of $\struct {T, \circ}$ +}} +{{eqn | r = \map {f_e} x \circ \map f x + | c = {{Defof|Constant Mapping}} $f_e$ +}} +{{eqn | r = \map {\paren {f \oplus f_e} } x + | c = {{Defof|Pointwise Operation}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Structure Induced by Abelian Group Operation is Abelian Group} +Tags: Abelian Groups, Pointwise Operations + +\begin{theorem} +Let $\struct {G, \circ}$ be an [[Definition:Abelian Group|abelian group]] whose [[Definition:Identity Element|identity]] is $e$. +Let $S$ be a [[Definition:Set|set]]. +Let $\struct {G^S, \oplus}$ be the [[Definition:Induced Structure|structure on $G^S$ induced]] by $\circ$. +Then $\struct {G^S, \oplus}$ is an [[Definition:Abelian Group|abelian group]]. +\end{theorem} + +\begin{proof} +From [[Structure Induced by Group Operation is Group]], $\struct {G^S, \oplus}$ is a [[Definition:Group|group]]. +From [[Structure Induced by Commutative Operation is Commutative]], so is the [[Definition:Pointwise Operation|pointwise operation it induces]] on $G^S$. +Hence the result. +{{qed}} +[[Category:Abelian Groups]] +[[Category:Pointwise Operations]] +gffz66j74cdmaqb4seldfypi6jcr7vr +\end{proof}<|endoftext|> +\section{Power Set with Union is Commutative Monoid} +Tags: Examples of Monoids, Set Union, Power Set + +\begin{theorem} +Let $S$ be a [[Definition:Set|set]] and let $\powerset S$ be its [[Definition:Power Set|power set]]. +Then $\struct {\powerset S, \cup}$ is a [[Definition:Commutative Monoid|commutative monoid]] whose [[Definition:Identity Element|identity]] is $\O$. +The only [[Definition:Invertible Element|invertible element]] of this [[Definition:Algebraic Structure|structure]] is $\O$. +Thus (except in the [[Definition:Degenerate Case|degenerate case]] $S = \O$) $\struct {\powerset S, \cup}$ cannot be a [[Definition:Group|group]]. +\end{theorem} + +\begin{proof} +From [[Power Set is Closed under Union]]: +:$\forall A, B \in \powerset S: A \cup B \in \powerset S$ +From [[Set System Closed under Union is Commutative Semigroup]], $\struct {\powerset S, \cup}$ is a [[Definition:Commutative Semigroup|commutative semigroup]]. +From [[Identity of Power Set with Union]], $\O$ acts as the [[Definition:Identity Element|identity]] of $\struct {\powerset S, \cup}$. +It remains to be shown that only $\O$ has an [[Definition:Inverse Element|inverse]]: +For $T \subseteq S$ to have an [[Definition:Inverse Element|inverse]] under $\cup$, we require $T^{-1} \cup T = \O$. +From this it follows that $T = \O = T^{-1}$. +The result follows by definition of [[Definition:Commutative Monoid|commutative monoid]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Power Set with Intersection is Commutative Monoid} +Tags: Examples of Monoids, Power Set, Set Intersection + +\begin{theorem} +Let $S$ be a [[Definition:Set|set]] and let $\powerset S$ be its [[Definition:Power Set|power set]]. +Then $\struct {\powerset S, \cap}$ is a [[Definition:Commutative Monoid|commutative monoid]] whose [[Definition:Identity Element|identity]] is $S$. +The only [[Definition:Invertible Element|invertible element]] of this [[Definition:Algebraic Structure|structure]] is $S$. +Thus (except in the [[Definition:Degenerate Case|degenerate case]] $S = \O$) $\struct {\powerset S, \cap}$ cannot be a [[Definition:Group|group]]. +\end{theorem} + +\begin{proof} +From [[Power Set is Closed under Intersection]]: +:$\forall A, B \in \powerset S: A \cap B \in \powerset S$ +From [[Set System Closed under Intersection is Commutative Semigroup]], $\struct {\powerset S, \cap}$ is a [[Definition:Commutative Semigroup|commutative semigroup]]. +From [[Identity of Power Set with Intersection]], we have that $S$ acts as the [[Definition:Identity Element|identity]]. +It remains to be shown that only $S$ has an [[Definition:Inverse Element|inverse]]: +For $T \subseteq S$ to have an [[Definition:Inverse Element|inverse]] under $\cap$, we require $T^{-1} \cap T = S$. +From this it follows that $T = S = T^{-1}$. +The result follows by definition of [[Definition:Commutative Monoid|commutative monoid]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Diagonal Relation is Universally Compatible} +Tags: Compatible Relations + +\begin{theorem} +The [[Definition:Diagonal Relation|diagonal relation]] $\Delta_S$ on a [[Definition:Set|set]] $S$ is [[Definition:Universally Compatible Relation|universally compatible]] with every [[Definition:Binary Operation|operation]] on $S$. +\end{theorem} + +\begin{proof} +{{questionable|It needs to be pointed out that the definition of [[Definition:Universally Compatible Relation|universally compatible relation]] may not be correct as such. I will address it when I work up to that chapter in my book.}} +Let $\struct {S, \circ}$ be any [[Definition:Algebraic Structure|algebraic structure]]. +{{begin-eqn}} +{{eqn | o = + | r = x_1 \Delta_S x_2 \land y_1 \Delta_S y_2 + | c = +}} +{{eqn | o = \leadsto + | r = x_1 = x_2 \land y_1 = y_2 + | c = {{Defof|Diagonal Relation}} +}} +{{eqn | o = \leadsto + | r = x_1 \circ y_1 = x_2 \circ y_2 + | c = (consequence of equality) +}} +{{eqn | o = \leadsto + | r = \paren {x_1 \circ y_1} \mathrel {\Delta_S} \paren {x_2 \circ y_2} + | c = {{Defof|Diagonal Relation}} +}} +{{end-eqn}} +{{qed}} +[[Category:Compatible Relations]] +pug8w8n3as389p9zzsr5bp1d4o64vbt +\end{proof}<|endoftext|> +\section{Trivial Relation is Universally Congruent} +Tags: Congruence Relations, Trivial Relation + +\begin{theorem} +The [[Definition:Trivial Relation|trivial relation]] $\mathcal R = S \times S$ on a [[Definition:Set|set]] $S$ is [[Definition:Universally Congruent|universally congruent]] with every [[Definition:Closed Algebraic Structure|closed operation]] on $S$. +\end{theorem} + +\begin{proof} +Let $\left({S, \circ}\right)$ be any [[Definition:Algebraic Structure|algebraic structure]] which is [[Definition:Closed Algebraic Structure|closed]] for $\circ$. +By definition of [[Definition:Trivial Relation|trivial relation]], $x \in S \land y \in S \implies x \mathcal R y$. So: +{{begin-eqn}} +{{eqn | l=x_1, x_2, y_1, y_2 + | o=\in + | r=S + | c= +}} +{{eqn | ll=\implies + | l=x_1 \circ y_1, x_2 \circ y_2 + | o=\in + | r=S + | c=Definition of [[Definition:Closed Algebraic Structure|closed algebraic structure]] +}} +{{eqn | ll=\implies + | l=\left({x_1 \circ y_1}\right) + | o=\mathcal R + | r=\left({x_2 \circ y_2}\right) + | c=Definition of [[Definition:Trivial Relation|Trivial Relation]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Equivalence Relation is Congruence for Constant Operation} +Tags: Constant Operation, Congruence Relations + +\begin{theorem} +Every [[Definition:Equivalence Relation|equivalence relation]] is a [[Definition:Congruence Relation|congruence relation]] for the [[Definition:Constant Operation|constant operation]]. +\end{theorem} + +\begin{proof} +Let $c \in S$. +By the definition of the [[Definition:Constant Operation|constant operation]]: +:$\forall x, y \in S: x \sqbrk c y = c$ +Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on $S$. +Every [[Definition:Equivalence Relation|equivalence relation]] is [[Definition:Reflexive Relation|reflexive]], so: +:$c \mathrel \RR c$ +So: +{{begin-eqn}} +{{eqn | l = x_1 \mathrel \RR x_2 + | o = \land + | r = y_1 \mathrel \RR y_2 + | c = +}} +{{eqn | ll= \leadsto + | l = \paren {x_1 \sqbrk c y_1} + | o = \RR + | r = \paren {x_2 \sqbrk c y_2} + | c = [[True Statement is implied by Every Statement]] +}} +{{end-eqn}} +Hence the result. +{{Qed}} +\end{proof}<|endoftext|> +\section{Quotient Structure is Well-Defined} +Tags: Quotient Structures + +\begin{theorem} +Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]]. +Let $\RR$ be a [[Definition:Congruence Relation|congruence relation]] on $\struct {S, \circ}$. +Let $S / \RR$ be the [[Definition:Quotient Set|quotient set of $S$ by $\RR$]]. +Let $\circ_\RR$ be the [[Definition:Operation Induced on Quotient Set|operation induced on $S / \RR$ by $\circ$]]. +Then $\circ_\RR$ is a [[Definition:Well-Defined Operation|well-defined operation]] in the [[Definition:Quotient Structure|quotient structure]] $\struct {S / \RR, \circ_\RR}$. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \eqclass {x_1} \RR = \eqclass {x_2} \RR + | o = \land + | r = \eqclass {y_1} \RR = \eqclass {y_2} \RR + | c = +}} +{{eqn | ll= \leadsto + | l = x_1 \mathop \RR x_2 + | o = \land + | r = y_1 \mathop \RR y_2 + | c = {{Defof|Equivalence Class}} +}} +{{eqn | ll= \leadsto + | l = \paren {x_1 \circ y_1} + | o = \RR + | r = \paren {x_2 \circ y_2} + | c = {{Defof|Congruence Relation}} +}} +{{eqn | ll= \leadsto + | l = \eqclass {x_1 \circ y_1} \RR + | r = \eqclass {x_2 \circ y_2} \RR + | c = {{Defof|Equivalence Class}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Quotient Structure on Subset Product} +Tags: Quotient Structures, Subset Products + +\begin{theorem} +Let $\left({S, \circ}\right)$ be an [[Definition:Algebraic Structure|algebraic structure]]. +Let $\mathcal R$ be a [[Definition:Congruence Relation|congruence]] for $\circ$ on $S$. +Then: +:$\forall X, Y \in S / \mathcal R: X \circ_\mathcal P Y \subseteq X \circ_\mathcal R Y$ +where: +: $S / \mathcal R$ is the [[Definition:Quotient Set|quotient of $S$ by $\mathcal R$]] +: $\circ_\mathcal P$ is the [[Definition:Subset Product|operation induced on $\mathcal P \left({S}\right)$ by $\circ$]] +: $\circ_\mathcal R$ is the [[Definition:Operation Induced on Quotient Set|operation induced on $S / \mathcal R$ by $\circ$]] +\end{theorem} + +\begin{proof} +By definition of [[Definition:Subset Product|subset product]]: +:$X \circ_\mathcal P Y = \left\{{x \circ y: x \in X, y \in Y}\right\}$ +Thus: +:$X \circ_\mathcal R Y = \left\{{x \circ y: x \in X, y \in Y}\right\} \cup \left\{{x \circ y: x \in \left[\!\left[{X}\right]\!\right]_\mathcal R, y \in \left[\!\left[{Y}\right]\!\right]_\mathcal R}\right\}$ +The result follows from [[Subset of Union]]. +{{qed}} +\end{proof}<|endoftext|> +\section{External Direct Product Closure} +Tags: Direct Products, External Direct Products + +\begin{theorem} +Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be [[Definition:Algebraic Structure|algebraic structures]]. +Let $\struct {S \times T, \circ}$ be the [[Definition:External Direct Product|external direct product]] of $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$. +Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be [[Definition:Closed Algebraic Structure|closed]]. +Then $\struct {S \times T, \circ}$ is also [[Definition:Closed Algebraic Structure|closed]]. +\end{theorem} + +\begin{proof} +Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be [[Definition:Closed Algebraic Structure|closed]]. +Let $\tuple {s_1, t_1} \in S \times T$ and $\tuple {s_2, t_2} \in S \times T$. +Then: +{{begin-eqn}} +{{eqn | l = \tuple {s_1, t_1} \circ \tuple {s_2, t_2} + | r = \tuple {s_1 \circ_1 s_2, t_1 \circ_1 t_2} + | c = {{Defof|External Direct Product}} +}} +{{eqn | o = \in + | r = S \times T + | c = as $S$ and $T$ are [[Definition:Closed Algebraic Structure|closed]]: $s_1 \circ_1 s_2 \in S, t_1 \circ_1 t_2 \in T$ +}} +{{end-eqn}} +demonstrating that $\struct {S \times T, \circ}$ is [[Definition:Closed Algebraic Structure|closed]]. +{{qed}} +[[Category:Direct Products]] +[[Category:External Direct Products]] +snavudfreayq56xw6zpqcy7nt5h6dok +\end{proof}<|endoftext|> +\section{External Direct Product Associativity} +Tags: Direct Products, External Direct Products + +\begin{theorem} +Let $\struct {S \times T, \circ}$ be the [[Definition:External Direct Product|external direct product]] of the two [[Definition:Algebraic Structure|algebraic structures]] $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$. +Let $\circ_1$ and $\circ_2$ be [[Definition:Associative Operation|associative]]. +Then $\circ$ is also [[Definition:Associative Operation|associative]]. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \paren {\tuple {s_1, t_1} \circ \tuple {s_2, t_2} } \circ \tuple {s_3, t_3} + | r = \tuple {\paren {s_1 \circ_1 s_2} \circ_1 s_3, \paren {t_1 \circ_2 t_2} \circ_2 t_3} + | c = +}} +{{eqn | r = \tuple {s_1 \circ_1 s_2 \circ_1 s_3, t_1 \circ_2 t_2 \circ_2 t_3} + | c = {{Defof|Associative Operation}} +}} +{{eqn | r = \tuple {s_1 \circ_1 \paren {s_2 \circ_1 s_3}, t_1 \circ_2 \paren {t_2 \circ_2 t_3} } + | c = +}} +{{eqn | r = \tuple {s_1, t_1} \circ \paren {\tuple {s_2, t_2} \circ \tuple {s_3, t_3} } + | c = +}} +{{end-eqn}} +Thus $\circ$ is [[Definition:Associative Operation|associative]]. +{{qed}} +\end{proof}<|endoftext|> +\section{External Direct Product of Semigroups} +Tags: Direct Products, Semigroups + +\begin{theorem} +The [[Definition:External Direct Product|external direct product]] of two [[Definition:Semigroup|semigroups]] is itself a [[Definition:Semigroup|semigroup]]. +\end{theorem} + +\begin{proof} +* From [[External Direct Product Closure]], the [[Definition:External Direct Product|external direct product]] of two closed [[Definition:Algebraic Structure|algebraic structures]] is [[Definition:Closure (Abstract Algebra)|closed]]. +* From [[External Direct Product Associativity]], the [[Definition:External Direct Product|external direct product]] of two [[Definition:Associative Algebraic Structure|associative algebraic structures]] is [[Definition:Associative Operation|associative]]. +The result follows. +{{qed}} +[[Category:Direct Products]] +[[Category:Semigroups]] +hfejmyry60s1o3wdbvteyfdeqxcglrt +\end{proof}<|endoftext|> +\section{External Direct Product Commutativity} +Tags: External Direct Products + +\begin{theorem} +Let $\left({S \times T, \circ}\right)$ be the [[Definition:External Direct Product|external direct product]] of the two [[Definition:Algebraic Structure|algebraic structures]] $\left({S, \circ_1}\right)$ and $\left({T, \circ_2}\right)$. +Let $\circ_1$ and $\circ_2$ be [[Definition:Commutative Operation|commutative operations]]. +Then $\circ$ is also a [[Definition:Commutative Operation|commutative operation]]. +\end{theorem} + +\begin{proof} +Let $\circ_1$ and $\circ_2$ be [[Definition:Commutative Operation|commutative operations]]. +{{begin-eqn}} +{{eqn | l = \left({s_1, t_1}\right) \circ \left({s_2, t_2}\right) + | r = \left({s_1 \circ_1 s_2, t_1 \circ_2 t_2}\right) + | c = +}} +{{eqn | r = \left({s_2 \circ_1 s_1, t_2 \circ_2 t_1}\right) + | c = +}} +{{eqn | r = \left({s_2, t_2}\right) \circ \left({s_1, t_1}\right) + | c = +}} +{{end-eqn}} +Thus $\circ$ is [[Definition:Commutative Operation|commutative]]. +{{qed}} +\end{proof}<|endoftext|> +\section{External Direct Product Identity} +Tags: Direct Products, External Direct Products + +\begin{theorem} +Let $\struct {S \times T, \circ}$ be the [[Definition:External Direct Product|external direct product]] of the two [[Definition:Monoid|monoids]] $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$. +Let: +: $e_S$ be the [[Definition:Identity Element|identity]] for $\struct {S, \circ_1}$ +and: +: $e_T$ be the [[Definition:Identity Element|identity]] for $\struct {T, \circ_2}$. +Then $\tuple {e_S, e_T}$ is the [[Definition:Identity Element|identity]] for $\struct {S \times T, \circ}$. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \tuple {s, t} \circ \tuple {e_S, e_T} + | r = \tuple {s \circ_1 e_S, t \circ_2 e_T} + | c = +}} +{{eqn | r = \tuple {s, t} + | c = +}} +{{eqn | l = \tuple {e_S, e_T} \circ \tuple {s, t} + | r = \tuple {e_S \circ_1 s, e_T \circ_2 t} = \left({s, t}\right) + | c = +}} +{{eqn | r = \tuple {s, t} + | c = +}} +{{end-eqn}} +Thus $\tuple {e_S, e_T}$ is the [[Definition:Identity Element|identity]] of $\struct {S \times T, \circ}$. +{{qed}} +\end{proof}<|endoftext|> +\section{External Direct Product Inverses} +Tags: External Direct Products + +\begin{theorem} +Let $\struct {S \times T, \circ}$ be the [[Definition:External Direct Product|external direct product]] of the two [[Definition:Monoid|monoids]] $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$. +Let: +:$s^{-1}$ be an [[Definition:Inverse Element|inverse]] of $s \in \struct {S, \circ_1}$ +and: +:$t^{-1}$ be an [[Definition:Inverse Element|inverse]] of $t \in \struct {T, \circ_2}$. +Then $\tuple {s^{-1}, t^{-1} }$ is an [[Definition:Inverse Element|inverse]] of $\tuple {s, t} \in \struct {S \times T, \circ}$. +\end{theorem} + +\begin{proof} +Let: +:$e_S$ be the [[Definition:Identity Element|identity]] for $\struct {S, \circ_1}$ +and: +:$e_T$ be the [[Definition:Identity Element|identity]] for $\struct {T, \circ_2}$. +Also let: +:$s^{-1}$ be the [[Definition:Inverse Element|inverse]] of $s \in \struct {S, \circ_1}$ +and +:$t^{-1}$ be the [[Definition:Inverse Element|inverse]] of $t \in \struct {T, \circ_2}$. +Then: +{{begin-eqn}} +{{eqn | l = \tuple {s, t} \circ \tuple {s^{-1}, t^{-1} } + | r = \tuple {s \circ_1 s^{-1}, t \circ_2 t^{-1} } + | c = +}} +{{eqn | r = \tuple {e_S, e_T} + | c = +}} +{{eqn | r = \tuple {s^{-1} \circ_1 s, t^{-1} \circ_2 t} + | c = +}} +{{eqn | r = \tuple {s^{-1}, t^{-1} } \circ \tuple {s, t} + | c = +}} +{{end-eqn}} +Thus the [[Definition:Inverse Element|inverse]] of $\tuple {s, t}$ is $\tuple {s^{-1}, t^{-1} }$. +{{qed}} +\end{proof}<|endoftext|> +\section{Morphism Property Preserves Closure} +Tags: Homomorphisms + +\begin{theorem} +Let $\phi: \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({T, *_1, *_2, \ldots, *_n}\right)$ be a [[Definition:Mapping|mapping]] from one [[Definition:Algebraic Structure|algebraic structure]] $\left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$ to another $\left({T, *_1, *_2, \ldots, *_n}\right)$. +Let $\circ_k$ have the [[Definition:Morphism Property|morphism property]] under $\phi$ for some operation $\circ_k$ in $\left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$. +Then the following properties hold: +: If $S' \subseteq S$ is [[Definition:Closed Algebraic Structure|closed]] under $\circ_k$, then $\phi \left[{S'}\right]$ is [[Definition:Closed Algebraic Structure|closed]] under $*_k$ +: If $T' \subseteq T$ is [[Definition:Closed Algebraic Structure|closed]] under $*_k$, then $\phi^{-1} \left[{T'}\right]$ is [[Definition:Closed Algebraic Structure|closed]] under $\circ_k$ +where $\phi \left[{S'}\right]$ denotes the [[Definition:Image of Subset under Mapping|image of $S'$]]. +\end{theorem} + +\begin{proof} +Suppose that $\circ_k$ has the [[Definition:Morphism Property|morphism property]] under $\phi$. +Suppose that $S' \subseteq S$ is [[Definition:Closed Algebraic Structure|closed]] under $\circ_k$. +Thus: +:$s_1, s_2 \in S' \implies s_1 \circ_k s_2 \in S'$ +Similarly, suppose that $T' \subseteq T$ is [[Definition:Closed Algebraic Structure|closed]] under $*_k$. +Thus: +:$t_1, t_2 \in T' \implies t_1 *_k t_2 \in T'$ +First we prove that $\phi \left[{S'}\right]$ is closed under $*_k$: +{{begin-eqn}} +{{eqn | l = t_1, t_2 + | o = \in + | r = \phi \left[{S'}\right] +}} +{{eqn | ll= \implies + | l = \exists s_1 \in S': t_1 + | r = \phi \left({s_1}\right) + | c = Definition of [[Definition:Image of Subset under Mapping|Image of Subset]] +}} +{{eqn | ll= \land + | l = \exists s_2 \in S': t_2 + | r = \phi \left({s_2}\right) +}} +{{eqn | ll= \implies + | l = t_1 *_k t_2 + | r = \phi \left({s_1}\right) *_k \phi \left({s_2}\right) + | c = +}} +{{eqn | r = \phi \left({s_1 \circ_k s_2}\right) + | c = Definition of [[Definition:Morphism Property|Morphism Property]] +}} +{{eqn | ll= \implies + | l = t_1 *_k t_2 + | o = \in + | r = \phi \left[{S'}\right] + | c = $S'$ is [[Definition:Closed Algebraic Structure|closed]] under $\circ$ +}} +{{end-eqn}} +Then we prove that $\phi^{-1} \left({T'}\right)$ is [[Definition:Closed Algebraic Structure|closed]] under $\circ_k$: +{{begin-eqn}} +{{eqn | l = s_1, s_2 + | o = \in + | r = \phi^{-1} \left[{T'}\right] +}} +{{eqn | ll= \implies + | l = \phi \left({s_1}\right), \phi \left({s_2}\right) + | o = \in + | r = T' + | c = Definition of [[Definition:Inverse Mapping|Inverse Mapping]] +}} +{{eqn | l = \phi \left({s_1}\right) *_k \phi \left({s_2}\right) + | o = \in + | r = T' + | c = +}} +{{eqn | ll= \implies + | l = \phi \left({s_1 \circ_k s_2}\right) + | o = \in + | r = T' + | c = Definition of [[Definition:Morphism Property|Morphism Property]] +}} +{{eqn | ll= \implies + | l = s_1 \circ_k s_2 + | o = \in + | r = \phi^{-1} \left[{T'}\right] + | c = Definition of [[Definition:Inverse Mapping|Inverse Mapping]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Morphism Property Preserves Cancellability} +Tags: Homomorphisms + +\begin{theorem} +Let: +: $\phi: \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({T, *_1, *_2, \ldots, *_n}\right)$ +be a [[Definition:Mapping|mapping]] from one [[Definition:Algebraic Structure|algebraic structure]]: +: $\left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$ +to another: +: $\left({T, *_1, *_2, \ldots, *_n}\right)$ +Let $\circ_k$ have the [[Definition:Morphism Property|morphism property]] under $\phi$ for some operation $\circ_k$ in $\left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$. +Then if an [[Definition:Element|element]] $a \in S$ is either [[Definition:Left Cancellable Element|left cancellable]] or [[Definition:Right Cancellable Element|right cancellable]] under $\circ_k$, then $\phi \left({a}\right)$ is correspondingly [[Definition:Left Cancellable Element|left cancellable]] or [[Definition:Right Cancellable Element|right cancellable]] under $*_k$. +Thus, the [[Definition:Morphism Property|morphism property]] is seen to preserve [[Definition:Cancellable Element|cancellability]]. +\end{theorem} + +\begin{proof} +We need to demonstrate the following properties: +* If $a \in S$ has the property that: +:$\forall x, y \in S: x \circ_k a = y \circ_k a \implies x = y$ +then: +:$\forall x, y \in S: \phi \left({x}\right) *_k \phi \left({a}\right) = \phi \left({y}\right) *_k \phi \left({a}\right) \implies \phi \left({x}\right) = \phi \left({y}\right)$ +* If $a \in S$ has the property that: +:$\forall x, y \in S: a \circ_k x = a \circ_k y \implies x = y$ +then: +:$\forall x, y \in S: \phi \left({a}\right) *_k \phi \left({x}\right) = \phi \left({a}\right) *_k \phi \left({y}\right) \implies \phi \left({x}\right) = \phi \left({y}\right)$ +* To show the first of the properties above: +{{begin-eqn}} +{{eqn | o= + | r=\left({x \circ_k a = y \circ_k a \implies x = y}\right) + | c= +}} +{{eqn | o=\implies + | r=\left({\phi \left({x \circ_k a}\right) = \phi \left({y \circ_k a}\right) \implies \phi \left({x}\right) = \phi \left({y}\right)}\right) + | c=[[Definition:Mapping|Mappings]] are [[Definition:Many-to-One Relation|many-to-one by definition]] +}} +{{eqn | o=\implies + | r=\left({\phi \left({x}\right) *_k \phi \left({a}\right) = \phi \left({y}\right) *_k \phi \left({a}\right) \implies \phi \left({x}\right) = \phi \left({y}\right)}\right) + | c=[[Definition:Morphism Property|Morphism Property]] +}} +{{end-eqn}} +and thus [[Definition:Left Cancellable Element|left cancellability]] is demonstrated. +* And the second is like it, namely this: +{{begin-eqn}} +{{eqn | o= + | r=\left({a \circ_k x = a \circ_k y \implies x = y}\right) + | c= +}} +{{eqn | o=\implies + | r=\left({\phi \left({a \circ_k x}\right) = \phi \left({a \circ_k y}\right) \implies \phi \left({x}\right) = \phi \left({y}\right)}\right) + | c=[[Definition:Mapping|Mappings]] are [[Definition:Many-to-One Relation|many-to-one by definition]] +}} +{{eqn | o=\implies + | r=\left({\phi \left({a}\right) *_k \phi \left({x}\right) = \phi \left({a}\right) *_k \phi \left({y}\right) \implies \phi \left({x}\right) = \phi \left({y}\right)}\right) + | c=[[Definition:Morphism Property|Morphism Property]] +}} +{{end-eqn}} +and thus [[Definition:Right Cancellable Element|right cancellability]] is demonstrated. +{{qed}} +[[Category:Homomorphisms]] +renh9lnsqjymefimq6eqfxvw23fgjy7 +\end{proof}<|endoftext|> +\section{Quotient Mapping on Structure is Canonical Epimorphism} +Tags: Epimorphisms, Quotient Structures + +\begin{theorem} +Let $\mathcal R$ be a [[Definition:Congruence Relation|congruence relation]] on an [[Definition:Algebraic Structure|algebraic structure]] $\struct {S, \circ}$. +Then the [[Definition:Quotient Mapping|quotient mapping]] from $\struct {S, \circ}$ to the [[Definition:Quotient Structure|quotient structure]] $\struct {S / \mathcal R, \circ_\mathcal R}$ is an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]]: +:$q_\mathcal R: \struct {S, \circ} \to \struct {S / \mathcal R, \circ_\mathcal R}: \forall x, y \in S: \map {q_\mathcal R} {x \circ y} = \map {q_\mathcal R} x \circ_\mathcal R \map {q_\mathcal R} y$ +This is sometimes called the '''canonical epimorphism''' from $\struct {S, \circ}$ to $\struct {S / \mathcal R, \circ_\mathcal R}$. +\end{theorem} + +\begin{proof} +The [[Definition:Quotient Mapping|quotient mapping]] $q_\mathcal R: S \to S / \mathcal R$ is the [[Quotient Mapping is Surjection|canonical surjection]] from $S$ to $S / \mathcal R$. +Next we show that this is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]]: +{{begin-eqn}} +{{eqn | l = \map {q_\mathcal R} x + | r = \eqclass x {\mathcal R} + | c = {{Defof|Quotient Mapping}} +}} +{{eqn | l = \map {q_\mathcal R} y + | r = \eqclass y {\mathcal R} + | c = {{Defof|Quotient Mapping}} +}} +{{eqn | l = \map {q_\mathcal R} {x \circ y} + | r = \eqclass {x \circ y} {\mathcal R} + | c = {{Defof|Quotient Mapping}} +}} +{{eqn | l = \eqclass {x \circ y} {\mathcal R} + | r = \eqclass x {\mathcal R} \circ_\mathcal R \eqclass y {\mathcal R} + | c = {{Defof|Operation Induced on Quotient Set}} +}} +{{eqn | ll= \leadsto + | l = \map {q_\mathcal R} {x \circ y} + | r = \map {q_\mathcal R} x \circ_\mathcal R \map {q_\mathcal R} y + | c = +}} +{{end-eqn}} +Thus the [[Definition:Morphism Property|morphism property]] is shown to hold. +So the [[Definition:Quotient Mapping|quotient mapping]] $q_\mathcal R: \struct {S, \circ} \to \struct {S / \mathcal R, \circ_\mathcal R}$ has been shown to be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] which is a [[Definition:Surjection|surjection]], and is thus an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]]. +{{Qed}} +\end{proof}<|endoftext|> +\section{Restriction of Homomorphism to Image is Epimorphism} +Tags: Epimorphisms + +\begin{theorem} +Let $S$ and $T$ be [[Definition:Algebraic Structure|algebraic structures]]. +Let $\phi: S \to T$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]]. +Then a [[Definition:Surjective Restriction|surjective restriction]] of $\phi$ can be produced by limiting the [[Definition:Codomain of Mapping|codomain]] of $\phi$ to its [[Definition:Image of Mapping|image]] $\operatorname{Im} \left({\phi}\right)$. +\end{theorem} + +\begin{proof} +Let $\phi: S \to T$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]]. +Let $\operatorname{Im} \left({\phi}\right) = T'$ +By [[Morphism Property Preserves Closure]], $T'$ is [[Definition:Closed Algebraic Structure|closed]]. +From [[Restriction of Mapping to Image is Surjection]], $\phi \to \operatorname{Im} \left({\phi}\right)$ is a [[Definition:Surjection|surjection]]. +Thus $\phi: S \to T$ is an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]]. +Therefore, by suitably restricting the