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Ionic bond

Ionic bonding occurs when an atom or molecule completely transfers electrons to another atom or molecule. This happens because the valence electron of what becomes the cation is attracted to what becomes the anion. This transfer of electrons causes electrostatic attraction; the receiving atom or molecule develops a negative charge to become an anion, and the transferring atom or molecule a positive charge to become a cation. The atoms and/or molecules, now having opposite charges, are attracted to each other, thus forming an ionic bond. Some examples of compounds with ionic bonds are NaCl (sodium chloride or table salt), and CaCO3 (calcium carbonate). Such compounds are called ionic compounds, as opposed to covalent or molecular compounds which have no ionic bonds at all.

The charge of a cation or anion is denoted as a superscript after the formula of the ion. A + sign denotes a positive charge resulting from loss of electrons. A - sign denotes a negative charge resulting from gain of electrons. The number of electrons missing or gained is written before the + or - sign to indicate the degree of charge; when a + or - sign is seen alone in the superscript, the charge 1+ or 1- respectively is implied. For example, the cation of NaCl is Na+, while the anion of CaCO3 is CO32-.

Typically, metals form the cation and nonmetals the anion, but there are exceptions. In ammonium chloride (NH4Cl), for example, the cation is NH4+, which is completely made of nonmentals.

Examples
When neutral single atoms of Na and Cl come together, the sodium atom may transfer a negatively-charged electron to the chlorine (see animation below). This leaves a positive charge on the sodium, and a negative on the chlorine, creating Na+ and Cl−. These charged atoms are called ions, and they will be attracted towards each other, forming an ion pair of sodium chloride. This attraction of oppositely-charged ions is called an ionic bond.

When a molecule picks up or loses one or more electrons, the result is a polyatomic ion. CO32- is a polyatomic ion consisting of a central carbon atom bonded to three oxygen atoms. Its overall charge is -2, having two more electrons than the total of what the carbon and oxygen atoms would normally have.

Two ionic compounds can switch their cations and anions in a double replacement or metathesis reaction. For example, solutions of barium nitrate and sodium sulfate can be mixed to precipitate solid barium sulfate: Ba(NO3)2+Na2SO4 → BaSO4+2NaNO3.

A free element can replace another element or polyatomic ion as the cation or anion of an ionic compound in a single replacement reaction: 2Fe2O3+2Al→Al2O3+2Fe 2NaBr+Cl2→2NaCl+Br2 In either case, ions that do not change in the formation of new ionic compounds are spectator ions; the Na+ ion in the latter example remains unchanged except being ionically bonded to a different anions.

Properties of ionic compounds
Because of the strong binding of the crystal lattice of most ionic compounds, and well-packing, they tend to have very high boiling and melting points. Ionic compounds do not conduct electricity while dry, but when the ions are separated in solution, they conduct electricity well. Ionic compounds tend to dissolve in substances with polar covalent bonds, such as water. Most do not dissolve in non-polar compounds like (CH4). For what is meant by "polar" or "non-polar", see covalent bond.

Differing between ionic and covalent bonds
Although ionic and covalent bonds are generally considered separately, there exists substantial grey zone between them. Beryllium chloride (BeCl2), which one might expect to be ionic, actually has significant covalent character.

In general, although metals tend to form ionic bonds with most nonmetals, the electronegativity differences between the bonding elements is what determines the degree of covalency of a bond. Electronegativity, for our purposes here, can be defined loosely as the affinity of the atom for the electron pair in a bond with another atom. Fluorine is the most electronegative element, assigned a dimensionless electronegativity value of 4.0. Caesium is the least electronegative non-radioactive element (francium may be less electronegative but is highly radioactive), with a value of .7. The smaller the electronegativity difference, the greater likelihood of a covalent bond.

Likewise, there is only a certain degree in which metals, especially transition metals, can completely lose electrons. For example, ruthenium tetroxide (RuO4) is highly covalent, even though oxygen is much more electronegative than ruthenium, because even oxygen cannot completely take all 8 of the electrons ruthenium uses to bond. Lacking the electronegativity for this, oxygen must settle for double covalent bonds. On the other hand, ruthenium dioxide (RuO2) is, for the most part, ionic, as only 4 electrons have to be contributed to bonding by ruthenium.

title: Ionic bond


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susto

Noun

 * 1) An illness in Latin American cultures, a condition of fear and chronic somatic suffering stemming from one's own or others' emotional trauma.

Etymology 1
.

Noun

 * 1) fright, scare
 * 2) fear, terror

Etymology
. Several theories exist. Possibly an expressive creation, or from a verb, from (though this is unlikely), or perhaps from. Compare 🇨🇬.

Noun

 * 1) fright

source: wikimedia

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: susto


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ascolano

Adjective

 * 1)  of the Ascoli Piceno football/soccer team
 * 1)  of the Ascoli Piceno football/soccer team

Noun

 * 1) a variety of olive which originated in the Marche region of Italy
 * 1) a variety of olive which originated in the Marche region of Italy

title: ascolano


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Ziarat, Nishapur

Ziarat (, also Romanized as Zīārat and Zeyārat) is a village in Sarvelayat Rural District, Sarvelayat District, Nishapur County, Razavi Khorasan Province, Iran. At the 2006 census, its population was 82, in 20 families.

title: Ziarat, Nishapur


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Talk:Dave Wottle

Tendonitis prior to Olympic Games
Some unsourced claims appear in the Web that Dave Wottle had tendonitis which prevented him from preparing to the Games properly, and that might be a reason for his incredibly slow start in the 800 m race. A 1972 NYT article seems to confirm the health condition, but it doesn't explain or confirm a correlation to the race. I suppose we can't add info based on such a weak source...? --CiaPan (talk) 20:03, 4 December 2022 (UTC)
 * 1972 Olympic 800m Final (Hi Quality) https://www.youtube.com/watch?v=5LHid-nC45k
 * https://www.reddit.com/r/Damnthatsinteresting/comments/zcbwq6/dave_wottle/
 * As I recall from memory, without looking at sources, Wottle ran a similar come from behind race in the Olympic preliminaries to qualify for the final. I'd also consider the New York Times link as a strong source.Skilgis1900 (talk) 23:14, 5 December 2022 (UTC)

title: Talk:Dave Wottle


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Talk:David Hart (actor)

Untitled
David is a very good actor. It was great to see him and Alan working together again! — Preceding unsigned comment added by <IP_ADDRESS> (talk) 16:55, 27 April 2012 (UTC)

title: Talk:David Hart (actor)


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CACS/Authors

This Wikibook is a the result of a collaborative effort by a number or contributors. All of them are gratefully acknowledged, and thanked for their good work.

The only ones identified here are those who have chosen to share the blame for its shortcomings. You can contact any of them through their Wikibook user talk page or by eMail. --- Contributors in alphabetic order are:
 * LouI - just a start up of the work. (instigator, nag, and critic).

title: CACS/Authors


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Wikibooks:Please do not bite the newcomers

Newcomers are prospective members of the Wikibooks community and therefore a valuable resource. We must treat newcomers with kindness and patience—nothing scares potentially valuable contributors away faster than a hostile atmosphere. Newcomers, by definition, lack knowledge about the way we do things. We were all newcomers once, and most of us were enthusiastic, but lacked knowledge of how the project worked, and how to find that information. Many of us still consider ourselves newcomers after months (or years) of contributing.

Even if you never were a typical newcomer—perhaps you were just super careful to read up on everything before even making slight contributions—understand that we do encourage contributors (yes, also newcomers) to be bold and keep in mind the value of others in the community. Pride for your successful contributions to Wikibooks is hardly cause for looking down on enthusiastic newcomers.

Correcting mistakes
When newcomers make mistakes, try to correct them and politely point out some avenues for reading up on best common practices. Also consider leaving a welcome note on the newcomer's talk page to let them know that they are welcome here (there are some useful ones available on Templates/User messages).

New contributors are often hesitant to make changes, especially major ones, such as NPOVing and moving. Rather than being annoyed by their timidness, kindly encourage them to be bold in updating pages. Avoid hostile comments about the newcomer's ignorance. Be very careful with the wording of your edit summaries as these cannot be reworded later.

When you give advice to new contributors, tone the rhetoric down even a few notches from the usual mellow discourse that dominates Wikibooks. Try to make the newcomer feel genuinely welcome, not as though they must win your approval in order to be granted membership into an exclusive club.

Stepping back and mentoring
If you find these frustrating, don't correct them and just let it pass. Newcomers that stick around are usually the most diligent in hunting down all their early mistakes once they get to know the project. The best way to stop that is to drive them away before they learn.

Remember that writing a book is hard to do right. There are going to be a number of experiments that go way beyond even just learning how to edit. Newcomers need a mentor far more than a critic. If you see common mistakes being made, instead of complaining about recurring issues, consider supplementing or improving our documentation for future newcomers; help by making it more accessible and point newcomers to it.

Final thoughts
Stay on the safe side by always assuming that newcomers sincerely want to contribute positively (a fair assumption since most do). Just because they do not know much about our standard editing practices doesn't mean they aren't willing to learn. Give them a chance!


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Discrete Mathematics/Number theory

Introduction
'Number theory' is a large encompassing subject in its own right. Here we will examine the key concepts of number theory.

Unlike real analysis and calculus which deals with the dense set of real numbers, number theory examines mathematics in discrete sets, such as N or Z. If you are unsure about sets, you may wish to revisit ../Set theory/.

Number Theory, the study of the integers, is one of the oldest and richest branches of mathematics. Its basic concepts are those of divisibility, prime numbers, and integer solutions to equations -- all very simple to understand, but immediately giving rise to some of the best known theorems and biggest unsolved problems in mathematics. The Theory of Numbers is also a very interdisciplinary subject. Ideas from combinatorics (the study of counting), algebra, and complex analysis all find their way in, and eventually become essential for understanding parts of number theory. Indeed, the greatest open problem in all mathematics, the Riemann Hypothesis, is deeply tied into Complex Analysis. But never fear, just start right into Elementary Number Theory, one of the warmest invitations to pure mathematics, and one of the most surprising areas of applied mathematics!

