TITLE: Congruency and Congruent Classes
QUESTION [5 upvotes]: so studying for my midterm on Tuesday (intro to abstract algebra). The topics on the exam are Division Algorithm, Divisibility, Prime Numbers, FTA, Congruency, Congruent Classes and very brief introduction to rings. 
I was reading a few theorems about Congruency and have a couple of questions. 
I want to know what a "congruent class" is. My notes say "the congruence class of a modulo n" is a set: 
$ \left\{ \text{all } b \in \mathbb{Z} | b \equiv a \pmod{n} \right\}  $ which is also saying
$ \left\{ \text{all } a + kn \in \mathbb{Z}  | k \in \mathbb{Z} \right\}  $
okay so got that. I just wrote it for some people who might need a refreshed (it is a 3rd undergrad course after all). 
So in my notes our professor has a following example:
$\left[ 60 \right]_{17} = \left[ 43\right]_{17}$ 
1) so the way I figured this out is that to check if they are equivalent, we subtract 60-43 and see if that is a multiple of n = 17. Is this how you can check if they are equal classes?  If not, is there a better way to do so? 
2) A certain theorem states: Let $n \in \mathbb{Z}_+; a, b \in \mathbb{Z}$ and $gcd(a,n) = d$ then $[a]x=[b]$ has exactly $d$ solutions. My question here is that is x a congruence class or a random integer? What is x and how do I solve for it?
3) Is it true that if we are in $\mathbb{Z}_{12}$ then $[7]x=[11]$ can be rewritten as $ 7x \equiv 11 \pmod{12}$? If so, would finding the solution be similar to solution in this question
Thankyou. I am just very confused about congurency and stuff. I understand the theorems but I am hoping someone would give me an "easy" explanation of what is going on. I still don't know the difference between circle plus and regular plus except that circle plus has to satisfy certain axioms. Am I right?

REPLY [4 votes]: Okay, let's start from the beginning. You define the congruence class of $a$ as:
\begin{equation*}
[a]_n = \{ b\in\mathbb{Z}\mid a\equiv a\pmod{n}\}.
\end{equation*}
Now, what does $a\equiv b\pmod{n}$ mean? It means that $n$ divides $a-b$, or equivalently, that $a$ and $b$ leave the same remainder when you divide them by $n$. Considering the last of the conditions, notice that:

Since each number leaves the same remainder as itself when divided by $n$, $a\equiv a \pmod{n}$ for all $a$.
If $a$ leaves the same remainder as $b$ when divided by $n$, then $b$ leaves the same remainder as $a$ when divided by $n$. So if $a\equiv b\pmod{n}$, then $b\equiv a \pmod{n}$.
If $a$ leaves the same remainder as $b$, and $b$ leaves the same remainder as $c$, then $a$ and $c$ leave the same remainder as well. So if $a\equiv b\pmod{n}$ and $b\equiv c\pmod{n}$, then $a\equiv c\pmod{n}$.

That means that:
\begin{equation*}
[a]_{n} = [b]_n\Longleftrightarrow a\equiv b\pmod{n}.
\end{equation*}
Why? Well, suppose the congruence classes are the same. Since $a\in[a]_n=[b]_n$, that means that $b\equiv a \pmod{n}$ (since $a\in[b]_n$), so $a\equiv b\pmod {n}$. That proves that if the classes are equal, then $a\equiv b\pmod{n}$.
Conversely, suppose that $a\equiv b\pmod{n}$. How do we prove that $[a]_{n}=[b]_{n}$? Since they are sets, the usual way is to show that each is contained in the other. If $c\in[a]_n$, then $a\equiv c\pmod{n}$ by definition; since we also have $b\equiv a\pmod{n}$ (since $a\equiv b\pmod{n}$ is our assumption), then $b\equiv c\pmod{n}$, so $c\in[b]_n$. Therefore, every that is in $[a]_{n}$ is also in $[b]_n$, so $[a]_n\subseteq [b]_n$. For the converse inclusion, if $c\in[b]_n$, then $b\equiv c\pmod{n}$, and since we also have $a\equiv b\pmod{n}$, then $a\equiv c \pmod{n}$ so $c\in[a]_n$. Therefore, $[b]_n\subseteq [a]_n$. Since we have the two inclusions, we conclude that $[a]_n=[b]_n$.