codomain of a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]], it is possible to regard it as an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]]. +{{qed}} +[[Category:Epimorphisms]] +stsi7puz9734bf4mk55ir5amqnele7l +\end{proof}<|endoftext|> +\section{Epimorphism Preserves Associativity} +Tags: Epimorphisms + +\begin{theorem} +Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure|algebraic structures]]. +Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]]. +Let $\circ$ be an [[Definition:Associative Operation|associative operation]]. +Then $*$ is also an [[Definition:Associative Operation|associative operation]]. +\end{theorem} + +\begin{proof} +Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]] in which $\circ$ is [[Definition:Associative Operation|associative]]. +Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]]. +As an epimorphism is [[Definition:Surjection|surjective]], it follows that: +:$\forall u, v, w \in T: \exists x, y, z \in S: \map \phi x = u, \, \map \phi y = v, \, \map \phi z = w$ +So: +{{begin-eqn}} +{{eqn | l = \paren {u * v} * w + | r = \paren {\map \phi x * \map \phi y} * \map \phi z + | c = [[Definition:Surjection|as $\phi$ is a surjection]] +}} +{{eqn | r = \map \phi {x \circ y} * \map \phi z + | c = {{Defof|Morphism Property}} +}} +{{eqn | r = \map \phi {\paren {x \circ y} \circ z} + | c = {{Defof|Morphism Property}} +}} +{{eqn | r = \map \phi {x \circ \paren {y \circ z} } + | c = [[Definition:Associative Operation|Associativity of $\circ$]] +}} +{{eqn | r = \map \phi x * \map \phi {y \circ z} + | c = {{Defof|Morphism Property}} +}} +{{eqn | r = \map \phi x * \paren {\map \phi y * \map \phi z} + | c = {{Defof|Morphism Property}} +}} +{{eqn | r = u * \paren {v * w} + | c = by definition as above +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Epimorphism Preserves Semigroups} +Tags: Epimorphisms, Semigroups + +\begin{theorem} +Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be [[Definition:Algebraic Structure|algebraic structures]]. +Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]]. +Let $\left({S, \circ}\right)$ be a [[Definition:Semigroup|semigroup]]. +Then $\left({T, *}\right)$ is also a [[Definition:Semigroup|semigroup]]. +\end{theorem} + +\begin{proof} +As $\left({S, \circ}\right)$ is a [[Definition:Semigroup|semigroup]], then by definition it is [[Definition:Closed Algebraic Structure|closed]]. +As $\phi$ is an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]], it is by definition [[Definition:Surjection|surjective]]. +That is: +:$T = \phi \left[{S}\right]$ +where $\phi \left[{S}\right]$ denotes the [[Definition:Image of Subset under Mapping|image of $S$ under $\phi$]]. +From [[Morphism Property Preserves Closure]] it follows that $\left({T, *}\right)$ is [[Definition:Closed Algebraic Structure|closed]]. +As $\left({S, \circ}\right)$ is a [[Definition:Semigroup|semigroup]], then by definition $\circ$ is [[Definition:Associative Operation|associative]]. +From [[Epimorphism Preserves Associativity]], $*$ is therefore also [[Definition:Associative Operation|associative]]. +So: +: $\left({T, *}\right)$ is [[Definition:Closed Algebraic Structure|closed]] +and: +: $*$ is [[Definition:Associative Operation|associative]]. +Therefore, by definition, $\left({T, *}\right)$ is a [[Definition:Semigroup|semigroup]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Homomorphism Preserves Subsemigroups} +Tags: Homomorphisms, Subsemigroups + +\begin{theorem} +Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Semigroup|semigroups]]. +Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]]. +Let $S'$ be a [[Definition:Subsemigroup|subsemigroup]] of $S$. +Then $\phi \paren {S'}$ is a [[Definition:Subsemigroup|subsemigroup]] of $T$. +\end{theorem} + +\begin{proof} +By [[Restriction of Homomorphism to Image is Epimorphism]], $\phi$ is an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]] onto its [[Definition:Image of Mapping|image]] . +Then by [[Epimorphism Preserves Semigroups]], it follows that the [[Definition:Image of Mapping|image]] of $S'$ is a [[Definition:Semigroup|semigroup]]. +The result follows. +{{Qed}} +[[Category:Homomorphisms]] +[[Category:Subsemigroups]] +jw5hpjmdnguazj8xyel7nwh0pvngfyu +\end{proof}<|endoftext|> +\section{Epimorphism Preserves Commutativity} +Tags: Epimorphisms + +\begin{theorem} +Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be [[Definition:Algebraic Structure|algebraic structures]]. +Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]]. +Let $\circ$ be a [[Definition:Commutative Operation|commutative operation]]. +Then $*$ is also a [[Definition:Commutative Operation|commutative operation]]. +\end{theorem} + +\begin{proof} +Let $\left({S, \circ}\right)$ be an [[Definition:Algebraic Structure|algebraic structure]] in which $\circ$ is [[Definition:Commutative Operation|commutative]]. +Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]]. +As an epimorphism is [[Definition:Surjection|surjective]], it follows that: +:$\forall u, v \in T: \exists x, y \in S: \phi \left({x}\right) = u, \phi \left({y}\right) = v$ +So: +{{begin-eqn}} +{{eqn | l = u * v + | r = \phi \left({x}\right) * \phi \left({y}\right) + | c = $\phi$ is a [[Definition:Surjection|surjection]] +}} +{{eqn | r = \phi \left({x \circ y}\right) + | c = [[Definition:Morphism Property|Morphism Property]] +}} +{{eqn | r = \phi \left({y \circ x}\right) + | c = [[Definition:Commutative Operation|Commutativity]] of $\circ$ +}} +{{eqn | r = \phi \left({y}\right) * \phi {\left({x}\right)} + | c = [[Definition:Morphism Property|Morphism Property]] +}} +{{eqn | r = v * u + | c = by definition as above +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Epimorphism Preserves Identity} +Tags: Epimorphisms, Identity Elements + +\begin{theorem} +Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure|algebraic structures]]. +Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]]. +Let $\struct {S, \circ}$ have an [[Definition:Identity Element|identity element]] $e_S$. +Then $\struct {T, *}$ has the [[Definition:Identity Element|identity element]] $\map \phi {e_S}$. +\end{theorem} + +\begin{proof} +Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]] in which $\circ$ has an [[Definition:Identity Element|identity]] $e_S$. +Then: +:$\forall x \in S: x \circ e_S = x = e_S \circ x$ +Thus: +{{begin-eqn}} +{{eqn | l = \map \phi x + | r = \map \phi {x \circ e_S} + | c = $e_S$ is an [[Definition:Identity Element|identity]] for $\circ$ +}} +{{eqn | r = \map \phi x * \map \phi {e_S} + | c = [[Definition:Morphism Property|Morphism property]] of $\circ$ under $\phi$ +}} +{{end-eqn}} +and: +{{begin-eqn}} +{{eqn | l = \map \phi x + | r = \map \phi {e_S \circ x} + | c = $e_S$ is an [[Definition:Identity Element|identity]] for $\circ$ +}} +{{eqn | r = \map \phi {e_S} * \map \phi x + | c = [[Definition:Morphism Property|Morphism property]] of $\circ$ under $\phi$ +}} +{{end-eqn}} +The result follows because every [[Definition:Element of Set|element]] $y \in T$ is of the form $\map \phi x$ with $x \in S$. +{{qed}} +\end{proof}<|endoftext|> +\section{Epimorphism Preserves Inverses} +Tags: Epimorphisms, Inverse Elements + +\begin{theorem} +Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be [[Definition:Algebraic Structure|algebraic structures]]. +Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]]. +Let $\left({S, \circ}\right)$ have an [[Definition:Identity Element|identity]] $e_S$. +Let $x^{-1}$ be an [[Definition:Inverse Element|inverse element]] of $x$ for $\circ$. +Then $\phi \left({x^{-1}}\right)$ is an [[Definition:Inverse Element|inverse element]] of $\phi \left({x}\right)$ for $*$. +That is: +: $\phi \left({x^{-1}}\right) = \left({\phi \left({x}\right)}\right)^{-1}$ +\end{theorem} + +\begin{proof} +Let $\left({S, \circ}\right)$ be an [[Definition:Algebraic Structure|algebraic structure]] in which $\circ$ has an [[Definition:Identity Element|identity element]] $e_S$. +From [[Epimorphism Preserves Identity]], it follows that $\left({T, *}\right)$ also has an [[Definition:Identity Element|identity element]], which is $\phi \left({e_S}\right)$. +Let $y$ be an [[Definition:Inverse Element|inverse]] of $x$ in $\left({S, \circ}\right)$. +Then: +{{begin-eqn}} +{{eqn | l = \phi \left({x}\right) * \phi \left({y}\right) + | r = \phi \left({x \circ y}\right) + | c = +}} +{{eqn | r = \phi \left({e_S}\right) + | c = +}} +{{eqn | r = \phi \left({y \circ x}\right) + | c = +}} +{{eqn | r = \phi \left({y}\right) * \phi \left({x}\right) + | c = +}} +{{end-eqn}} +So $\phi \left({y}\right)$ is an [[Definition:Inverse Element|inverse]] of $\phi \left({x}\right)$ in $\left({T, *}\right)$. +{{qed}} +\end{proof}<|endoftext|> +\section{Homomorphism with Cancellable Codomain Preserves Identity} +Tags: Homomorphisms, Identity Elements + +\begin{theorem} +Let $\struct{S, \circ}$ and $\struct{T, *}$ be [[Definition:Algebraic Structure|algebraic structures]]. +Let $\phi: \struct{S, \circ} \to \struct{T, *}$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]]. +Let $\struct{S, \circ}$ have an [[Definition:Identity Element|identity]] $e_S$. +Let $\struct{T, *}$ have an [[Definition:Identity Element|identity]] $e_T$. +Let every element of $\struct{T, *}$ be [[Definition:Cancellable Element|cancellable]]. +Then $\map \phi {e_S}$ is the [[Definition:Identity Element|identity]] $e_T$. +\end{theorem} + +\begin{proof} +Let $\struct{S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]] in which $\circ$ has an [[Definition:Identity Element|identity]] $e_S$. +Let $\struct{T, *}$ be an [[Definition:Algebraic Structure|algebraic structure]] in which $*$ has an [[Definition:Identity Element|identity]] $e_T$. +Let $\struct{T, *}$ be such that every element is [[Definition:Cancellable Element|cancellable]]. +Let $\phi: \struct{S, \circ} \to \struct{T, *}$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]]. +Every element of $\struct{T, *}$ is [[Definition:Cancellable Element|cancellable]]. +Suppose there is an [[Definition:Idempotent Element|idempotent element]] in $\struct{T, *}$ +So from [[Identity is only Idempotent Cancellable Element]], it must be the [[Definition:Identity Element|identity]] $e_T$. +Thus: +{{begin-eqn}} +{{eqn | l = \map \phi {e_S} * \map \phi {e_S} + | r = \map \phi {e_S \circ e_S} + | c = by the [[Definition:Morphism Property|morphism property]] of $\circ$ under $\phi$ +}} +{{eqn | r = \map \phi {e_S} + | c = as $e_S$ is the [[Definition:Identity Element|identity]] of $\left({S, \circ}\right)$ +}} +{{end-eqn}} +So $\map \phi {e_S}$ is [[Definition:Idempotent Element|idempotent]] in $\struct{T, *}$ and the result follows. +{{qed}} +\end{proof}<|endoftext|> +\section{Homomorphism with Identity Preserves Inverses} +Tags: Homomorphisms, Inverse Elements + +\begin{theorem} +Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be [[Definition:Algebraic Structure|algebraic structures]]. +Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]]. +Let $\left({S, \circ}\right)$ have an [[Definition:Identity Element|identity]] $e_S$. +Let $\left({T, *}\right)$ also have an identity $e_T = \phi \left({e_S}\right)$. +If $x^{-1}$ is an [[Definition:Inverse Element|inverse]] of $x$ for $\circ$, then $\phi \left({x^{-1}}\right)$ is an [[Definition:Inverse Element|inverse]] of $\phi \left({x}\right)$ for $*$. +That is: +: $\phi \left({x^{-1}}\right) = \left({\phi \left({x}\right)}\right)^{-1}$ +\end{theorem} + +\begin{proof} +Let $\left({S, \circ}\right)$ be an [[Definition:Algebraic Structure|algebraic structure]] in which $\circ$ has an [[Definition:Identity Element|identity]] $e_S$. +Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]]. +Let $\left({T, *}\right)$ be an [[Definition:Algebraic Structure|algebraic structure]] in which $*$ has an [[Definition:Identity Element|identity]] $e_T = \phi \left({e_S}\right)$. +{{begin-eqn}} +{{eqn | l = \phi \left({x}\right) * \phi \left({x^{-1} }\right) + | r = \phi \left({x \circ x^{-1} }\right) + | c = [[Definition:Morphism Property|Morphism property]] of $\phi$ +}} +{{eqn | r = \phi \left({e_S}\right) = e_T + | c = +}} +{{eqn | r = \phi \left({x^{-1} \circ x}\right) + | c = +}} +{{eqn | r = \phi \left({x^{-1} }\right) * \phi \left({x}\right) + | c = [[Definition:Morphism Property|Morphism property]] of $\phi$ +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Homomorphism to Group Preserves Identity} +Tags: Homomorphisms, Group Theory + +\begin{theorem} +Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]]. +Let $\struct {T, *}$ be a [[Definition:Group|group]]. +Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]]. +Let $\struct {S, \circ}$ have an [[Definition:Identity Element|identity]] $e_S$. +Then: +:$\map \phi {e_S} = e_T$ +\end{theorem} + +\begin{proof} +By hypothesis, $\struct {T, *}$ is a [[Definition:Group|group]]. +By the [[Cancellation Laws]], all [[Definition:Element|elements]] of $T$ are [[Definition:Cancellable Element|cancellable]]. +Thus [[Homomorphism with Cancellable Codomain Preserves Identity]] can be applied. +{{Qed}} +\end{proof}<|endoftext|> +\section{Homomorphism of External Direct Products} +Tags: Homomorphisms, Cartesian Product + +\begin{theorem} +Let: +: $\left({S_1 \times S_2, \circ}\right)$ be the [[Definition:External Direct Product|external direct product]] of two [[Definition:Algebraic Structure|algebraic structures]] $\left({S_1, \circ_1}\right)$ and $\left({S_2, \circ_2}\right)$ +: $\left({T_1 \times T_2, *}\right)$ be the [[Definition:External Direct Product|external direct product]] of two [[Definition:Algebraic Structure|algebraic structures]] $\left({T_1, *_1}\right)$ and $\left({T_2, *_2}\right)$ +: $\phi_1$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] from $\left({S_1, \circ_1}\right)$ onto $\left({T_1, *_1}\right)$ +: $\phi_2$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] from $\left({S_2, \circ_2}\right)$ onto $\left({T_2, *_2}\right)$. +Then the [[Definition:Mapping|mapping]] $\phi_1 \times \phi_2: \left({S_1 \times S_2, \circ}\right) \to \left({T_1 \times T_2, *}\right)$ defined as: +:$\left({\phi_1 \times \phi_2}\right) \left({\left({x, y}\right)}\right) = \left({\phi_1 \left({x}\right), \phi_2 \left({y}\right)}\right)$ +is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] from $\left({S_1 \times S_2, \circ}\right)$ to $\left({T_1 \times T_2, *}\right)$. +\end{theorem} + +\begin{proof} +Let $\left({x_1, x_2}\right), \left({y_1, y_2}\right) \in S_1 \circ S_2$. +Then: +{{begin-eqn}} +{{eqn | l = \left({\phi_1 \times \phi_2}\right) \left({\left({x_1, x_2}\right) \circ \left({y_1, y_2}\right)}\right) + | r = \left({\phi_1 \times \phi_2}\right) \left({\left({x_1 \circ_1 y_1, x_2 \circ_2 y_2}\right)}\right) + | c = +}} +{{eqn | r = \left({\phi_1 \left({x_1 \circ_1 y_1}\right), \phi_2 \left({x_2 \circ_2 y_2}\right)}\right) + | c = +}} +{{eqn | r = \left({\phi_1 \left({x_1}\right) \ast_1 \phi_1 \left({y_1}\right), \phi_2 \left({x_2}\right) \ast_2 \phi_2 \left({y_2}\right)}\right) + | c = +}} +{{eqn | r = \left({\phi_1 \left({x_1}\right), \phi_2 \left({x_2}\right)}\right) \ast \left({\phi_1 \left({y_1}\right), \phi_2 \left({y_2}\right)}\right) + | c = +}} +{{eqn | r = \left({\phi_1 \times \phi_2}\right) \left({\left({x_1, x_2}\right)}\right) \ast \left({\phi_1 \times \phi_2}\right) \left({\left({y_1, y_2}\right)}\right) + | c = +}} +{{end-eqn}} +thus demonstrating the [[Definition:Morphism Property|morphism property]]. +{{Qed}} +[[Category:Homomorphisms]] +[[Category:Cartesian Product]] +sg0mmocq3mnm7li4x7qx66ymostexp0 +\end{proof}<|endoftext|> +\section{Monomorphism Image is Isomorphic to Domain} +Tags: Monomorphisms + +\begin{theorem} +The [[Definition:Image of Mapping|image]] of a [[Definition:Monomorphism (Abstract Algebra)|monomorphism]] is [[Definition:Isomorphism (Abstract Algebra)|isomorphic]] to its [[Definition:Domain of Mapping|domain]]. +That is, if $\phi: S_1 \to S_2$ is a [[Definition:Monomorphism (Abstract Algebra)|monomorphism]], then: +:$\phi: S_1 \to \Img \phi$ +is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]]. +\end{theorem} + +\begin{proof} +Let $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$ be [[Definition:Closed Algebraic Structure|closed algebraic structures]]. +Let $\phi$ be a [[Definition:Monomorphism (Abstract Algebra)|monomorphism]] from $\struct {S_1, \circ_1}$ to $\struct {S_2, \circ_2}$. +Let $T = \Img \phi$ be the [[Definition:Image of Mapping|image]] of $\phi$. +By [[Morphism Property Preserves Closure]], $\struct {T, \circ_2}$ is [[Definition:Closed Algebraic Structure|closed]]. +As $\phi$ is a [[Definition:Monomorphism (Abstract Algebra)|monomorphism]], it is by definition an [[Definition:Injection|injective]] [[Definition:Homomorphism (Abstract Algebra)|homomorphism]]. +From [[Restriction of Injection is Injection]], $\phi: S_1 \to \Img \phi$ is an [[Definition:Injection|injection]]. +From [[Restriction of Mapping to Image is Surjection]], $\phi: S_1 \to \Img \phi$ is a [[Definition:Surjection|surjection]]. +Thus $\phi \to \Img \phi$ is by definition a [[Definition:Bijection|bijection]]. +Thus $\phi: S_1 \to \Img \phi$ is a [[Definition:Bijection|bijective]] [[Definition:Homomorphism (Abstract Algebra)|homomorphism]]. +Hence, by definition, $\phi: S_1 \to \Img \phi$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Inverse of Algebraic Structure Isomorphism is Isomorphism} +Tags: Isomorphisms + +\begin{theorem} +Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be [[Definition:Algebraic Structure|algebraic structures]]. +Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be a [[Definition:Mapping|mapping]]. +Then $\phi$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] iff $\phi^{-1}: \left({T, *}\right) \to \left({S, \circ}\right)$ is also an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]]. +\end{theorem} + +\begin{proof} +Let $\phi$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]]. +Then by definition $\phi$ is a [[Definition:Bijection|bijection]]. +Thus $\exists \phi^{-1}$ such that $\phi^{-1}$ is also a [[Definition:Bijection|bijection]] from [[Bijection iff Inverse is Bijection]]. +That is: +:$\exists \phi^{-1}: \left({T, *}\right) \to \left({S, \circ}\right)$ +It follows that: +{{begin-eqn}} +{{eqn | ll=\forall s \in S, t \in T: + | o= + | r=\phi \left({s}\right) = t \iff \phi^{-1} \left({t}\right) = s + | c=[[Inverse Element of Bijection]] +}} +{{eqn | o=\implies + | r=\phi \left({s_1 \circ s_2}\right) = t_1 * t_2 + | c=[[Definition:Morphism Property|Morphism Property]] +}} +{{eqn | o=\implies + | r=\phi^{-1} \left({t_1 * t_2}\right) = s_1 \circ s_2 = \phi^{-1} \left({t_1}\right) \circ \phi^{-1} \left({t_2}\right) + | c=[[Inverse Element of Bijection]] +}} +{{end-eqn}} +So $\phi^{-1}: \left({T, *}\right) \to \left({S, \circ}\right)$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]]. +$\phi^{-1}$ is also (from above) a [[Definition:Bijection|bijection]]. +Thus, by definition, $\phi^{-1}$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]]. +Let $\phi^{-1}: \left({T, *}\right) \to \left({S, \circ}\right)$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]]. +Applying the same result as above in reverse, we have that $\left({\phi^{-1}}\right)^{-1}: \left({S, \circ}\right) \to \left({T, *}\right)$ is also an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]]. +But by [[Inverse of Inverse of Bijection]]: +: $\left({\phi^{-1}}\right)^{-1} = \phi$ +and hence the result. +{{qed}} +\end{proof}<|endoftext|> +\section{Isomorphism is Equivalence Relation} +Tags: Isomorphisms, Equivalence Relations + +\begin{theorem} +[[Definition:Isomorphism (Abstract Algebra)|Isomorphism]] is an [[Definition:Equivalence Relation|equivalence]] on a [[Definition:Set|set]] of [[Definition:Magma|magmas]]. +This result applies to all [[Definition:Magma|magmas]]: [[Definition:Ring (Abstract Algebra)|rings]], [[Definition:Group|groups]], [[Definition:R-Algebraic Structure|R-algebraic structures]] etc. +\end{theorem} + +\begin{proof} +To prove a [[Definition:Relation|relation]] is an [[Definition:Equivalence Relation|equivalence]], we need to prove it is [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]]. +So, checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: +=== Reflexive === +From [[Identity Mapping is Automorphism]], the [[Definition:Identity Mapping|identity mapping]] $I_S: S \to S$ is an [[Definition:Automorphism (Abstract Algebra)|automorphism]], which is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] from a [[Definition:Magma|magma]] onto itself. +So $S \cong S$, and [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is seen to be [[Definition:Reflexive Relation|reflexive]]. +{{Qed|lemma}} +=== Symmetric === +If $S \cong T$, then $\exists \phi: S \to T$ such that $\phi$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]]. +Thus, from [[Inverse of Algebraic Structure Isomorphism is Isomorphism]], $\phi^{-1}: T \to S$ is also an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]], and $T \cong S$. +Thus we have shown that $S \cong T \implies T \cong S$, so $\cong$ is [[Definition:Symmetric Relation|symmetric]]. +{{Qed|lemma}} +=== Transitive === +Let $S_1 \cong S_2$, and $S_2 \cong S_3$. +We can define these two [[Definition:Isomorphism (Abstract Algebra)|isomorphisms]]: +: $\phi: S_1 \to S_2$ +: $\psi: S_2 \to S_3$ +From [[Composite of Isomorphisms in Algebraic Structure is Isomorphism]], we have that $\psi \circ \phi$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] between $S_1$ and $S_3$. +Thus we have shown that $\cong$ is [[Definition:Transitive Relation|transitive]]. +{{Qed|lemma}} +Thus [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]], and therefore an [[Definition:Equivalence Relation|equivalence]]. +{{Qed}} +\end{proof} + +\begin{proof} +{{Finish|Needs filling out.}} +* [[Definition:Reflexive Relation|Reflexive]]: +Follows directly from [[Identity Mapping is Order Isomorphism]]. +* [[Definition:Symmetric Relation|Symmetric]]: +Follows directly from [[Inverse of Order Isomorphism is Order Isomorphism]]. +* [[Definition:Transitive Relation|Transitive]]: +Follows directly from [[Composite of Order Isomorphisms is Order Isomorphism]]. +{{qed}} +\end{proof} + +\begin{proof} +An [[Definition:Ordered Set|ordered set]] is a [[Definition:Relational Structure|relational structure]] where [[Definition:Order Isomorphism|order isomorphism]] is a special case of [[Definition:Relation Isomorphism|relation isomorphism]]. +The result follows directly from [[Relation Isomorphism is Equivalence Relation]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Projection is Epimorphism} +Tags: Epimorphisms, Projections + +\begin{theorem} +Let $\struct {S, \circ}$ be the [[Definition:External Direct Product|external direct product]] of the [[Definition:Algebraic Structure|algebraic structures]] $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$. +Then: +:$\pr_1$ is an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]] from $\struct {S, \circ}$ to $\struct {S_1, \circ_1}$ +:$\pr_2$ is an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]] from $\struct {S, \circ}$ to $\struct {S_2, \circ_2}$ +where $\pr_1$ and $\pr_2$ are the [[Definition:First Projection|first]] and [[Definition:Second Projection|second projection]] respectively of $\struct {S, \circ}$. +\end{theorem} + +\begin{proof} +From [[Projection is Surjection]], $\pr_1$ and $\pr_2$ are [[Definition:Surjection|surjections]]. +We now need to show they are [[Definition:Homomorphism (Abstract Algebra)|homomorphisms]]. +Let $s, t \in \struct {S, \circ}$ where $s = \tuple {s_1, s_2}$ and $t = \tuple {t_1, t_2$. +Then: +{{begin-eqn}} +{{eqn | l = \map {\pr_1} {s \circ t} + | r = \map {\pr_1} {\tuple {s_1, s_2} \circ \tuple {t_1, t_2} } + | c = +}} +{{eqn | r = \map {\pr_1} {\tuple {s_1 \circ_1 t_1, s_2 \circ_2 t_2} } + | c = +}} +{{eqn | r = s_1 \circ_j t_1 + | c = +}} +{{eqn | r = \map {\pr_1} s \circ_1 \map {\pr_1} t + | c = +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | l = \map {\pr_2} {s \circ t} + | r = \map {\pr_2} {\tuple {s_1, s_2} \circ \tuple {t_1, t_2} } + | c = +}} +{{eqn | r = \map {\pr_2} {\tuple {s_1 \circ_1 t_1, s_2 \circ_2 t_2} } + | c = +}} +{{eqn | r = s_2 \circ_j t_2 + | c = +}} +{{eqn | r = \map {\pr_2} s \circ_2 \map {\pr_2} t + | c = +}} +{{end-eqn}} +Thus the [[Definition:Morphism Property|morphism property]] is demonstrated for both $\pr_1$ and $\pr_2$. +{{Qed}} +\end{proof}<|endoftext|> +\section{Quotient Theorem for Epimorphisms} +Tags: Epimorphisms, Isomorphisms, Quotient Mappings, Quotient Structures, Named Theorems + +\begin{theorem} +Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure|algebraic structures]]. +Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]]. +Let $\mathcal R_\phi$ be the [[Definition:Equivalence Relation Induced by Mapping|equivalence induced by $\phi$]]. +Let $S / \mathcal R_\phi$ be the [[Definition:Quotient Set|quotient of $S$ by $\mathcal R_\phi$]]. +Let $q_{\mathcal R_\phi}: S \to S / \mathcal R_\phi$ be the [[Definition:Quotient Mapping|quotient mapping induced by $\mathcal R_\phi$]]. +Let $\struct {S / \mathcal R_\phi}, {\circ_{\mathcal R_\phi} }$ be the [[Definition:Quotient Structure|quotient structure defined by $\mathcal R_\phi$]]. +Then: +:The [[Definition:Equivalence Relation Induced by Mapping|induced equivalence]] $\mathcal R_\phi$ is a [[Definition:Congruence Relation|congruence relation]] for $\circ$ +:There is one and only one [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] $\psi: \struct {S / \mathcal R_\phi}, {\circ_{\mathcal R_\phi} } \to \struct {T, *}$ which satisfies $\psi \bullet q_{\mathcal R_\phi} = \phi$. +where, in order not to cause notational confusion, $\bullet$ is used as the symbol to denote [[Definition:Composition of Mappings|composition of mappings]]. +\end{theorem} + +\begin{proof} +=== [[Equivalence Induced by Epimorphism is Congruence Relation|Proof of Congruence Relation]] === +{{:Equivalence Induced by Epimorphism is Congruence Relation}}{{qed|lemma}} +=== [[Unique Isomorphism from Quotient Mapping to Epimorphism Domain|Proof of Unique Isomorphism]] === +{{:Unique Isomorphism from Quotient Mapping to Epimorphism Domain}} +\end{proof}<|endoftext|> +\section{Homomorphism on Induced Structure} +Tags: Homomorphisms, Pointwise Operations + +\begin{theorem} +Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]]. +Let $\struct {T, \oplus}$ be a [[Definition:Commutative Semigroup|commutative semigroup]]. +Let $T^S$ be the [[Definition:Set of All Mappings|set of all mappings]] from $S$ to $T$. +Let $f$ and $g$ be [[Definition:Homomorphism (Abstract Algebra)|homomorphisms]] from $S$ into $T$. +Let $f \oplus' g$ be the [[Definition:Pointwise Operation|pointwise operation on $T^S$ induced by $\oplus$]]. +Then $f \oplus' g$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] from $\struct {S, \circ}$ into $\struct {T, \oplus}$. +\end{theorem} + +\begin{proof} +Let $\struct {T, \oplus}$ be a [[Definition:Commutative Semigroup|commutative semigroup]]. +Let $x, y \in S$. +Then: +{{begin-eqn}} +{{eqn | l = \map {\paren {f \oplus' g} } {x \circ y} + | r = \map f {x \circ y} \oplus \map g {x \circ y} + | c = {{Defof|Pointwise Operation}} +}} +{{eqn | r = \paren {\map f x \oplus \map f y} \oplus \paren {\map g x \oplus \map g y} + | c = $f$ and $g$ are [[Definition:Homomorphism (Abstract Algebra)|homomorphisms]] +}} +{{eqn | r = \paren {\map f x \oplus \map g x} \oplus \paren {\map f y \oplus \map g y} + | c = $\oplus$ is [[Definition:Commutative Operation|Commutative]] and [[Definition:Associative Operation|Associative]] +}} +{{eqn | r = \map {\paren {f \oplus' g} } x \oplus \map {\paren {f \oplus' g} } Y + | c = {{Defof|Pointwise Operation}} +}} +{{end-eqn}} +{{qed}} +Notice that for this to work, $\oplus$ needs to be both [[Definition:Associative Operation|associative]] ''and'' [[Definition:Commutative Operation|commutative]]. +\end{proof}<|endoftext|> +\section{Inverse Mapping in Induced Structure} +Tags: Homomorphisms, Group Theory + +\begin{theorem} +Let $\left({S, \circ}\right)$ be an [[Definition:Algebraic Structure|algebraic structure]]. +Let $\left({T, \oplus}\right)$ be an [[Definition:Abelian Group|abelian group]]. +Let $f$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] from $S$ into $T$. +Let $f^*$ be the [[Induced Structure Inverse|induced structure inverse of $f$]]. +Then $f^*$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] from $\left({S, \circ}\right)$ into $\left({T, \oplus}\right)$. +\end{theorem} + +\begin{proof} +Let $\left({T, \oplus}\right)$ be an [[Definition:Abelian Group|abelian group]]. +Let $x, y \in S$. +Then: +{{begin-eqn}} +{{eqn | l = f^* \left({x \circ y}\right) + | r = \left({f \left({x \circ y}\right)}\right)^{-1} + | c = [[Induced Structure Inverse]] +}} +{{eqn | r = \left({f \left({x}\right) \oplus f \left({y}\right)}\right)^{-1} + | c = $f$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] +}} +{{eqn | r = \left({f \left({y}\right) \oplus f \left({x}\right)}\right)^{-1} + | c = [[Definition:Commutative Operation|Commutativity]] of $\oplus$ +}} +{{eqn | r = \left({f \left({x}\right)}\right)^{-1} \oplus \left({f \left({y}\right)}\right)^{-1} + | c = [[Inverse of Group Product]] +}} +{{eqn | r = f^* \left({x}\right) \oplus f^* \left({y}\right) + | c = [[Induced Structure Inverse]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Set of Homomorphisms is Subgroup of All Mappings} +Tags: Homomorphisms, Subgroups + +\begin{theorem} +Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]]. +Let $\struct {T, \oplus}$ be an [[Definition:Abelian Group|abelian group]]. +Let $\struct {T^S, \oplus}$ be the [[Definition:Induced Structure|algebraic structure on $T^S$ induced by $\oplus$]]. +Then the set of all [[Definition:Homomorphism (Abstract Algebra)|homomorphisms]] from $\struct {S, \circ}$ into $\struct {T, \oplus}$ is a [[Definition:Subgroup|subgroup]] of $\struct {T^S, \oplus}$. +\end{theorem} + +\begin{proof} +Let $H$ be the set of all [[Definition:Homomorphism (Abstract Algebra)|homomorphisms]] from $\struct {S, \circ}$ into $\struct {T, \oplus}$. +=== {{GroupAxiom|0}} === +From [[Homomorphism on Induced Structure]], we have that $\forall f, g \in H: f \oplus g$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] from $\struct {S, \circ}$ into $\struct {T, \oplus}$. +Hence $\struct {H, \oplus}$ is [[Definition:Closed Algebraic Structure|closed]]. +=== {{GroupAxiom|3}} === +From [[Inverse Mapping in Induced Structure]], if $g$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] then its [[Induced Structure Inverse]] $g^*$ is one also. +Thus $g \in H \implies g^* \in H$. +So by the [[Two-Step Subgroup Test]], $\struct {H, \oplus}$ is a [[Definition:Subgroup|subgroup]] of $\struct {T^S, \oplus}$. +{{qed}} +\end{proof}<|endoftext|> +\section{Transplanting Theorem} +Tags: Isomorphisms, Named Theorems + +\begin{theorem} +Let $\left({S, \circ}\right)$ be an [[Definition:Algebraic Structure|algebraic structure]]. +Let $f: S \to T$ be a [[Definition:Bijection|bijection]]. +Then there exists [[Definition:Unique|one and only one]] [[Definition:Binary Operation|operation]] $\oplus$ such that $f: \left({S, \circ}\right) \to \left({T, \oplus}\right)$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]]. +The [[Definition:Binary Operation|operation]] $\oplus$ is defined by: +:$\forall x, y \in T: x \oplus y = f \left({f^{-1} \left({x}\right) \circ f^{-1} \left({y}\right)}\right)$ +The [[Definition:Binary Operation|operation]] $\oplus$ is called the '''transplant of $\circ$ under $f$'''. +\end{theorem} + +\begin{proof} +=== Existence === +To show that $\oplus$ as defined above exists: +Let $u, v \in S$, and let $x = f \left({u}\right), y = f \left({v}\right)$. +Then as $f$ is a [[Definition:Bijection|bijection]], $u = f^{-1} \left({x}\right), v = f^{-1} \left({y}\right)$. +Thus: +{{begin-eqn}} +{{eqn | l = f \left({u \circ v}\right) + | r = f \left({f^{-1} \left({x}\right) \circ f^{-1} \left({y}\right)}\right) + | c = Definition of $x$ and $y$ +}} +{{eqn | r = f \left({f^{-1} \left({x}\right)}\right) \oplus f \left({f^{-1} \left({y}\right)}\right) + | c = [[Definition:Morphism Property|Morphism Property]] +}} +{{eqn | r = x \oplus y + | c = [[Inverse of Inverse of Bijection]] +}} +{{eqn | r = f \left({u}\right) \oplus f \left({v}\right) + | c = Definition of $x$ and $y$ +}} +{{end-eqn}} +It is seen that $f$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] as required. +{{qed|lemma}} +=== Uniqueness === +Let $f$ be the [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] whose existence has been proven [[Transplanting Theorem#Existence|above]]. +Thus: +{{begin-eqn}} +{{eqn | l = f \left({f^{-1} \left({x}\right) \circ f^{-1} \left({y}\right)}\right) + | r = f \left({f^{-1} \left({x}\right)}\right) \oplus f \left({f^{-1} \left({y}\right)}\right) + | c = [[Definition:Morphism Property|Morphism Property]] +}} +{{eqn | r = \left({f \circ f^{-1} \left({x}\right)}\right) \oplus \left({f \circ f^{-1} \left({y}\right)}\right) + | c = Definition of [[Definition:Composition of Mappings|Composite Mapping]] +}} +{{eqn | r = \left({I_T \left({x}\right)}\right) \oplus \left({I_T \left({y}\right)}\right) + | c = [[Bijection has Left and Right Inverse]] +}} +{{eqn | r = x \oplus y + | c = Definition of [[Definition:Identity Mapping|Identity Mapping]] +}} +{{end-eqn}} +So, if $\oplus$ is an [[Definition:Binary Operation|operation]] on $T$ such that $f$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] from $\left({S, \circ}\right) \to \left({T, \oplus}\right)$, then $\oplus$ must be defined as by this theorem, and there can be no other such [[Definition:Binary Operation|operations]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Exists Bijection to a Disjoint Set} +Tags: Bijections + +\begin{theorem} +Let $S$ and $T$ be [[Definition:Set|sets]]. +Then there exists a [[Definition:Bijection|bijection]] from $T$ onto a set $T'$ [[Definition:Disjoint Sets|disjoint]] from $S$. +\end{theorem} + +\begin{proof} +Consider the set: +:$X = \left\{{y \in S: \left({\exists x \in T: \left({x, y}\right) \in S}\right)}\right\}$ +That is, $X$ consists of all the [[Definition:Element|elements]] of $S$ which are the [[Definition:Coordinate|second coordinate]] of some [[Definition:Ordered Pair|ordered pair]] which ''also'' happens to be in $S$ and whose [[Definition:Coordinate|first coordinate]] is in $T$. +From [[Exists Element Not in Set]], we have that $\exists z: z \notin X$. +Now consider the [[Definition:Cartesian Product|cartesian product]] $T' = T \times \left\{{z}\right\}$. +Suppose that $p \in T'$. +Then $p = \left({c, z}\right)$ where $c \in T$. +Suppose $\left({c, z}\right) \in S$. +Then that would mean that $z \in X$. +But we have specifically selected $z$ such that $z \notin X$. +So $p = \left({c, z}\right) \notin S$. +Thus we have that $T' \cap S = \varnothing$. +There is an obvious [[Definition:Bijection|bijection]] $g: T \to T'$: +:$\forall t \in T: g \left({t}\right) = \left({t, z}\right)$ +and hence the result. +{{qed}} +\end{proof}<|endoftext|> +\section{Embedding Theorem} +Tags: Isomorphisms, Named Theorems + +\begin{theorem} +Let: +:$(1): \quad \left({T_2, \oplus_2}\right)$ be a [[Definition:Submagma|submagma]] of $\left({S_2, *_2}\right)$ +:$(2): \quad f: \left({T_1, \oplus_1}\right) \to \left({T_2, \oplus_2}\right)$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] +then there exists: +:$(1): \quad$ a [[Definition:Magma|magma]] $\left({S_1, *_1}\right)$ which [[Definition:Submagma|algebraically contains]] $\left({T_1, \oplus_1}\right)$ +:$(2): \quad g: \left({S_1, *_1}\right) \to \left({S_2, *_2}\right)$ where $g$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] which [[Definition:Extension of Mapping|extends]] $f$. +{{Expand|Add the corollary that extends this theorem to structures with two operations.}} +\end{theorem} + +\begin{proof} +In order to reduce notational confusion, the [[Definition:Composition of Mappings|composition of two mappings]] $f$ and $g$ is denoted $f \bullet g$ instead of the more usual $f \circ g$. +There are two cases to consider: when $T_1$ and $S_2$ are [[Definition:Disjoint Sets|disjoint]], and when they are not. +Suppose $T_1$ and $S_2$ are [[Definition:Disjoint Sets|disjoint]]. +Let $S_1$ be the [[Definition:Set|set]] $T_1 \cup \left({S_2 \setminus T_2}\right)$. +Then we can define the [[Definition:Mapping|mapping]] $h: S_2 \to S_1$ as: +:$\forall x \in S_2: h \left({x}\right) = \begin{cases} +x & : x \in S_2 \setminus T_2 \\ +f^{-1} \left({x}\right) & : x \in T_2 +\end{cases}$ +Because $T_1$ and $S_2$ are [[Definition:Disjoint Sets|disjoint]], $h$ can be seen to be a [[Definition:Bijection|bijection]]. +What $h$ is doing is that it effectively "slots" $T_1$ into the gap in $S_2$ that was taken up by the removed $T_2$. +We are going to show that $\left({T_1, \oplus_1}\right)$ is [[Definition:Submagma|embedded]] in $\left({S_1, *_1}\right)$. +The [[Definition:Binary Operation|operation]] $*_1$ is defined as the [[Transplanting Theorem|transplant of $*_2$ under $h$]]. +Thus by the [[Transplanting Theorem]]: +:$\forall x, y \in S_1: x *_1 y = h \left({h^{-1} \left({x}\right) *_2 h^{-1} \left({y}\right)}\right)$ +Let $x, y \in T_1$. +Then, from the definition of $h$ above: +:$h^{-1} \left({x}\right) = \left({f^{-1} \left({x}\right)}\right)^{-1} = f \left({x}\right)$ +and similarly: +:$h^{-1} \left({y}\right) = f \left({y}\right)$ +So: +{{begin-eqn}} +{{eqn | l = x *_1 y + | r = h \left({f \left({x}\right) *_2 f \left({y}\right)}\right) + | c = [[Transplanting Theorem]] +}} +{{eqn | r = h \left({f \left({x}\right) \oplus_2 f \left({y}\right)}\right) + | c = as $*_2$ is [[Definition:Submagma|embedded]] in $\oplus_2$ +}} +{{eqn | r = h \left({f \left({x \oplus_1 y}\right)}\right) + | c = [[Definition:Morphism Property|Morphism Property of $\oplus_1$ under $f$]] +}} +{{eqn | r = f^{-1} \left({f \left({x \oplus_1 y}\right)}\right) + | c = Definition of $h$ on elements of $T_1$ +}} +{{eqn | r = x \oplus_1 y + | c = [[Inverse of Inverse of Bijection]] +}} +{{end-eqn}} +proving that $x \oplus_1 y$ is [[Definition:Closed Algebraic Structure|closed]]. +Thus: +: $\left({T_1, \oplus_1}\right) \subseteq \left({S_1, *_1}\right)$ +that is, $\left({T_1, \oplus_1}\right)$ is [[Definition:Submagma|embedded]] in $\left({S_1, *_1}\right)$. +Now, let $g = h^{-1}$. +By the definition of $*_1$, $g$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] from $\left({S_1, *_1}\right)$ onto $\left({S_2, *_2}\right)$. +Then: +{{begin-eqn}} +{{eqn | l = x \in T_1 + | o = \implies + | r = f \left({x}\right) \in T_2 + | c = by definition of $f$ +}} +{{eqn | o = \implies + | r = h \left({f \left({x}\right)}\right) = f^{-1} \left({f \left({x}\right)}\right) = x + | c = by definition of $h$ +}} +{{end-eqn}} +Thus: +{{begin-eqn}} +{{eqn | l = g \left({x}\right) + | r = g \left({h \left({f \left({x}\right)}\right)}\right) + | c = from above +}} +{{eqn | r = h^{-1} \left({h \left({f \left({x}\right)}\right)}\right) + | c = by definition of $g$ +}} +{{eqn | r = f \left({x}\right) + | c = [[Inverse of Inverse of Bijection]] +}} +{{end-eqn}} +So $g$ is an [[Definition:Extension of Mapping|extension]] of $f$. +We have therefore proved that the embedding theorem holds when $T_1$ and $S_2$ are [[Definition:Disjoint Sets|disjoint]]. +{{qed|lemma}} +Next, suppose $T_1$ and $S_2$ are ''not'' [[Definition:Disjoint Sets|disjoint]]. +We can use [[Exists Bijection to a Disjoint Set]] to assume the existence of a [[Definition:Bijection|bijection]] $k: S_2 \to S_3$ such that $S_3 \cap T = \varnothing$. +Let $*'$ be the [[Transplanting Theorem|transplant of $*$ under $k$]]. +Let: +:$T_3 = \left\{{k \left({x}\right): x \in T_2}\right\}$ +Then: +{{begin-eqn}} +{{eqn | l = x, y \in T_3 + | o = \implies + | r = k^{-1} \left({x}\right), k^{-1} \left({y}\right) \in T_2 + | c = by definition of $k$ +}} +{{eqn | o = \implies + | r = k^{-1} \left({x}\right) * k^{-1} \left({y}\right) \in T_2 + | c = by definition of $*$ in $T_2$ +}} +{{eqn | o = \implies + | r = x *' y = k \left({k^{-1} \left({x}\right) * k^{-1} \left({y}\right)}\right) \in T_3 + | c = [[Transplanting Theorem]] +}} +{{end-eqn}} +So $T_3$ is closed under $*'$. +Let $*' {\restriction_{T_3} }$ be the [[Definition:Operation Induced by Restriction|operation on $T_3$ induced by $*'$]]. +Then $\left({T_3, *' {\restriction_{T_3} } }\right)$ is embedded in $\left({S_3, *'}\right)$. +Thus $k \circ f$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] from $\left({T_1, \oplus_1}\right)$ onto $\left({T_3, *' {\restriction_{T_3} } }\right)$. +$S_3$ has been constructed so as to be [[Definition:Disjoint Sets|disjoint]] from $T_1$. +In such a case it has already been shown that the embedding theorem holds. +That is: +:$(1): \quad$ there exists a [[Definition:Magma|magma]] $\left({S_1, *_1}\right)$ [[Definition:Submagma|containing $\left({T_1, \oplus_1}\right)$ algebraically]] +:$(2): \quad$ there exists an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] $g_1$ from $\left({S_1, *_1}\right)$ to $\left({S_3, \oplus_1}\right)$ [[Definition:Extension of Mapping|extending]] $k \bullet f$. +Let $g = k^{-1} \bullet g_1$. +Let $x \in T$. +Then: +{{begin-eqn}} +{{eqn | l = g \left({x}\right) + | r = k^{-1} \left({g_1 \left({x}\right)}\right) + | c = definition of $g$ +}} +{{eqn | r = k^{-1} \left({k \left({f \left({x}\right)}\right)}\right) + | c = as $g_1$ extends $k \bullet f$ +}} +{{eqn | r = f \left({x}\right) + | c = [[Inverse of Inverse of Bijection]] +}} +{{end-eqn}} +and so $g$ [[Definition:Extension of Mapping|extends]] $f$. +As $k$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] from $\left({S_2, *}\right)$ onto $\left({S_3, *'}\right)$, then: +:$(1): \quad g$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] from $\left({S_1, *_1}\right)$ onto $\left({S_2, *}\right)$ +:$(2): \quad g$ [[Definition:Extension of Mapping|extends]] $f$. +Hence the result. +{{qed}} +{{Proofread}} +\end{proof}<|endoftext|> +\section{Quotient Structure of Inverse Completion} +Tags: Inverse Completions, Quotient Structures + +\begin{theorem} +Let $\left({T, \circ'}\right)$ be an [[Definition:Inverse Completion|inverse completion]] of a [[Definition:Commutative Semigroup|commutative semigroup]] $\left({S, \circ}\right)$, where $C$ is the set of [[Definition:Cancellable Element|cancellable elements]] of $S$. +Let $f: S \times C: T$ be the [[Definition:Mapping|mapping]] defined as: +:$\forall x \in S, y \in C: f \left({x, y}\right) = x \circ' y^{-1}$ +Then the mapping $g: \left({S \times C}\right) / \mathcal R_f \to T$ defined by $g \left({\left[\!\left[{x, y}\right]\!\right]_{\mathcal R_f}}\right) = x \circ' y^{-1}$, +where $\left({S \times C}\right) / \mathcal R_f$ is a [[Definition:Quotient Structure|quotient structure]], is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]]. +\end{theorem} + +\begin{proof} +$T$ is [[Definition:Commutative Semigroup|commutative]], from [[Inverse Completion is Commutative Semigroup]]. +The mapping $f \left({x, y}\right) = x \circ' y^{-1}$ is an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]] from the [[Definition:Cartesian Product|cartesian product]] of $\left({S, \circ}\right)$ and $\left({C, \circ \restriction_C}\right)$ onto $\left({T, \circ'}\right)$. +By the [[Quotient Theorem for Epimorphisms]], the proof follows. +{{Explain|The mapping $f \left({x, y}\right) {{=}} x \circ' y^{-1}$ is an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]] from the [[Definition:Cartesian Product|cartesian product]] of $\left({S, \circ}\right)$ and $\left({C, \circ \restriction_C}\right)$ onto $\left({T, \circ'}\right)$.}} +{{Proofread}} +[[Category:Inverse Completions]] +[[Category:Quotient Structures]] +5bw3dopvx32rcfb6a9lyp5eqyjb3m8v +\end{proof}<|endoftext|> +\section{Quotient Mapping of Inverse Completion} +Tags: Inverse Completions + +\begin{theorem} +Let $\struct {T, \circ'}$ be an [[Definition:Inverse Completion|inverse completion]] of a [[Definition:Commutative Semigroup|commutative semigroup]] $\struct {S, \circ}$, where $C$ is the set of [[Definition:Cancellable Element|cancellable elements]] of $S$. +Let $f: S \times C: T$ be the [[Definition:Mapping|mapping]] defined as: +:$\forall x \in S, y \in C: \map f {x, y} = x \circ' y^{-1}$ +Let $\RR_f$ be the [[Definition:Equivalence Relation Induced by Mapping|equivalence relation induced by $f$]]. +Then: +:$\tuple {x_1, y_1} \mathrel {\RR_f} \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$ +\end{theorem} + +\begin{proof} +By the definition of $\RR_f$: +:$\tuple {x_1, y_1} \mathrel {\RR_f} \tuple {x_2, y_2} \iff x_1 \circ' y_1^{-1} = x_2 \circ' y_2^{-1}$ +Now: +{{begin-eqn}} +{{eqn | l = x_1 \circ' y_1^{-1} + | r = x_2 \circ' y_2^{-1} + | c = +}} +{{eqn | ll= \leadsto + | l = x_1 \circ' y_1^{-1} \circ' y_1 \circ' y_2 + | r = x_2 \circ' y_2^{-1} \circ' y_1 \circ' y_2 + | c = Elements $y_1$ and $y_2$ are [[Definition:Cancellable Element|cancellable]] +}} +{{eqn | ll= \leadsto + | l = x_1 \circ' y_2 + | r = x_2 \circ' y_1 + | c = {{Defof|Inverse Element}}, and [[Definition:Commutative Operation|commutativity]] of $\circ'$ +}} +{{eqn | ll= \leadsto + | l = x_1 \circ y_2 + | r = x_2 \circ y_1 + | c = [[Definition:Extension of Operation|$\circ'$ extends $\circ$]] +}} +{{end-eqn}} +which leads us to: +:$\tuple {x_1, y_1} \mathrel {\RR_f} \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$ +{{qed}} +[[Category:Inverse Completions]] +2om0id39oam1sk3ljj5keesgt7lf0w0 +\end{proof}<|endoftext|> +\section{Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup} +Tags: Inverse Completions + +\begin{theorem} +Let $\left({S, \circ}\right)$ be a [[Definition:Commutative Semigroup|commutative semigroup]]. +Let $\left ({C, \circ}\right) \subseteq \left({S, \circ}\right)$ be the [[Definition:Subsemigroup|subsemigroup]] of [[Definition:Cancellable Element|cancellable elements]] of $\left({S, \circ}\right)$. +Let $\left({T, \circ'}\right)$ be an [[Definition:Inverse Completion|inverse completion]] of $\left({S, \circ}\right)$. +Then: +:$S \circ' C^{-1}$ is a [[Definition:Commutative Semigroup|commutative semigroup]] +where $S \circ' C^{-1}$ is the [[Definition:Subset Product|subset product]] of $S$ with $C^{-1}$ under $\circ'$ in $T$. +\end{theorem} + +\begin{proof} +Note that by definition of [[Definition:Inverse Completion|inverse completion]], $\left({T, \circ'}\right)$ is a [[Definition:Semigroup|semigroup]]. +Thus $\circ'$ is [[Definition:Associative Operation|associative]]. +First it is demonstrated that $S \circ' C^{-1}$ is a [[Definition:Semigroup|semigroup]]. +Let $x, z \in S$. +Let $y, w \in C$. +Then: +{{begin-eqn}} +{{eqn | l = \left({x \circ' y^{-1} }\right) \circ' \left({z \circ' w^{-1} }\right) + | r = x \circ' \left({y^{-1} \circ' z}\right) \circ' w^{-1} + | c = [[Definition:Associative Operation|Associativity of $\circ'$]] +}} +{{eqn | r = x \circ' \left({z \circ' y^{-1} }\right) \circ' w^{-1} + | c = [[Commutation with Inverse in Monoid]] +}} +{{eqn | r = \left({x \circ' z}\right) \circ' \left({y^{-1} \circ' w^{-1} }\right) + | c = [[Definition:Associative Operation|Associativity of $\circ'$]] +}} +{{eqn | r = \left({x \circ' z}\right) \circ' \left({w \circ' y}\right)^{-1} + | c = [[Inverse of Product in Associative Structure]] +}} +{{eqn | r = \left({x \circ z}\right) \circ' \left({w \circ y}\right)^{-1} + | c = [[Definition:Extension of Operation|$\circ'$ extends $\circ$]] +}} +{{end-eqn}} +Thus: +:$\left({x \circ z}\right)\circ' \left({w \circ y}\right)^{-1} \in S \circ' C^{-1}$ +proving that $S \circ' C^{-1}$ is [[Definition:Closed Algebraic Structure|closed]]. +Therefore by [[Subsemigroup Closure Test]]: +:$S \circ' C^{-1}$ is a [[Definition:Subsemigroup|subsemigroup]] of $\left({T, \circ'}\right)$ +and thus a [[Definition:Semigroup|semigroup]]. +{{qed|lemma}} +It remains to be shown that $\circ'$ is a [[Definition:Commutative Operation|commutative operation]]. +Let $\left({x \circ' y^{-1} }\right)$ and $\left({z \circ' w^{-1} }\right)$ be two arbitrary [[Definition:Element|elements]] of $S \circ' C^{-1}$. +By [[Element Commutes with Product of Commuting Elements]], $x, y, z, w$ all [[Definition:Commute|commute]] with each other under $\circ$. +As $\circ'$ is an [[Definition:Extension of Operation|extension]] of $\circ$, it follows that $x, y, z, w$ also all [[Definition:Commute|commute]] with each other under $\circ'$. +Then: +{{begin-eqn}} +{{eqn | l = \left({x \circ' y^{-1} }\right) \circ' \left({z \circ' w^{-1} }\right) + | r = x \circ' \left({y^{-1} \circ' z}\right) \circ' w^{-1} + | c = [[Definition:Associative Operation|Associativity of $\circ'$]] +}} +{{eqn | r = x \circ' \left({z \circ' y^{-1} }\right) \circ' w^{-1} + | c = [[Commutation with Inverse in Monoid]] +}} +{{eqn | r = \left({x \circ' z}\right) \circ' \left({y^{-1} \circ' w^{-1} }\right) + | c = [[Definition:Associative Operation|Associativity of $\circ'$]] +}} +{{eqn | r = \left({z \circ' x}\right) \circ' \left({w^{-1} \circ' y^{-1} }\right) + | c = [[Commutation of Inverses in Monoid]] +}} +{{eqn | r = z \circ' \left({x \circ' w^{-1} }\right) \circ' y^{-1} + | c = [[Definition:Associative Operation|Associativity of $\circ'$]] +}} +{{eqn | r = z \circ' \left({w^{-1} \circ' x}\right) \circ' y^{-1} + | c = [[Commutation with Inverse in Monoid]] +}} +{{eqn | r = \left({z \circ' w^{-1} }\right) \circ' \left({x \circ' y^{-1} }\right) + | c = [[Definition:Associative Operation|Associativity of $\circ'$]] +}} +{{end-eqn}} +So $x \circ' y^{-1}$ commutes with $z \circ' w^{-1}$. +It follows by definition that $S \circ' C^{-1}$ is a [[Definition:Commutative Semigroup|commutative semigroup]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Inverse Completion of Commutative Semigroup is Inverse Completion of Itself} +Tags: Inverse Completions + +\begin{theorem} +Let $\left({S, \circ}\right)$ be a [[Definition:Commutative Semigroup|commutative semigroup]]. +Let $\left ({C, \circ}\right) \subseteq \left({S, \circ}\right)$ be the [[Definition:Subsemigroup|subsemigroup]] of [[Definition:Cancellable Element|cancellable]] elements of $\left({S, \circ}\right)$. +Let $\left({T, \circ'}\right)$ be an [[Definition:Inverse Completion|inverse completion]] of $\left({S, \circ}\right)$. +Then $\left({T, \circ'}\right)$ is its own [[Definition:Inverse Completion|inverse completion]]. +\end{theorem} + +\begin{proof} +Let $x \circ' y^{-1}$ be [[Definition:Cancellable Element|cancellable]] for $\circ'$, where $x \in S$ and $y \in C$. +We have that $y$ is [[Definition:Invertible Element|invertible]] for $\circ'$. +So by [[Invertible Element of Associative Structure is Cancellable]]: +:$y$ is [[Definition:Cancellable Element|cancellable]] for $\circ'$. +Now by definition of [[Definition:Inverse Element|inverse element]]: +:$x = \left({x \circ' y^{-1}}\right) \circ' y$ +Thus $x$ is also [[Definition:Cancellable Element|cancellable]] for $\circ'$. +By [[Cancellable Elements of Semigroup form Subsemigroup]], $x$ is [[Definition:Cancellable Element|cancellable]] for $\circ$. +So: +: $x \in S \implies x \in C$ +Thus $x$ is [[Definition:Invertible Element|invertible]] for $\circ'$. +Hence by [[Inverse of Product in Associative Structure]]: +: $x \circ' y$ is [[Definition:Invertible Element|invertible]] for $\circ'$. +So every [[Definition:Cancellable Element|cancellable element]] of $\left({T, \circ'}\right)$ is [[Definition:Invertible Element|invertible]]. +So, by definition, $\left({T, \circ'}\right)$ is its own [[Definition:Inverse Completion|inverse completion]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Identity of Inverse Completion of Commutative Monoid} +Tags: Inverse Completions + +\begin{theorem} +Let $\left({S, \circ}\right)$ be a [[Definition:Commutative Monoid|commutative monoid]] whose [[Definition:Identity Element|identity]] is $e$. +Let $\left ({C, \circ}\right) \subseteq \left({S, \circ}\right)$ be the [[Definition:Subsemigroup|subsemigroup]] of [[Definition:Cancellable Element|cancellable elements]] of $\left({S, \circ}\right)$. +Let $\left({T, \circ'}\right)$ be an [[Definition:Inverse Completion|inverse completion]] of $\left({S, \circ}\right)$. +Then $e \in T$ is the [[Definition:Identity Element|identity]] for $\circ'$. +\end{theorem} + +\begin{proof} +Let $e$ be the [[Definition:Identity Element|identity]] for $\circ$. +Let $e = x \circ' y^{-1}$, where $x \in S, y \in C$. +Then: +{{begin-eqn}} +{{eqn | l = y + | r = e \circ y + | c = Definition of [[Definition:Identity Element|Identity]] +}} +{{eqn | r = \left({x \circ' y^{-1} }\right) \circ' y + | c = Definition of [[Definition:Inverse Element|Inverse]] +}} +{{eqn | r = x \circ' \left({y^{-1} \circ' y}\right) + | c = $\circ'$ is [[Definition:Associative Operation|associative]] +}} +{{eqn | r = x \circ' e + | c = Definition of [[Definition:Inverse Element|Inverse]] +}} +{{end-eqn}} +Thus $e = y^{-1} \circ' y$, and $y^{-1} \circ' y$ is the identity for $\circ'$. +{{qed}} +\end{proof}<|endoftext|> +\section{Inverse Completion of Commutative Semigroup is Abelian Group} +Tags: Inverse Completions, Abelian Groups + +\begin{theorem} +Let $\struct {S, \circ}$ be a [[Definition:Commutative Semigroup|commutative semigroup]] +Let all the [[Definition:Element|elements]] of $\struct {S, \circ}$ be [[Definition:Cancellable Element|cancellable]]. +Then an [[Definition:Inverse Completion|inverse completion]] of $\struct {S, \circ}$ is an [[Definition:Abelian Group|abelian group]]. +\end{theorem} + +\begin{proof} +Let $\struct {S, \circ}$ be a [[Definition:Commutative Semigroup|commutative semigroup]], all of whose [[Definition:Element|elements]] are [[Definition:Cancellable Element|cancellable]]. +Let $\struct {T, \circ'}$ be an [[Definition:Inverse Completion|inverse completion]] of $\struct {S, \circ}$. +From [[Inverse Completion is Commutative Monoid]]: +:$\struct {T, \circ'} = \struct {S \circ' S^{-1}, \circ'}$ +has been shown to be a [[Definition:Commutative Monoid|commutative monoid]]. +Taking the [[Definition:Group Axioms|group axioms]] in turn: +=== {{GroupAxiom|0|nolink}} === +As $\struct {T, \circ'}$ is a [[Definition:Commutative Semigroup|commutative semigroup]], it is by definition [[Definition:Closed Algebraic Structure|closed]]. +{{qed|lemma}} +=== {{GroupAxiom|1|nolink}} === +As $\struct {T, \circ'}$ is a [[Definition:Commutative Semigroup|commutative semigroup]], it is by definition [[Definition:Associative Operation|associative]]. +{{qed|lemma}} +=== {{GroupAxiom|2|nolink}} === +Let $x \in S$. +Then $x^{-1} \in S^{-1}$ by definition. +As $\struct {T, \circ'} = \struct {S \circ' S^{-1}, \circ'}$ is [[Definition:Closed Algebraic Structure|closed]]: +:$x \circ' x^{-1} \in T$ +This holds for all $x \in S$. +As $\struct {T, \circ'}$ is a [[Definition:Commutative Semigroup|commutative semigroup]]: +:$\exists e \in T: \forall x \in S: x \circ' x^{-1} = e^{-1} = x \circ' x$ +Thus $\struct {T, \circ'}$ has an [[Definition:Identity Element|identity element]]. +{{qed|lemma}} +=== {{GroupAxiom|3|nolink}} === +Every [[Definition:Element|element]] of $S$ has an [[Definition:Inverse Element|inverse]] in $S^{-1}$. +Therefore every [[Definition:Element|element]] of $S$ and $S^{-1}$ is [[Definition:Invertible Element|invertible]]. +From [[Inverse of Product in Associative Structure]], every [[Definition:Element|element]] of $S \circ' S^{-1}$ is therefore also [[Definition:Invertible Element|invertible]]. +Thus every element of $T$ has an [[Definition:Inverse Element|inverse]]. +{{qed|lemma}} +All the [[Definition:Group Axioms|group axioms]] are thus seen to be fulfilled, and so $\struct {T, \circ'}$ is a [[Definition:Group|group]]. +[[Definition:Commutative Operation|Commutativity]] of $\circ$ has been demonstrated. +Hence the result, by definition of [[Definition:Abelian Group|abelian group]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Taylor's Theorem} +Tags: Calculus, Taylor's Theorem + +\begin{theorem} +Every [[Definition:Smooth Real Function|infinitely differentiable function]] can be approximated by a [[Definition:Series|series]] of [[Definition:Real Polynomial Function|polynomials]]. +\end{theorem}<|endoftext|> +\section{Inverse Completion Theorem} +Tags: Inverse Completions, Semigroups, Named Theorems + +\begin{theorem} +Every [[Definition:Commutative Semigroup|commutative semigroup]] containing [[Definition:Cancellable Element|cancellable elements]] admits an [[Definition:Inverse Completion|inverse completion]]. +\end{theorem} + +\begin{proof} +Let $\left({S, \circ}\right)$ be a [[Definition:Commutative Semigroup|commutative semigroup]] which has [[Definition:Cancellable Element|cancellable elements]]. +From [[Construction of Inverse Completion]], we can construct an [[Definition:Inverse Completion|inverse completion]] $\left({T', \oplus'}\right)$ of $\left({S', \oplus'}\right)$, which is an [[Definition:Isomorphism (Abstract Algebra)|isomorphic]] copy of $S$ under the mapping $\psi: S \to S'$. +By the [[Embedding Theorem]], there exists a [[Definition:Semigroup|semigroup]] $\left({T, \circ}\right)$ which contains $\left({S, \circ}\right)$, and an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] $\Psi$ from $\left({T, \circ}\right)$ to $\left({T', \oplus'}\right)$ which extends $\psi$. +Thus $T$ is an [[Definition:Inverse Completion|inverse completion]] of $S$. +{{Qed}} +\end{proof}<|endoftext|> +\section{Extension Theorem for Homomorphisms} +Tags: Homomorphisms, Monomorphisms, Semigroups, Inverse Completions, Named Theorems + +\begin{theorem} +Let $\struct {S, \circ}$ be a [[Definition:Commutative Semigroup|commutative semigroup]] with [[Definition:Cancellable Element|cancellable elements]] +Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the [[Definition:Subsemigroup|subsemigroup]] of all [[Definition:Cancellable Element|cancellable elements]] of $S$ +Let $\struct {S', \circ'}$ be an [[Definition:Inverse Completion|inverse completion]] of $\struct {S, \circ}$ +Let $\phi$ be a [[Definition:Semigroup Homomorphism|(semigroup) homomorphism]] from $\struct {S, \circ}$ into a [[Definition:Semigroup|semigroup]] $\struct {T, *}$ such that $\map \phi y$ is [[Definition:Invertible Element|invertible]] for all $y \in C$. +Then: +:$(1): \quad$ There is one and only one [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] $\psi$ from $\struct {S', \circ'}$ into $\struct {T, *}$ [[Definition:Extension of Mapping|extending]] $\phi$ +:$(2): \quad \forall x \in S, y \in C: \map \psi {x \circ' y^{-1} } = \map \phi * \paren {\map \phi y}^{-1}$ +:$(3): \quad$ If $\phi$ is a [[Definition:Monomorphism (Abstract Algebra)|monomorphism]], then so is $\psi$. +\end{theorem} + +\begin{proof} +It is proved that $\struct {C, \circ}$ is a [[Definition:Subsemigroup|subsemigroup]] of $\struct {S, \circ}$ by [[Cancellable Elements of Semigroup form Subsemigroup]]. +=== Proof of at most one such homomorphism === +To show there is at most one such [[Definition:Homomorphism (Abstract Algebra)|homomorphism]]: +Let $\psi$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] from $\struct {S', \circ'}$ into $\struct {T, *}$ [[Definition:Extension of Mapping|extending]] $\phi$. +It will be shown that for each [[Definition:Element|element]] in the [[Definition:Domain of Mapping|domain]] of $\psi$ , its [[Definition:Image of Element under Mapping|image]] is uniquely determined by $\phi$. +It will therefore follow that $\psi$ is the only such [[Definition:Extension of Mapping|extension]] of $\phi$. +Now $\map \phi y$ is [[Definition:Invertible Element|invertible]] and hence [[Definition:Cancellable Element|cancellable]] for $*$ by [[Invertible Element of Monoid is Cancellable]]. +As $\psi$ is an [[Definition:Extension of Mapping|extension]] of $\phi$, it follows that $\map \psi y$ is also [[Definition:Cancellable Element|cancellable]] for $*$. +Consider: +:$\map \psi {x \circ' y^{-1} } * \map \psi y$ +for arbitrary $x \in S, y \in C$. +We have: +{{begin-eqn}} +{{eqn | l = \map \psi {x \circ' y^{-1} } * \map \psi y + | r = \map \psi {x \circ' y^{-1} \circ' y} + | c = {{Defof|Morphism Property}} +}} +{{eqn | r = \map \psi x + | c = {{Defof|Inverse Element}} +}} +{{eqn | r = \map \phi x + | c = $\psi$ [[Definition:Extension of Mapping|extends]] $\phi$ +}} +{{eqn | r = \map \phi x * \paren {\map \phi y}^{-1} * \map \phi y + | c = [[Definition:By Hypothesis|by hypothesis]]: $\map \phi y$ is [[Definition:Invertible Element|invertible]] in $\struct {T, *}$ +}} +{{eqn | r = \map \phi x * \paren {\map \phi y}^{-1} * \map \psi y + | c = $\psi$ [[Definition:Extension of Mapping|extends]] $\phi$ +}} +{{end-eqn}} +So as $\map \psi y$ is [[Definition:Cancellable Element|cancellable]] for $*$: +:$(1): \quad \map \psi {x \circ' y^{-1} } = \map \phi x * \paren {\map \phi y}^{-1}$ +From [[Inverse Completion is Commutative Semigroup]]: +:$S' = S \circ' C^{-1}$ +thus $(1)$ holds for all $x \in S, y \in C$. +So, there is at most one [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] extending $\phi$. +{{Qed}} +=== Proof of at least one such homomorphism === +By [[Restriction of Mapping to Image is Surjection]], $\phi: S \to \phi \sqbrk S$ is a [[Definition:Surjection|surjection]]. +Therefore, by definition, $\phi$ is an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]]. +We have that $\struct {S, \circ}$ is a [[Definition:Commutative Semigroup|commutative semigroup]]. +By [[Epimorphism Preserves Semigroups]] and [[Epimorphism Preserves Commutativity]], $\phi \sqbrk {\struct {S, \circ} }$ is also a [[Definition:Commutative Semigroup|commutative semigroup]]. +As $S \subseteq T$ it follows that $\phi \sqbrk {\struct {S, \circ} }$ is a [[Definition:Subsemigroup|subsemigroup]] of $\struct {T, *}$. +By [[Commutation with Inverse in Monoid]], every [[Definition:Element|element]] of $\phi \sqbrk S$ [[Definition:Commute|commutes]] with every [[Definition:Element|element]] of $\paren {\phi \sqbrk C}^{-1}$. +Thus it follows that the [[Definition:Mapping|mapping]] $\psi: S' \to T$ defined by: +:$\forall x \in S, y \in C: \map \psi {x \circ' y^{-1} } = \map \phi x * \paren {\map \phi y}^{-1}$ +is a well-defined [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] extending $\phi$. +{{Qed}} +=== Proof of Monomorphism === +Let $\phi$ be a [[Definition:Monomorphism (Abstract Algebra)|monomorphism]]. +Then: +:$\forall x, y \in S: \map \phi x = \map \phi y \implies x = y$ +Now let $x \circ y^{-1}, z \circ w^{-1} \in S'$ such that $\map \psi {x \circ y^{-1} } = \map \psi {z \circ w^{-1} }$. +Then: +{{begin-eqn}} +{{eqn | l = \map \psi {x \circ' y^{-1} } + | r = \map \psi {z \circ' w^{-1} } + | c = {{Defof|Morphism Property}} +}} +{{eqn | ll= \leadsto + | l = \map \psi x * \map \psi {y^{-1} } + | r = \map \psi z * \map \psi {w^{-1} } + | c = +}} +{{eqn | ll= \leadsto + | l = \map \psi x * \map \psi w + | r = \map \psi z * \map \psi y + | c = +}} +{{eqn | ll= \leadsto + | l = \map \phi x * \map \phi w + | r = \map \phi z * \map \phi y + | c = $\psi$ is an [[Definition:Extension of Mapping|extension]] of $\phi$ +}} +{{eqn | ll= \leadsto + | l = \map \phi {x \circ w} + | r = \map \phi {z \circ y} + | c = {{Defof|Morphism Property}} +}} +{{eqn | ll= \leadsto + | l = x \circ w + | r = z \circ y + | c = $\phi$ is a [[Definition:Monomorphism (Abstract Algebra)|monomorphism]] +}} +{{eqn | ll= \leadsto + | l = x \circ' y^{-1} + | r = z \circ' w^{-1} + | c = +}} +{{end-eqn}} +Thus $\psi$ is a [[Definition:Monomorphism (Abstract Algebra)|monomorphism]] if $\phi$ is. +{{Qed}} +\end{proof}<|endoftext|> +\section{Extension Theorem for Isomorphisms} +Tags: Isomorphisms, Inverse Completions, Semigroups, Named Theorems + +\begin{theorem} +Let the following conditions be fulfilled: +: Let $\left({S, \circ}\right)$ be a [[Definition:Commutative Semigroup|commutative semigroup]] with [[Definition:Cancellable Element|cancellable elements]] +: Let $\phi$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] from $\left({S, \circ}\right)$ into a [[Definition:Semigroup|semigroup]] $\left({T, *}\right)$ +: Let $\left({S', \circ'}\right)$ be an [[Definition:Inverse Completion|inverse completion]] of $\left({S, \circ}\right)$ +: Let $\left({T', \circ'}\right)$ be an [[Definition:Inverse Completion|inverse completion]] of $\left({T, \circ}\right)$. +Then there is a unique [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] $\phi': S' \to T'$ extending $\phi$. +\end{theorem} + +\begin{proof} +Let $C$ be the [[Definition:Subsemigroup|subsemigroup]] of [[Definition:Cancellable Element|cancellable elements]] of $S$. +It is proved that this is a semigroup by [[Cancellable Elements of Semigroup form Subsemigroup]]. +The set of [[Definition:Cancellable Element|cancellable elements]] of $T$ is $\phi \left({C}\right)$. +By the [[Extension Theorem for Homomorphisms]], there is: +: A unique [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] $\phi'$ from $S'$ into $T'$ extending $\phi$ +and: +: A unique [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] $\psi$ from $T'$ into $S'$ extending $\phi^{-1}$. +Therefore, $\psi \circ \phi'$ is an [[Definition:Endomorphism|endomorphism]] of $S'$ whose [[Definition:Restriction of Mapping|restriction]] to $S$ is the [[Definition:Inclusion Mapping|inclusion mapping]] $\iota_S: S \to S'$: +:$\forall x \in S: \iota \left({x}\right) = x$ +From [[Inclusion Mapping is Monomorphism]], $\iota_S$ is a [[Definition:Monomorphism (Abstract Algebra)|monomorphism]] from $S$ into $S'$. +But by the [[Extension Theorem for Homomorphisms]], the [[Definition:Identity Mapping|identity]] [[Definition:Automorphism (Abstract Algebra)|automorphism]] of $S'$ is the only [[Definition:Endomorphism|endomorphism]] of $S'$ extending $\iota_S$ from $S$ into $S'$, so: +:$\psi \circ \phi' = I_{S'}$ +Similarly: +:$\phi' \circ \psi = I_{T'}$ +So by [[Bijection iff Left and Right Inverse]], $\phi'$ is a [[Definition:Bijection|bijection]] and therefore an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]]. +{{Qed}} +\end{proof}<|endoftext|> +\section{Inverse Completion is Unique} +Tags: Inverse Completions, Semigroups + +\begin{theorem} +An [[Definition:Inverse Completion|inverse completion]] of a [[Definition:Commutative Semigroup|commutative semigroup]] is unique up to [[Definition:Isomorphism (Abstract Algebra)|isomorphism]]. +\end{theorem} + +\begin{proof} +Let $T$ and $T'$ both be [[Definition:Inverse Completion|inverse completions]] of a [[Definition:Commutative Semigroup|commutative semigroup]] $S$ having [[Definition:Cancellable Element|cancellable elements]]. +Then from the [[Extension Theorem for Isomorphisms]], there is a unique [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] $\phi: T \to T'$ satisfying $\forall x \in S: \phi \left({x}\right) = x$. +Hence the result. +{{Qed}} +\end{proof}<|endoftext|> +\section{Extension Theorem for Distributive Operations} +Tags: Inverse Completions, Semigroups, Distributive Operations + +\begin{theorem} +Let $\struct {R, *}$ be a [[Definition:Commutative Semigroup|commutative semigroup]], all of whose [[Definition:Element|elements]] are [[Definition:Cancellable Element|cancellable]]. +Let $\struct {T, *}$ be an [[Definition:Inverse Completion|inverse completion]] of $\struct {R, *}$. +Let $\circ$ be an [[Definition:Binary Operation|operation]] on $R$ which [[Definition:Distributive Operation|distributes]] over $*$. +Then: +:$(1): \quad$ There is a unique [[Definition:Binary Operation|operation]] $\circ'$ on $T$ which [[Definition:Distributive Operation|distributes]] over $*$ in $T$ and induces on $R$ the [[Definition:Binary Operation|operation]] $\circ$ +:$(2): \quad$ If $\circ$ is [[Definition:Associative Operation|associative]], then so is $\circ'$ +:$(3): \quad$ If $\circ$ is [[Definition:Commutative Operation|commutative]], then so is $\circ'$ +:$(4): \quad$ If $e$ is an [[Definition:Identity Element|identity]] for $\circ$, then $e$ is also an [[Definition:Identity Element|identity]] for $\circ'$ +:$(5): \quad$ Every element [[Definition:Cancellable Element|cancellable]] for $\circ$ is also [[Definition:Cancellable Element|cancellable]] for $\circ'$. +\end{theorem} + +\begin{proof} +[[Definition:By Hypothesis|By hypothesis]], all the [[Definition:Element|elements]] of $\struct {R, *}$ are [[Definition:Cancellable Element|cancellable]]. +Thus [[Inverse Completion of Commutative Semigroup is Abelian Group]] can be applied. +So $\struct {T, *}$ is an [[Definition:Abelian Group|abelian group]]. +{{explain|Where is the above used?}} +=== Existence of Distributive Operation === +For each $m \in R$, we define $\lambda_m: R \to T$ as: +:$\forall x \in R: \map {\lambda_m} x = m \circ x$ +Then: +{{begin-eqn}} +{{eqn | l = \map {\lambda_m} {x * y} + | r = m \circ \paren {x * y} + | c = +}} +{{eqn | r = \paren {m \circ x} * \paren {m \circ y} + | c = as $\circ$ [[Definition:Distributive Operation|distributes]] over $*$ +}} +{{eqn | r = \map {\lambda_m} x * \map {\lambda_m} y + | c = +}} +{{end-eqn}} +So $\lambda_m$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] from $\struct {R, *}$ into $\struct {T, *}$. +Now, by the [[Extension Theorem for Homomorphisms]], every [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] from $\struct {R, *}$ into $\struct {T, *}$ is the [[Definition:Restriction of Mapping|restriction]] to $R$ of a unique [[Definition:Endomorphism|endomorphism]] of $\struct {T, *}$. +We have just shown that $\lambda_m$ is such a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]]. +Therefore there exists a unique [[Definition:Endomorphism|endomorphism]] $\lambda'_m: T \to T$ which [[Definition:Extension of Mapping|extends]] $\lambda_m$. +Now: +{{begin-eqn}} +{{eqn | lo= \forall m, n, z \in R: + | l = \map {\lambda_{m * n} } z + | r = \paren {m * n} \circ z + | c = Definition of $\lambda$ +}} +{{eqn | r = \paren {m \circ z} * \paren {n \circ z} + | c = [[Definition:Distributive Operation|Distributivity of $\circ$ over $*$]] +}} +{{eqn | r = \map {\lambda_m} z * \map {\lambda_n} z + | c = Definition of $\lambda$ +}} +{{eqn | r = \map {\paren {\lambda_m * \lambda_n} } z + | c = Here $*$ is the [[Definition:Induced Structure|operation induced on $T^T$ by $*$]] +}} +{{end-eqn}} +By [[Homomorphism on Induced Structure]], $\lambda'_m * \lambda'_n$ is an [[Definition:Endomorphism|endomorphism]] of $\struct {T, *}$ that, as we have just seen, coincides on $R$ with $\lambda'_{m * n}$. +Hence $\lambda'_{m * n} = \lambda'_m * \lambda'_n$. +Similarly, for each $z \in T$, we define $\rho_z: R \to T$ as: +:$\forall m \in R: \map {\rho_z} m = \map {\lambda'_m} z$ +Then: +{{begin-eqn}} +{{eqn | l = \map {\rho_z} {m * n} + | r = \map {\lambda'_{m * n} } z + | c = Definition of $\rho_z$ +}} +{{eqn | r = \map {\lambda'_m} z * \map {\lambda'_n} z + | c = Behaviour of $\lambda'_{m * n}$ +}} +{{eqn | r = \map {\rho_z} m * \map {\rho_z} n + | c = Definition of $\rho_z$ +}} +{{end-eqn}} +Therefore $\rho_z$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] from $\struct {R, *}$ into $\struct {T, *}$. +Consequently there exists a unique endomorphism $\rho'_z: T \to T$ extending $\rho_z$. +{{begin-eqn}} +{{eqn | lo= \forall y, z \in T, m \in R: + | l = \map {\rho_{y * z} } m + | r = \map {\lambda'_m} {y * z} + | c = Definition of $\rho_{y * z}$ +}} +{{eqn | r = \map {\lambda'_m} y * \map {\lambda'_m} z + | c = $\lambda'_m$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] +}} +{{eqn | r = \map {\paren {\rho_y * \rho_z} } m + | c = Definition of $\rho_y$ and $\rho_z$ +}} +{{end-eqn}} +By [[Homomorphism on Induced Structure]], $\rho'_y * \rho'_z$ is an [[Definition:Endomorphism|endomorphism]] on $\struct {T, *}$ that coincides (as we have just seen) with $\rho'_{y * z}$ on $R$. +Hence $\rho'_{y * z} = \rho'_y * \rho'_z$. +Now we define an operation $\circ'$ on $T$ by: +:$\forall x, y \in T: x \circ' y = \map {\rho'_y} x$ +Now suppose $x, y \in R$. Then: +{{begin-eqn}} +{{eqn | l = x \circ' y + | r = \map {\rho_y} x + | c = as $\rho'_y = \rho_y$ on $R$ +}} +{{eqn | r = \map {\lambda_x} y + | c = Definition of $\rho_x$ +}} +{{eqn | r = x \circ y + | c = Definition of $\lambda_x$ +}} +{{end-eqn}} +so $\circ'$ is an extension of $\circ$. +Next, let $x, y, z \in T$. Then: +{{begin-eqn}} +{{eqn | l = \paren {x * y} \circ' z + | r = \map {\rho'_z} {x * y} + | c = +}} +{{eqn | r = \map {\rho'_z} x * \map {\rho'_z} y + | c = +}} +{{eqn | r = x \circ' z * y \circ' z + | c = +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | l = x \circ' \paren {y * z} + | r = \map {\rho'_{y * z} } x + | c = +}} +{{eqn | r = \map {\rho'_y} x * \map {\rho'_z} x + | c = +}} +{{eqn | r = x \circ' y * x \circ' z + | c = +}} +{{end-eqn}} +So $\circ'$ is [[Definition:Distributive Operation|distributive]] over $*$. +{{qed}} +=== Uniqueness of Distributive Operation === +To show that $\circ'$ is unique, let $\circ_1$ be any operation on $T$ [[Definition:Distributive Operation|distributive]] over $*$ that induces $\circ$ on $R$. +Since $\circ'$ and $\circ_1$ both [[Definition:Distributive Operation|distribute]] over $*$, for every $m \in R$, the mappings: +{{explain|Should this be $y \in T$? If not, how does it define a $T$-endomorphism?}} +{{begin-eqn}} +{{eqn | l = y \mapsto m \circ_1 y + | o = , + | r = y \in R + | c = +}} +{{eqn | l = y \mapsto m \circ' y + | o = , + | r = y \in R + | c = +}} +{{end-eqn}} +are [[Definition:Endomorphism|endomorphisms]] of $\struct {T, *}$ that coincide on $R$ so must be the same mapping. +Therefore $\forall m \in R, y \in T: m \circ_1 y = m \circ' y$. +Similarly, for every $y \in T$, the mappings: +{{begin-eqn}} +{{eqn | l = x \mapsto x \circ_1 y + | o = , + | r = x \in T + | c = +}} +{{eqn | l = x \mapsto x \circ' y + | o = , + | r = x \in T + | c = +}} +{{end-eqn}} +are endomorphisms of $\struct {T, *}$ that coincide on $R$ by what we have just proved, so must be the same mapping. +Hence $\forall x, y \in T: x \circ_1 y = x \circ' y$. +Thus $\circ'$ is the only operation on $T$ which extends $\circ$ and distributes over $*$. +{{qed}} +=== Proof of Associativity === +Suppose $\circ$ is [[Definition:Associative Operation|associative]]. +As $\circ'$ distributes over $*$, for all $n, p \in R$, the mappings: +{{begin-eqn}} +{{eqn | l = x \mapsto \paren {x \circ' n} \circ' p + | o = , + | r = x \in T + | c = +}} +{{eqn | l = x \mapsto x \circ' \paren {n \circ' p} + | o = , + | r = x \in T + | c = +}} +{{end-eqn}} +are endomorphisms of $\struct {T, *}$ that coincide on $R$ by the associativity of $\circ$ and hence are the same mapping. +Therefore: +:$\forall x \in T, n, p \in R: \paren {x \circ' n} \circ' p = x \circ' \paren {n \circ' p}$ +Similarly, for all $x \in T, p \in R$, the mappings: +{{begin-eqn}} +{{eqn | l = y \mapsto \paren {x \circ' y} \circ' p + | o = , + | r = y \in T + | c = +}} +{{eqn | l = y \mapsto x \circ' \paren {y \circ' p} + | o = , + | r = y \in T + | c = +}} +{{end-eqn}} +are endomorphisms of $\struct {T, *}$ that coincide on $R$ by what we have proved and hence are the same mapping. +Therefore: +:$\forall x, y \in T, p \in R: \paren {x \circ' y} \circ' p = x \circ' \paren {y \circ' p}$ +Finally, for all $x, y \in T$, the mappings: +{{begin-eqn}} +{{eqn | l = z \mapsto \paren {x \circ' y} \circ' z + | o = , + | r = z \in T + | c = +}} +{{eqn | l = z \mapsto x \circ' \paren {y \circ' z} + | o = , + | r = z \in T + | c = +}} +{{end-eqn}} +are endomorphisms of $\struct {T, *}$ that coincide on $R$ by what we have proved and hence are the same mapping. +Therefore $\circ'$ is [[Definition:Associative Operation|associative]]. +{{qed}} +=== Proof of Commutativity === +Suppose $\circ$ is [[Definition:Commutative Operation|commutative]]. +As $\circ'$ distributes over $*$, for all $n \in R$, the mappings: +{{begin-eqn}} +{{eqn | l = x \mapsto x \circ' n + | o = , + | r = x \in T + | c = +}} +{{eqn | l = x \mapsto n \circ' x + | o = , + | r = x \in T + | c = +}} +{{end-eqn}} +are endomorphisms of $\struct {T, *}$ that coincide on $R$ by the commutativity of $\circ$ and hence are the same mapping. +Therefore $\forall x \in T, n \in R: x \circ' n = n \circ' x$. +Finally, for all $y \in T$, the mappings: +{{begin-eqn}} +{{eqn | l = z \mapsto z \circ' y + | o = , + | r = z \in T + | c = +}} +{{eqn | l = z \mapsto y \circ' z + | o = , + | r = z \in T + | c = +}} +{{end-eqn}} +are endomorphisms of $\struct {T, *}$ that coincide on $R$ by what we have proved and hence are the same mapping. +Therefore $\circ'$ is [[Definition:Commutative Operation|commutative]]. +{{qed}} +=== Proof of Identity === +Let $e$ be the [[Definition:Identity Element|identity element]] of $\struct {R, \circ}$. +Then the restrictions to $R$ of the endomorphisms $\lambda_e: x \mapsto e \circ' x$ and $\rho_e: x \mapsto x \circ' e$ of $\struct {T, *}$ are [[Definition:Monomorphism (Abstract Algebra)|monomorphisms]]. +But then $\lambda_e$ and $\rho_e$ are [[Definition:Monomorphism (Abstract Algebra)|monomorphisms]] by the [[Extension Theorem for Homomorphisms]], so $e$ is the [[Definition:Identity Element|identity element]] of $T$. +{{explain|How does the above explain why $e$ is the identity element of $T$?}} +{{qed}} +=== Proof of Cancellability === +To prove that every element of $R$ [[Definition:Cancellable Element|cancellable]] for $\circ$ is also [[Definition:Cancellable Element|cancellable]] for $\circ'$: +Let $a$ be an element of $R$ cancellable for $\circ$. +Then the restrictions to $R$ of the endomorphisms $\lambda_a: x \mapsto a \circ' x$ and $\rho_a: x \mapsto x \circ' a$ of $\struct {T, *}$ are [[Definition:Monomorphism (Abstract Algebra)|monomorphisms]]. +But then $\lambda_a$ and $\rho_a$ are [[Definition:Monomorphism (Abstract Algebra)|monomorphisms]] by the [[Extension Theorem for Homomorphisms]], so $a$ is [[Definition:Cancellable Element|cancellable]] for $\circ'$. +{{qed}} +\end{proof}<|endoftext|> +\section{Monoid is not Empty} +Tags: Monoids + +\begin{theorem} +A [[Definition:Monoid|monoid]] cannot be [[Definition:Empty Set|empty]]. +\end{theorem} + +\begin{proof} +Let $\left({S, \circ}\right)$ be a the [[Definition:Monoid|monoid]]. +By [[Definition:Monoid|definition]]: +:[[Definition:Identity Element|Identity]]: $\exists e_S \in S: \forall a \in S: a \circ e_S = a = e_S \circ a$ +So a monoid must ''at least'' have an [[Definition:Identity Element|identity]]. +Therefore $e_S \in S$ and so $S$ is not the [[Definition:Empty Set|empty set]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Group is not Empty} +Tags: Group Theory + +\begin{theorem} +A [[Definition:Group|group]] cannot be [[Definition:Empty Set|empty]]. +\end{theorem} + +\begin{proof} +A [[Definition:Group|group]] is defined as a [[Definition:Monoid|monoid]] for which every [[Definition:Element|element]] has an [[Definition:Inverse (Abstract Algebra)|inverse]]. +Thus, as a group is already a [[Definition:Monoid|monoid]], it must ''at least'' have an [[Definition:Identity (Abstract Algebra)|identity]], therefore can not be empty. +{{qed}} +\end{proof}<|endoftext|> +\section{Identity is only Idempotent Element in Group} +Tags: Group Theory, Idempotence, Identity Elements, Identity is only Idempotent Element in Group + +\begin{theorem} +Every [[Definition:Group|group]] has [[Definition:Exactly One|exactly one]] [[Definition:Idempotent Element|idempotent element]]: the [[Definition:Identity Element|identity]]. +\end{theorem} + +\begin{proof} +The [[Identity Element is Idempotent]]. +From the [[Cancellation Laws]], all [[Definition:Group|group]] [[Definition:Element|elements]] are [[Definition:Cancellable Element|cancellable]]. +The result follows from [[Identity is only Idempotent Cancellable Element]]. +{{qed}} +\end{proof} + +\begin{proof} +Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$. +Let $x \in G$ such that $x \circ x = x$. +{{begin-eqn}} +{{eqn | l = e + | r = x \circ x^{-1} + | c = {{GroupAxiom|3}} +}} +{{eqn | r = \paren {x \circ x} \circ x^{-1} + | c = [[Definition:By Hypothesis|by hypothesis]]: $x \circ x = x$ +}} +{{eqn | r = x \circ \paren {x \circ x^{-1} } + | c = {{GroupAxiom|1}} +}} +{{eqn | r = x \circ e + | c = {{GroupAxiom|3}} +}} +{{eqn | r = x + | c = {{GroupAxiom|2}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Group Product Identity therefore Inverses} +Tags: Group Theory, Group Product Identity therefore Inverses + +\begin{theorem} +Let $g$ and $h$ be [[Definition:Element|elements]] of a [[Definition:Group|group]] $G$ whose [[Definition:Identity Element|identity element]] is $e$. +Then if either: +:$g h = e$ +or: +:$h g = e$ +it follows that: +:$g = h^{-1}$ +and: +:$h = g^{-1}$ +\end{theorem}<|endoftext|> +\section{Self-Inverse Elements Commute iff Product is Self-Inverse} +Tags: Group Theory, Commutativity + +\begin{theorem} +Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. +Let $x, y \in \struct {G, \circ}$, such that $x$ and $y$ are [[Definition:Self-Inverse Element|self-inverse]]. +Then $x$ and $y$ [[Definition:Commute|commute]] {{iff}} $x \circ y$ is also [[Definition:Self-Inverse Element|self-inverse]]. +\end{theorem} + +\begin{proof} +Let the [[Definition:Identity Element|identity element]] of $\struct {G, \circ}$ be $e_G$. +=== Necessary Condition === +Let $x$ and $y$ commute. +Then: +{{begin-eqn}} +{{eqn | l = \paren {x \circ y} \circ \paren {x \circ y} + | r = x \circ \paren {y \circ x} \circ y + | c = $\circ$ is [[Definition:Associative Operation|associative]] +}} +{{eqn | r = x \circ \paren {x \circ y} \circ y + | c = $x$ and $y$ [[Definition:Commute|commute]] +}} +{{eqn | r = \paren {x \circ x} \circ \paren {y \circ y} + | c = $\circ$ is [[Definition:Associative Operation|associative]] +}} +{{eqn | r = e_G \circ e_G + | c = $x$ and $y$ are [[Definition:Self-Inverse Element|self-inverse]] +}} +{{eqn | r = e_G + | c = {{Defof|Identity Element}} +}} +{{end-eqn}} +Thus $\paren {x \circ y} \circ \paren {x \circ y} = e_G$, proving that $x \circ y$ is [[Definition:Self-Inverse Element|self-inverse]]. +{{qed|lemma}} +=== Sufficient Condition === +Now, suppose that $x \circ y$ is [[Definition:Self-Inverse Element|self-inverse]]. +We already have that $x$ and $y$ are [[Definition:Self-Inverse Element|self-inverse]]. +Thus: +:$\paren {x \circ x} \circ \paren {y \circ y} = e_G \circ e_G = e_G$ +Because $x \circ y$ is [[Definition:Self-Inverse Element|self-inverse]], we have: +:$\paren {x \circ y} \circ \paren {x \circ y} = e_G$ +Thus: +{{begin-eqn}} +{{eqn | l = \paren {x \circ y} \circ \paren {x \circ y} + | r = \paren {x \circ x} \circ \paren {y \circ y} = e_G + | c = +}} +{{eqn | ll= \leadsto + | l = x \circ \paren {y \circ x} \circ y + | r = x \circ \paren {x \circ y} \circ y + | c = $\circ$ is [[Definition:Associative Operation|associative]] +}} +{{eqn | ll= \leadsto + | l = y \circ x + | r = x \circ y + | c = [[Cancellation Laws]] +}} +{{end-eqn}} +So $x$ and $y$ [[Definition:Commute|commute]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Power Set of Group under Induced Operation is Semigroup} +Tags: Group Theory, Semigroups, Power Set, Subset Products, Power Set of Group under Induced Operation is Semigroup + +\begin{theorem} +Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. +Let $\struct {\powerset G, \circ_\mathcal P}$ be the [[Definition:Algebraic Structure|algebraic structure]] consisting of the [[Definition:Power Set|power set]] of $G$ and the [[Definition:Subset Product|operation induced on $\powerset G$ by $\circ$]]. +Then $\struct {\powerset G, \circ_\mathcal P}$ is a [[Definition:Semigroup|semigroup]]. +\end{theorem} + +\begin{proof} +We need to prove [[Definition:Closed Algebraic Structure|closure]] and [[Definition:Associative Operation|associativity]]. +=== Closure === +Let $\struct {G, \circ}$ be a [[Definition:Group|group]], and let $A, B \subseteq G$. +{{begin-eqn}} +{{eqn | o = + | r = \forall a \in A, b \in B: a \circ b \in G + | c = +}} +{{eqn | o = \leadsto + | r = A \circ B \subseteq G + | c = {{Defof|Subset Product}} +}} +{{eqn | o = \leadsto + | r = A \circ B \in \powerset G + | c = {{Defof|Power Set}} +}} +{{end-eqn}} +Thus $\struct {\powerset G, \circ_\PP}$ is [[Definition:Closed Algebraic Structure|closed]]. +{{qed|lemma}} +=== Associativity === +It follows from [[Subset Product within Semigroup is Associative]] that $\circ_\PP$ is [[Definition:Associative Operation|associative]] in $\struct {\powerset G, \circ_\PP}$. +{{qed|lemma}} +Thus $\struct {\powerset G, \circ_\PP}$ is a [[Definition:Semigroup|semigroup]]. +{{qed}} +\end{proof} + +\begin{proof} +By definition a [[Definition:Group|group]] is also a [[Definition:Semigroup|semigroup]]. +The result then follows from [[Power Set of Semigroup under Induced Operation is Semigroup]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Inverse of Product of Subsets of Group} +Tags: Group Theory, Subset Products, Inverse of Product + +\begin{theorem} +Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. +Let $X, Y \subseteq G$. +Then: +:$\paren {X \circ Y}^{-1} = Y^{-1} \circ X^{-1}$ +where $X^{-1}$ is the [[Definition:Inverse of Subset of Group|inverse]] of $X$. +\end{theorem} + +\begin{proof} +First, note that: +{{begin-eqn}} +{{eqn | o = + | r = x \in X, y \in Y + | c = +}} +{{eqn | o = \leadsto + | r = x^{-1} \in X^{-1}, y^{-1} \in Y^{-1} + | c = {{Defof|Inverse of Subset of Group}} +}} +{{eqn | o = \leadsto + | r = y^{-1} \circ x^{-1} \in Y^{-1} \circ X^{-1} + | c = {{Defof|Subset Product}} +}} +{{end-eqn}} +Now: +{{begin-eqn}} +{{eqn | l = x \circ y + | o = \in + | r = X \circ Y + | c = {{Defof|Subset Product}} +}} +{{eqn | ll= \leadsto + | l = \paren {x \circ y}^{-1} + | o = \in + | r = \paren {X \circ Y}^{-1} + | c = {{Defof|Inverse of Subset of Group}} +}} +{{eqn | ll= \leadsto + | l = y^{-1} \circ x^{-1} + | o = \in + | r = \paren {X \circ Y}^{-1} + | c = [[Inverse of Group Product]] +}} +{{eqn | ll= \leadsto + | l = Y^{-1} \circ X^{-1} + | o = \subseteq + | r = \paren {X \circ Y}^{-1} + | c = {{Defof|Subset}} +}} +{{end-eqn}} +By a similar argument we see that $\paren {X \circ Y}^{-1} \subseteq Y^{-1} \circ X^{-1}$. +Hence the result. +{{qed}} +\end{proof} + +\begin{proof} +=== Superset === +We will show that $\forall z \in Y^{-1} \circ X^{-1}: z \in \left({X \circ Y}\right)^{-1}$, from which: +:$Y^{-1} \circ X^{-1} \subseteq \left({X \circ Y}\right)^{-1}$ +Let $z \in Y^{-1} \circ X^{-1}$. +By the definition of [[Definition:Subset Product|subset product]]: +:$\exists x' \in X^{-1}, y' \in Y^{-1}: z = y' \circ x'$ +Then by [[Inverse of Group Product]]: +:$(2)\quad z^{-1} = x'^{-1} \circ y'^{-1}$ +By the definition of [[Definition:Inverse of Subset of Group|inverse of subset]]: +:$x'^{-1} \in X$ and $y'^{-1} \in Y$ +By the definition of [[Definition:Subset Product|subset product]]: +:$x'^{-1} \circ y'^{-1} \in X \circ Y$ +Thus by $(2)$: +:$z^{-1} \in X \circ Y$ +By the definition of [[Definition:Inverse of Subset of Group|inverse of subset]]: +:$z \in \left({X \circ Y}\right)^{-1}$ +=== Subset === +We will show that $\forall z \in \left({X \circ Y}\right)^{-1}: z \in Y^{-1} \circ X^{-1}$, from which: +:$\left({X \circ Y}\right)^{-1} \subseteq Y^{-1} \circ X^{-1}$ +Let $z \in \left({X \circ Y}\right)^{-1}$. +By the definition of [[Definition:Inverse of Subset of Group|inverse of subset]]: +:$z^{-1} \in X \circ Y$ +From [[Inverse of Inverse of Subset of Group]]: +:$z^{-1} \in \left({ X^{-1} }\right)^{-1} \circ \left({ Y^{-1} }\right)^{-1}$ +Thus by the superset proof above: +:$z^{-1} \in \left({ Y^{-1} \circ X^{-1} }\right)^{-1}$ +From the definition of [[Definition:Inverse of Subset of Group|inverse of subset]]: +:$z \in Y^{-1} \circ X^{-1}$ +{{qed}} +\end{proof}<|endoftext|> +\section{Regular Representations in Group are Permutations} +Tags: Group Theory, Regular Representations + +\begin{theorem} +Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. +Let $a \in G$ be any [[Definition:Element|element]] of $G$. +Then the [[Definition:Left Regular Representation|left regular representation]] $\lambda_a$ and the [[Definition:Right Regular Representation|right regular representation]] $\rho_a$ are [[Definition:Permutation|permutations]] of $G$. +\end{theorem} + +\begin{proof} +This follows directly from the fact that all [[Definition:Element|elements]] of a [[Definition:Group|group]] are by definition [[Definition:Invertible Element|invertible]]. +Therefore the result [[Regular Representation of Invertible Element is Permutation]] applies. +{{Qed}} +\end{proof}<|endoftext|> +\section{Set Equivalence of Regular Representations} +Tags: Group Theory, Regular Representations + +\begin{theorem} +If $S$ is a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of a [[Definition:Group|group]] $G$, then: +:$\card {a \circ S} = \card S = \left|{S \circ a}\right|$ +That is, $a \circ S$, $S$ and $S \circ a$ are [[Definition:Set Equivalence|equivalent]]: $a \circ S \sim S \sim S \circ a$. +\end{theorem} + +\begin{proof} +Follows immediately from the fact that both the [[Definition:Left Regular Representation|left]] and [[Definition:Right Regular Representation|right regular representation]] are [[Regular Representations in Group are Permutations|permutations]], and therefore [[Definition:Bijection|bijections]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Composition of Regular Representations} +Tags: Semigroups, Regular Representations + +\begin{theorem} +Let $\left({S, *}\right)$ be a [[Definition:Semigroup|semigroup]]. +Let $\lambda_x, \rho_x$ be the [[Definition:Left Regular Representation|left]] and [[Definition:Right Regular Representation|right]] regular representations of $\left({S, *}\right)$ with respect to $x$. +Let $\lambda_x \circ \lambda_y$, $\rho_x \circ \rho_y$ etc. be defined as the [[Definition:Composition of Mappings|composition of the mappings]] $\lambda_x$ and $\lambda_y$ etc. +Then $\forall x, y \in S$, the following results hold: +\end{theorem}<|endoftext|> +\section{All Elements Self-Inverse then Abelian} +Tags: Abelian Groups + +\begin{theorem} +Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. +Suppose that every [[Definition:Element|element]] of $G$ is [[Definition:Self-Inverse Element|self-inverse]]. +Then $G$ is [[Definition:Abelian Group|abelian]]. +\end{theorem} + +\begin{proof} +Every [[Definition:Element|element]] of $G$ is [[Definition:Self-Inverse Element|self-inverse]], that is: +:$\forall x \in G: x \circ x = e$ +In particular, for all $x, y \in G$: +:$\paren {x \circ y} \circ \paren {x \circ y} = e$ +that is, $x \circ y$ is also [[Definition:Self-Inverse Element|self-inverse]]. +From [[Self-Inverse Elements Commute iff Product is Self-Inverse]], it follows that: +:$\forall x, y \in G: y \circ x = x \circ y$ +that is: $x$ and $y$ [[Definition:Commute|commute]] for all $x, y \in G$. +Hence $G$ is an [[Definition:Abelian Group|abelian group]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Commutation Property in Group} +Tags: Group Theory, Commutativity + +\begin{theorem} +Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. +Then $x$ and $y$ [[Definition:Commute|commute]] {{iff}} $x \circ y \circ x^{-1} = y$. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = x \circ y + | r = y \circ x +}} +{{eqn | ll= \iff + | l = \paren {x \circ y} \circ x^{-1} + | r = \paren {y \circ x} \circ x^{-1} + | c = [[Cancellation Laws]] +}} +{{eqn | ll= \iff + | l = x \circ y \circ x^{-1} + | r = y \circ \paren {x \circ x^{-1} } + | c = {{Defof|Associative Operation}} +}} +{{eqn | ll= \iff + | l = x \circ y \circ x^{-1} + | r = y \circ e + | c = {{Defof|Inverse Element}} +}} +{{eqn | ll= \iff + | l = x \circ y \circ x^{-1} + | r = y + | c = {{Defof|Identity Element}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Identity Mapping is Automorphism} +Tags: Automorphisms, Identity Mappings + +\begin{theorem} +The [[Definition:Identity Mapping|identity mapping]] $I_S: \left({S, \circ}\right) \to \left({S, \circ}\right)$ on the [[Definition:Algebraic Structure|algebraic structure]] $\left({S, \circ}\right)$ is an [[Definition:Automorphism (Abstract Algebra)|automorphism]]. +Its [[Definition:Image of Mapping|image]] is $S$. +\end{theorem} + +\begin{proof} +By definition, an [[Definition:Automorphism (Abstract Algebra)|automorphism]] is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] from an [[Definition:Algebraic Structure|algebraic structure]] onto itself. +An [[Definition:Isomorphism (Abstract Algebra)|isomorphism]], in turn, is a [[Definition:Bijection|bijective]] [[Definition:Homomorphism (Abstract Algebra)|homomorphism]]. +From [[Identity Mapping is Bijection]], the [[Definition:Identity Mapping|identity mapping]] $I_S: S \to S$ on the [[Definition:Set|set]] $S$ is a [[Definition:Bijection|bijection]] from $S$ onto itself. +Now we need to show it is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]]. +Let $x, y \in S$. Then: +{{begin-eqn}} +{{eqn | l=I_S \left({x \circ y}\right) + | r=x \circ y + | c=[[Definition:Identity Mapping|Definition of Identity Mapping]] +}} +{{eqn | r=I_S \left({x}\right) \circ I_S \left({y}\right) + | c=[[Definition:Identity Mapping|Definition of Identity Mapping]] +}} +{{end-eqn}} +Thus $I_S \left({x \circ y}\right) = I_S \left({x}\right) \circ I_S \left({y}\right)$ and the [[Definition:Morphism Property|morphism property]] holds, proving that $I_S: S \to S$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]]. +As $I_S$ is a [[Definition:Bijection|bijection]], its [[Definition:Image of Mapping|image]] is $S$. +{{Qed}} +\end{proof}<|endoftext|> +\section{Group Homomorphism of Product with Inverse} +Tags: Group Homomorphisms + +\begin{theorem} +Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a [[Definition:Group Homomorphism|group homomorphism]]. +Then: +:$(1): \quad \forall x, y \in G: \map \phi {x \circ y^{-1} } = \map \phi x * \paren {\map \phi y}^{-1}$ +:$(2): \quad \forall x, y \in G: \map \phi {y^{-1} \circ x} = \paren {\map \phi y}^{-1} * \map \phi x$ +\end{theorem} + +\begin{proof} +Result $(1)$: +{{begin-eqn}} +{{eqn | l = \map \phi {x \circ y^{-1} } * \map \phi y + | r = \map \phi {\paren {x \circ y^{-1} } \circ y} + | c = {{Defof|Group Homomorphism}} +}} +{{eqn | r = \map \phi x + | c = +}} +{{eqn | ll= \leadsto + | l = \map \phi {x \circ y^{-1} } * e_H + | r = \map \phi x * \paren {\map \phi y}^{-1} + | c = multiplying on the right by $\paren {\map \phi y}^{-1}$ +}} +{{end-eqn}} +Result $(2)$: +{{begin-eqn}} +{{eqn | l = \map \phi y * \map \phi {y^{-1} \circ x} + | r = \map \phi {y \circ \paren {y^{-1} \circ x} } + | c = {{Defof|Group Homomorphism}} +}} +{{eqn | r = \map \phi x + | c = +}} +{{eqn | ll= \leadsto + | l = e_H * \map \phi {y^{-1} \circ x} + | r = \paren {\map \phi y}^{-1} * \map \phi x + | c = multiplying on the left by $\paren {\map \phi y}^{-1}$ +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Mapping to Square is Endomorphism iff Abelian} +Tags: Abelian Groups, Group Endomorphisms + +\begin{theorem} +Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. +Let $\phi: G \to G$ be defined as: +:$\forall g \in G: \map \phi g = g \circ g$ +Then $\struct {G, \circ}$ is [[Definition:Abelian Group|abelian]] {{iff}} $\phi$ is a [[Definition:Group Endomorphism|(group) endomorphism]]. +\end{theorem} + +\begin{proof} +=== Necessary Condition === +Let $\struct {G, \circ}$ be an [[Definition:Abelian Group|abelian group]]. +Let $a, b \in G$ be arbitrary. +Then: +{{begin-eqn}} +{{eqn | l = \map \phi {a \circ b} + | r = \paren {a \circ b} \circ \paren {a \circ b} + | c = Definition of $\phi$ +}} +{{eqn | r = a \circ \paren {b \circ a} \circ b + | c = {{GroupAxiom|1}} +}} +{{eqn | r = a \circ \paren {a \circ b} \circ b + | c = {{Defof|Abelian Group}}: [[Definition:Commutative Operation|Commutativity]] +}} +{{eqn | r = \paren {a \circ a} \circ \paren {b \circ b} + | c = {{GroupAxiom|1}} +}} +{{eqn | r = \map \phi a \circ \map \phi b + | c = Definition of $\phi$ +}} +{{end-eqn}} +As $a$ and $b$ are arbitrary, the above holds for all $a, b \in G$. +Thus $\phi$ is a [[Definition:Group Homomorphism|group homomorphism]] from $G$ to $G$. +So by definition, $\phi$ is a [[Definition:Group Endomorphism|group endomorphism]]. +{{qed|lemma}} +=== Sufficient Condition === +Let $\phi: G \to G$ as defined above be a [[Definition:Group Endomorphism|group endomorphism]]. +Then: +{{begin-eqn}} +{{eqn | lo= \forall a, b \in G: + | l = \map \phi {a \circ b} + | r = \map \phi a \circ \map \phi b + | c = {{Defof|Group Endomorphism}} +}} +{{eqn | ll= \leadsto + | lo= \forall a, b \in G: + | l = \paren {a \circ b} \circ \paren {a \circ b} + | r = \paren {a \circ a} \circ \paren {b \circ b} + | c = Definition of $\phi$ +}} +{{eqn | ll= \leadsto + | lo= \forall a, b \in G: + | l = a \circ \paren {b \circ a} \circ b + | r = a \circ \paren {a \circ b} \circ b + | c = {{GroupAxiom|1}} +}} +{{eqn | ll= \leadsto + | lo= \forall a, b \in G: + | l = b \circ a + | r = a \circ b + | c = [[Cancellation Laws]] +}} +{{end-eqn}} +Thus, by definition, $G$ is an [[Definition:Abelian Group|abelian group]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Induced Group Product is Homomorphism iff Commutative} +Tags: Group Homomorphisms + +\begin{theorem} +Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. +Let $H_1, H_2$ be [[Definition:Subgroup|subgroups]] of $G$. +Let $\phi: H_1 \times H_2 \to G$ be defined such that: +:$\forall \tuple {h_1, h_2} \in H_1 \times H_2: \map \phi {h_1, h_2} = h_1 \circ h_2$ +Then $\phi$ is a [[Definition:Group Homomorphism|homomorphism]] {{iff}} every element of $H_1$ [[Definition:Commute|commutes]] with every element of $H_2$. +\end{theorem} + +\begin{proof} +We have $\tuple {h_1, h_2} \circ \tuple {k_1, k_2} = \tuple {h_1 \circ k_1, h_2 \circ k_2}$ by definition of [[Definition:Group Direct Product|group direct product]]. +=== Necessary Condition === +Let $\phi$ be a [[Definition:Group Homomorphism|homomorphism]]. +Then: +{{begin-eqn}} +{{eqn | l = \map \phi {\tuple {h_1, h_2} \circ \tuple {k_1, k_2} } + | r = \map \phi {h_1, h_2} \circ \map \phi {k_1, k_2} + | c = +}} +{{eqn | ll= \leadsto + | l = \map \phi {h_1 \circ k_1, h_2 \circ k_2} + | r = \tuple {h_1 \circ h_2} \circ \tuple {k_1 \circ k_2} + | c = +}} +{{eqn | ll= \leadsto + | l = h_1 \circ k_1 \circ h_2 \circ k_2 + | r = h_1 \circ h_2 \circ k_1 \circ k_2 + | c = +}} +{{eqn | ll= \leadsto + | l = k_1 \circ h_2 + | r = h_2 \circ k_1 + | c = +}} +{{end-eqn}} +This follows whatever $k_1$ and $h_2$ are. +So in order for $\phi$ to be a [[Definition:Group Homomorphism|homomorphism]], every element of $H_1$ must [[Definition:Commute|commute]] with every element of $H_2$. +{{qed|lemma}} +=== Sufficient Condition === +Let every element of $H_1$ [[Definition:Commute|commute]] with every element of $H_2$. +Let $\tuple {h_1, h_2}, \tuple {k_1, k_2} \in H_1 \times H_2$. +Then: +{{begin-eqn}} +{{eqn | l = \map \phi {\tuple {h_1, h_2} \circ \tuple {k_1, k_2} } + | r = \map \phi {h_1 \circ k_1, h_2 \circ k_2} + | c = +}} +{{eqn | r = h_1 \circ k_1 \circ h_2 \circ k_2 + | c = +}} +{{eqn | r = h_1 \circ h_2 \circ k_1 \circ k_2 + | c = $k_1$ [[Definition:Commute|commutes]] with $h_2$ +}} +{{eqn | r = \map \phi {h_1, h_2} \circ \map \phi {k_1, k_2} + | c = +}} +{{end-eqn}} +Thus $\phi$ is shown to be a [[Definition:Group Homomorphism|homomorphism]]. +{{Qed}} +[[Category:Group Homomorphisms]] +ahllokut0j0saikd89t07n91h4lhxgg +\end{proof}<|endoftext|> +\section{Isomorphism of Abelian Groups} +Tags: Abelian Groups, Group Isomorphisms + +\begin{theorem} +Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a [[Definition:Group Isomorphism|group isomorphism]]. +Then $\struct {G, \circ}$ is [[Definition:Abelian Group|abelian]] {{iff}} $\struct {H, *}$ is [[Definition:Abelian Group|abelian]]. +\end{theorem} + +\begin{proof} +We have that [[Isomorphism Preserves Commutativity]]. +Thus: +:$\forall x, y \in G: x \circ y = y \circ x \implies \map \phi x * \map \phi y = \map \phi y * \map \phi x$ +Thus if $G$ is [[Definition:Abelian Group|abelian]], so is $H$. +As $\phi^{-1}: H \to G$ is [[Inverse of Algebraic Structure Isomorphism is Isomorphism|also an isomorphism]], it is clear that if $H$ is [[Definition:Abelian Group|abelian]], then so is $G$. +{{qed}} +\end{proof}<|endoftext|> +\section{Opposite Group is Group} +Tags: Opposite Groups + +\begin{theorem} +Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. +Let $\struct {G, *}$ be the [[Definition:Opposite Group|opposite group]] to $G$. +Then $\struct {G, *}$ is a [[Definition:Group|group]]. +\end{theorem} + +\begin{proof} +=== $\text G 0$: Closure === +$\struct {G, *}$ is [[Definition:Closed Algebraic Structure|closed]]: +:$b \circ a \in G \implies a * b \in G$ +{{qed|lemma}} +=== $\text G 1$: Associativity === +$*$ is [[Definition:Associative Operation|associative]] on $G$: +{{begin-eqn}} +{{eqn | l = a * \paren {b * c} + | r = \paren {c \circ b} \circ a + | c = Definition of $*$ +}} +{{eqn | r = c \circ \paren {b \circ a} + | c = [[Definition:Associative Operation|Associativity]] of $\circ$ +}} +{{eqn | r = \paren {a * b} * c + | c = Definition of $*$ +}} +{{end-eqn}} +{{qed|lemma}} +=== $\text G 2$: Identity === +Let $e$ be the [[Definition:Identity Element|identity]] of $\struct {G, \circ}$: +{{begin-eqn}} +{{eqn | l = a * e + | r = e \circ a = a +}} +{{eqn | l = e * a + | r = a \circ e = a +}} +{{end-eqn}} +Thus $e$ is the [[Definition:Identity Element|identity]] of $\struct {G, *}$. +{{qed|lemma}} +=== $\text G 3$: Inverses === +Let the [[Definition:Inverse Element|inverse]] of $a \in \struct {G, \circ}$ be $a^{-1}$: +{{begin-eqn}} +{{eqn | l = a * a^{-1} + | r = a^{-1} \circ a = e +}} +{{eqn | l = a^{-1} * a + | r = a \circ a^{-1} = e +}} +{{end-eqn}} +Thus $a^{-1}$ is the [[Definition:Inverse Element|inverse]] of $a \in \struct {G, *}$ +{{qed|lemma}} +So all the [[Definition:Group Axioms|group axioms]] are satisfied, and $\struct {G, *}$ is a [[Definition:Group|group]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Group Example: x inv c y} +Tags: Examples of Groups + +\begin{theorem} +Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. +Let $c \in G$. +We define a new [[Definition:Group Operation|operation]] $*$ on $G$ as: +:$\forall x, y \in G: x * y = x \circ c^{-1} \circ y$ +Then $\struct {G, *}$ is a [[Definition:Group|group]]. +\end{theorem} + +\begin{proof} +=== $G \, 0$: Closure === +Let $x, y \in G$. +Then: +:$\forall x * y = x \circ c^{-1} \circ y \in G$ as $c^{-1} \in G$ +thus demonstrating that $\struct {G, *}$ is [[Definition:Closed Algebraic Structure|closed]]. +{{qed|lemma}} +=== $G \, 1$: Associativity === +Let $x, y, z \in G$. +{{begin-eqn}} +{{eqn | l = x * \paren {y * z} + | r = x \circ c^{-1} \circ \paren {y \circ c^{-1} \circ z} + | c = Definition of $*$ +}} +{{eqn | r = \paren {x \circ c^{-1} \circ y} \circ c^{-1} \circ z + | c = [[Definition:Associative Operation|Associativity]] of $\circ$ +}} +{{eqn | r = \paren {x * y} * z + | c = Definition of $*$ +}} +{{end-eqn}} +thus demonstrating that $\struct {G, *}$ is [[Definition:Associative Algebraic Structure|associative]]. +{{qed|lemma}} +=== $G \, 2$: Identity === +Let $x \in G$. +{{begin-eqn}} +{{eqn | l = x * c + | r = x \circ c^{-1} \circ c = x +}} +{{eqn | l = c * x + | r = c \circ c^{-1} \circ x = x +}} +{{end-eqn}} +So $c$ serves as the [[Definition:Identity Element|identity]] in $\struct {G, *}$. +{{qed|lemma}} +=== $G \, 3$: Inverses === +Let $x \in G$. +We need to find $y \in G$ such that $x * y = c \implies x \circ c^{-1} \circ y = c$. +{{begin-eqn}} +{{eqn | l = x * y + | r = c + | c = +}} +{{eqn | ll= \leadsto + | l = x \circ c^{-1} \circ y + | r = c + | c = +}} +{{eqn | ll= \leadsto + | l = c^{-1} \circ y + | r = x^{-1} \circ c + | c = +}} +{{eqn | ll= \leadsto + | l = y + | r = c \circ x^{-1} \circ c + | c = +}} +{{end-eqn}} +Thus the [[Definition:Inverse Element|inverse]] of $x$ under the operation $*$ is $c \circ x^{-1} \circ c$ where $x^{-1}$ is the [[Definition:Inverse Element|inverse]] of $x$ under $\circ$. +{{qed|lemma}} +All criteria have been demonstrated to be fulfilled, and so $\struct {G, *}$ is a [[Definition:Group|group]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Symmetric Difference on Power Set forms Abelian Group} +Tags: Abelian Groups, Symmetric Difference, Power Set + +\begin{theorem} +Let $S$ be a [[Definition:Set|set]] such that $\O \subset S$ (that is, $S$ is [[Definition:Non-Empty Set|non-empty]]). +Let $A * B$ be defined as the [[Definition:Symmetric Difference|symmetric difference]] between $A$ and $B$. +Let $\powerset S$ denote the [[Definition:Power Set|power set]] of $S$. +Then the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\powerset S, *}$ is an [[Definition:Abelian Group|abelian group]]. +\end{theorem} + +\begin{proof} +From [[Power Set is Closed under Symmetric Difference]], we have that $\struct {\powerset S, *}$ is [[Definition:Closed Algebraic Structure|closed]]. +The result follows directly from [[Set System Closed under Symmetric Difference is Abelian Group]]. +{{Qed}} +\end{proof}<|endoftext|> +\section{Group is Subgroup of Itself} +Tags: Subgroups + +\begin{theorem} +Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. +Then: +:$\struct {G, \circ} \le \struct {G, \circ}$ +That is, a [[Definition:Group|group]] is always a [[Definition:Subgroup|subgroup]] of itself. +\end{theorem} + +\begin{proof} +By [[Set is Subset of Itself]], we have that: +:$G \subseteq G$ +Thus $\struct {G, \circ}$ is a [[Definition:Group|group]] which is a [[Definition:Subset|subset]] of $\struct {G, \circ}$, and therefore a [[Definition:Subgroup|subgroup]] of $\struct {G, \circ}$. +{{Qed}} +\end{proof}<|endoftext|> +\section{Identity of Subgroup} +Tags: Subgroups, Identity Elements + +\begin{theorem} +Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$. +Let $H$ be a [[Definition:Subgroup|subgroup]] of [[Definition:Group|group]] $G$. +Then the [[Definition:Identity Element|identity]] of $H$ is also $e$. +\end{theorem} + +\begin{proof} +From the [[Definition:Group|definition]], a group is a [[Definition:Monoid|monoid]]. +Also, [[Cancellation Laws|all of its elements are cancellable]]. +The result then follows directly from the result for monoids each of whose [[Definition:Element|elements]] are [[Definition:Cancellable Element|cancellable]]: [[Identity of Cancellable Monoid is Identity of Submonoid]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Inverses in Subgroup} +Tags: Subgroups + +\begin{theorem} +Let $G$ be a [[Definition:Group|group]]. +Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$. +Then for each $h \in H$, the [[Definition:Inverse Element|inverse]] of $h$ in $H$ is the same as the [[Definition:Inverse Element|inverse]] of $h$ in $G$. +\end{theorem} + +\begin{proof} +Let $h \in H$. +Let: +: $h'$ be the [[Definition:Inverse Element|inverse]] of $h$ in $H$ +: $h^{-1}$ be the [[Definition:Inverse Element|inverse]] of $h$ in $G$. +From [[Identity of Subgroup]], $h \circ h' = e$. +From [[Inverse in Group is Unique]], it follows that $h' = h^{-1}$. +{{qed}} +\end{proof}<|endoftext|> +\section{Subgroup of Abelian Group is Abelian} +Tags: Abelian Groups, Subgroups + +\begin{theorem} +A [[Definition:Subgroup|subgroup]] of an [[Definition:Abelian Group|abelian group]] is itself [[Definition:Abelian Group|abelian]]. +\end{theorem} + +\begin{proof} +Follows directly from [[Restriction of Commutative Operation is Commutative]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Intersection of Subgroups is Subgroup} +Tags: Subgroups, Set Intersection + +\begin{theorem} +The [[Definition:Set Intersection|intersection]] of two [[Definition:Subgroup|subgroups]] of a [[Definition:Group|group]] is itself a [[Definition:Subgroup|subgroup]] of that [[Definition:Group|group]]: +:$\forall H_1, H_2 \le \struct {G, \circ}: H_1 \cap H_2 \le G$ +It also follows that $H_1 \cap H_2 \le H_1$ and $H_1 \cap H_2 \le H_2$. +\end{theorem} + +\begin{proof} +Let $H = H_1 \cap H_2$ where $H_1, H_2 \le \struct {G, \circ}$. +Then: +{{begin-eqn}} +{{eqn | o = + | r = a, b \in H + | c = +}} +{{eqn | o = \leadsto + | r = a, b \in H_1 \land a, b \in H_2 + | c = {{Defof|Set Intersection}} +}} +{{eqn | o = \leadsto + | r = a \circ b^{-1} \in H_1 \land a \circ b^{-1} \in H_2 + | c = [[Definition:Group|Group Properties]] +}} +{{eqn | o = \leadsto + | r = a \circ b^{-1} \in H + | c = {{Defof|Set Intersection}} +}} +{{eqn | o = \leadsto + | r = H \le G + | c = [[One-Step Subgroup Test]] +}} +{{end-eqn}} +As $H \subseteq H_1$ and $H \subseteq H_2$, the other results follow directly. +{{qed}} +\end{proof}<|endoftext|> +\section{Union of Subgroups} +Tags: Subgroups, Set Union, Union of Subgroups + +\begin{theorem} +Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. +Let $H, K \le G$ be [[Definition:Subgroup|subgroups]] of $G$. +Let neither $H \subseteq K$ nor $K \subseteq H$. +Then $H \cup K$ is ''not'' a [[Definition:Subgroup|subgroup]] of $G$. +\end{theorem} + +\begin{proof} +As neither $H \subseteq K$ nor $K \subseteq H$, it follows from [[Set Difference with Superset is Empty Set]] that neither $H \setminus K = \O$ nor $K \setminus H = \O$. +So, let $h \in H \setminus K, k \in K \setminus H$. +Thus, $h \notin K, k \notin H$. +If $\struct {H \cup K, \circ}$ is a [[Definition:Group|group]], then it must be [[Definition:Closed Algebraic Structure|closed]]. +If $\struct {H \cup K, \circ}$ is [[Definition:Closed Algebraic Structure|closed]], then $h \circ k \in H \cup K \implies h \circ k \in H \lor h \circ k \in K$. +If $h \circ k \in H$ then $h^{-1} \circ h \circ k \in H \implies k \in H$. +If $h \circ k \in K$ then $h \circ k \circ k^{-1} \in K \implies h \in K$. +So $h \circ k$ can be in neither $H$ nor $K$. +Therefore $\struct {H \cup K, \circ}$ is not [[Definition:Closed Algebraic Structure|closed]]. +Therefore $H \cup K$ is not a [[Definition:Subgroup|subgroup]] of $G$. +{{qed}} +\end{proof}<|endoftext|> +\section{Elements of Group with Equal Images under Homomorphisms form Subgroup} +Tags: Group Homomorphisms + +\begin{theorem} +Let $\struct {G, \circ}$ and $\struct {H, *}$ be [[Definition:Group|groups]]. +Let $f: G \to H$ and $g: G \to H$ be [[Definition:Group Homomorphism|group homomorphisms]]. +Then the [[Definition:Set|set]]: +:$S = \set {x \in G: \map f x = \map g x}$ +is a [[Definition:Subgroup|subgroup]] of $G$. +\end{theorem} + +\begin{proof} +Let the [[Definition:Identity Element|identities]] of $\struct {G, \circ}$ and $\struct {H, *}$ be $e_G$ and $e_H$ respectively. +By [[Homomorphism to Group Preserves Identity]]: +:$\map f {e_G} = \map g {e_G} = e_H$ +Thus $e_G \in S$, and so $S \ne \O$. +Similarly, from [[Homomorphism to Group Preserves Inverses]], $x \in S \implies x^{-1} \in S$. +Let $x, y \in S$. Then: +{{begin-eqn}} +{{eqn | l = \map f {x \circ y} + | r = \map f x * \map f y + | c = [[Definition:Morphism Property|Morphism Property]] +}} +{{eqn | r = \map g x * \map g y + | c = Definition of $S$ +}} +{{eqn | r = \map g {x \circ y} + | c = [[Definition:Morphism Property|Morphism Property]] +}} +{{end-eqn}} +Thus $x \circ y \in S$. +So, by the [[Two-Step Subgroup Test]]: +:$S \le G$ +{{Qed}} +\end{proof}<|endoftext|> +\section{Product of Subgroup with Itself} +Tags: Subgroups, Subset Products + +\begin{theorem} +Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. +Then: +: $\forall H \le \struct {G, \circ}: H \circ H = H$ +\end{theorem} + +\begin{proof} +From [[Magma Subset Product with Self]], we have: +:$H \circ H \subseteq H$ +Let $e$ be the [[Definition:Identity Element|identity]] of $G$. +By [[Identity of Subgroup]], it is also the [[Definition:Identity Element|identity]] of $H$. +So: +{{begin-eqn}} +{{eqn | l = h + | o = \in + | r = H + | c = +}} +{{eqn | ll= \leadsto + | l = e \circ h + | o = \in + | r = H \circ H + | c = +}} +{{eqn | ll= \leadsto + | l = h + | o = \in + | r = H \circ H + | c = +}} +{{eqn | ll= \leadsto + | l = H + | o = \subseteq + | r = H \circ H + | c = +}} +{{end-eqn}} +Hence the result from the [[Definition:Set Equality|definition of set equality]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Inverse of Subgroup} +Tags: Subgroups, Subset Products + +\begin{theorem} +Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. +Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$. +Then: +:$H^{-1} = H$ +where $H^{-1}$ is the [[Definition:Inverse of Subset of Group|inverse]] of $H$. +\end{theorem} + +\begin{proof} +As $H$ is a [[Definition:Subgroup|subgroup]] of $G$: +:$\forall h \in H: h^{-1} \in H$ +The result follows. +\end{proof}<|endoftext|> +\section{Subset Product of Subgroups} +Tags: Subgroups, Subset Products, Subset Product of Subgroups + +\begin{theorem} +Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. +Let $H, K$ be [[Definition:Subgroup|subgroups]] of $G$. +Then $H \circ K$ is a [[Definition:Subgroup|subgroup]] of $G$ {{iff}} $H$ and $K$ are [[Definition:Permutable Subgroups|permutable]]. +That is: +$H \circ K$ is a [[Definition:Subgroup|subgroup]] of $G$ {{iff}}: +:$H \circ K = K \circ H$ +where $H \circ K$ denotes [[Definition:Subset Product|subset product]]. +\end{theorem}<|endoftext|> +\section{Group Homomorphism Preserves Subgroups} +Tags: Subgroups, Group Homomorphisms + +\begin{theorem} +Let $\struct {G_1, \circ}$ and $\struct {G_2, *}$ be [[Definition:Group|groups]]. +Let $\phi: \struct {G_1, \circ} \to \struct {G_2, *}$ be a [[Definition:Group Homomorphism|group homomorphism]]. +Then: +:$H \le G_1 \implies \phi \sqbrk H \le G_2$ +where: +:$\phi \sqbrk H$ denotes the [[Definition:Image of Subset under Mapping|image of $H$ under $\phi$]] +:$\le$ denotes [[Definition:Subgroup|subgroup]]. +That is, [[Definition:Group Homomorphism|group homomorphism]] preserves [[Definition:Subgroup|subgroups]]. +\end{theorem} + +\begin{proof} +Let $H \le G_1$. +First note that from [[Null Relation is Mapping iff Domain is Empty Set]]: +:$H \ne \O \implies \phi \sqbrk H \ne \O$ +and so $\phi \sqbrk H$ is not [[Definition:Empty Set|empty]]. +Next, let $x, y \in \phi \sqbrk H$. +Then: +:$\exists h_1, h_2 \in H: x = \map \phi {h_1}, y = \map \phi {h_2}$ +From the definition of [[Definition:Group Homomorphism|Group Homomorphism]], we have: +:$\map \phi {h_1^{-1} \circ h_2} = x^{-1} * y$ +Since $H$ is a [[Definition:Subgroup|subgroup]]: +:$h_1^{-1} \circ h_2 \in H$ +Hence: +:$x^{-1} * y \in \phi \sqbrk H$ +Thus from the [[One-Step Subgroup Test]]: +:$\phi \sqbrk H \le G_2$ +{{qed}} +\end{proof}<|endoftext|> +\section{Image of Group Homomorphism is Subgroup} +Tags: Subgroups, Group Homomorphisms + +\begin{theorem} +Let $\phi: G_1 \to G_2$ be a [[Definition:Group Homomorphism|group homomorphism]]. +Then: +:$\Img \phi \le G_2$ +where $\le$ denotes the relation of being a [[Definition:Subgroup|subgroup]]. +\end{theorem} + +\begin{proof} +This is a special case of [[Group Homomorphism Preserves Subgroups]], where we set $H = G_1$. +{{Qed}} +\end{proof}<|endoftext|> +\section{Conjugacy is Equivalence Relation} +Tags: Conjugacy + +\begin{theorem} +[[Definition:Conjugate of Group Element|Conjugacy of group elements]] is an [[Definition:Equivalence Relation|equivalence relation]]. +\end{theorem} + +\begin{proof} +Checking each of the criteria for an [[Definition:Equivalence Relation|equivalence relation]] in turn: +=== Reflexive === +:$\forall x \in G: e_G \circ x = x \circ e_G \implies x \sim x$ +Thus [[Definition:Conjugate of Group Element|conjugacy of group elements]] is [[Definition:Reflexive Relation|reflexive]]. +{{qed|lemma}} +=== Symmetric === +{{begin-eqn}} +{{eqn | o = + | r = x \sim y + | c = +}} +{{eqn | o = \leadsto + | r = a \circ x = y \circ a + | c = {{Defof|Conjugate of Group Element}} +}} +{{eqn | o = \leadsto + | r = a \circ x \circ a^{-1} = y + | c = {{Defof|Group}} +}} +{{eqn | o = \leadsto + | r = a^{-1} \circ y = x \circ a^{-1} + | c = {{Defof|Group}} +}} +{{eqn | o = \leadsto + | r = y \sim x + | c = {{Defof|Conjugate of Group Element}} +}} +{{end-eqn}} +Thus [[Definition:Conjugate of Group Element|conjugacy of group elements]] is [[Definition:Symmetric Relation|symmetric]]. +{{qed|lemma}} +=== Transitive === +{{begin-eqn}} +{{eqn | o = + | r = x \sim y, y \sim z + | c = +}} +{{eqn | o = \leadsto + | r = a_1 \circ x = y \circ a_1, a_2 \circ y = z \circ a_2 + | c = {{Defof|Conjugate of Group Element}} +}} +{{eqn | o = \leadsto + | r = a_2 \circ a_1 \circ x = a_2 \circ y \circ a_1 = z \circ a_2 \circ a_1 + | c = {{Defof|Group}} +}} +{{eqn | o = \leadsto + | r = x \sim z + | c = {{Defof|Conjugate of Group Element}} +}} +{{end-eqn}} +Thus [[Definition:Conjugate of Group Element|conjugacy of group elements]] is [[Definition:Transitive Relation|transitive]]. +{{qed|lemma}} +All criteria are satisfied, and so [[Definition:Conjugate of Group Element|conjugacy of group elements]] is shown to be an [[Definition:Equivalence Relation|equivalence relation]]. +{{Qed}} +\end{proof}<|endoftext|> +\section{Kernel of Group Homomorphism is Subgroup} +Tags: Subgroups, Group Homomorphisms + +\begin{theorem} +The [[Definition:Kernel of Group Homomorphism|kernel of a group homomorphism]] is a [[Definition:Subgroup|subgroup]] of its [[Definition:Domain of Mapping|domain]]: +:$\map \ker \phi \le \Dom \phi$ +\end{theorem} + +\begin{proof} +Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a [[Definition:Group Homomorphism|group homomorphism]]. +From [[Homomorphism to Group Preserves Identity]], $\map \phi {e_G} = e_H$, so $e_G \in \map \ker \phi$. +Therefore $\map \ker \phi \ne \O$. +Let $x, y \in \map \ker \phi$, so that $\map \phi x = \map \phi y = e_H$. +Then: +{{begin-eqn}} +{{eqn | l = \map \phi {x^{-1} \circ y} + | r = \map \phi {x^{-1} } * \map \phi y + | c = {{Defof|Morphism Property}} +}} +{{eqn | r = \paren {\map \phi x}^{-1} * \map \phi y + | c = [[Homomorphism with Identity Preserves Inverses]] +}} +{{eqn | r = e_T^{-1} * e_T + | c = as $x, y \in \map \ker \phi$ +}} +{{eqn | r = e_T + | c = {{Defof|Identity Element}} +}} +{{end-eqn}} +So $x^{-1} \circ y \in \map \ker \phi$, and from the [[One-Step Subgroup Test]], $\map \ker \phi \le S$. +{{qed}} +\end{proof}<|endoftext|> +\section{Commutative Semigroup is Entropic Structure} +Tags: Semigroups, Entropic Structures, Commutativity + +\begin{theorem} +A [[Definition:Commutative Semigroup|commutative semigroup]] is an [[Definition:Entropic Structure|entropic structure]]. +\end{theorem} + +\begin{proof} +Let $\struct {S, \circ}$ be a [[Definition:Commutative Semigroup|commutative semigroup]]. +Let $a, b, c, d \in S$. +Then: +{{begin-eqn}} +{{eqn | l = \paren {a \circ b} \circ \paren {c \circ d} + | r = a \circ \paren {b \circ \paren {c \circ d} } + | c = [[Definition:Associative Operation|Associativity]] of $\circ$ +}} +{{eqn | r = a \circ \paren {\paren {b \circ c} \circ d} + | c = [[Definition:Associative Operation|Associativity]] of $\circ$ +}} +{{eqn | r = a \circ \paren {\paren {c \circ b} \circ d} + | c = [[Definition:Commutative Operation|Commutativity]] of $\circ$ +}} +{{eqn | r = a \circ \paren {c \circ \paren {b \circ d} } + | c = [[Definition:Associative Operation|Associativity]] of $\circ$ +}} +{{eqn | r = \paren {a \circ c} \circ \paren {b \circ d} + | c = [[Definition:Associative Operation|Associativity]] of $\circ$ +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Abelian Group Induces Entropic Structure} +Tags: Abelian Groups, Entropic Structures + +\begin{theorem} +Let $\struct {G, \circ}$ be an [[Definition:Abelian Group|abelian group]]. +Let the operation $*$ be defined on $G$ such that: +:$\forall x, y \in G: x * y = x \circ y^{-1}$ +Then $\struct {G, *}$ is an [[Definition:Entropic Structure|entropic structure]]. +\end{theorem} + +\begin{proof} +We need to prove that $\forall a, b, c, d \in G: \paren {a * b} * \paren {c * d} = \paren {a * c} * \paren {b * d}$. +So: +{{begin-eqn}} +{{eqn | l = \paren {a * b} * \paren {c * d} + | r = \paren {a \circ b^{-1} } \circ \paren {c \circ d^{-1} }^{-1} + | c = Definition of $*$ +}} +{{eqn | r = \paren {a \circ b^{-1} } \circ \paren {\paren {d^{-1} }^{-1} \circ c^{-1} } + | c = [[Inverse of Group Product]] +}} +{{eqn | r = \paren {a \circ b^{-1} } \circ \paren {d \circ c^{-1} } + | c = [[Inverse of Group Inverse]] +}} +{{eqn | r = \paren {a \circ c^{-1} } \circ \paren {d \circ b^{-1} } + | c = {{Defof|Abelian Group}} +}} +{{eqn | r = \paren {a \circ c^{-1} } \circ \paren {\paren {d^{-1} }^{-1} \circ b^{-1} } + | c = [[Inverse of Group Inverse]] +}} +{{eqn | r = \paren {a \circ c^{-1} } \circ \paren {b \circ d^{-1} }^{-1} + | c = [[Inverse of Group Product]] +}} +{{eqn | r = \paren {a * c} * \paren {b * d} + | c = Definition of $*$ +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Cancellable Semiring with Unity is Additive Semiring} +Tags: Semirings + +\begin{theorem} +Let $\struct {S, *, \circ}$ be a [[Definition:Cancellable Semiring|cancellable semiring]] with [[Definition:Unity of Semiring|unity]] $1_S$. +Then the [[Definition:Distributand|distributand]] $*$ is [[Definition:Commutative Operation|commutative]]. +That is to say, $\struct {S, *, \circ}$ is also an [[Definition:Additive Semiring|additive semiring]]. +\end{theorem} + +\begin{proof} +Let $\struct {S, *, \circ}$ be a [[Definition:Semiring (Abstract Algebra)|semiring]], all of whose [[Definition:Element|elements]] of $S$ are [[Definition:Cancellable Element|cancellable]] for $*$. +We expand the expression $\paren {a * b} \circ \paren {c * d}$ using the [[Definition:Distributive Operation|distributive]] law in two ways: +{{begin-eqn}} +{{eqn | l = \paren {a * b} \circ \paren {c * d} + | r = \paren {\paren {a * b} \circ c} * \paren {\paren {a * b} \circ d} + | c = +}} +{{eqn | r = \paren {a \circ c} * \paren {b \circ c} * \paren {a \circ d} * \paren {b \circ d} + | c = +}} +{{end-eqn}} +{{begin-eqn}} +{{eqn | l = \paren {a * b} \circ \paren {c * d} + | r = \paren {a \circ \paren {c * d} } * \paren {b \circ \paren {c * d} } + | c = +}} +{{eqn | r = \paren {a \circ c} * \paren {a \circ d} * \paren {b \circ c} * \paren {b \circ d} + | c = +}} +{{end-eqn}} +So, by the fact that all [[Definition:Element|elements]] of $\struct {S, *}$ are [[Definition:Cancellable Element|cancellable]] (and thus, [[Definition:A Fortiori|a fortiori]], are $a \circ c$ and $b \circ d$), we have: +{{begin-eqn}} +{{eqn | l = \paren {a \circ c} * \paren {b \circ c} * \paren {a \circ d} * \paren {b \circ d} + | r = \paren {a \circ c} * \paren {a \circ d} * \paren {b \circ c} * \paren {b \circ d} + | c = +}} +{{eqn | ll= \leadsto + | l = \paren {b \circ c} * \paren {a \circ d} + | r = \paren {a \circ d} * \paren {b \circ c} + | c = +}} +{{end-eqn}} +As this is true for all $a, b, c, d \in S$, it is true in particular if $c = d = 1_S$. +Thus it is clear that $b * a = a * b$, which is exactly to say that $*$ is [[Definition:Commutative Operation|commutative]]. +Hence the result, by definition of [[Definition:Additive Semiring|additive semiring]]. +{{Qed}} +[[Category:Semirings]] +gwveb67h30ny0u227ct67e0iioxn4vn +\end{proof}<|endoftext|> +\section{Ring is not Empty} +Tags: Ring Theory + +\begin{theorem} +A [[Definition:Ring (Abstract Algebra)|ring]] cannot be [[Definition:Empty Set|empty]]. +\end{theorem} + +\begin{proof} +In a ring $\struct {R, +, \circ}$, $\struct {R, +}$ forms a [[Definition:Group|group]]. +From [[Group is not Empty]], the [[Definition:Group|group]] $\struct {R, +}$ contains at least the [[Definition:Identity Element|identity]], so cannot be [[Definition:Empty Set|empty]]. +So every ring $\struct {R, +, \circ}$ contains at least the [[Definition:Identity Element|identity]] for [[Definition:Ring Addition|ring addition]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Ring Product with Zero} +Tags: Ring Theory + +\begin{theorem} +Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$. +Then: +:$\forall x \in R: 0_R \circ x = 0_R = x \circ 0_R$ +That is, the [[Definition:Ring Zero|zero]] is a [[Definition:Zero Element|zero element]] for the [[Definition:Ring Product|ring product]], thereby justifying its name. +\end{theorem} + +\begin{proof} +Because $\struct {R, +, \circ}$ is a [[Definition:Ring (Abstract Algebra)|ring]], $\struct {R, +}$ is a [[Definition:Group|group]]. +Since $0_R$ is the [[Definition:Identity Element|identity]] in $\struct {R, +}$, we have $0_R + 0_R = 0_R$. +From the [[Cancellation Laws]], all [[Definition:Group|group]] [[Definition:Element|elements]] are [[Definition:Cancellable Element|cancellable]], so every [[Definition:Element|element]] of $\struct {R, +}$ is [[Definition:Cancellable Element|cancellable]] for $+$. +Thus: +{{begin-eqn}} +{{eqn | l = x \circ \paren {0_R + 0_R} + | r = x \circ 0_R + | c = {{Defof|Ring Zero}} +}} +{{eqn | ll= \leadsto + | l = \paren {x \circ 0_R} + \paren {x \circ 0_R} + | r = x \circ 0_R + | c = {{Ring-axiom|D}} +}} +{{eqn | ll= \leadsto + | l = \paren {x \circ 0_R} + \paren {x \circ 0_R} + | r = \paren {x \circ 0_R} + 0_R + | c = {{Defof|Ring Zero}} +}} +{{eqn | ll= \leadsto + | l = x \circ 0_R + | r = 0_R + | c = [[Cancellation Laws]] +}} +{{end-eqn}} +Next: +{{begin-eqn}} +{{eqn | l = \paren {0_R + 0_R} \circ x + | r = 0_R \circ x + | c = {{Defof|Ring Zero}} +}} +{{eqn | ll= \leadsto + | l = \paren {0_R \circ x} + \paren {0_R \circ x} + | r = 0_R \circ x + | c = {{Ring-axiom|D}} +}} +{{eqn | ll= \leadsto + | l = \paren {0_R \circ x} + \paren {0_R \circ x} + | r = 0_R + \paren {0_R \circ x} + | c = {{Defof|Ring Zero}} +}} +{{eqn | ll= \leadsto + | l = 0_R \circ x + | r = 0_R + | c = [[Cancellation Laws]] +}} +{{end-eqn}} +{{Qed}} +\end{proof}<|endoftext|> +\section{Product with Ring Negative} +Tags: Ring Theory + +\begin{theorem} +Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]]. +Then: +:$\forall x, y \in \struct {R, +, \circ}: \paren {-x} \circ y = -\paren {x \circ y} = x \circ \paren {-y}$ +where $\paren {-x}$ denotes the [[Definition:Ring Negative|negative]] of $x$. +\end{theorem} + +\begin{proof} +We have: +{{begin-eqn}} +{{eqn | l = \paren {x + \paren {-x} } \circ y + | r = 0_R \circ y + | c = {{Defof|Ring Zero}} +}} +{{eqn | r = 0_R + | c = [[Ring Product with Zero]] +}} +{{eqn | ll= \leadstoandfrom + | l = \paren {x \circ y} + \paren {\paren {-x} \circ y} + | r = 0_R + | c = {{Ring-axiom|D}} +}} +{{end-eqn}} +So from {{GroupAxiom|3}} as applied to $\struct {R, +}$: +:$\paren {-x} \circ y = -\paren {x \circ y}$ +The proof that $x \circ \paren {-y} = -\paren {x \circ y}$ follows identical lines. +{{qed}} +\end{proof}<|endoftext|> +\section{Unity of Ring is Unique} +Tags: Rings with Unity + +\begin{theorem} +A [[Definition:Ring (Abstract Algebra)|ring]] can have no more than one [[Definition:Unity of Ring|unity]]. +\end{theorem} + +\begin{proof} +Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]]. +If $\struct {R, \circ}$ has an [[Definition:Identity (Abstract Algebra)|identity]], then it is a [[Definition:Monoid|monoid]]. +From [[Identity of Monoid is Unique]], it follows that such an identity is unique. +{{qed}} +\end{proof}<|endoftext|> +\section{Trivial Ring from Abelian Group} +Tags: Trivial Rings, Abelian Groups + +\begin{theorem} +Any [[Definition:Abelian Group|abelian group]] $\struct {G, +}$ may be turned into a [[Definition:Trivial Ring|trivial ring]] by defining the [[Definition:Ring Product|ring product]] to be: +:$\forall x, y \in G: x \circ y = e_G$ +\end{theorem} + +\begin{proof} +Follows directly from the definition of a [[Definition:Trivial Ring|trivial ring]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Null Ring iff Zero and Unity Coincide} +Tags: Ring Theory + +\begin{theorem} +The [[Definition:Null Ring|null ring]] is the only [[Definition:Ring (Abstract Algebra)|ring]] in which the [[Definition:Unity of Ring|unity]] and [[Definition:Ring Zero|zero]] coincide. +\end{theorem} + +\begin{proof} +The single element of the [[Definition:Null Ring|null ring]] serves as an [[Definition:Identity Element|identity]] for both of the operations. +So, in this particular ring, the [[Definition:Unity of Ring|unity]] and the [[Definition:Ring Zero|zero]] are the same [[Definition:Element|element]]. +A [[Definition:Non-Null Ring|non-null ring]] contains some non-zero element $a$. +Since $1_R \circ a = a \ne 0_R = 0_R \circ a$, then $1_R \ne 0_R$. +So if a [[Definition:Ring (Abstract Algebra)|ring]] is [[Definition:Non-Null Ring|non-null]], its [[Definition:Unity of Ring|unity]] cannot be [[Definition:Ring Zero|zero]]. +So a ring is [[Definition:Null Ring|null]] {{iff}} its [[Definition:Unity of Ring|unity]] is also its [[Definition:Ring Zero|zero]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Unity is Unit} +Tags: Rings with Unity + +\begin{theorem} +The [[Definition:Unity of Ring|unity]] in a [[Definition:Ring (Abstract Algebra)|ring]] is a [[Definition:Unit of Ring|unit]]. +\end{theorem} + +\begin{proof} +Let $\left({R, +, \circ}\right)$ be a [[Definition:Ring with Unity|ring with unity]] $1_R$. +From [[Identity is Self-Inverse]]: +:$1_R^{-1} = 1_R \in R \implies 1_R \in U_R$ +{{qed}} +[[Category:Rings with Unity]] +1kgq0vcsz8xqaswu08wx5qsyj7zb6a3 +\end{proof}<|endoftext|> +\section{Unity and Negative form Subgroup of Units} +Tags: Rings with Unity, Subgroups + +\begin{theorem} +Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]]. +Then: +:$\struct {\set {1_R, -1_R}, \circ} \le U_R$ +That is, the [[Definition:Set|set]] consisting of the [[Definition:Unity of Ring|unity]] and its [[Definition:Ring Negative|negative]] forms a [[Definition:Subgroup|subgroup]] of the [[Definition:Group of Units of Ring|group of units]]. +\end{theorem} + +\begin{proof} +From [[Unity is Unit]]: +:$1_R \in U_R$ +It remains to be shown that $-1_R \in U_R$. +From {{Ring-axiom|A}}, $\struct {R, +}$ is an [[Definition:Abelian Group|abelian group]]. +Therefore: +:$1_R \in R \implies -1_R \in R$. +From [[Product of Ring Negatives]]: +:$-1_R \circ -1_R = 1_R \circ 1_R = 1_R$ +Thus $-1_R$ has a [[Definition:Ring Product Inverse|ring product inverse]] (itself) and therefore $-1_R \in U_R$. +We have: +{{begin-eqn}} +{{eqn | l = 1_R \circ 1_R + | r = 1_R + | c = +}} +{{eqn | l = 1_R \circ -1_R + | r = -1_R + | c = +}} +{{eqn | l = -1_R \circ 1_R + | r = -1_R + | c = +}} +{{eqn | l = -1_R \circ -1_R + | r = 1_R + | c = +}} +{{end-eqn}} +This exhausts all the ways we can form a [[Definition:Ring Product|ring product]] between $1_R$ and $-1_R$. +That is: +:$\forall x, y \in \set {1_R, -1_R}: x \circ y \in \set {1_R, -1_R}$ +Hence the result from the [[Two-Step Subgroup Test]]. +{{qed}} +[[Category:Rings with Unity]] +[[Category:Subgroups]] +4xfqsv3mwxeuhx44xgisxjrxwdhk8k6 +\end{proof}<|endoftext|> +\section{Negative of Product Inverse} +Tags: Rings with Unity + +\begin{theorem} +Let $\left({R, +, \circ}\right)$ be a [[Definition:Ring with Unity|ring with unity]]. +Let $z \in U_R$, where $U_R$ is the [[Definition:Set|set]] of [[Definition:Unit of Ring|units]]. +Then: +:$\left({- z}\right)^{-1} = - \left({z^{-1}}\right)$. +where $z^{-1}$ is the [[Definition:Ring Product Inverse|ring product inverse]] of $z$. +\end{theorem} + +\begin{proof} +Let the [[Definition:Unity of Ring|unity]] of $\left({R, +, \circ}\right)$ be $1_R$. +{{begin-eqn}} +{{eqn | l = \left({- \left({z^{-1} }\right)}\right) \circ \left({- z}\right) + | r = z^{-1} \circ z + | c = [[Product of Ring Negatives]] +}} +{{eqn | r = 1_R + | c = [[Definition:Inverse Element|Inverse under $\circ$]] +}} +{{eqn | r = z \circ z^{-1} + | c = [[Definition:Inverse Element|Inverse under $\circ$]] +}} +{{eqn | r = \left({- z}\right) \circ \left({- \left({z^{-1} }\right)}\right) + | c = [[Product of Ring Negatives]] +}} +{{end-eqn}} +Thus: +: $\left({- z}\right)^{-1} = - \left({z^{-1}}\right)$ +{{qed}} +\end{proof}<|endoftext|> +\section{Product of Negative with Product Inverse} +Tags: Rings with Unity + +\begin{theorem} +Let $\left({R, +, \circ}\right)$ be a [[Definition:Ring with Unity|ring with unity]]. +Let $z \in U_R$, where $U_R$ is the [[Definition:Set|set]] of [[Definition:Unit of Ring|units]]. +Then: +: $(1): \quad \forall x \in R: -\left({x \circ z^{-1}}\right) = \left({- x}\right) \circ z^{-1} = x \circ \left({\left({- z}\right)^{-1}}\right)$ +: $(2): \quad \forall x \in R: -\left({z^{-1} \circ x}\right) = z^{-1} \circ \left({- x}\right) = \left({\left({- z}\right)^{-1}}\right) \circ x$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | n=1 + | l=-\left({x \circ z^{-1} }\right) + | r=\left({- x}\right) \circ z^{-1} + | c=[[Product with Ring Negative]] +}} +{{eqn | r=x \circ \left({- \left({z^{-1} }\right)}\right) + | c=[[Product with Ring Negative]] +}} +{{eqn | r=x \circ \left({\left({- z}\right)^{-1} }\right) + | c=[[Negative of Product Inverse]] +}} +{{end-eqn}} +{{qed|lemma}} +{{begin-eqn}} +{{eqn | n=2 + | l=-\left({z^{-1} \circ x}\right) + | r=z^{-1} \circ \left({- x}\right) + | c=[[Product with Ring Negative]] +}} +{{eqn | r=\left({-\left({z^{-1} }\right)}\right) \circ x + | c=[[Product with Ring Negative]] +}} +{{eqn | r=\left({\left({- z}\right)^{-1} }\right) \circ x + | c=[[Negative of Product Inverse]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Negative of Division Product} +Tags: Commutative Algebra + +\begin{theorem} +Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]]. +Let $\struct {U_R, \circ}$ be the [[Definition:Group of Units of Ring|group of units]] of $\struct {R, +, \circ}$. +Then: +:$\displaystyle \forall x \in R: -\frac x z = \frac {-x} z = \frac x {-z}$ +where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is the [[Definition:Division Product|division product]] of $x$ by $z$. +\end{theorem} + +\begin{proof} +Follows directly from [[Product of Negative with Product Inverse]] and the definition of [[Definition:Division Product|division product]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Addition of Division Products} +Tags: Commutative Algebra + +\begin{theorem} +Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]]. +Let $\struct {U_R, \circ}$ be the [[Definition:Group of Units of Ring|group of units]] of $\struct {R, +, \circ}$. +Let $a, c \in R, b, d \in U_R$. +Then: +:$\dfrac a b + \dfrac c d = \dfrac {a \circ d + b \circ c} {b \circ d}$ +where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is, $x$ [[Definition:Division Product|divided by]] $z$. +The [[Definition:Binary Operation|operation]] $+$ is [[Definition:Well-Defined Operation|well-defined]]. +That is: +:$\dfrac a b = \dfrac {a'} {b'}, \dfrac c d = \dfrac {c'} {d'} \implies \dfrac a b + \dfrac c d = \dfrac {a'} {b'} + \dfrac {c'} {d'}$ +{{questionable|This is an existing operation and thus already well-defined. See talk.