Divisibility
Note that in R, Q, and C, we can divide freely, except by zero. This property is often known as closure -- the quotient of two rationals is again a rational, etc.. However, if we move to performing mathematics purely in a set such as Z, we come into difficulty. This is because, in the integers, the result of a division of two integers might not be another integer. For example, we can of course divide 6 by 2 to get 3, but we cannot divide 6 by 5, because the fraction 6/5 is not in the set of integers.

However we can introduce a new relation where division is defined. We call this relation divisibility, and if $$\frac{b}{a}$$ is an integer, we say:
 * $$a$$ divides $$b$$
 * $$a$$ is a factor of $$b$$
 * $$b$$ is a multiple of $$a$$
 * $$b$$ is divisible by $$a$$

Formally, if there exists an integer $$q$$ such that $$b = qa$$ then we say that $$a$$ divides $$b$$ and write $$a \mid b$$. If $$a$$ does not divide $$b$$ then we write $$a \nmid b$$:

Proposition. The following are basic consequences of this definition. Let a, b, and c be integers:
 * (a) If a|b then a|(bc).
 * (b) If a|b and b|c, then a|c.
 * (c) If a|b and a|c, then for any integers x and y, a|(xb+yc) -- in other words a divides any linear combination of its multiples.
 * (d) If both a|b and b|a, then a = b or a = -b.
 * (e) If c is not 0, then a|b is equivalent to ca|cb.

Quotient and divisor theorem
For any integer n and any k > 0, there is a unique q and r such that:

 * n = qk + r (with 0 ≤ r < k)

Here n is known as dividend.

We call q the quotient, r the remainder, and k the divisor.

It is probably easier to recognize this as division by the algebraic re-arrangement:

 * n/k = q + r/k (0 ≤ r/k < 1)

Modular arithmetic
What can we say about the numbers that divide another? Pick the number 8 for example. What is the remainder on dividing 1 by 8? Using the division theorem above

 * 0 = 8*0 + 0
 * 1 = 8*0 + 1
 * 2 = 8*0 + 2
 * 8 = 8*1 + 0
 * 9 = 8*1 + 1
 * 10 = 8 * 1 + 2
 * and so on
 * and so on
 * and so on

We have a notation for the remainders, and can write the above equations as

 * 0 mod 8 = 0
 * 1 mod 8 = 1
 * 2 mod 8 = 2
 * 3 mod 8 = 3
 * 4 mod 8 = 4
 * 5 mod 8 = 5
 * 6 mod 8 = 6
 * 7 mod 8 = 7
 * 8 mod 8 = 0
 * 9 mod 8 = 1
 * 10 mod 8 = 2

We can also write
 * 1 ≡ 1 (mod 8)
 * 2 ≡ 2 (mod 8)
 * 3 ≡ 3 (mod 8)
 * 4 ≡ 4 (mod 8)
 * 5 ≡ 5 (mod 8)
 * 6 ≡ 6 (mod 8)
 * 7 ≡ 7 (mod 8)
 * 8 ≡ 0 (mod 8)
 * 9 ≡ 1 (mod 8)
 * 10 ≡ 2 (mod 8)

These notations are all short for

 * a = 8k+r for some integer k.

So x ≡ 1 (mod 8), for example, is the same as saying

 * x = 8k+1

Observe that the remainder here, in comparing it with the division algorithm is 1. x ≡ 1 (mod 8) asks what numbers have the remainder 1 on division by 8? Clearly the solutions are x=8×0+1, 8×1+1,... = 1, 9, ...

Often the entire set of remainders on dividing by n - which we say modulo n - are interesting to look at. We write this set Zn. Note that this set is finite. The remainder on dividing 9 by 8 is 1 - the same as dividing 1 by 8. So in a sense 9 is really "the same as" 1. In fact, the relation "≡"

 * x ≡ y iff x mod n = y mod n.

is an equivalence relation. We call this relation congruence. Note that the equivalence classes defined by congruence are precisely the elements of Zn.

We can find some number a modulo n (or we say a congruent to n) by finding its decomposition using the division algorithm.

Addition, subtraction, and multiplication work in Zn - for example 3 + 6 (mod 8) = 9 (mod 8) = 1 (mod 8). The numbers do look strange but they follow many normal properties such as commutativity and associativity.

If we have a number greater than n we often reduce it modulo n first - again using the division algorithm. For example if we want to find 11+3 mod 8, its often easier to calculate 3 + 3 (mod 8) rather than reducing 14 mod 8. A trick that's often used is that, say, if we have 6 + 7 (mod 8) we can use negative numbers instead so the problem becomes -2 + -1 = -3 = 5 (mod 8).

We often use the second notation when we want to look at equations involving numbers modulo some n. For example, we may want to find a number x such that
 * 3x ≡ 5 (mod 8)

We can find solutions by trial substitution (going through all the numbers 0 through 7), but what if the moduli are very large? We will look at a more systematic solution later.

Note: we often say that we are working in Zn and use equals signs throughout. Familiarize yourself with the three ways of writing modular equations and expressions.

The Consistency of Modular Arithmetic
Let $$n \geq 2$$ denote an arbitrary base. Given an arbitrary integer $$x$$, the sequence of integers $$..., x-2n, x-n, x, x+n, x+2n, ...$$ are all congruent to each other modulo $$n$$:

$$ ... \equiv x-2n \equiv x-n \equiv x \equiv x+n \equiv x+2n \equiv ... \;\;(\text{mod}\; n) $$

In modular arithmetic, two integers $$x$$ and $$y$$ that are congruent modulo $$n$$ ($$x \equiv y \;\;(\text{mod}\; n)$$) both "represent" the same quantity from $$\Z_n$$. It should be possible to substitute an arbitrary integer $$x$$ in place of integer $$y$$ provided that $$x \equiv y \;\;(\text{mod}\; n)$$.

This means that:
 * Given arbitrary integers $$x_1, x_2$$ and $$y_1, y_2$$, if $$x_1 \equiv x_2 \;\;(\text{mod}\; n)$$ and $$y_1 \equiv y_2 \;\;(\text{mod}\; n)$$, then $$x_1 + y_1 \equiv x_2 + y_2 \;\;(\text{mod}\; n)$$.

 * Given arbitrary integers $$x_1, x_2$$ and $$y_1, y_2$$, if $$x_1 \equiv x_2 \;\;(\text{mod}\; n)$$ and $$y_1 \equiv y_2 \;\;(\text{mod}\; n)$$, then $$x_1y_1 \equiv x_2y_2 \;\;(\text{mod}\; n)$$.

Number Bases
Converting between various number bases is one of the most tedious processes in mathematics.

The numbers that are generally used in transactions are all in base-10. This means that there are 10 digits that are used to describe a number. These ten digits are {0,1,2,3,4,5,6,7,8,9}.

Similarly, base-4 has 4 digits {0,1,2,3} and base-2 has two digits {0,1}. Base two is sometimes referred to as Binary.

There are also bases greater than 10. For these bases, it is customary to use letters to represent digits greater than 10. An example is Base-16 (Hexadecimal). The digits used in this base are {0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F}.

In order to convert between number bases, it is critical that one knows how to divide numbers and find remainders.

To convert from decimal to another base one must simply start dividing by the value of the other base, then dividing the result of the first division and overlooking the remainder, and so on until the base is larger than the result (so the result of the division would be a zero). Then the number in the desired base is the remainders read from end to start.

The following shows how to convert a number (105) which is in base-10 into base-2.

Answer : 1101001

After finishing this process, the remainders are taken and placed in a row (from bottom to top) after the final quotient (1101001, in this example) is shown as the base-2 equivalent of the number 105.

To sum up the process, simply take the original number in base 10, and divide that number repeatedly, keeping track of the remainders, until the quotient becomes less than the numerical value of the base.

This works when converting any number from base-10 to any base. If there are any letters in the base digits, then use the letters to replace any remainder greater than 9. For example, writing 11(of base-10) in base 14.

Answer: B

As 11 is a single remainder, it is written as a single digit. Following the pattern {10=A, 11=B, 12=C...35=Z}, write it as B. If you were to write "11" as the answer, it would be wrong, as "11" Base-14 is equal to 15 in base-10!

In order to convert from a number in any base back to base ten, the following process should be used:

Take the number 3210 (in base-10). In the units place (100), there is a 0. In the tens place (101), there is a 1. In the hundreds place (102), there is a 2. In the thousands place (103), there is a 3.

The formula to find the value of the above number is:

3×103 + 2×102 + 1×101 + 0×100 = 3000 + 200 + 10 + 0 = 3210.

The process is similar when converting from any base to base-10. For example, take the number 3452 (in base-6). In the units place (60), there is a 2. In the sixths place (61) there is a 5. In the thirty-sixths place (62), there is a 4. In the 216th place (63), there is a 3.

The formula to find the value of the above number (in base-10) is:

3×63 + 4×62 + 5×61 + 2×60 = 648 + 144 + 30 + 2 = 824.

The value of 3452 (base-6) is 824 in base-10.

A more efficient algorithm is to add the left-most digit and multiply by the base, and repeat with the next digit and so on.

((3 * 6 + 4) * 6 + 5) * 6 + 2 = 824

The processes to convert between number bases may seem difficult at first, but become easy if one practices often.

Prime numbers
Prime numbers are the building blocks of the integers. A prime number is a positive integer greater than one that has only two divisors: 1, and the number itself. For example, 17 is prime because the only positive integers that divide evenly into it are 1 and 17. The number 6 is not a prime since more than two divisors 1, 2, 3, 6 divide 6. Also, note that 1 is not a prime since 1 has only one divisor.

Some prime numbers
The prime numbers as a sequence begin
 * 2, 3, 5, 7, 11, 13, 17, 19, 23, ...

Euclid's Proof that There are Infinitely Many Primes
The Greek mathematician Euclid gave the following elegant proof that there are an infinite number of primes. It relies on the fact that all non-prime numbers --- composites --- have a unique factorization into primes.

Euclid's proof works by contradiction: we will assume that there are a finite number of primes, and show that we can derive a logically contradictory fact.