We also have the following, which is perhaps more surprising:
\begin{equation*}
[a]_n\cap[b]_n\neq\emptyset\Longleftrightarrow [a]_n=[b]_n.
\end{equation*}
That is: the only way the class of $a$ and the class of $b$ have anything in common is if they are identical. Why? Well, if they are identical they certainly have nonempty intersection, since the class of $a$ always includes at least $a$. And conversely, if $c\in[a]_n\cap[b]_n$, then $a\equiv c\pmod{n}$ and $b\equiv c\pmod{n}$, from which we conclude that $a\equiv b\pmod{n}$, so $[a]_n=[b]_n$ by what we just finished proving.
On to your questions:
(1) How do we check if $[a]_{17} = [b]_{17}$? Precisely the way you did it: the classes are equal if and only if $a\equiv b \pmod{17}$. How do you check if $a\equiv b\pmod{17}$? By checking to see if $17$ divides $a-b$. So how do you check in general whether $[a]_n=[b]_n$? You check to see if $a-b$ is a multiple of $n$. If it is, then the classes are equal. If $a-b$ is not a multiple of $n$, then they are not equal, and in fact they are disjoint.
There are other ways: for example, if you could show that $[60]_{17}$ and $[43]_{17}$ have any element in common, then you would be able to conclude that they are equal. Sometimes it may be simpler to see that $[a]_n$ and $[b]_n$ have some element in common than to check if $n$ divides $a-b$; but the standard way of checking is to see whether $n$ divides $a-b$, just as you did.
(2) What you have is an equation in which $x$ is an unknown. You are really looking for all solutions to the congruence
\begin{equation*}
ax \equiv b \pmod {n}
\end{equation*}
That is, all integers $x$ that make the congruence true. For example, $x=3$ is a solution to
\begin{equation*}
2x \equiv 1 \pmod{5}
\end{equation*}
because $(2)(3)= 6\equiv 1 \pmod{5}$. In fact, because $3$ is a solution, so is $3+5k$ for any integer $k$; that is, any element in $[3]_5$ is a solution, since $3$ is a solution. So the solutions will actually be a collection of congruence classes.
Moreover, if we replace $2$ in the equation with $7$, so that we are trying to solve $7x\equiv 1 \pmod{5}$, then anything which was a solution to $2x\equiv 1\pmod{5}$ is still a solution, and any solution to the first one is still a solution to the new congruence; why? because $2x\equiv 7x\pmod{5}$ for all $x$, as $2x-7x = -5(x)$ is always a multiple of $5$. So we can replace $2$ with any element of $[2]_5$ and not change the solutions. Likewise, we can replace $1$ with any element of $[1]_5$ and not change the solutions. So it's almost as if instead of trying to solve the single congruence
\begin{equation*}
2x \equiv 1 \pmod{5}
\end{equation*}
we are trying to solve the equation
\begin{equation*}
[2]_5 x = [1]_5
\end{equation*}
So that's why you have written $[a]x=[b]$. That really means the congruence $ax\equiv b \pmod{n}$.
What you wrote, however, is incorrect. You are missing a clause: the congruence will have $d$ solutions if $d$ divides $b$. Otherwise, it's not going to have any. For an easy example, consider the congruence $2x\equiv 1 \pmod{4}$. Then $\gcd(2,4)=2$, but there are no solutions to the congruence, because $2x$ is always even, so $2x-1$ is always odd, so $4$ never divides $2x-1$. 