}} +\end{theorem} + +\begin{proof} +First we demonstrate the operation has the specified property: +{{begin-eqn}} +{{eqn | l = \frac a b + \frac c d + | r = a \circ b^{-1} + c \circ d^{-1} + | c = {{Defof|Division Product}} +}} +{{eqn | r = a \circ b^{-1} \circ d \circ d^{-1} + c \circ d^{-1} \circ b \circ b^{-1} + | c = {{Defof|Inverse Element}} and {{Defof|Identity Element}} under $\circ$ +}} +{{eqn | r = \paren {a \circ d} \circ \paren {d^{-1} \circ b^{-1} } + \paren {b \circ c} \circ \paren {d^{-1} \circ b^{-1} } + | c = {{Defof|Commutative Operation}} +}} +{{eqn | r = \paren {a \circ d + b \circ c} \circ \paren {b \circ d}^{-1} + | c = {{Defof|Distributive Operation}} $\circ$ over $+$ +}} +{{eqn | r = \frac {a \circ d + b \circ c} {b \circ d} + | c = {{Defof|Division Product}} +}} +{{end-eqn}} +Notice that this works only if $\struct {R, +, \circ}$ is [[Definition:Commutative Ring|commutative]]. +Now we show that $+$ is [[Definition:Well-Defined Operation|well-defined]]. +Let $a, c, a', c' \in D, b, d, b', d' \in D^*$ such that $\dfrac a b = \dfrac {a'} {b'}$ and $\dfrac c d = \dfrac {c'} {d'}$. +Then: +{{begin-eqn}} +{{eqn | l = a \circ b' + | r = \paren {a \circ b^{-1} } \circ b \circ b' + | c = +}} +{{eqn | r = \paren {a' \circ \paren {b'}^{-1} } \circ b \circ b' + | c = as $a / b = a' / b'$ +}} +{{eqn | r = a' \circ b + | c = +}} +{{end-eqn}} +Similarly, $c \circ d' = c' \circ d$. +Hence: +{{begin-eqn}} +{{eqn | l = \paren {\paren {a' \circ d'} + \paren {c' \circ b'} } \circ b \circ d + | r = a' \circ b \circ d' \circ d + c' \circ d \circ b' \circ b + | c = +}} +{{eqn | r = a \circ b' \circ d' \circ d + c \circ d' \circ b' \circ b + | c = +}} +{{eqn | r = \paren {\paren {a \circ d} + \paren {c \circ b} } \circ b' \circ d' + | c = +}} +{{end-eqn}} +Thus: +{{begin-eqn}} +{{eqn | l = \frac {\paren {a' \circ d'} + \paren {c' \circ b'} } {b' \circ d'} + | r = \paren {\paren {\paren {a' \circ d'} + \paren {c' \circ b'} } \circ b \circ d} \circ \paren {b^{-1} \circ d^{-1} \circ \paren {b'}^{-1} \circ \paren {d'}^{-1} } + | c = +}} +{{eqn | r = \paren {\paren {\paren {a \circ d} + \paren {c \circ b} } \circ b' \circ d'} \circ \paren {b^{-1} \circ d^{-1} \circ \paren {b'}^{-1} \circ \paren {d'}^{-1} } + | c = +}} +{{eqn | r = \paren {\paren {a \circ d} + \paren {c \circ b} } \circ b^{-1} \circ d^{-1} + | c = +}} +{{eqn | r = \frac {\paren {a \circ d} + \paren {c \circ b} } {b \circ d} + | c = +}} +{{end-eqn}} +showing that $+$ is indeed [[Definition:Well-Defined Operation|well-defined]]. +{{Qed}} +\end{proof}<|endoftext|> +\section{Equality of Division Products} +Tags: Commutative Algebra + +\begin{theorem} +Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]]. +Let $\struct {U_R, \circ}$ be the [[Definition:Group of Units of Ring|group of units]] of $\struct {R, +, \circ}$. +Let $a, b \in R, c, d \in U_R$. +Then: +:$\dfrac a c = \dfrac b d \iff a \circ d = b \circ c$ +where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is, $x$ [[Definition:Division Product|divided by]] $z$. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \frac a c + | r = \frac b d + | c = +}} +{{eqn | ll= \leadstoandfrom + | l = a \circ c^{-1} + | r = b \circ d^{-1} + | c = {{Defof|Division Product}} +}} +{{eqn | ll= \leadstoandfrom + | l = a \circ c^{-1} \circ c \circ d + | r = b \circ d^{-1} \circ c \circ d + | c = {{Defof|Cancellable Element}} of $U_R$ +}} +{{eqn | ll= \leadstoandfrom + | l = \paren {a \circ d} \circ \paren {c^{-1} \circ c} + | r = \paren {b \circ c} \circ \paren {d^{-1} \circ d} + | c = {{Defof|Commutative Operation}} +}} +{{eqn | ll= \leadstoandfrom + | l = a \circ d + | r = b \circ c + | c = {{Defof|Identity Element}} and {{Defof|Inverse Element}} +}} +{{end-eqn}} +{{qed}} +{{refactor|Construct a proof based on the below}} +Alternatively, a proof can be built using [[Addition of Division Products]]. +\end{proof}<|endoftext|> +\section{Product of Division Products} +Tags: Commutative Algebra + +\begin{theorem} +Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]]. +Let $\struct {U_R, \circ}$ be the [[Definition:Group of Units of Ring|group of units]] of $\struct {R, +, \circ}$. +Let $a, b \in R, c, d \in U_R$. +Then: +:$\dfrac a c \circ \dfrac b d = \dfrac {a \circ b} {c \circ d}$ +where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is, $x$ [[Definition:Division Product|divided by]] $z$. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \frac a c \circ \frac b d + | r = a \circ c^{-1} \circ b \circ d^{-1} + | c = {{Defof|Division Product}} +}} +{{eqn | r = \paren {a \circ b} \circ \paren {d^{-1} \circ c^{-1} } + | c = {{Defof|Commutative Operation}} +}} +{{eqn | r = \paren {a \circ b} \circ \paren {c \circ d}^{-1} + | c = [[Inverse of Product]] +}} +{{eqn | r = \frac {a \circ b} {c \circ d} + | c = {{Defof|Division Product}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Inverse of Division Product} +Tags: Commutative Algebra + +\begin{theorem} +Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]]. +Let $\struct {U_R, \circ}$ be the [[Definition:Group of Units of Ring|group of units]] of $\struct {R, +, \circ}$. +Let $a, b \in U_R$. +Then: +:$\paren {\dfrac a b}^{-1} = \dfrac {1_R} {\paren {a / b}} = \dfrac b a$ +where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is, $x$ [[Definition:Division Product|divided by]] $z$. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \frac {1_R} {\paren {a / b} } + | r = 1_R / \paren {a \circ b^{-1} } + | c = {{Defof|Division Product}} +}} +{{eqn | r = 1_R \circ \paren {a \circ b^{-1} }^{-1} + | c = {{Defof|Division Product}} +}} +{{eqn | r = \paren {a \circ b^{-1} }^{-1} + | c = {{Defof|Identity Element}} of $\circ$ +}} +{{eqn | r = \paren {\frac a b}^{-1} + | c = {{Defof|Division Product}} +}} +{{eqn | r = \paren {a \circ b^{-1} }^{-1} + | c = {{Defof|Division Product}} +}} +{{eqn | r = \paren {b^{-1} }^{-1} \circ a^{-1} + | c = [[Inverse of Group Product]] +}} +{{eqn | r = b \circ a^{-1} + | c = [[Inverse of Group Inverse]] +}} +{{eqn | r = \frac b a + | c = {{Defof|Division Product}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Zero Product with Proper Zero Divisor is with Zero Divisor} +Tags: Ring Theory + +\begin{theorem} +Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]]. +Let $x \in R$ be a [[Definition:Proper Zero Divisor|proper zero divisor]] of $R$. +Then: +:$\paren {x \divides 0_R} \land \paren {x \circ y = 0_R} \land \paren {y \ne 0_R} \implies y \divides 0_R$ +That is, if $x$ is a [[Definition:Proper Zero Divisor|proper zero divisor]], then whatever non-[[Definition:Ring Zero|zero]] [[Definition:Element|element]] you form the product with it by to get [[Definition:Ring Zero|zero]] must itself be a [[Definition:Zero Divisor of Ring|zero divisor]]. +\end{theorem} + +\begin{proof} +Follows directly from the definition of [[Definition:Proper Zero Divisor|proper zero divisor]]. +If $y \ne 0_R$ and $x \circ y = 0_R$ and $x \in R^*$ (which is has to be if it's a proper zero divisor), then all the criteria of being a [[Definition:Zero Divisor of Ring|zero divisor]] are fulfilled by $y$. +{{qed}} +[[Category:Ring Theory]] +otdlkr5hues1586vxgj40zmmh38byr4 +\end{proof}<|endoftext|> +\section{Unit Not Zero Divisor} +Tags: Rings with Unity + +\begin{theorem} +A [[Definition:Unit of Ring|unit of a ring]] is not a [[Definition:Zero Divisor of Ring|zero divisor]]. +\end{theorem} + +\begin{proof} +Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Unity of Ring|unity]] is $1_R$ and whose [[Definition:Ring Zero|zero]] is $0_R$. +Let $x$ be a [[Definition:Unit of Ring|unit]] of $\struct {R, +, \circ}$. +{{AimForCont}} $x$ is such that: +:$x \circ y = 0_R, y \ne 0_R$ +Then: +{{begin-eqn}} +{{eqn | l = \paren {x^{-1} \circ x} \circ y + | r = 0_R + | c = +}} +{{eqn | ll= \leadsto + | l = 1_R \circ y + | r = 0_R + | c = {{Defof|Inverse Element}} +}} +{{eqn | ll= \leadsto + | l = 1_R + | r = 0_R + | c = as $y \ne 0_R$ +}} +{{end-eqn}} +From this [[Definition:Contradiction|contradiction]] it follows that $x$ cannot have such a property. +Thus by [[Proof by Contradiction]] $x$ is not a [[Definition:Zero Divisor of Ring|zero divisor]]. +{{qed}} +[[Category:Rings with Unity]] +jjh16erxnmb4k3yno1on1yjw5d2unfe +\end{proof}<|endoftext|> +\section{Zero Divisor Product is Zero Divisor} +Tags: Ring Theory + +\begin{theorem} +The [[Definition:Ring Product|ring product]] of a [[Definition:Zero Divisor of Ring|zero divisor]] with any ring element is a [[Definition:Zero Divisor of Ring|zero divisor]]. +\end{theorem} + +\begin{proof} +Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]]. +Let $x \divides 0_R$ in $R$. +Then: +{{begin-eqn}} +{{eqn | lo= \exists y \in R, y \ne 0: + | l = x \circ y + | r = 0_R + | c = {{Defof|Zero Divisor of Ring}} +}} +{{eqn | ll= \leadsto + | lo= \forall z \in R: + | l = z \circ \paren {x \circ y} + | r = z \circ 0_R +}} +{{eqn | r = 0_R + | c = Property of [[Definition:Ring Zero|Ring Zero]] +}} +{{eqn | ll= \leadsto + | lo= \forall z \in R: + | l = \paren {z \circ x} \circ y + | r = 0_R + | c = [[Definition:Associative Operation|Associativity of $\circ$]] +}} +{{end-eqn}} +So $z \circ x \divides 0_R$ in $R$. +The same thing happens if we form the product $\paren {x \circ y} \circ z$. +{{Qed}} +[[Category:Ring Theory]] +gid9x63l0fomedgnaji3f26ntjao24n +\end{proof}<|endoftext|> +\section{Product is Zero Divisor means Zero Divisor} +Tags: Ring Theory + +\begin{theorem} +If the [[Definition:Ring Product|ring product]] of two [[Definition:Element|elements]] of a [[Definition:Ring (Abstract Algebra)|ring]] is a [[Definition:Zero Divisor of Ring|zero divisor]], then one of the two elements must be a [[Definition:Zero Divisor of Ring|zero divisor]]. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = \paren {x \circ y} + | o = \divides + | r = 0_R +}} +{{eqn | ll= \leadsto + | l = \exists z \divides 0_R \in R: \paren {x \circ y} \circ z + | r = 0_R, x \ne 0_R, y \ne 0_R + | c = {{Defof|Zero Divisor of Ring}} +}} +{{eqn | ll= \leadsto + | l = x \circ \paren {y \circ z} + | r = 0_R + | c = {{Ring-axiom|M1}} +}} +{{eqn | ll= \leadsto + | l = x \divides 0_R + | o = \lor + | r = \paren {y \circ z} \divides 0_R + | c = {{Defof|Zero Divisor of Ring}} +}} +{{eqn | ll= \leadsto + | l = x \divides 0_R + | o = \lor + | r = y \divides 0_R + | c = [[Zero Product with Proper Zero Divisor is with Zero Divisor]] applies, as $z \divides 0_R$ +}} +{{end-eqn}} +{{qed}} +[[Category:Ring Theory]] +3dh13ct8ok76uxfnchcs2qkr33s2on2 +\end{proof}<|endoftext|> +\section{Ring Element is Zero Divisor iff not Cancellable} +Tags: Ring Theory + +\begin{theorem} +Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] which is [[Definition:Non-Null Ring|not null]]. +Let $z \in R^*$. +Then $z$ is a [[Definition:Zero Divisor of Ring|zero divisor]] {{iff}} $z$ is not [[Definition:Cancellable Element|cancellable]] for $\circ$. +\end{theorem} + +\begin{proof} +=== Sufficient Condition === +If $z \circ x = 0_R$ or $x \circ z = 0_R$ for some $x \in R^*$, then $z$ can not be [[Definition:Cancellable Element|cancellable]] for $\circ$ since $z \circ 0_R = 0_R = 0_R \circ z$. +=== Necessary Condition === +Let $z \circ x = z \circ y$ where $x \ne y$. +Then: +{{begin-eqn}} +{{eqn | l = z \circ \paren {x + \paren {-y} } + | r = z \circ x + z \circ \paren {-y} + | c = [[Definition:Distributive Operation|$\circ$ distributes over $+$]] +}} +{{eqn | r = z \circ x + \paren {-z \circ y} + | c = [[Product with Ring Negative]] +}} +{{eqn | r = 0_R + | c = as $z \circ x = z \circ y$ +}} +{{end-eqn}} +But $x \ne y$, so $x + \paren {-y} \ne 0$. +Thus $z$ is a [[Definition:Zero Divisor of Ring|zero divisor]]. +Similarly if $x \circ z = y \circ z$ where $x \ne y$. +{{Qed}} +\end{proof}<|endoftext|> +\section{Ring Less Zero is Semigroup for Product iff No Proper Zero Divisors} +Tags: Ring Theory, Semigroups + +\begin{theorem} +Let $\struct {R, +, \circ}$ be a [[Definition:Non-Null Ring|non-null]] [[Definition:Ring (Abstract Algebra)|ring]]. +Then $R$ has no [[Definition:Zero Divisor of Ring|zero divisors]] {{iff}} $\struct {R^*, \circ}$ is a [[Definition:Semigroup|semigroup]]. +\end{theorem} + +\begin{proof} +=== Necessary Condition === +Let $\struct {R, +, \circ}$ be a [[Definition:Non-Null Ring|non-null]] [[Definition:Ring (Abstract Algebra)|ring]] with no [[Definition:Zero Divisor of Ring|zero divisors]]. +The set $R^* = R \setminus \set {0_R} \ne \O$ as $\struct {R, +, \circ}$ is [[Definition:Non-Null Ring|non-null]]. +All elements of $R$ are not [[Definition:Zero Divisor of Ring|zero divisors]], and therefore are [[Definition:Cancellable Element|cancellable]]. +$\struct {R, +, \circ}$ is [[Definition:Closed Algebraic Structure|closed]] under $\circ$, from the fact that there are no [[Definition:Zero Divisor of Ring|zero divisors]], and also that $\struct {R, \circ}$ is also closed. +From [[Restriction of Associative Operation is Associative]], [[Definition:Ring Product|ring product]] is [[Definition:Associative Operation|associative]] on $\struct {R^*, +, \circ}$, as it is [[Definition:Associative Operation|associative]] on $\struct {R, +, \circ}$. +Thus $\struct {R^*, \circ}$ is a [[Definition:Semigroup|semigroup]]. +{{qed|lemma}} +=== Sufficient Condition === +Let $\struct {R^*, \circ}$ be a [[Definition:Semigroup|semigroup]]. +Then: +:$\neg \exists x, y \in R^*: x \circ y \notin R^*$ +Thus: +:$\neg \exists x, y \in R^*: x \circ y = 0_R$ +and the result follows. +{{qed}} +\end{proof}<|endoftext|> +\section{Idempotent Elements of Ring with No Proper Zero Divisors} +Tags: Ring Theory + +\begin{theorem} +Let $\left({R, +, \circ}\right)$ be a [[Definition:Non-Null Ring|non-null]] [[Definition:Ring (Abstract Algebra)|ring]] with no [[Definition:Proper Zero Divisor|(proper) zero divisors]]. +Let $x \in R$. +Then: +:$x \circ x = x \iff x \in \left\{{0_R, 1_R}\right\}$ +That is, the only [[Definition:Element|elements]] of $R$ that are [[Definition:Idempotent Element|idempotent]] are [[Definition:Ring Zero|zero]] and [[Definition:Unity of Ring|unity]]. +\end{theorem} + +\begin{proof} +We have $0_R \circ 0_R = 0_R$, so that sorts out one [[Definition:Element|element]]. +Let $R^*$ be the [[Definition:Ring Less Zero|ring $R$ without the zero]]: $R^* = R \setminus \left\{{0_R}\right\}$. +By [[Ring Element is Zero Divisor iff not Cancellable]], all elements of $R^*$ that are not [[Definition:Zero Divisor|zero divisors]] are [[Definition:Cancellable Element|cancellable]]. +Therefore all elements of $R^*$ are [[Definition:Cancellable Element|cancellable]]. +Suppose $x \circ x = x$ where $x \ne 0_R$. +Then $x \circ x = x = x \circ 1_R$. +As $x$ is [[Definition:Cancellable Element|cancellable]], the result follows. +{{Qed}} +\end{proof}<|endoftext|> +\section{Non-Zero Elements of Division Ring form Group} +Tags: Division Rings, Group Theory + +\begin{theorem} +Let $\struct {R, +, \circ}$ be a [[Definition:Division Ring|division ring]]. +Then $\struct {R^*, \circ}$ is a [[Definition:Group|group]]. +\end{theorem} + +\begin{proof} +A [[Definition:Division Ring|division ring]] by definition is a [[Definition:Ring with Unity|ring with unity]], and therefore not [[Definition:Null Ring|null]]. +A [[Definition:Division Ring|division ring]] by definition has no [[Definition:Zero Divisor of Ring|zero divisors]], so $\struct {R^*, \circ}$ is a [[Ring Less Zero is Semigroup for Product iff No Proper Zero Divisors|semigroup]]. +$1_R \in \struct {R^*, \circ}$ and so the [[Definition:Identity Element|identity]] of $\circ$ is in $\struct {R^*, \circ}$. +By the definition of a [[Definition:Division Ring|division ring]], each element of $\struct {R^*, \circ}$ is a [[Definition:Unit of Ring|unit]], and therefore has a unique [[Definition:Inverse Element|inverse]] in $\struct {R^*, \circ}$. +Thus $\struct {R^*, \circ}$ is a [[Definition:Semigroup|semigroup]] with an [[Definition:Identity Element|identity]] and [[Definition:Inverse Element|inverses]] and so is a [[Definition:Group|group]]. +{{qed}} +[[Category:Division Rings]] +[[Category:Group Theory]] +7qcq9nzppvyldguhqwc62adejiwizur +\end{proof}<|endoftext|> +\section{Null Ring and Ring Itself Subrings} +Tags: Subrings + +\begin{theorem} +In any [[Definition:Ring (Abstract Algebra)|ring]] $R$, the [[Definition:Null Ring|null ring]] and $R$ itself are [[Definition:Subring|subrings]] of $R$. +\end{theorem}<|endoftext|> +\section{Subring Test} +Tags: Subrings + +\begin{theorem} +Let $S$ be a [[Definition:Subset|subset]] of a [[Definition:Ring (Abstract Algebra)|ring]] $\struct {R, +, \circ}$. +Then $\struct {S, +, \circ}$ is a [[Definition:Subring|subring]] of $\struct {R, +, \circ}$ {{iff}} these all hold: +:$(1): \quad S \ne \O$ +:$(2): \quad \forall x, y \in S: x + \paren {-y} \in S$ +:$(3): \quad \forall x, y \in S: x \circ y \in S$ +\end{theorem} + +\begin{proof} +=== Necessary Condition === +If $S$ is a [[Definition:Subring|subring]] of $\struct {R, +, \circ}$, the conditions hold by virtue of the [[Definition:Ring Axioms|ring axioms]] as applied to $S$. +=== Sufficient Condition === +Conversely, suppose the conditions hold. We check that the [[Definition:Ring Axioms|ring axioms]] hold for $S$. +:$(1): \quad$ '''{{Ring-axiom|A}}:''' By $1$ and $2$ above, it follows from: +:::: the [[One-Step Subgroup Test]] +::::[[Subgroup of Abelian Group is Abelian]] +:::that $\struct {S, +}$ is an [[Definition:Abelian Group|abelian]] [[Definition:Subgroup|subgroup]] of $\struct {R, +}$, and therefore an [[Definition:Abelian Group|abelian group]]. +:$(2): \quad$ '''{{Ring-axiom|M0}}:''' From $3$, $\struct {S, \circ}$ is [[Definition:Closed Algebraic Structure|closed]]. +:$(3): \quad$ '''{{Ring-axiom|M1}}:''' From [[Restriction of Associative Operation is Associative]], $\circ$ is [[Definition:Associative Operation|associative]] on $R$, therefore also associative on $S$. +:$(4): \quad$ '''{{Ring-axiom|D}}:''' From [[Restriction of Operation Distributivity]], $\circ$ [[Definition:Distributive Operation|distributes]] over $+$ for the whole of $R$, therefore for $S$ also. +So $\struct {S, +, \circ}$ is a [[Definition:Ring (Abstract Algebra)|ring]], and therefore a [[Definition:Subring|subring]] of $\struct {R, +, \circ}$. +{{qed}} +\end{proof}<|endoftext|> +\section{Subdomain Test} +Tags: Integral Domains, Subrings + +\begin{theorem} +Let $S$ be a [[Definition:Subset|subset]] of an [[Definition:Integral Domain|integral domain]] $\struct {R, +, \circ}$. +Then $\struct {S, +\restriction_S, \circ \restriction_S}$ is a [[Definition:Subdomain|subdomain]] of $\struct {R, +, \circ}$ {{iff}} these conditions hold: +:$(1): \quad$ $\struct {S, + \restriction_S, \circ \restriction_S}$ is a [[Definition:Subring|subring]] of $\struct {R, +, \circ}$ +:$(2): \quad$ The [[Definition:Unity of Ring|unity]] of $R$ is also in $S$, that is $1_R = 1_S$. +\end{theorem} + +\begin{proof} +By [[Idempotent Elements of Ring with No Proper Zero Divisors]], it follows that the [[Definition:Unity of Ring|unity]] of a [[Definition:Subdomain|subdomain]] is the [[Definition:Unity of Ring|unity]] of the [[Definition:Integral Domain|integral domain]] it's a [[Definition:Subdomain|subdomain]] of. +{{qed}} +\end{proof}<|endoftext|> +\section{Centralizer of Ring Subset is Subring} +Tags: Ring Theory, Subrings + +\begin{theorem} +Let $S$ be a [[Definition:Subset|subset]] of a [[Definition:Ring (Abstract Algebra)|ring]] $\struct {R, +, \circ}$ +Then $\map {C_R} S$, the [[Definition:Centralizer of Ring Subset|centralizer]] of $S$ in $R$, is a [[Definition:Subring|subring]] of $R$. +\end{theorem} + +\begin{proof} +Certainly $0_R \in \map {C_R} S$ as $0_R$ [[Definition:Commute|commutes]] (trivially) with all [[Definition:Element|elements]] of $R$. +Suppose $x, y \in \map {C_R} S$. +Then: +{{begin-eqn}} +{{eqn | l = \forall s \in S: s \circ \paren {x + \paren {-y} } + | r = s \circ x + s \circ \paren {-y} + | c = [[Definition:Distributive Operation|Distributivity]] of $\circ$ over $+$ +}} +{{eqn | r = x \circ s + \paren {-y} \circ s + | c = [[Definition:Centralizer of Ring Subset|$x$ and $y$ are in $\map {C_R} S$]] +}} +{{eqn | r = \paren {x + \paren {-y} } \circ s + | c = [[Definition:Distributive Operation|Distributivity]] of $\circ$ over $+$ +}} +{{end-eqn}} +So: +:$x + \paren {-y} \in \map {C_R} s$ +Suppose $x, y \in \map {C_R} S$ again. +Then from [[Element Commutes with Product of Commuting Elements]]: +:$x \circ y \in \map {C_R} S$ +Thus all the conditions are fulfilled for [[Subring Test]], and $\map {C_R} S$ is a [[Definition:Subring|subring]] of $R$. +{{qed}} +\end{proof}<|endoftext|> +\section{Center of Ring is Commutative Subring} +Tags: Ring Theory, Subrings + +\begin{theorem} +The [[Definition:Center of Ring|center]] $\map Z R$ of a ring $R$ is a [[Definition:Commutative Ring|commutative]] [[Definition:Subring|subring]] of $R$. +\end{theorem} + +\begin{proof} +Follows directly from the definition of [[Definition:Center of Ring|center]] and [[Centralizer of Ring Subset is Subring]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Ideal is Subring} +Tags: Ideal Theory + +\begin{theorem} +Let $\left({R, +, \circ}\right)$ be a [[Definition:Ring (Abstract Algebra)|ring]], and let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$. +Then $J$ is a [[Definition:Subring|subring]] of $R$. +\end{theorem} + +\begin{proof} +This follows directly from the definition of an [[Definition:Ideal of Ring|ideal]] and [[Subring Test]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Ring is Ideal of Itself} +Tags: Ideal Theory + +\begin{theorem} +Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]]. +Then $R$ is an [[Definition:Ideal of Ring|ideal]] of $R$. +\end{theorem} + +\begin{proof} +From [[Null Ring and Ring Itself Subrings]], $\struct {R, +, \circ}$ is a [[Definition:Subring|subring]] of $\struct {R, +, \circ}$. +Also: +:$\forall x, y \in \struct {R, +, \circ}: x \circ y \in R$ +thus fulfilling the [[Definition:Ideal of Ring|condition]] for $R$ to be an [[Definition:Ideal of Ring|ideal]] of $R$. +{{qed}} +\end{proof}<|endoftext|> +\section{Ideal of Unit is Whole Ring} +Tags: Ideal Theory + +\begin{theorem} +Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]]. +Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$. +If $J$ contains a [[Definition:Unit of Ring|unit]] of $R$, then $J = R$. +\end{theorem} + +\begin{proof} +Let $u \in J$, where $u \in U_R$. +Also by definition, we have $u^{-1} \in U_R$. +Let $x \in R$. +{{begin-eqn}} +{{eqn | o = + | r = x \in R + | c = +}} +{{eqn | o = \leadsto + | r = x \circ u^{-1} \in R + | c = as $R$ is [[Definition:Closed Algebraic Structure|closed]] +}} +{{eqn | o = \leadsto + | r = \paren {x \circ u^{-1} } \circ u \in J + | c = {{Defof|Ideal of Ring}} +}} +{{eqn | o = \leadsto + | r = x \in J + | c = Ring properties: $u \circ u^{-1} = 1_R$ +}} +{{end-eqn}} +Thus $R \subseteq J$. +As $J \subseteq R$ by definition, it follows that $J = R$. +{{Qed}} +\end{proof}<|endoftext|> +\section{Test for Ideal} +Tags: Ideal Theory + +\begin{theorem} +Let $J$ be a [[Definition:Subset|subset]] of a [[Definition:Ring (Abstract Algebra)|ring]] $\struct {R, +, \circ}$. +Then $J$ is an [[Definition:Ideal of Ring|ideal]] of $\struct {R, +, \circ}$ {{iff}} these all hold: +: $(1): \quad J \ne \O$ +: $(2): \quad \forall x, y \in J: x + \paren {-y} \in J$ +: $(3): \quad \forall j \in J, r \in R: r \circ j \in J, j \circ r \in J$ +\end{theorem} + +\begin{proof} +=== Necessary Condition === +Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $\struct {R, +, \circ}$. +Then conditions $(1)$ to $(3)$ hold by virtue of the [[Definition:Ring Axioms|ring axioms]] and $J$ being an [[Definition:Ideal of Ring|ideal]]. +{{qed|lemma}} +=== Sufficient Condition === +Suppose conditions $(1)$ to $(3)$ hold. +As $r \in J \implies r \in R$, if $(3)$ holds for $J$, then $J$ is [[Definition:Closed Algebraic Structure|closed]] under $\circ$ and condition $(3)$ of [[Subring Test]] holds. +Thus, $J$ is a [[Definition:Subring|subring]] of $R$. +As $(3)$ defines the condition for $J$, being a [[Definition:Subring|subring]], to be an [[Definition:Ideal of Ring|ideal]], the result holds. +So $J$ is an [[Definition:Ideal of Ring|ideal]] of $\struct {R, +, \circ}$. +{{qed}} +\end{proof}<|endoftext|> +\section{Epimorphism Preserves Rings} +Tags: Ring Theory, Ring Epimorphisms + +\begin{theorem} +Let $\struct {R_1, +_1, \circ_1}$ be a [[Definition:Ring (Abstract Algebra)|ring]], and $\struct {R_2, +_2, \circ_2}$ be a [[Definition:Closed Algebraic Structure|closed algebraic structure]]. +Let $\phi: R_1 \to R_2$ be an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]]. +Then $\struct {R_2, +_2, \circ_2}$ is a [[Definition:Ring (Abstract Algebra)|ring]]. +\end{theorem} + +\begin{proof} +From [[Epimorphism Preserves Groups]], we have that if $\struct {R_1, +_1}$ is a [[Definition:Group|group]] then so is $\struct {R_2, +_2}$. +From [[Epimorphism Preserves Semigroups]], we have that if $\struct {R_1, \circ_1}$ is a [[Definition:Semigroup|semigroup]] then so is $\struct {R_2, \circ_2}$. +From [[Epimorphism Preserves Distributivity]], we have that if $\circ_1$ distributes over $+_1$ then $\circ_2$ distributes over $+_2$. +So it follows from the definition of a [[Definition:Ring (Abstract Algebra)|ring]] that if $\struct {R_1, +_1, \circ_1}$ is a [[Definition:Ring (Abstract Algebra)|ring]] then so is $\struct {R_2, +_2, \circ_2}$. +{{qed}} +\end{proof}<|endoftext|> +\section{Ring Homomorphism of Addition is Group Homomorphism} +Tags: Ring Homomorphisms, Group Homomorphisms + +\begin{theorem} +Let $\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$ be a [[Definition:Ring Homomorphism|ring homomorphism]]. +Then $\phi: \left({R_1, +_1}\right) \to \left({R_2, +_2}\right)$ is a [[Definition:Group Homomorphism|group homomorphism]]. +\end{theorem} + +\begin{proof} +From the definition of a [[Definition:Ring (Abstract Algebra)|ring]], both $\left({R_1, +_1}\right)$ and $\left({R_2, +_2}\right)$ are [[Definition:Abelian Group|abelian groups]]. +The result follows from the definitions of [[Definition:Ring Homomorphism|ring homomorphism]] and [[Definition:Group Homomorphism|group homomorphism]]. +{{Qed}} +[[Category:Ring Homomorphisms]] +[[Category:Group Homomorphisms]] +0vdhqve9vlm1czgtl0xvsx794hmny89 +\end{proof}<|endoftext|> +\section{Element of Integral Domain Divides Zero} +Tags: Integral Domains + +\begin{theorem} +Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$. +Then every [[Definition:Element|element]] of $D$ is a [[Definition:Divisor of Ring Element|divisor]] of [[Definition:Ring Zero|zero]]: +:$\forall x \in D: x \divides 0_D$ +\end{theorem} + +\begin{proof} +By definition, an [[Definition:Integral Domain|integral domain]] is a [[Definition:Ring (Abstract Algebra)|ring]]. +So, from [[Ring Product with Zero]]: +:$\forall x \in D: 0_D = x \circ 0_D$. +The result follows from the definition of [[Definition:Divisor of Ring Element|divisor]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Unity Divides All Elements} +Tags: Integral Domains, Unity Divides All Elements, Divisibility + +\begin{theorem} +Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Unity of Ring|unity]] is $1_D$. +Then [[Definition:Unity of Ring|unity]] is a [[Definition:Divisor of Ring Element|divisor]] of every [[Definition:Element|element]] of $D$: +:$\forall x \in D: 1_D \divides x$ +Also: +:$\forall x \in D: -1_D \divides x$ +\end{theorem} + +\begin{proof} +The element $1_D$ is the [[Definition:Unity of Ring|unity]] of $\struct {D, +, \circ}$, and so: +:$1_D \in D: x = 1_D \circ x$ +Similarly, from [[Product of Ring Negatives]]: +:$-1_D \in D: x = \paren {-1_D} \circ \paren {-x}$ +The result follows from the definition of [[Definition:Divisor of Ring Element|divisor]]. +{{qed}} +\end{proof} + +\begin{proof} +This is a special case of [[Unit of Integral Domain divides all Elements]], as [[Unity is Unit]]. +Furthermore, from [[Unity and Negative form Subgroup of Units]] we also have that $-1_D$ is a [[Definition:Unit of Ring|unit of $D$]]. +Hence the result. +{{qed}} +\end{proof}<|endoftext|> +\section{Element of Integral Domain is Divisor of Itself} +Tags: Integral Domains + +\begin{theorem} +Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Unity of Ring|unity]] is $1_D$. +Then every [[Definition:Element|element]] of $D$ is a [[Definition:Divisor of Ring Element|divisor]] of itself: +:$\forall x \in D: x \divides x$ +\end{theorem} + +\begin{proof} +Follows directly from the definition of [[Definition:Divisor of Ring Element|divisor]]: +:$\forall x \in D: \exists 1_D \in D: x = 1_D \circ x$ +{{qed}} +[[Category:Integral Domains]] +8xrney68i9349jnuparynsqc0h08i4k +\end{proof}<|endoftext|> +\section{Unit of Integral Domain divides all Elements} +Tags: Integral Domains + +\begin{theorem} +Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Unity of Ring|unity]] is $1_D$. +Let $\struct {U_D, \circ}$ be the [[Definition:Group of Units of Ring|group of units]] of $\struct {D, +, \circ}$. +Then: +:$\forall x \in D: \forall u \in U_D: u \divides x$ +That is, every [[Definition:Unit of Ring|unit]] of $D$ is a [[Definition:Divisor of Ring Element|divisor]] of every [[Definition:Element|element]] of $D$. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | lo= \forall x \in D, u \in U_D: + | l = x + | r = u \circ \paren {u^{-1} \circ x} + | c = {{Defof|Unit of Ring}} +}} +{{eqn | ll= \leadsto + | l = u + | o = \divides + | r = x + | c = {{Defof|Divisor of Ring Element}} +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Divisor of Unit is Unit} +Tags: Integral Domains + +\begin{theorem} +Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Unity of Ring|unity]] is $1_D$. +Let $\struct {U_D, \circ}$ be the [[Definition:Group of Units of Ring|group of units]] of $\struct {D, +, \circ}$. +Then: +:$x \in D, u \in U_D: x \divides u \implies x \in U_D$ +That is, if $x$ is a [[Definition:Divisor of Ring Element|divisor]] of a [[Definition:Unit of Ring|unit]], $x$ must itself be a [[Definition:Unit of Ring|unit]]. +\end{theorem} + +\begin{proof} +Let $x \in D, u \in U_D$ such that $x \divides u$. +By [[Definition:Divisor of Ring Element|definition]]: +:$\exists t \in D: u = t \circ x$ +Thus: +:$1_D = u^{-1} \circ u = u^{-1} \circ t \circ x$ +Also, as $D$ is an [[Definition:Integral Domain|integral domain]] and hence a [[Definition:Commutative Ring|commutative ring]], we have: +:$u^{-1} \circ t \circ x = 1_D = x \circ u^{-1} \circ t$ +It follows by definition that $x$ is a [[Definition:Unit of Ring|unit]], as it has an [[Definition:Inverse Element|inverse]], namely $u^{-1} \circ t$. +{{qed}} +\end{proof}<|endoftext|> +\section{Associatehood is Equivalence Relation} +Tags: Integral Domains, Equivalence Relations, Associates + +\begin{theorem} +Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$ and whose [[Definition:Unity of Ring|unity]] is $1_D$. +Let $\sim$ be the [[Definition:Relation|relation]] defined on $D$ as: +$\forall x, y \in D: x \sim y$ {{iff}} $x$ is an [[Definition:Associate in Integral Domain|associate]] of $y$ +Then $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]]. +\end{theorem} + +\begin{proof} +Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: +=== Reflexive === +Clearly $x \divides x$ as $x = 1_D \circ x$, so $x \sim x$. +Thus $\sim$ is [[Definition:Reflexive Relation|reflexive]]. +{{qed|lemma}} +=== Symmetric === +By the definition: +: $x \sim y \iff x \divides y \land y \divides x \iff y \sim x$ +Thus $\sim$ is [[Definition:Symmetric Relation|symmetric]]. +{{qed|lemma}} +=== Transitive === +{{begin-eqn}} +{{eqn | o = + | r = x \sim y \land y \sim z + | c = +}} +{{eqn | o = \leadsto + | r = \paren {x \divides y \land y \divides x} \land \paren {y \divides z \land z \divides y} + | c = {{Defof|Associate/Integral Domain|index = 1|Associate}} +}} +{{eqn | o = \leadsto + | r = \paren {x \divides y \land y \divides z} \land \paren {z \divides y \land y \divides x} + | c = [[Rule of Association]] and [[Rule of Commutation]] +}} +{{eqn | o = \leadsto + | r = x \divides z \land z \divides x + | c = [[Divisor Relation in Integral Domain is Transitive]] +}} +{{eqn | o = \leadsto + | r = x \sim z + | c = {{Defof|Associate/Integral Domain|index = 1|Associate}} +}} +{{end-eqn}} +Thus $\sim$ is [[Definition:Transitive Relation|transitive]]. +{{qed|lemma}} +All criteria are fulfilled for $\sim$ to be an [[Definition:Equivalence Relation|equivalence relation]]. +Hence the result. +{{Qed}} +\end{proof}<|endoftext|> +\section{Divisor Relation in Integral Domain is Transitive} +Tags: Integral Domains + +\begin{theorem} +Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]]. +Let $x, y, z \in D$. +Then: +:$x \divides y \land y \divides z \implies x \divides z$ +\end{theorem} + +\begin{proof} +Let $x \divides y \land y \divides z$. +Then from the definition of [[Definition:Divisor of Ring Element|divisor]], we have: +:$x \divides y \iff \exists s \in D: y = s \circ x$ +:$y \divides z \iff \exists t \in D: z = t \circ y$ +Then: +:$z = t \circ \paren {s \circ x} = \paren {t \circ s} \circ x$ +Thus: +:$\exists \paren {t \circ s} \in D: z = \paren {t \circ s} \circ x$ +and the result follows. +{{Qed}} +\end{proof}<|endoftext|> +\section{Integers form Unique Factorization Domain} +Tags: Integral Domains, Integers, Factorization, Unique Factorization Domains + +\begin{theorem} +The [[Integers form Integral Domain|integers]] $\struct {\Z, +, \times}$ form a [[Definition:Unique Factorization Domain|unique factorization domain]]. +\end{theorem} + +\begin{proof} +Follows directly from: +:$(1) \quad$ the [[Fundamental Theorem of Arithmetic|fundamental theorem of arithmetic]] +:$(2) \quad$ the fact that [[Integers form Integral Domain|$\struct {\Z, +, \times}$ is an integral domain]] +:$(3) \quad$ the definitions [[Definition:Complete Factorization|complete factorization]] and [[Definition:Equivalent Factorizations|equivalent factorizations]]. +{{qed}} +[[Category:Integral Domains]] +[[Category:Integers]] +[[Category:Factorization]] +[[Category:Unique Factorization Domains]] +32n4jwtruzn4meukioxmgtt2arwfda5 +\end{proof}<|endoftext|> +\section{Trivial Ordering Compatibility in Boolean Ring} +Tags: Boolean Rings, Order Theory + +\begin{theorem} +Let $\struct {S, +, \circ}$ be a [[Definition:Boolean Ring|Boolean ring]]. +Then the [[Definition:Trivial Ordering|trivial ordering]] is the only [[Definition:Ordering|ordering]] on $S$ [[Definition:Relation Compatible with Operation|compatible]] with both its [[Definition:Binary Operation|operations]]. +\end{theorem} + +\begin{proof} +That the [[Definition:Trivial Ordering|trivial ordering]] is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$ and $*$ follows from [[Trivial Ordering is Universally Compatible]]. +Conversely, suppose that $\preceq$ is a [[Definition:Ordering|ordering]] [[Definition:Relation Compatible with Operation|compatible]] with $\circ$ and $*$. +{{proof wanted}} +\end{proof}<|endoftext|> +\section{Smallest Field is Field} +Tags: Field Theory + +\begin{theorem} +The [[Definition:Ring (Abstract Algebra)|ring]] $\struct {\set {0_R, 1_R}, +, \circ}$ is the smallest [[Definition:Algebraic Structure|algebraic structure]] which is a [[Definition:Field (Abstract Algebra)|field]]. +\end{theorem} + +\begin{proof} +The [[Definition:Null Ring|null ring]], which contains one [[Definition:Element|element]], is ''not'' a [[Definition:Field (Abstract Algebra)|field]] as it is [[Definition:Null Ring|trivial]]. +Therefore any [[Definition:Field (Abstract Algebra)|field]] must contain at least two [[Definition:Element|elements]]. +For $\struct {\set {0_R, 1_R}, +, \circ}$ to be a [[Definition:Field (Abstract Algebra)|field]]: +:$\struct {\set {0_R, 1_R}, +}$ must be an [[Definition:Abelian Group|abelian group]]. +This is fulfilled as this is the [[Definition:Parity Group|parity group]]. +:$\struct {\set {0_R, 1_R}, \circ}$ must be a [[Definition:Commutative Ring|commutative]] [[Definition:Division Ring|division ring]]. +This is fulfilled, as $\struct {\set {0_R, 1_R}^*, \circ} = \struct {\set {1_R}, \circ}$ is the [[Definition:Trivial Group|trivial group]]. +:$\circ$ needs to [[Definition:Distributive Operation|distribute]] over $+$. This follows directly from [[Ring Product with Zero]] and the behaviour of the [[Definition:Identity Element|identity element]] in a [[Definition:Group|group]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Field is Integral Domain} +Tags: Field Theory, Integral Domains, Field is Integral Domain + +\begin{theorem} +Every [[Definition:Field (Abstract Algebra)|field]] is an [[Definition:Integral Domain|integral domain]]. +\end{theorem} + +\begin{proof} +This result follows directly from: +:[[Field has no Proper Zero Divisors]] +:By definition, that a [[Definition:Field (Abstract Algebra)|field]] is a [[Definition:Commutative Ring|commutative ring]]. +{{qed}} +\end{proof} + +\begin{proof} +Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$. +Suppose $\exists x, y \in F: x \circ y = 0_F$. +Suppose $x \ne 0_F$. +Then, by the definition of a [[Definition:Field (Abstract Algebra)|field]], $x^{-1}$ exists in $F$ and: +:$y = 1_F \circ y = x^{-1} \circ x \circ y = x^{-1} \circ 0_F = 0_F$. +Otherwise $x = 0_F$. +So if $x \circ y = 0_F$, either $x = 0_F$ or $y = 0_F$ as we were to show. +{{qed}} +\end{proof}<|endoftext|> +\section{Finite Integral Domain is Galois Field} +Tags: Galois Fields, Integral Domains, Finite Integral Domain is Galois Field + +\begin{theorem} +A [[Definition:Finite Integral Domain|finite integral domain]] is a [[Definition:Galois Field|Galois field]]. +\end{theorem} + +\begin{proof} +Let $R$ be a [[Definition:Finite Integral Domain|finite integral domain]] whose [[Definition:Unity of Ring|unity]] is $1$ and whose [[Definition:Ring Zero|zero]] is $0$. +Let $a \in R$ such that $a \ne 0$. +We wish to show that $a$ has a [[Definition:Ring Product Inverse|product inverse]] in $R$. So consider the [[Definition:Mapping|map]] $f: R \to R$ defined by $f: x \mapsto a x$. +We first show that the [[Definition:Kernel of Ring Homomorphism|kernel]] of $f$ is trivial. +Consider that: +:$\map \ker f = \set {x \in R: \map f x = 0} = \set {x \in R: a x = 0}$ +Since $R$ is an [[Definition:Integral Domain|integral domain]], it has no [[Definition:Proper Zero Divisor|proper zero divisors]] and thus $a x = 0$ means that $a = 0$ or $x = 0$. +Since, by definition, $a \ne 0$, then it must be true that $x = 0$. +Therefore, $\map \ker f = \set 0$ and so $f$ is [[Definition:Injection|injective]]. +Next, the [[Pigeonhole Principle]] gives us that an [[Definition:Injection|injective mapping]] from a [[Definition:Finite Set|finite set]] onto itself is [[Definition:Surjection|surjective]]. +Since $R$ is [[Definition:Finite Set|finite]], the [[Definition:Mapping|mapping]] $f$ is [[Definition:Surjection|surjective]]. +Finally, since $f$ is [[Definition:Surjection|surjective]] and $1 \in R$, we have: +:$\exists x \in R: \map f x = a x = 1$ +So this $x$ is the [[Definition:Ring Product Inverse|product inverse]] of $a$ and the proof is complete. +{{qed}} +\end{proof} + +\begin{proof} +Let $R$ be a [[Definition:Finite Integral Domain|finite integral domain]] with [[Definition:Unity of Ring|unity]] $1$ and [[Definition:Ring Zero|zero]] $0$. +Let $R^*$ denote $R \setminus \set 0$, the set $R$ without the [[Definition:Ring Zero|zero]]. +As $R$ is [[Definition:Finite Set|finite]], we may enumerate the elements of $R$ as: +:$x_0 = 0, x_1 = 1, x_2, x_3, \ldots, x_n$ +Let $R^*$. +Consider the elements: +:$x_k x_1, x_k x_2, \ldots, x_k x_n$ +which are members of $R$ by [[Definition:Closed Algebraic Structure|closure]] of the [[Definition:Ring Product|ring product]]. +Since $R$ is an [[Definition:Integral Domain|integral domain]], by definition it has no [[Definition:Proper Zero Divisor|proper zero divisors]]. +Therefore all of $x_k x_1, x_k x_2, \ldots, x_k x_n$ are elements of $R^*$. +Recall: +:$x_k x_p = x_k x_q \implies x_p = x_q$ +by the [[Cancellation Law of Ring Product of Integral Domain]]. +So the list of products consists of $n$ distinct elements and is therefore equal to the whole $R^*$. +Since $1 \in R^*$, we have that $x_k x_{k^*} = 1$ for some $k^*$. +Therefore, there exists an [[Definition:Ring Product Inverse|inverse]] for every non-[[Definition:Zero Ring|zero]] [[Definition:Element|element]], so $R$ is a [[Definition:Field (Abstract Algebra)|field]]. +As the [[Definition:Underlying Set|underlying set]] of $R$ is [[Definition:Finite Set|finite]], $R$ is by definition a [[Definition:Galois Field|Galois field]]. +{{qed}} +\end{proof} + +\begin{proof} +An [[Definition:Integral Domain|integral domain]] is by definition a [[Definition:Ring (Abstract Algebra)|ring]] which has no [[Definition:Proper Zero Divisor|proper zero divisors]]. +By definition, a [[Definition:Galois Field|Galois field]] is a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Underlying Set|underlying set]] is [[Definition:Finite Set|finite]]. +The result follows from [[Finite Ring with No Proper Zero Divisors is Field]]. +{{qed}} +\end{proof} + +\begin{proof} +{{AimForCont}} $\struct {D, +, \circ}$ is a [[Definition:Finite Set|finite]] [[Definition:Integral Domain|integral domain]] which is not a [[Definition:Field (Abstract Algebra)|field]]. +From [[Non-Field Integral Domain has Infinite Number of Ideals]], $\struct {D, +, \circ}$ has an [[Definition:Infinite Set|infinite number]] of [[Definition:Distinct Elements|distinct]] [[Definition:Ideal of Ring|ideals]]. +But this [[Definition:Contradiction|contradicts]] the assertion that $\struct {D, +, \circ}$ is [[Definition:Finite Set|finite]]. +Hence the result by [[Proof by Contradiction]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Field is Subfield of Itself} +Tags: Subfields + +\begin{theorem} +Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]]. +Then $\struct {F, +, \circ}$ is a [[Definition:Subfield|subfield]] of $\struct {F, +, \circ}$. +\end{theorem} + +\begin{proof} +$F$ is a [[Definition:Field (Abstract Algebra)|field]] and $F \subseteq F$ from [[Set is Subset of Itself]]. +{{Qed}} +\end{proof}<|endoftext|> +\section{Center of Division Ring is Subfield} +Tags: Division Rings, Subfields + +\begin{theorem} +Let $\struct {K, +, \circ}$ be an [[Definition:Division Ring|division ring]]. +Then $\map Z K$, the [[Definition:Center of Ring|center]] of $K$, is a [[Definition:Subfield|subfield]] of $K$. +\end{theorem} + +\begin{proof} +For $\map Z K$ to be a [[Definition:Subfield|subfield]] of $K$, it needs to be a [[Definition:Division Ring|division ring]] that is [[Definition:Commutative Ring|commutative]]. +Thus the result follows directly from [[Center of Ring is Commutative Subring]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Ideals of Field} +Tags: Field Theory, Ideal Theory, Commutative Algebra + +\begin{theorem} +Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$. +Then $\struct {R, +, \circ}$ is a [[Definition:Field (Abstract Algebra)|field]] {{iff}} the only [[Definition:Ideal of Ring|ideals]] of $\struct {R, +, \circ}$ are $\struct {R, +, \circ}$ and $\set {0_R}$. +\end{theorem} + +\begin{proof} +=== Necessary Condition === +Let $\struct {R, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]]. +The result follows from [[Field has 2 Ideals]]. +{{qed|lemma}} +=== Sufficient Condition === +Suppose that the only [[Definition:Ideal of Ring|ideals]] of $\struct {R, +, \circ}$ are $\struct {R, +, \circ}$ and $\set {0_R}$. +The result follows from [[Commutative and Unitary Ring with 2 Ideals is Field]] +{{Qed}} +\end{proof}<|endoftext|> +\section{Epimorphism from Division Ring to Ring} +Tags: Ring Theory, Division Rings, Ring Epimorphisms + +\begin{theorem} +Let $\left({K, +, \circ}\right)$ be a [[Definition:Division Ring|division ring]] whose [[Definition:Ring Zero|zero]] is $0_K$. +Let $\left({R, +, \circ}\right)$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$. +Let $\phi: K \to R$ be a [[Definition:Ring Epimorphism|ring epimorphism]]. +Then one of the following applies: +:$(1): \quad R$ is a [[Definition:Null Ring|null ring]] +:$(2): \quad R$ is a [[Definition:Division Ring|division ring]] and $\phi$ is a [[Definition:Ring Isomorphism|ring isomorphism]]. +\end{theorem} + +\begin{proof} +We have that the [[Kernel of Ring Epimorphism is Ideal|kernel of $K$ is an ideal]]. +From [[Ideals of Division Ring]], $\ker \left({K}\right)$ must therefore either be $0_K$ or $K$. +* If $\ker \left({K}\right) = 0_K$, then by [[Ring Epimorphism with Trivial Kernel is Isomorphism]] $\phi$ is an [[Definition:Ring Isomorphism|ring isomorphism]], thus making $R$ a [[Definition:Division Ring|division ring]] like $K$. +* If $\ker \left({K}\right) = K$, then $\forall x \in K: \phi \left({K}\right) = 0_R$. +As $\phi$ is an [[Definition:Ring Epimorphism|epimorphism]], it is [[Definition:Surjection|surjective]] and therefore $R = \left\{{0_R}\right\}$. +{{qed}} +\end{proof}<|endoftext|> +\section{Congruence Class Modulo Subgroup is Coset} +Tags: Cosets + +\begin{theorem} +Let $G$ be a [[Definition:Group|group]]. +Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$. +\end{theorem}<|endoftext|> +\section{Cosets are Equivalent} +Tags: Cosets, Cosets are Equivalent + +\begin{theorem} +All [[Definition:Left Coset|left cosets]] of a [[Definition:Group|group]] $G$ with respect to a [[Definition:Subgroup|subgroup]] $H$ are [[Definition:Set Equivalence|equivalent]]. +That is, any two [[Definition:Left Coset|left cosets]] are [[Definition:Bijection|in one-to-one correspondence]]. +The same applies to [[Definition:Right Coset|right cosets]]. +As a special case of this: +:$\forall x \in G: \order {x H} = \order H = \order {H x}$ +where $H$ is a [[Definition:Subgroup|subgroup]] of $G$. +\end{theorem} + +\begin{proof} +Let us set up [[Definition:Mapping|mappings]] $\theta: H \to H x$ and $\phi: H x \to H$ as follows: +:$\forall u \in H: \map \theta u = u x$ +:$\forall v \in H x: \map \phi v = v x^{-1}$ +From [[Element in Right Coset iff Product with Inverse in Subgroup]]: +:$v \in H x \implies v x^{-1} \in H$ +Now: +:$\forall v \in H x: \theta \circ \map \phi v = v x^{-1} x = v$ +:$\forall u \in H: \phi \circ \map \theta u = u x x^{-1} = u$ +Thus $\theta \circ \phi = I_{H x}$ and $\phi \circ \theta = I_H$ are [[Definition:Identity Mapping|identity mappings]]. +So $\theta = \phi^{-1}$: both are [[Definition:Bijection|bijections]] and one is the [[Definition:Inverse Mapping|inverse]] of the other. +This establishes, for each $x \in G$, the [[Definition:Set Equivalence|set equivalence]] between $H$ and $H x$: +:$H \simeq H x$ +In particular, for any $x, y \in G$, it follows from $H x \simeq H$ and $H y \simeq H$ that: +:$H x \simeq H y$ +by [[Set Equivalence is Equivalence Relation]]. +Similarly, we can set up [[Definition:Mapping|mappings]] $\alpha: H \to x H$ and $\beta: x H \to H$ as follows: +:$\forall u \in H: \map \alpha u = x u$ +:$\forall v \in x H: \map \beta v = x^{-1} v$ +Analogous to above reasoning gives $\alpha = \beta^{-1}$ which establishes $H \simeq x H$. +Also similarly, $x H \simeq y H$ for all $x, y \in G$. +Hence the result. +{{qed}} +\end{proof} + +\begin{proof} +Follows directly from [[Set Equivalence of Regular Representations]]. +{{qed}} +\end{proof}<|endoftext|> +\section{Left and Right Coset Spaces are Equivalent} +Tags: Cosets, Left and Right Coset Spaces are Equivalent + +\begin{theorem} +Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. +Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$. +Let: +:$x H$ denote the [[Definition:Left Coset|left coset]] of $H$ by $x$ +:$H y$ denote the [[Definition:Right Coset|right coset]] of $H$ by $y$. +Then: +:$\order {\set {x H: x \in G} } = \order {\set {H y: y \in G} }$ +\end{theorem} + +\begin{proof} +Let there be exactly $r$ different [[Definition:Left Coset|left cosets]] of $H$ in $G$. +Let a complete repetition-free list of these [[Definition:Left Coset|left cosets]] be: +:$a_1 H, a_2 H, a_3 H \ldots, a_r H: a_1, a_2, \ldots, a_r \in G$ +From [[Left Coset Space forms Partition]], every element of $G$ is contained in exactly one of the [[Definition:Left Coset|left cosets]]. +Let $x \in G$. +Then, for $1 \le i \le r$: +{{begin-eqn}} +{{eqn | l = x + | o = \in + | r = H a_i^{-1} + | c = +}} +{{eqn | ll= \iff + | l = x \paren {a_i^{-1} }^{-1} + | o = \in + | r = H + | c = [[Element in Right Coset iff Product with Inverse in Subgroup]] +}} +{{eqn | ll= \iff + | l = x a_i + | o = \in + | r = H + | c = [[Inverse of Group Inverse]] +}} +{{eqn | ll= \iff + | l = \paren {x^{-1} }^{-1} a_i + | o = \in + | r = H + | c = [[Inverse of Group Inverse]] +}} +{{eqn | ll= \iff + | l = x^{-1} + | o = \in + | r = a_i H + | c = [[Element in Left Coset iff Product with Inverse in Subgroup]] +}} +{{end-eqn}} +Since $x^{-1} \in a_i H$ is true for precisely one value of $i$, it follows that $x \in H a_i^{-1}$ is also true for precisely that value of $i$. +So there are exactly $r$ different [[Definition:Right Coset|right cosets]] of $H$ in $G$, and a complete repetition-free list of these is: +:$H a_1^{-1}, H a_2^{-1}, H a_3^{-1}, \ldots, H a_r^{-1}$ +The result follows. +{{qed}} +\end{proof} + +\begin{proof} +Let $G$ be a [[Definition:Group|group]] and let $H \le G$. +Consider the [[Definition:Mapping|mapping]] $\phi$ from the [[Definition:Left Coset Space|left coset space]] to the [[Definition:Right Coset Space|right coset space]] defined as: +:$\forall g \in G: \map \phi {g H} = H g^{-1}$ +We need to show that $\phi$ is a [[Definition:Bijection|bijection]]. +First we need to show that $\phi$ is [[Definition:Well-Defined Mapping|well-defined]]. +That is, that $a H = b H \implies \map \phi {a H} = \map \phi {b H}$. +Suppose $a H = b H$. +{{begin-eqn}} +{{eqn | l = a H = b H + | o = \iff + | r = a^{-1} b \in H + | c = [[Left Cosets are Equal iff Product with Inverse in Subgroup]] +}} +{{eqn | l = H a^{-1} = H b^{-1} + | o = \iff + | r = a^{-1} \paren {b^{-1} }^{-1} \in H + | c = [[Right Cosets are Equal iff Product with Inverse in Subgroup]] +}} +{{end-eqn}} +But $a^{-1} \paren {b^{-1} }^{-1} = a^{-1} b \in H$ as $a H = b H$. +So $H a^{-1} = H b^{-1}$ and $\phi$ is [[Definition:Well-Defined Mapping|well-defined]]. +Next we show that $\phi$ is [[Definition:Injection|injective]]: +Suppose $\exists x, y \in G: \map \phi {x H} = \map \phi {y H}$. +Then $H x^{-1} = H y^{-1}$, so $x^{-1} = e_G x^{-1} = h y^{-1}$ for some $h \in H$. +Thus $h = x^{-1} y \implies h^{-1} = y^{-1} x$. +As $H$ is a [[Definition:Subgroup|subgroup]], $h^{-1} \in H$. +Thus: +: $y^{-1} x \in H$ +So by [[Left Cosets are Equal iff Product with Inverse in Subgroup]]: +:$x H = y H$ +Thus $\phi$ is [[Definition:Injection|injective]]. +Next we show that $\phi$ is [[Definition:Surjection|surjective]]: +Let $H x$ be a [[Definition:Right Coset|right coset]] of $H$ in $G$. +Since $x = \paren {x^{-1} }^{-1}$, $H x = \map \phi {x^{-1} H}$ and so $\phi$ is [[Definition:Surjection|surjective]]. +Thus $\phi$ constitutes a [[Definition:Bijection|bijection]] from the [[Definition:Left Coset Space|left coset space]] to the [[Definition:Right Coset Space|right coset space]], and the result follows. +{{qed}} +\end{proof}<|endoftext|> +\section{Cosets in Abelian Group} +Tags: Cosets, Abelian Groups + +\begin{theorem} +Let $G$ be an [[Definition:Abelian Group|abelian group]]. +Then every [[Definition:Right Coset|right coset]] modulo $H$ is a [[Definition:Left Coset|left coset]] modulo $H$. +That is: +:$\forall x \in G: x H = H x$ +In an [[Definition:Abelian Group|abelian group]], therefore, we can talk about '''[[Definition:Congruence Modulo Subgroup|congruence modulo $H$]]''' and not worry about whether it is [[Definition:Left Congruence Modulo Subgroup|left]] or [[Definition:Right Congruence Modulo Subgroup|right]]. +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | o = + | r = \forall x, y \in G: x^{-1} y = y x^{-1} + | c = +}} +{{eqn | o = \leadsto + | r = \paren {x \equiv^l y \pmod H \iff y \equiv^r x \pmod H} + | c = {{Defof|Congruence Modulo Subgroup}} +}} +{{eqn | o = \leadsto + | r = \paren {x \equiv^l y \pmod H \iff x \equiv^r y \pmod H} + | c = [[Congruence Modulo Subgroup is Equivalence Relation]], therefore [[Definition:Symmetric Relation|Symmetric]] +}} +{{end-eqn}} +{{qed}} +\end{proof}<|endoftext|> +\section{Conjugate of Set by Identity} +Tags: Conjugacy + +\begin{theorem} +Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$. +Let $S \subseteq G$. +Then the [[Definition:Conjugate of Group Subset|conjugate of $S$ by $e$]] is $S$: +:$S^e = S$ +\end{theorem} + +\begin{proof} +{{begin-eqn}} +{{eqn | l = S^e + | r = \set {y \in G: \exists x \in S: y = e \circ x \circ e^{-1} } + | c = {{Defof|Conjugate of Group Subset}} +}} +{{eqn | r = \set {y \in G: \exists x \in S: y = x} + | c = {{Defof|Identity Element}} +}} +{{eqn | r = S + | c= +}} +{{end-eqn}} +{{Qed}} +\end{proof}<|endoftext|> +\section{Conjugate of Set by Group Product} +Tags: Conjugacy + +\begin{theorem} +Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. +Let $S \subseteq G$. +Let $S^a$ denote the [[Definition:Conjugate of Group Subset|$G$-conjugate of $S$ by $a$]] as: +:$S^a := \set {y \in G: \exists x \in S: y = a \circ x \circ a^{-1} } = a \circ S \circ a^{-1}$ +Then: +:$\paren {S^a}^b = S^{b \circ a}$ +\end{theorem} + +\begin{proof} +$S^a$ is defined as $a \circ S \circ a^{-1}$ from the definition of the [[Definition:Conjugate of Group Subset|conjugate of a set]]. +From the definition of [[Definition:Subset Product with Singleton|subset product with a singleton]], this can be seen to be the same thing as: +:$S^a = \set a \circ S \circ \set {a^{-1} }$. +Thus we can express $\paren {S^a}^b$ as $b \circ \paren {a \circ S \circ a^{-1} } \circ b^{-1}$, and understand that the {{RHS}} refers to [[Definition:Subset Product|subset products]]. +From [[Subset Product within Semigroup is Associative]] (which applies because $\circ$ is [[Definition:Associative Operation|associative]]), it then follows directly that: +{{begin-eqn}} +{{eqn | l = \paren {S^a}^b + | r = b \circ \paren {a \circ S \circ a^{-1} } \circ b^{-1} + | c = from above +}} +{{eqn | r = \paren {b \circ a} \circ S \circ \paren {a^{-1} \circ b^{-1} } + | c = [[Subset Product within Semigroup is Associative]] +}} +{{eqn | r = \paren {b \circ a} \circ S \circ \paren {b \circ a}^{-1} + | c = [[Inverse of Group Product]] +}} +{{eqn | r = S^{b \circ a} + | c = {{Defof|Conjugate of Group Subset}} +}} +{{end-eqn}} +{{Qed}} +=== Proof for Alternative Definition === +Using the same preliminary argument as above, we then follow: +{{begin-eqn}} +{{eqn | l = \paren {S^a}^b + | r = b^{-1} \circ \paren {a^{-1} \circ S \circ a} \circ b + | c = from above +}} +{{eqn | r = \paren {b^{-1} \circ a^{-1} } \circ S \circ \paren {a \circ b} + | c = [[Subset Product within Semigroup is Associative]] +}} +{{eqn | r = \paren {a \circ b}^{-1} \circ S \circ \paren {a \circ b} + | c = [[Inverse of Group Product]] +}} +{{eqn | r = S^{a \circ b} + | c = {{Defof|Conjugate (Group Theory)/Subset/Also defined as|Conjugate of Group Subset}} +}} +{{end-eqn}} +{{Qed}} +\end{proof} \ No newline at end of file