So here we go. First, we assume that that there are a finite number of primes:

 * p1, p2, ..., pn

Now consider the number M defined as follows:

 * M = 1 + p1 *  p2 * ... *  pn

There are two important --- and ultimately contradictory --- facts about the number M:

 * 1) It cannot be prime because pn is the biggest prime (by our initial assumption), and M is clearly bigger than pn. Thus, there must be some prime p that divides M.
 * 2) It is not divisible by any of the numbers p1, p2, ..., pn.   Consider what would happen if you tried to divide M by any of the primes in the list p1, p2, ..., pn.  From the definition of M, you can tell that you would end up with a remainder of 1. That means that p --- the prime that divides M --- must be bigger than any of p1, ..., pn.

Thus, we have shown that M is divisible by a prime p that is not on the finite list of all prime. And so there must be an infinite number of primes.

These two facts imply that M must be divisible by a prime number bigger than pn. Thus, there cannot be a biggest prime.

Note that this proof does not provide us with a direct way to generate arbitrarily large primes, although it always generates a number which is divisible by a new prime. Suppose we know only one prime: 2. So, our list of primes is simply p1=2. Then, in the notation of the proof, M=1+2=3. We note that M is prime, so we add 3 to the list. Now, M = 1 +2 *3 = 7. Again, 7 is prime. So we add it to the list. Now, M = 1+2*3*7 = 43: again prime. Continuing in this way one more time, we calculate M = 1+2*3*7*43 = 1807 =13*139. So we see that M is not prime.

Viewed another way: note that while 1+2, 1+2*3, 1+2*3*5, 1+2*3*5*7, and 1+2*3*5*7*11 are prime, 1+2*3*5*7*11*13=30031=59*509 is not.

Testing for primality
There are a number of simple and sophisticated primality tests. We will consider some simple tests here. In upper-level courses we will consider some faster and more sophisticated methods to test whether a number is prime.

Inspection
The most immediate and simple test to eliminate a number n as a prime is to inspect the units digit or the last digit of a number.

If the number n ends in an even number 0, 2, 4, 6, 8 we can show that number n cannot be a prime. For example, take n = 87536 = 8753(10) + 6. Since 10 is divisible by 2 and 6 is divisible by 2 then 87536 must be divisible by 2. In general, any even number can be expressed in the form n = a*10 + b, where b = 0, 2, 4, 6, 8. Since 10 is divisible by 2 and b is divisible by 2 then n = a*10 + b is divisible by 2. Consequently, any number n which ends in an even number such as 7777732 or 8896 is divisible by 2 so n is not a prime.

In a similar type of argument, if a number n ends in a 5 we can show the number n cannot be a prime. If the last digit of n, call it b, is a 5 we can express n in the form n = a*10 + b, where b = 5. Since 10 is divisible by 5 and b = 5 is divisible by 5 then n = a*10 + b is divisible by 5. Hence, any number n which ends in a 5 such as 93475 is divisible by 5 so n is not a prime.

Thus, if a number greater than 5 is a prime it must end with either a 1, 3, 7, or 9. Note that this does not mean all numbers that end in a 1, 3, 7, or 9 are primes. For example, while the numbers 11, 23, 37, 59 are primes, the numbers 21 = 3*7, 33 = 3*11, 27 = 3*9, 39 = 3*13 are not primes. Consequently, if a number ends in a 1, 3, 7, or 9 we have to test further.

Trial Division Method
To test if a number n that ends in a 1, 3, 7, or 9 is prime, we could simply try the smallest prime number and try to divide it in n. If that doesn't divide, we would take the next largest prime number and try again etc. Certainly, if we took all primes numbers in this manner that were less than n and we could not divide n then we would be justified in saying n is prime. However, it can be shown that you don't have to take all primes smaller than n to test if n is prime. We can stop earlier by using the Trial Division Method.

The justification of the Trial Division Method is if a number n has no divisors less than or equal to $$\sqrt{n}$$ then n must be a prime. We can show this by contradiction. Let us assume n has no divisors less than or equal to $$\sqrt{n}$$. If n is not a prime, there must be two numbers a and b such that $$a*b = n$$. In particular, by our assumption $$\sqrt{n} < a $$ and $$\sqrt{n} < b $$. But then $$n = \sqrt{n}\sqrt{n} < a*b = n$$. Since a number can not be greater than itself the number n must be a prime.

Trial Division Method is a method of primality testing that involves taking a number n and then sequentially dividing it by primes up to $$\sqrt{n}$$.

For example, is 113 prime? $$\sqrt{113}$$ is approximately 10.63... We only need to test whether 2, 3, 5, 7 divide 113 cleanly (leave no remainder, i.e., the quotient is an integer).
 * 113/2 is not an integer since the last digit is not even.
 * 113/3 (=37.666...) is not an integer.
 * 113/5 is not an integer since the last digit does not end in a 0 or 5.
 * 113/7 (=16.142857...) is not an integer.

So we need not look at any more primes such as 11, 13, 17 etc. less than 113 to test, since 2, 3, 5, 7 does not divide 113 cleanly, 113 is prime.

Notice that after rejecting 2 and 3 as a divisor, we next considered the next prime number 5 and not the next number 4. We know not to consider 4 because we know 2 does not divide 113. If 2 cannot divide 113 then certainly 4 cannot because if 4 divided 113 and since 2 divides 4 then 2 would divide 113. So we only use the next cheapest available prime to test not the next consecutive number.

If we test 91 we get,

 * 91/2 is not an integer since that last digit is not even.
 * 91/3= (30.333) is not an integer.
 * 91/5= is not an integer since the last digit does not end in a 0 or 5.
 * 91/7=13 is an integer

So we know since 7 divides 91, 91 is not a prime.

Trial division is normally only used for relatively small numbers due to its inefficiency. However this technique has the two advantages that firstly once we have tested a number we know for sure that it is prime and secondly if a number is not prime it also gives us the number's factors.

To obtain a few small primes, it may be best to use the Sieve of Eratosthenes than to test each number sequentially using trial division. The Sieve of Eratosthenes method is basically a process of finding primes by elimination. We start by taking a list of consecutive numbers say 1 to 100. Cross out the number 1 because the number is not prime. Take the next least uncrossed off number which is 2 and circle it. Now cross out all multiples of 2 on the list. Next take the next least uncircled number which is 3. Circle the number 3 and cross out all multiples of 3. The next least uncircled number should be 5 since 4 is a multiple of 2 and should have been crossed off. Circle the number 5 and cross out all multiples of 5. The next least uncircled number should be a 7 since 6 is a multiple of 2. Circle the 7 and mark off all multiples of 7. Now the next uncrossed off number should be 11 since 8,9,10 is a multiple of 2, 3, and 2. If we continue in this manner what is left is the circled numbers which are primes. But notice we can actually stop now and circle all the unmarked numbers after crossing off multiples of 7 because of the result that since $$\sqrt{100} = 10$$ any number less than 100 which is not prime must be divisible by 2, 3, 5, or 7.

The Fundamental Theorem of Arithmetic
The Fundamental Theorem of Arithmetic is an important theorem relating to the factorization of numbers. The theorem basically states that every positive integer can be written as the product of prime numbers in a unique way (ignoring reordering the product of prime numbers).

In particular, The Fundamental Theorem of Arithmetic means any number such as 1,943,032,663 is either a prime or can be factored into a product of primes. If a number such as 1,943,032,663 can be factored into primes such as 11×13×17×19×23×31×59 it is futile to try to find another different combination of prime numbers that will also give you the same number.

To make the theorem work even for the number 1, we think of 1 as being the product of zero prime numbers.

More formally,
 * For all n∈N
 * n=p1p2p3...
 * where the pi are all prime numbers, and can be repeated.

Here are some examples.
 * 4 = 2 × 2 = 22
 * 12 = 2 × 2 × 3 = 22 × 3
 * 11 = 11.

A proof of the Fundamental Theorem of Arithmetic will be given after Bezout's identity has been established.

LCM and GCD
Two characteristics we can determine between two numbers based on their factorizations are the lowest common multiple, the LCM and greatest common divisor, the GCD (also greatest common factor, GCF)

LCM
The lowest common multiple, or the least common multiple, for two numbers a and b is the smallest number designated by LCM(a,b) that is divisible by both the number a and the number b. We can find LCM(a,b) by finding the prime factorization of a and b and choosing the maximum power for each prime factor.

In another words, if the number a factors to $$p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_n^{\alpha_n}$$, and the number b factors to $$p_1^{\beta_1}p_2^{\beta_2}\cdots p_n^{\beta_n}$$, then LCM(a,b) = $$p_1^{\gamma_1}p_2^{\gamma_2}\cdots p_n^{\gamma_n}$$ where $$\gamma_i = Maximum (\alpha_i, \beta_i)$$ for i = 1 to n.

An example, let us see the process on how we find lowest common multiple for 5500 and 450 which happens to be 49500. First, we find the prime factorization for 5500 and 450 which is
 * 5500=22 53 11
 * 450=2 32 52

Notice the different primes we came up for both the number 5500 and the number 450 are 2, 3, 5, and 11. Now let us express 5500 and 450 completely in a product of these primes raised to the appropriate power.
 * 5500=22 53 11 = 22 30 53 111
 * 450=2 32 52 = 21 32 52 110

The LCM(5500,450) is going to be in the form 2? 3? 5? 11?. All we now have to do is find what the powers of each individual prime will be.

So now we compare the power of each prime for 5500 and 450. Let us consider the powers of the first prime 2. In the number 5500, the prime 2 is raised to the second power and in the number 450, prime 2 is raised to the first power. Since the maximum between 2 and 1 for the power of the prime 2 is 2, we use 2 for the power of the prime 2.

Now let us consider the powers of the prime 3. In the number 5500, the prime 3 is raised to the zero power and in the number 450 the prime 3 is raised to the second power. Since the maximum between 0 and 2 for the power of the prime 3 is 2, we use 2 for the power of the prime 3.

Similarly, let us consider the powers of the next prime 5. In the number 5500, the prime 5 is raised to the third power and in the number 450 the prime 5 is raised to the second power. Since the maximum between 3 and 2 for the power of the prime 5 is 3, we use 3 for the power of the prime 5.

Finally, let us consider the powers of the prime 11, the last prime on our list. In the number 5500, the prime 11 is raised to the first power and in the number 450 the prime 11 is raised to the zero power. Since the maximum between 1 and 0 for the power of the prime 11 is 1, we use 1 for the power of the last prime 11.