Now, how do you solve a congruence $ax\equiv b\pmod{n}$ when $\gcd(a,n)$ divides $b$? let $d=\gcd(a,n)$. Then we can write $a=da'$, $b=db'$, and $n=dn'$. Then $ax\equiv b\pmod{n}$ if and only if $n$ divides $ax-b$. But $ax-b = d(a'x - b')$, and $dn'$ divides $d(a'x-b')$ if and only if $n'$ divides $a'x-b'$, if and only if $a'x\equiv b'\pmod{n'}$. So if we can solve a congruence when $\gcd(a,n)=1$, then we can solve any congruence where $\gcd(a,n)$ divides $b$
So, how do you solve a system $ax \equiv b\pmod{n}$ when $\gcd(a,n)=1$ (which will of course divide $n$)? Since $\gcd(a,n)=1$, then we can write $n$ as a linear combination of $a$ and $n$, $1=ar+ns$ for some integers $r$ and $s$ (for example, using the Euclidean algorithm). Multiplying through by $b$ you get $b=a(rb)+ n(sb)$. That means that $n(sb) = a(rb)-b$, so $a(rb)\equiv b \pmod{n}$, which means that $x=rb$ is a solution. If $y$ is any other solution, then $ay\equiv b\pmod{n}$, and $ax\equiv b\pmod{n}$, so $ax\equiv ay\pmod{n}$, hence $n$ divides $ax-ay=a(x-y)$, and since $\gcd(a,n)=1$, then $n$ divides $x-y$; so if $y$ is any other solution, then $x\equiv y\pmod{n}$. So the only congruence class that is a solution is $[x]_n$. 
Now, what about general congruences? Suppose you have $[a]_nx=[b]_n$ and $\gcd(a,n)=d$ divides $b$. Write $a=da'$, $b=db'$, and $n=dn'$. Notice that $\gcd(a',n')=1$. Find a solution $x_0$ to $[a']_{n'}x=[b']_{n'}$. Then $n'$ divides $a'x_0 -b'$, so $dn'=n$ divides $d(a'x_0-b') = da'x_0-db' = ax_0-b$. So $x_0$ is also a solution to the original problem. If $y$ is any other solution, then as before we get that $n$ divides $a(x_0-y)$; so $dn'$ divides $da'(x_0-y)$, hence $n'$ divides $a'(x_0-y)$, and since $\gcd(a',n')=1$, then $n'$ divides $x_0 - y$; that is, $y\equiv x_0\pmod{n'}$, so $y=x_0+kn'$ for some $k$. Conversely, if $y = x_0+\ell n'$ for some $\ell$, then 
\begin{equation*}
ay = a(x_0+\ell n') = ax_0 + \ell an' = ax_0 + \ell (da')n' = ax_0 + \ell a'(dn') = ax_0 + (\ell a')n,
\end{equation*} 
which says that $ay \equiv ax_0 \pmod{n}$, so $y$ is also a solution. So what are the different solutions? Well, we have $x_0$, $x_0+n'$, $x_0+2n'$, $x_0+3n',\ldots, x_0+(d-1)n'$. All of these are not congruent to one another modulo $n$; the "next" one, however, will be $x_0+dn' = x_0+n$, which is congruent to $x_0$ modulo $n$. So the only distinct congruence classes that are solutions are $[x_0]_n$, $[x_0+n']_n,\ldots,[x_0+(d-1)n']_n$, giving you exactly $d$ distinct classes that are solutions to $[a]_nx=[b]_n$.
To see this in practice, take $n= 77$, $a=21$, and $b=14$. We want to solve the system $[21]_{77} x = [9]_{77}$ Since $\gcd(77,21)=7$ divides $b$, the system has solutions; in fact, it has $7$ different congruence classes modulo $77$ as solutions.  
So, we write $21 = 7\cdot 7$, $14 = 7\cdot 2$, $77 = 7\cdot 11$. And we first solve the system $[3]_{11}x=[2]_{11}$. To do this, we write $1$ as a linear combination of $3$ and $11$, like so: $1 = 3(4) - 11(1)$. Then multiply by $2$ to get $2 = 3(8)-11(2)$. So $x_0=8$ is a solution (indeed, $3(8)=24\equiv 2\pmod{11}$). Now going back to the original congruence, we take $x_0$ and we add multiples of $11$ (why $11$? Because $n=77$ is $7\cdot 11$, and $7$ is the $d$ from before) until we have our $7$ different solutions. So the solutions are: $[8]_{77}$, $[19]_{77}$, $[30]_{77}$, $[41]_{77}$, $[52]_{77}$, $[63]_{77}$, and $[74]_{77}$. You can verify that they all are solutions to $[21]_{77}x = [14]_{77}$. For example, $21\cdot 41 = 861$, and $[861]_{77}=[14]_{77}$ because $861-14 = 847 = 77\cdot 11$. 
(3) Yes: that is exactly the meaning of $[a]_nx=[b]_n$; it means $ax\equiv b\pmod{n}$. 
Hope this helps, despite the length.