Consequently, the product of our results is LCM(5500,450)=22 32 53 111 = 49500.

GCD
The greatest common divisor for two numbers a and b is the biggest number designated by GCD(a,b) that divides both the number a and the number b. In a similar process to finding LCM(a,b), we can find GCD(a,b) by finding the prime factorization of a and b but choosing the minimum power for each prime factor instead.

In other words, if the number a factors to $$p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_n^{\alpha_n}$$, and the number b factors to $$p_1^{\beta_1}p_2^{\beta_2}\cdots p_n^{\beta_n}$$, then GCD(a,b) = $$p_1^{\gamma_1}p_2^{\gamma_2}\cdots p_n^{\gamma_n}$$ where $$\gamma_i = Minimum (\alpha_i, \beta_i)$$ for i = 1 to n.

An example, let us see the process on how we find the greatest common divisor for 5500 and 450 which happens to be 50. First, we find the prime factorization for 5500 and 450 which is
 * 5500=22 53 11
 * 450=2 32 52

Notice the different primes we came up for both the number 5500 and the number 450 are 2, 3, 5, and 11. Now let us express 5500 and 450 completely in a product of these primes raised to the appropriate power.
 * 5500=22 53 11 = 22 30 53 111
 * 450=2 32 52 = 21 32 52 110

The GCD(5500,450) is going to be in the form 2? 3? 5? 11?. All we now have to do is find what the powers of each individual prime will be.

So now we compare the power of each prime for 5500 and 450. Let us consider the powers of the first prime 2. In the number 5500, the prime 2 is raised to the second power and in the number 4 50, prime 2 is raised to the first power. Since the minimum between 2 and 1 for the power of the prime 2 is 1, we use 1 for the power of the prime 2.

Now let us consider the powers of the prime 3. In the number 5500, the prime 3 is raised to the zero power and in the number 450 the prime 3 is raised to the second power. Since the minimum between 0 and 2 for the power of the prime 3 is 0, we use 0 for the power of the prime 3.

Similarly, let us consider the powers of the next prime 5. In the number 5500, the prime 5 is raised to the third power and in the number 450 the prime 5 is raised to the second power. Since the minimum between 3 and 2 for the power of the prime 5 is 2, we use 2 for the power of the prime 5.

Finally, let us consider the powers of the prime 11, the last prime on our list. In the number 5500, the prime 11 is raised to the first power and in the number 450 the prime 11 is raised to the zero power. Since the minimum between 1 and 0 for the power of the prime 11 is 0, we use 0 for the power of the last prime 11.

Consequently, the product of our results is GCD(5500,450)=21 30 52 110 = 50.

Properties

 * gcd(a,b)=gcd(b,a)
 * gcd(a,b) = gcd(b,q), where q is the remainder of a divided by b
 * gcd(a/d, b/d)=1, where d is gcd(a,b)

The Euclidean algorithm
The Euclidean algorithm is such that we can find the gcd of two numbers without finding the factorization*. The Euclidean algorithm consists of only addition and multiplication, and uses the above properties of gcd as its basis.

 * Factorization is a "hard" process, that is, to factor a number takes a long time depending on the length of the number. This is why later, when you need to find the gcd of a pair of numbers, you will most likely never factorize the numbers and use the properties of the primes but will use the Euclidean algorithm instead.

An example
We will see how this works by calculating gcd(458,44)

First, divide 458 by 44 and obtain the remainder:
 * 458 = 44 × 10 + 18

Now suppose that a number is a common divisor of 458 and 44. Then it must also be a divisor of 18. To see this, rearrange the above equation to:
 * 458 - 44×10 =18

When this equation is divided by a common divisor of 44 and 458, an integer is obtained on the left, and so must also be obtained on the right. This, by definition, means that the number is also a divisor of 18. By the same reasoning, any common divisor of 18 and 44 is also a divisor of 458. Since all of the common divisors of 458 and 44 are equal to common divisors of 44 and 18, then in particular the greatest common divisors are equal. So we have gcd(458,44)=gcd(44,18)

The next step in the algorithm is to divide 44 by 18 and find the remainder.
 * 44 = 18 × k + r
 * 44 = 18 × 2 + 8

Repeat this process; keep dividing the previous divisor by the previous remainder:
 * 18 = 8 × 2 + 2
 * 8 = 2 × 4 + 0

Our gcd is the last remainder before the zero, in this case, 2. This is because the reasoning that proved gcd(458,44)=gcd(44,18) applies at every step, so gcd(458,44)=gcd(44,18)=gcd(18,8)=gcd(8,2)=gcd(2,0)=2.

The Matrix Method
We can construct a matrix that provides an alternative method for calculating the greatest common divisor. In its general form, the matrix is

$$ \begin{bmatrix} 1 & 0 & b \\ 0 & 1 & a \\ \end{bmatrix} $$

Recall that one way to write the gcd of two numbers is as an integral linear combination. If we are finding the gcd, for example, we could represent it as as + bt, where a and b are the two numbers we are comparing, and s and t are integers. We also know that b = aq + r where r is the remainder upon division of b by a. After we subtract row 2 from row 1, we get

$$ \begin{bmatrix} 1 & -q_{1} & r_{1} \\ 0 & 1 & a \\ \end{bmatrix} $$

If r_2 is nonzero, we must continue the process; this time, subtracting row 1 from row 2. We continue this process until one of the  r's  has been reduced as far as possible. We now have our gcd. The numbers that are in that row, where the 1 and the 0 used to be, represent t and s, respectively.

Let us now look at a computational example.

$$ \begin{bmatrix} 1 & 0 & 99 \\  0 & 1 & 7 \\ \end{bmatrix} \rightsquigarrow \begin{bmatrix} 1 & 0 & 99 \\  0 & 14 & 98 \\   \end{bmatrix} \rightsquigarrow \begin{bmatrix} 1 & -14 & 1 \\  0 & 14 & 98 \\   \end{bmatrix} $$

We see that it would be trivial at this point to go any further; we would just end up with row-2 containing a zero where a used to be. So we look at row-1 and remember that the 1 represents our remainder, 1(=t) multiplies b and -14(=s) multiplies a such that

$$1 = 99 \times 1 + -14 \times 7 $$

This can be checked by the Euclidean algorithm that gcd(7,99)=1.

The extended Euclidean algorithm
What happens if we try and reverse the process of the Euclidean algorithm by substituting back? Back-substitution is rather tedious and prone to error, so here is a faster method.

Draw up a table with four columns, label these from left to right q, r, u, v. For convenience label a column i representing the step we're currently up to. Place a and b with the greater of these on top in the column r, and place 1s and 0s accordingly:

$$ \begin{matrix} i & q & r & u & v \\ -1 & . & b & 0 & 1\\ 0 & . & a & 1 & 0 \\ \end{matrix} $$

Now iterate downwards by taking the quotient of b/a and putting it in the next space in the q column, then of b-aq in the r column.

To update u and v, take
 * ui = ui-2-ui-1qi
 * vi = vi-2-vi-1qi

Indeed, you are looking for u and v such that au + bv = gcd (a,b). At some point, gcd (a,b) is in fact the remainder at the ith stage, so you might as well compute ui and vi such that aui + bvi = ri, at EACH stage.

Deriving the recurrences found above results from these three equations (the second equation is Euclid's algorithm's basic property, the other two are constraints we set to attain our desired goal):

 * aui-1 + bvi-1 = ri-1
 * ri-2 = qiri-1 + ri
 * aui + bvi = ri

The trick is to then appropriately express ri-2.

Stop writing when you obtain a 0 in the r column.

Finding then, gcd(450,44) (this is the same as gcd(44,450) )

$$ \begin{matrix} i & q  & r   & u  & v \\ -1 & . & 450 & 0  & 1 \\ 0  & .  & 44  & 1  & 0 \\ 1  & 10 & 10  & -10  & 1 \\ 2  & 4  & 4   & 41 & -4 \\ 3  & 2  & \mathbf{2}   & -92  & 9 \\ 4 & 2  & 0   & -  & - \\ \end{matrix} $$

The bold number is the gcd. Observe (9)×450+(-92)×44=2 Clearly these u and v are very special. What can we say about the general case?

Bezout's identity
In the above case we have 9×450+(-92)×44=gcd(450,44). So the greatest common divisor of 450 and 44 can be written as a linear combination of 450 and 44 with integer coefficients. This turns out to be true of any pair of integers. This result is known as "Bezout's Identity" and can be stated as follows:
 * For any pair of nonzero integers, a and b, there exists integers u and v such that
 * au+bv=gcd(a,b)

Proof
 * If a and b are any pair of integers, not both 0, then clearly there exist integers u and v such that au+bv is positive (just match the signs of u and v to those of a and b, respectively, for instance), and for all integer u and v, au+bv is also an integer (because the integers are closed under addition and multiplication). So there is a non-empty set of positive integers consisting of numbers of the form au+bv; call it S.  Since S is a non-empty set of positive integers, it must have a smallest element (this is the so called  Well-ordering principle).  Call this number d.  The claim is that $$d=gcd(a,b)$$.  Since d belongs to S, then

 * (1)$$d=ua + vb$$

 * for suitable u and v. Let n be any positive common divisor of a and b.  Since n divides the right side of (1), it also divides d.  So $$d=qn$$ for some integer $$q\geq 1$$.  So any common divisor of a and b is less than or equal to d.  Therefore, if d is a common divisor of a and b, it is the greatest one.  It only remains to show that d is in fact a common divisor.

 * To prove this, it will be shown that any element of S is divisible by d. This will prove that d is a common divisor of a and b because a and b are both elements of S  (because a = 1×a + 0×b and b = 0×a + 1×b). Let t be any element of S.  Then, by the division algorithm
 * $$t=qd+r$$

 * for some $$0\leq r<d$$.  If r is not 0, then it is in S.  This is because d and t are in S, so
 * $$r=t-qd=(u_1a+v_1b)-q(u_2a+v_2b)=(u_1-qu_2)a+(v_1-qv_2)b=u'a+v'b$$
 * But, since $$r<d$$ and d is the least element of S, this is impossible. The only other possibility is that r=0.  Therefore any element, t, of S is divisible by d.  Since this includes both a and b, d is a common divisor.  Combining this with the previous result establishes Bezout's Identity.

The numbers u and v can either be obtained using the tabular methods or back-substitution in the Euclidean Algorithm.

Proof of the Fundamental Theorem of Arithmetic
One use of Bezout's identity is in a proof of the Fundamental Theorem of Arithmetic. Before this is proven, two other results are needed: Lemma 1: If a prime number, p, divides a product of two integers, $$ab$$, then it must divide a or b (or both).
 * Proof: If p divides both a and b, there is nothing to prove. Assume p does not divide a.  If it can be proven under that assumption that p does divide b, the lemma will be proven.
 * Since p does not divide a, then gcd(a,p)=1 (because the only divisors of p are 1 and p, but p is not a common divisor). Therefore, by Bezout's Identity, there exist integers u and v such that
 * $$1=ua+vp$$
 * Multiply this equation by b to obtain:
 * $$b=uab+vbp$$
 * p divides both terms on the right hand side and, therefore, divides the left hand side. Hence, p divides b, as was to be shown.

Lemma 2: If a prime number, p, divides a product of integers, $$a_1a_2...a_n$$, then it must divide at least one of the factors.
 * Proof: The proof is by induction on n, the number of factors. The statement is true for n=2, by Lemma 1.  Assume the statement is true for n=k and consider a product of k+1 factors.  If p divides more than one of the factors, once again there is nothing to prove.  Assume that p does not divide any of the factors $$a_1, a_2, ... a_k$$.  It will be shown that p must divide $$a_{k+1}$$.  Since the statement is true for n=k, then since p does not divide any of the factors in $$a_1a_2...a_k$$, it must not divide the product (by Contrapositive).  Let $$a_1a_2...a_k=b$$.  Then $$a_1a_2...a_ka_{k+1}=ba_{k+1}$$.  The conclusion then follows by Lemma 1.

Fundamental Theorem of Arithmetic: Any positive integer, n, can be expressed as a product of primes. This product is unique up to the order in which the terms appear.
 * Proof: The proof of the first part of the theorem is by induction on n.  If n=1, it is the product of 0 primes.  Assume all positive integers less than n can be expressed as a product of primes.  If n is prime, then it is the product of 1 prime.  Assume n is not prime or 1.  Then $$n=ab$$,for some positive integers a and b both less than n.  Since a and b are both less than n, they can be written as a product of primes by the induciton hypothesis.  Combining these products gives n as a product of primes as required.
 * Now to prove the second part. Assume there are two prime factorizations of n,
 * $$n=p_1p_2...p_k=q_1q_2...q_s$$
 * $$p_1$$ divides the left side and so must also divide the right side. By Lemma 2, this means that $$p_1$$ must divide one of the $$q_i$$.  But these are all prime, so the only way $$p_1$$ can divide $$q_i$$ is if $$p_1=q_i$$ for some i.  Canceling $$p_1$$ from both sides of the equation forms another equation of the same form.  So it can likewise be proven that $$p_2=q_i$$ for some other i, and so on until all the factors on the left are exhausted.  After this, there must not be any factors remaining on the right side since it must equal 1.  This proves that any two prime factorizations consist of the same prime factors, as was to be shown.

Partitioning the Divisors of Products
The Fundamental Theorem of Arithmetic can also be derived from the following lemma:

Lemma: Given integers $$a$$, $$b$$, and $$c$$, if $$c$$ divides $$ab$$ (denoted by $$c|ab$$), then there exist integers $$m$$ and $$n$$ such that $$c = mn$$ and $$m|a$$ and $$n|b$$.

In other words, an integer that divides a product can itself be factored into a product where each factor divides the corresponding factor from $$ab$$. This means that no new primes are "created" when $$a$$ and $$b$$ are multiplied together.

This Lemma follows from the Fundamental Theorem of Arithmetic and Bezout's identity, but here a more direct proof will be given.

Proof: If any of $$a$$, $$b$$, or $$c$$ is $$0$$, then the Lemma is trivial. In addition, if any of $$a$$, $$b$$, or $$c$$ is negative, then if the Lemma holds for the absolute values $$|a|$$, $$|b|$$, and $$|c|$$, then it is trivial to show that the Lemma holds for $$a$$, $$b$$, and $$c$$. It will now be assumed that $$a$$, $$b$$, and $$c$$ are all strictly positive integers.

Form an $$a \times b$$ array $$S$$ of integers that has $$a$$ columns and $$b$$ rows. $$s_{(x,y)}$$ will denote the integer at column $$x$$ and row $$y$$. Fill the array by sweeping the array row by row starting with row 1, with each row swept starting from column 1. During this "raster" sweep of $$S$$, assign values to $$s_{(x,y)}$$ using the following cyclical pattern: $$1, 2, ..., c$$. In essence, $$s_{(x,y)}$$ is the unique integer from the range $$\{1, 2, \dots, c\}$$ such that $$s_{(x,y)} \equiv x + a(y-1) \;\;(\text{mod}\; c)$$. Since $$c|ab$$, it is the case that $$s_{(1,1)} = 1$$ and $$s_{(a,b)} = c$$.

As previously indicated, the "raster sweep" through array $$S$$ is a cyclical progression through the entries of $$S$$ where column index $$x$$ cycles around $$1, 2, ..., a$$, and every time $$x$$ transitions from $$a$$ to $$1$$, row index $$y$$ advances by one step around the cycle $$1, 2, ..., b$$.

In the image below, the grid $$S$$ where $$a = 14$$; $$b = 15$$; and $$c = 6$$ is depicted both explicitly and using a brickwork pattern.

Array $$S$$ can be endlessly replicated and used to form the infinite array $$S_\infty$$. For arbitrary integers $$u$$ and $$v$$, the block of entries in $$S_\infty$$ formed by columns $$a(u-1)+1$$ to $$a(u-1)+a$$ and rows $$(b(v-1)-(u-1))+1$$ to $$(b(v-1)-(u-1))+b$$ is a copy of $$S$$. For arbitrary integers $$x$$ and $$y$$, the entry $$s_{\infty,(x,y)}$$ of $$S_\infty$$ is the unique integer from the range $$\{1, 2, \dots, c\}$$ such that $$s_{\infty,(x,y)} \equiv x + a(y-1) \;\;(\text{mod}\; c)$$.

Given any column $$x$$ and row $$y$$, the entry $$s_{\infty,(x,y)}$$ of $$S_\infty$$ located at $$(x,y)$$ is the unique integer from $$\{1, 2, ..., c\}$$ such that $$s_{\infty,(x,y)} \equiv x + a(y-1) \;\;(\text{mod}\; c)$$. Given an arbitrary displacement column displacement $$\Delta x$$ and row displacement $$\Delta y$$, the difference $$s_{\infty,(x+\Delta x, y+\Delta y)}-s_{\infty,(x,y)}$$ is separated from $$\Delta x + a\Delta y$$ by a multiple of $$c$$. This gives the entries of $$S_\infty$$ the following symmetries:
 * Given column displacement $$\Delta x$$ and columns $$x_1, x_2$$, and row displacement $$\Delta y$$ and rows $$y_1, y_2$$, then $$s_{\infty,(x_1+\Delta x, y_1+\Delta y)}-s_{\infty,(x_1,y_1)} \equiv s_{\infty,(x_2+\Delta x, y_2+\Delta y)}-s_{\infty,(x_2,y_2)} \;\;(\text{mod}\; c)$$. Specifically if $$s_{\infty,(x_1+\Delta x, y_1+\Delta y)}-s_{\infty,(x_1,y_1)} = 0$$ then $$s_{\infty,(x_2+\Delta x, y_2+\Delta y)}-s_{\infty,(x_2,y_2)} = 0$$ since the difference between any two entries of $$S_\infty$$ is confined to the set $$\{-(c-1), -(c-2), ..., -1, 0, 1, ..., c-1\}$$.
 * Given columns $$x_1, x_2$$, and rows $$y_1, y_2$$, if $$s_{\infty,(x_1,y_1)} = s_{\infty,(x_2,y_2)}$$, then shifting $$S_\infty$$ by $$x_2-x_1$$ columns and $$y_2-y_1$$ rows does not change the entries of $$S_\infty$$.

The columns of $$S_\infty$$ that contain $$1$$ are spaced evenly due to the aforementioned symmetry. Let $$m$$ denote the smallest positive integer such that every $$m^\text{th}$$ column contains $$1$$.
 * The entries $$s_{\infty,(1,1)} = 1$$ and $$s_{\infty,(a+1,b)} = 1$$ so it must be the case that $$a$$ is an integer multiple of the period $$m$$: $$m|a$$.
 * The entries $$s_{\infty,(1,1)} = 1$$ and $$s_{\infty,(c+1,1)} = 1$$ so it must be the case that $$c$$ is an integer multiple of the period $$m$$: $$m|c$$.

The rows of $$S_\infty$$ eventually repeat (not allowing any column shifts) with a period of $$n$$. A row does not appear twice in a single cycle due to the symmetry of $$S_\infty$$. Row $$1$$ is identical to row $$b+1$$ so it must be the case that $$b$$ is an integer multiple of the period $$n$$: $$n|b$$.

It will now be proven that $$c = mn$$ by showing that a sub-block of $$S_\infty$$ that consists of $$m$$ columns and $$n$$ rows contains every integer from $$\{1, 2, ..., c\}$$ exactly once.

To clarify notation, given the column indices $$x_1, x_2$$ where $$x_1 \leq x_2$$, and the row indices $$y_1, y_2$$ where $$y_1 \leq y_2$$, then $$S_\infty([x_1,x_2]\times[y_1,y_2])$$ will denote the sub-block of $$S_\infty$$ consisting of columns $$x_1$$ through to $$x_2$$, and rows $$y_1$$ through to $$y_2$$.

Since a row can be uniquely determined from a single cell and the rows only repeat with a period of $$n$$, any block $$S_\infty([x,x]\times[y,y+(n-1)])$$ will contain exactly $$n$$ unique entries.

Columns that contain $$1$$ occur with a period of $$m$$. Due to the symmetry of $$S_\infty$$, given any integer $$s \in \{1, 2, ..., c\}$$, columns that contain $$s$$ occur with a period of $$m$$. Any block $$S_\infty([x,x+(m-1)]\times[y,y+(n-1)])$$ will contain exactly $$mn$$ unique entries.

Given any integer $$s \in \{1, 2, ..., c\}$$, integer $$s$$ will occur in every $$m^\text{th}$$ column, and within that column, in every $$n^\text{th}$$ cell. Any block $$S_\infty([x,x+(m-1)]\times[y,y+(n-1)])$$ will contain $$s$$. Any block $$S_\infty([x,x+(m-1)]\times[y,y+(n-1)])$$ will contain every integer from $$\{1, 2, ..., c\}$$ exactly once. So therefore $$c = mn$$. $$\Box$$.

Solving linear modular equations - back to Bezout
Bezout's identity above provides us with the key to solving equations in the form
 * ax ≡ b (mod m)

Coprime case - gcd(a, m) is 1
Consider the case where
 * ax ≡ b (mod m)

but with gcd(a, m)=1

Because of Bezout's identity
 * 1 = au+mv

When we calculate u, this number is special.

Say if we have the equation
 * 4x=11 (mod 21)

4 and 21 are coprime since gcd(4,21)=1. Now 1=4*16+(-3)*(21). Our u in this case is 16. Observe now that 4*16=64. 64 (mod 21) = 1. This number u is very special - it is known as the multiplicative inverse. It is the number u on multiplication by a gives 1 mod m. Bezout's identity on calculating gcd(a, m) will always give you the multiplicative inverse of a modulo m. The multiplicative inverse of a is often written a-1 but note that this does not mean 1/a since we have seen in the first sections that we can not always divide in the integers.

Note that in Zp there is one number without a multiplicative inverse - 0. It may be useful to exclude 0 when considering modular arithmetic, so instead of having to say Zp\{0} all the time, we merely write Zp*.

Now since we have the magic multiplicative inverse, our problem becomes relatively easy to solve. 4-1=16 in Z21 and now, on multiplying throughout by 16
 * x = 11 × 16 (mod 21)

(since 4×16=1 because 16 is 4's multiplicative inverse mod 21). 11×16=176 and using a calculator or using the division theorem we obtain
 * x = 8 (mod 21)

which is our solution! Verify - 8×4 = 32 = 11 (mod 21).

The general case
Consider the general case where
 * ax ≡ b (mod m)

with no restrictions on a, b and m.

Firstly we calculate gcd(a, m) again to obtain d. Now d is a divisor since the d in gcd means greatest common divisor. So we can now divide a and m - but what about b? Since we have calculated the gcd of a and m but not b we have no guarantees that d will divide b. This then becomes a condition that the equation has no solution.

Now we have reduced the problem to the previous coprime case because gcd(a/d, m/d)=1 with d as above. However we do not have 1 solution any more - this is true because we have reduced the solution to being x = c (mod m/d) and we must bring the solution back mod m. This will be come clearer in the examples.

Let's work through some examples.

Example 1. Solve 4x ≡ 3 (mod 20). Firstly, gcd(4, 20) = 4. 4 does not divide 3 and we have no solution.

Example 2. Solve 9x ≡ 6 (mod 15). gcd(9, 15) = 3 and 3 does divide 6 and we have 3 solutions.

Now, divide through by 3 to obtain
 * 3x ≡ 2 (mod 5)

gcd(3, 5) = 1 = 3 × 2 + -1 × 5 So the inverse of 3 mod 5 is 2. Now we obtain the solution
 * x ≡ 4 (mod 5)

Now in Z15 we must obtain the two extra solutions 9 and 14 mod 15 - 9 mod 5 = 4 and 14 mod 5 = 4.

Generally we can say that if we have the solution to the reduced equation x, the general solution is x+(m/d)k for k={0, 1, .., d-1}.

Chinese Remainder Theorem
Very often congruence relations are required to hold simultaneously. Given positive integer bases $$m_1$$ and $$m_2$$ and arbitrary integers $$a$$ and $$b$$ where $$0 \leq a < m_1$$ and $$0 \leq b < m_2$$, a common question is what integers $$x$$ satisfy the following congruence relations simultaneously:

$$\left\{\begin{array}{c} x \equiv a \;\;(\text{mod}\; m_1) \\ x \equiv b \;\;(\text{mod}\; m_2) \end{array}\right.$$

The Chinese Remainder Theorem dictates that when $$\text{gcd}(m_1, m_2) = 1$$, for any choice of $$a \in \Z_{m_1} = \{0, 1, ..., m_1-1\}$$ and $$b \in \Z_{m_2} = \{0, 1, ..., m_2-1\}$$ there exists a unique integer $$c \in \Z_{m_1m_2} = \{0, 1, ..., m_1m_2-1\}$$ such that:
 * the only integers $$x$$ that satisfy $$\left\{\begin{array}{c} x \equiv a \;\;(\text{mod}\; m_1) \\ x \equiv b \;\;(\text{mod}\; m_2) \end{array}\right.$$ simultaneously are those that satisfy $$x \equiv c \;\;(\text{mod}\; m_1m_2)$$

In essence, when $$m_1$$ and $$m_2$$ are coprime, there is a 1-to-1 correspondence between ordered pairs from $$\Z_{m_1} \times \Z_{m_2}$$ and the set $$\Z_{m_1m_2}$$.

Proof 1
Proof: To begin, observe that $$|\Z_{m_1m_2}| = m_1m_2$$ and $$|\Z_{m_1} \times \Z_{m_2}| = |\Z_{m_1}||\Z_{m_2}| = m_1m_2$$ so it is possible to pair each $$c \in \Z_{m_1m_2}$$ with a unique ordered pair $$\langle a, b \rangle \in \Z_{m_1} \times \Z_{m_2}$$ and vice-versa.

Given any $$c \in \Z_{m_1m_2}$$, integer $$c$$ can be reduced modulo $$m_1$$ to get an integer $$a \in \Z_{m_1}$$, and can be reduced modulo $$b$$ to get an integer $$b \in \Z_{m_2}$$. Integers $$a$$ and $$b$$ satisfy: $$\forall x \in \Z : (x \equiv c \;\;(\text{mod}\; m_1m_2)) \implies ((x \equiv a \;\;(\text{mod}\; m_1)) \;\text{and}\; (x \equiv b \;\;(\text{mod}\; m_2)))$$

It is not obvious that for any choice of $$\langle a, b\rangle \in \Z_{m_1} \times \Z_{m_2}$$ that there exists a unique $$c \in \Z_{m_1m_2}$$ such that $$\forall x \in \Z : ((x \equiv a \;\;(\text{mod}\; m_1)) \;\text{and}\; (x \equiv b \;\;(\text{mod}\; m_2))) \iff (x \equiv c \;\;(\text{mod}\; m_1m_2))$$

Let $$S$$ denote an infinite array with two rows indexed by $$y = 1, 2$$, and an infinite number of columns indexed by $$x \in \Z$$. $$s_{(x,y)}$$ will denote the entry of $$S$$ at column $$x$$ and row $$y$$. $$s_{(x,y)}$$ is the unique integer from $$\Z_y = \{0, 1, ..., m_y-1\}$$ where $$s_{(x,y)} \equiv x \;\;(\text{mod}\; m_y)$$.

Given column indices $$x_1, x_2$$ where $$x_1 \leq x_2$$, then $$S_1[x_1,x_2]; S_2[x_1,x_2]; S_{1,2}[x_1,x_2]$$ will respectively denote the sub-blocks formed by columns $$x_1$$ through to $$x_2$$ and row sets $$\{1\}; \{2\}; \{1,2\}$$.

Partition $$S$$ into the series of $$m_1 \times 2$$ blocks $$..., S_{1,2}[-2m_1,-m_1-1], S_{1,2}[-m_1,-1], S_{1,2}[0,m_1-1], S_{1,2}[m_1,2m_1-1], S_{1,2}[2m_1,3m_1-1], ...$$ where block $$k \in \Z$$ is $$S_{1,2}[km_1, (k+1)m_1-1]$$.

Row 1 of block $$k$$ will always be the sequence $$0, 1, ..., m_1-1$$. Row 2 of block $$k$$ can be uniquely determined by its first entry, $$s_{(km_1, 2)}$$ since $$s_{(km_1+r, 2)} \equiv s_{(km_1, 2)} + r \;\;(\text{mod}\; m_2)$$. The blocks only differ by row 2, and row 2 for each block is uniquely determined by its first entry $$s_{(km_1, 2)}$$.

$$s_{(km_1, 2)} \equiv km_1 \;\;(\text{mod}\; m_2)$$ and $$s_{((k+1)m_1, 2)} \equiv (k+1)m_1 \;\;(\text{mod}\; m_2)$$ so $$s_{((k+1)m_1, 2)} \equiv s_{(km_1, 2)} + m_1 \;\;(\text{mod}\; m_2)$$. This implies that the first entry of row 2 for the next block can be uniquely determined from the first entry of row 2 for the current block.

Since each block is uniquely determined from the previous block, the row 2 pattern for each block will repeat after a regular period of $$p$$ blocks: $$S_{1,2}[km_1, (k+1)m_1-1] = S_{1,2}[(k+p)m_1, (k+p+1)m_1-1]$$. Let set $$U \subseteq \Z_{m_2}$$ denote the total range of values attained by $$s_{(km_1,2)}$$. It is the case that $$p = |U|$$. Let $$d$$ be the minimum positive difference between any two elements from $$U$$. The cyclical nature of the elements from $$U$$ makes it easy to show that any element $$t \in U$$ is congruent to a multiple of $$d$$ modulo $$m_2$$. In essence: $$U = \{t| t \in \Z_{m_2} \;\text{and}\; \exists u \in \Z: t \equiv ud \;\;(\text{mod}\; m_2)\}$$. Since $$d$$ is the minimum positive difference between any two elements of $$U$$, both $$m_1$$ and $$m_2$$ are multiples of $$d$$ (in fact, $$m_2 = pd$$). Since $$\text{gcd}(m_1,m_2) = 1$$, it must be the case that $$d = 1$$. This implies that $$U = \Z_{m_2}$$ and that $$p = m_2$$. A total of $$m_2$$ blocks are encountered before any repetition occurs in array $$S$$, and therefore all $$m_1m_2$$ possible columns occur exactly once in a column period of $$m_1m_2$$ in array $$S$$. This establishes the Chinese Remainder Theorem. $$\Box$$

Proof 2
A second (more intuitive) proof can be derived by constructing a mesh to depict the space $$\mathbb{Z}_{m_1} \times \mathbb{Z}_{m_2}$$. This mesh is a rectangular array of points with $$m_1$$ columns and $$m_2$$ rows. The columns are indexed from left to right by $$0, 1, 2, ..., m_1-1$$, and the rows are indexed from bottom to top by $$0, 1, 2, ..., m_2-1$$. Most importantly, the mesh will "wrap around" in the horizontal and vertical dimensions. This means that moving to the right from column $$m_1-1$$ will return you to column $$0$$; moving to the left from column $$0$$ will send you to column $$m_1-1$$; moving up from row $$m_2-1$$ will return you to row $$0$$; and moving down from row $$0$$ will send you to row $$m_2-1$$.

The mesh has $$m_1m_2$$ points, and the horizontal and vertical coordinates of each point are the remainders from dividing an integer $$x$$ by $$m_1$$ and $$m_2$$ respectively. For convenience, given an arbitrary dividend $$x$$, $$x \;\text{mod}\; m_1$$ and $$x \;\text{mod}\; m_2$$ will denote the remainders after $$x$$ is divided by $$m_1$$ and $$m_2$$ respectively. If the dividend $$x$$ is incremented by $$1$$, then the coordinate formed by the remainders moves to the right by one step, and up by one step, wrapping around if necessary. $$x = 0$$ corresponds to the coordinate $$(0,0)$$. Increasing $$x$$ in steps of $$1$$ will trace a ray that originates from $$(0,0)$$ and moves one step to the right and one step up each interation, wrapping around if necessary. The images below give examples of this ray for $$m_1 = 6; m_2 = 5$$ and for $$m_1 = 6; m_2 = 4$$. In the images below, a copy of column 0 and row 0 appears at the right and top of the mesh respectively to illustrate the wrap around property. When $$m_1 = 6; m_2 = 4$$, $$m_1$$ and $$m_2$$ are not coprime and fail to satisfy the conditions of the Chinese remainder theorem.

The ray forms diagonal "stripes" in the mesh, and if these stripes are all equally spaced by $$1$$, then the ray passes through every point in the mesh exactly once proving that every remainder pair is possible and hence the Chinese remainder theorem. When $$m_1 = 6; m_2 = 4$$, $$m_1$$ and $$m_2$$ are not coprime and fail to satisfy the conditions of the Chinese remainder theorem, hence the ray does not hit every mesh point. The wrap around property of the mesh makes the mesh "symmetric" in both the horizontal and vertical dimensions, which means that if the wrap around "seams" were moved to any column and row where the ray passes through the lower left corner, then the ray is completely unchanged. This requires that the stripes be equally spaced. Let $$d$$ denote this equal spacing, and it will be shown that $$d|m_1$$ and $$d|m_2$$. The ray passes through row $$0$$ every $$d$$ steps to the right. The ray passes through $$(0,0)$$, and the wrap around property implies that moving $$m_1$$ steps to the right returns to this same intersection point. This can only occur if $$d|m_1$$. By a similar argument $$d|m_2$$. If $$m_1$$ and $$m_2$$ are coprime, then $$d = 1$$, the stripes are evenly spaced by $$1$$, every remainder pair is possible, and the Chinese remainder theorem is therefore true. $$\Box$$

title: Discrete Mathematics/Number theory


text:
Jakub Pokorný

Jakub Pokorný (born 11 September 1996) is a Czech professional footballer who plays as a defender for Sigma Olomouc in the Czech First League.

Club career
He started his professional career at Znojmo. He made his senior league debut for them on 28 September 2014 in their Czech National Football League 4–1 loss at Varnsdorf. He scored his first league goal on 27 March 2016 in Znojmo's Czech National Football League 3–1 loss at Hradec Králové. He moved to Baník Ostrava in 2016 and won promotion to the Czech First League with them the same season. On 30 August 2022, Pokorný joined Sigma Olomouc on a three-year contract.

International career
He represented the Czech Republic in the Under-20 and Under-21 youth categories.

title: Jakub Pokorný


text:
Janis Crystal Lipzin

Janis Crystal Lipzin (born 1945), is an American artist and educator, working with film, photography, video, audio, multi-media installations, and media performance. Lipzin is known for her work in many media and taught at the San Francisco Art Institute for over three decades. Lipzin's films offer a unique blend of rigorous conceptual structure, formal investigation, and sensual discovery. The Bladderwort Document is a haunting visual fantasia of her life on a farm in the 1970s; Trepanations is a droll meditation on social forces and women's appearance; and Seasonal Forces, Part One creates a fluid and immediate record of the cultural and seasonal changes in the rural landscape where she lives. She has been an active filmmaker since 1974, when she became attracted to using Super-8 cameras, in part because of their easy portability and flexibility to make changes to a film up to the moment of projection. Her more recent work incorporates both digital and analog film methods. wherein light and photo-chemistry collide and conspire to reveal aspects of our world deserving of more careful scrutiny. Her work blends an enduring interest in the volatility of nature and human events with a sympathy for alternative, hand-made methods that she interweaves with digital processes. Lipzin is based in Sonoma County, California.

Early life and education
Janis Crystal Lipzin was born in 1945, in the United States. Lipzin attended Ohio University, where she received a BFA; New York University where she studied painting; and the San Francisco Art Institute, where she received an MFA. She attended the University of Pittsburgh where she received an MSLS in library and information science.

Career
Her work has been recognized with exhibitions at the Museum of Modern Art, New York Film Society, Centre Pompidou (Paris), Venice Biennale, San Francisco Museum of Modern Art, the Tate Modern and other international venues. Lipzin's work was included in the Dreamlands: Immersive Cinema and Art 1905–2016 exhibition and the Color of Light exhibition at the Whitney Museum of American Art.

She taught at the San Francisco Art Institute from 1978 to 2009 where she served as Chair of the Film Department and before that directed the Film/Photography Program at Antioch College in Yellow Springs, Ohio.

Among her awards are fellowships, commissions and grants from the John Simon Guggenheim Memorial Foundation, National Endowment for the Arts, Center for Cultural Innovation, Ohio Arts Council, California Arts Council, Mission Eye and Ear,  and Echo Park Film Center, Los Angeles.

Lipzin's work is included in the Carnegie Museum of Art's collection Berkeley Art Museum, and the New Museum of Contemporary Art in New York.

title: Janis Crystal Lipzin


text:
Evian (disambiguation)

Evian is a brand of bottled spring water.

Evian or variation, may also refer to:

 * Évian-les-Bains (aka Évian) a commune in France
 * Évian Accords (1962) a treaty that ended the Algerian War of Independence
 * Évian Conference (1938) a conference to deal with Jewish refugees fleeing Germany
 * The Evian Group at IMD (formerly Evian Group) a group of governments and businesses committed to global markets that are open, inclusive, equitable, sustainable
 * Evian Resort Golf Club, Évian-les-Bains, France
 * The Evian Championship (golf) a women's golf championship in France
 * Thonon Évian Savoie F. C. (aka Évian) a soccer team located in Thonon-les-Bains, France

title: Evian (disambiguation)


text:
Wikiversity:Custodianship

A Wikiversity custodian is an experienced and trusted user who can protect, delete and restore pages as well as block users from editing. Custodians are the Wikiversity equivalent of administrators, also known as sysops, bibliotecarios (librarians), and moderators.

The English Wikiversity currently has custodianNaN ss (full list), including human and bot users.

Request assistance - List of custodians - Requests for custodianship

How does one become a custodian?
Any Wikiversity participant willing to do a lot of dull and boring work for the community can become a custodian. If you have a good editing record then you are likely to be trusted and granted the privileges and responsibilities of custodianship. If you are still interested in custodianship, here is the process it takes:

Deletion and undeletion of pages
Custodians can delete pages including images, categories, templates, etc. Deletion is subject to policy. Deletion requests may be submitted by any user at Requests for Deletion.

Deleting a page does not actually remove it from the database. It is merely invisible to non-custodians and can be restored at request, which may be submitted at Requests for Deletion. Page deletions and undeletions can be monitored by viewing the deletion log.

Custodians can also delete individual versions with the RevisionDelete feature. This is particularly useful to remove materials that violate copyrights or other applicable law.

Before you delete a page, read: Welcome templates. See also: Deletion policy.

Edit and move protection of pages
Custodians can protect pages to prevent editing. There are two types of page protection: semi-protection, which prevents anonymous and new users from editing, and full protection, which prevents all non-custodians from editing. A page can also be protected to prevent moving. Page protection can be lifted by any custodian upon request, which may be submitted at Request custodian action. Page protections and unprotections can be monitored by viewing the protection log.

Rollback
Custodians have a rollback tool to revert the last change or group of changes made to a page by the same user. There is no option to provide an edit summary when using this tool. Instead, a summary such as "" will be used automatically. This tool is primarily used to respond to obvious vandalism and spam. For other edit reversions, the rollback button should not be used and a good edit summary should be provided.

User blocks
Custodians can block users from editing by specifying a username, an IP address or range of addresses. Blocks can be temporary or permanent. Most frequently, blocking occurs in response to obvious and repeated vandalism. When blocking a user, a reason for blocking must be provided which is displayed in the block log. It is also possible to block IP addresses from creating new user accounts. See Blocking policy.

Editing MediaWiki namespace
Custodians are able to edit system messages (the standard texts that are used by the MediaWiki software), though site CSS and site JS editing requires interface administrator privileges.

Import
Custodians have access to the Import tool, to bring materials from Wikipedia, Wikibooks, Beta.Wikiversity, Wikiquote, and Wikisource.

How are custodians expected to act?
Custodians are supposed to follow the same principles as every other user, including being civil, assuming good faith, and understanding what Wikiversity is. They are expected to act professionally, and to respect policy and community consensus.

For a discussion of possible failure to act professionally, see a proposed recusal policy at Recusal.

Problems with custodians
If you have a question about an action (page deletion, page protection, user block, editing MediaWiki namespace pages, violation of Wikiversity policy or some other action that does serious damage to the project) by a Wikiversity custodian, the first thing to do is leave a message on that custodian's user discussion page. Custodians should always be able to explain how their actions support Wikiversity. If you cannot get satisfaction from discussion with the custodian, you can place comments, suggestions, complaints or questions at Custodian feedback. Try to resolve all custodian problems by discussion.

Custodians can lose their status for egregious violations of policies. Loss of custodianship involves a process that establishes community consensus. If a specific complaint is not resolved at Custodian feedback then a Community Review can be initiated to establish if there is community consensus in support for the custodianship of the custodian who is the subject of the unresolved complaint. At the end of the review, Wikiversity bureaucrats will review the discussion. If a bureaucrat decides that there is good reason for removal of a custodianship, that bureaucrat will go to the meta-wiki and request that stewards review the community discussion. If a steward agrees that the Wikiversity community has reached consensus about a problem custodian, then that steward can terminate the custodianship of the custodian.

If you have a question about an action of a probationary custodian, you should consult their mentor(s). Actions of a probationary custodian are the responsibility of their mentor(s) during their probationary period. Partially depending on agreements made in the probationary candidacy, and with respect for the 48 hour period to find a new mentor, a request to remove custodianship for a probationary custodian may often be made at Meta by any mentor or bureaucrat without need for community consensus. The local process to remove custodianship should only need to be used as a last resort, because of this alternative option.

Wikiversity

 * Blocking policy
 * No shrines for vandals
 * Policies
 * Revision deletion
 * How to be a Wikimedia sysop/Wikiversity

MediaWiki/Wikimedia

 * How to be a Wikimedia sysop
 * MediaWiki Administrator's Handbook

title: Wikiversity:Custodianship


text:
User:TheUnintendedGeek

Who I Am
I am a geek. Hence the name "TheUnintendedGeek." Now, I love the thought of space, extra terrestrials, and things based around science, physiology, biology, physics, and historical evolution. I'm also a fan of archaeology. I just like topics the average geek would like. I'm a simple person. But I love complexity. I am just over the age of thirteen (As of 2017). Any topic and/or question, I would love to answer, and try to provoke others to join in as well. Conversations will start, and they will be put to rest. I may become a well known person on here, as for I love editing, and commenting. Gaming is something I do a lot. Any question, comments, concerns, contact me at theunintendedgeek@gmail.com

title: User:TheUnintendedGeek


text:
Eiho Sato

Eiho Sato (佐藤 永芳) was a Japanese painter. His work was part of the painting event in the art competition at the 1936 Summer Olympics.

title: Eiho Sato


text:
1992 Girabola

The 1992 Girabola was the 14th season of top-tier football competition in Angola. C.D. Primeiro de Agosto were the defending champions.

The league comprised 16 teams, the bottom two of which were relegated.

Primeiro de Agosto were crowned champions, winning their 5th title, while Inter de Luanda and Benfica de Cabinda were relegated.

Amaral Aleixo of Sagrada Esperança finished as the top scorer with 20 goals.

Changes from the 1991 season
Relegated: Desportivo da Cuca, Desportivo de Saurimo Promoted: Desportivo da Nocal, Inter da Huíla

Top scorer

 * Amaral Aleixo

Champions
Squad: Barbosa, Bolefo, Corolla, Fuidimau, Ivo, Kiss, Loth, Mbila, Mwanza, Ndisso, Neto, Rabolé, Russo, Valentim, Velho, Vieira Dias, Zacarias, Zé Gordo Head coach: Dušan Condić

title: 1992 Girabola


text:
User:Sindrake/sandbox

Systemic Child Abuse Culture (Also known as Culturally Accepted Pedophilia) is a form of child abuse that is embedded as normal practice within society or organization. It can lead to such issues as grooming children, pedophilia, child abuse, human trafficking, and CPS (Child Protective Services) false reports.and the foster care system among other reports.

The Term Systemic Child Abuse Culture was coined by Jeremy Powell on the 22nd of July 2020 in an effort to gain assistance with his case against CPS involving fraudulent claims of neglect up to and including Educational Neglect (Homeschooling), Religious Neglect (Not attending church), and Poverty (Having a low income job). Jeremy Powell claims that the CPS caseworkers and supervisors abused their power in order to kidnap his children from their home on the 29th of April 2018. Upon further research discovered that Systemic Child Abuse Culture has been common for quite some time in order to groom children to think, act, dress, make decisions, and live in a certain way. This form of grooming comes from Hollywood via television, movies, and music.

Hollywood films such as Baby Burlesks 1930's, Little girl down the lane 1976, and Pretty baby 1978 are known to have images very suggestive in nature portraying children in very adult manner. Music circling the internet placing children in very adult situations, and television shows are particularly programmed to groom children to believe certain acts of adult nature are commonplace in order for the child to be more submissive to a controller or handler.

Systemic Child Abuse Culture has yet to be defined as it is only recently been discovered in large scale relating to human trafficking, CPS, and the foster care system linking all three in what could be considered conspiracy theory to some, but sources would reveal otherwise.

title: User:Sindrake/sandbox


text:
User:Arnelapolto

Arnel Manalo Apolto is a Filipino man born in September 26, 1992 his Mother is Mrs. Juliet Apolto and his Father is Alberto Apolto. He was born in manila at Amisola Hospital in Hermosa tondo manila, His siblings are Janet Apolto, Fatima Apolto, Jubert Apolto, John John Apolto.

title: User:Arnelapolto


text:
Talk:Discrete Mathematics/Introduction

Groups are usually considered introductory level, not advanced. Elementary group theory, I mean...

 * I think a consolidated and thorough treatment of group theory constitutes it being in the upper level section. However I might rearrange some things. Dysprosia 10:43, 21 Aug 2003 (UTC)

-

Groups are usually considered low level, but that doesn't necessarily mean they should be treated in the introductory part. Nonetheless, Groups probably shouldn't be called complex.

Anyway, what exactly is this introduction about? Is it an introduction
 * to the topic itself, i.e. discrete maths, or
 * to this book, i.e. the subset treated here, or
 * to this Wikibook, i.e. the way it's treated here?

 * A bit on the topic, and a bit on the Wikibook. Reworded intro. Dysprosia 07:29, 2 Feb 2004 (UTC)

title: Talk:Discrete Mathematics/Introduction


text:
Henry Ziegler

Henry Ziegler (February 10, 1889 – July 1968) was an American painter. His work was part of the painting event in the art competition at the 1936 Summer Olympics.

title: Henry Ziegler


text:
User:Bawolff

→ See Wikinews:User:Bawolff

I'm Bawolff. I'm mostly active at Wikinews (read it almost everyday as well as a admin and arbitrator there). However I also do do a little at 'pedia (~600 edits) and read Wikibooks occasionally. This sounds like an intreasting project with lots of potential, and I hope, over time, it will grow to the magnitude of the other Wikimedia projects.

title: User:Bawolff


text:
Asturian Left

Izquierda Asturiana (IAS; Asturian Left) is an Asturian nationalist leftist political party of Asturias, Spain.

History
Izquierda Asturiana was founded in 1992 by former members of the Unidá Nacionalista Asturiana. In the European Parliamentary elections of 2004, they made contact with Andecha Astur but could not agree on a platform, so they ran separate candidates.

Ideology
Among the party's objectives is the defense of the Asturian language, as well as the promotion of Asturian leftist politics without the influence of an outside party.

title: Asturian Left


text:
Category:Discoveries by Friedrich Tietjen

This is a category for astronomical objects discovered by German astronomer Friedrich Tietjen.

See parent category for proper sortkey usage instructions.


text:
Category:13th century in economic history

These articles pertain to economics, finance, or business in the 13th century, 1201–1300.


text:
Elizaveta Shanaeva

Elizaveta Tamerlanovna Shanaeva (Елизавета Тамерлановна Шанаева, born 30 May 2003) is a Russian competitive ice dancer. With her former skating partner, Devid Naryzhnyy, she is the 2020 World Junior bronze medalist and the 2019–20 Junior Grand Prix Final bronze medalist. She has also won three medals on the ISU Junior Grand Prix series, including gold medals at 2019 France and 2019 Russia.

Personal life
Elizaveta Tamerlanovna Shanaeva was born on 30 May 2003 in Alania, Russia.

Early years
Shanaeva began learning to skate in 2005. She skated with Sergei Semko before teaming up with Devid Naryzhnyy during the 2016–2017 season. They are coached by Irina Zhuk and Alexander Svinin.

Shanaeva/Naryzhnyy placed first at the 2017 Moscow Championship.

2018–2019 season
Shanaeva/Naryzhnyy received their first ISU Junior Grand Prix (JGP) assignments in the 2018–2019 season. They won silver medals at 2018 JGP event in Bratislava, Slovakia and placed 4th at 2018 JGP event in Yerevan, Armenia.

In November 2018, they won the junior gold medal at the 2018 Grand Prix of Bratislava. They placed fourth at the 2019 Russian Junior Championships.

2019–2020 season
Returning to the Junior Grand Prix, Shanaeva/Naryzhnyy won their first JGP gold medal in September at the 2019 JGP France in Courchevel. Three weeks later, they won a second gold medal at the 2019 JGP Russia in Chelyabinsk. With these results, they qualified for the 2019–20 ISU Junior Grand Prix Final in Torino. Shanaeva/Naryzhnyy placed third in the rhythm dance there, with her describing them as "quite happy" with their performance. They were also third in the free dance, despite Naryzhnyy missing a twizzle level, and won the bronze medal.

After winning the junior national title at the 2020 Russian Junior Championships, Shanaeva/Naryzhnyy were assigned to compete at the 2020 World Junior Championships in Tallinn, Estonia. First in the free dance, they won a small gold medal for the segment, becoming the only team to score above 70 points in the junior rhythm dance that season. Third in the free dance, they dropped to third place overall and won the bronze medal. Shanaeva said, "we got a lot of energy to show our maximum next season and to be ready to beat everyone."

2020–2021 season
After junior Russian test skates in August, both became ill with COVID-19, first Shanaeva and then Naryzhnyy. This caused them to miss the first half of the season, competing only in December, on the fifth stage of the domestic Cup of Russia series, but having to withdraw after the rhythm dance due to Naryzhnyy getting food poisoned.

At the beginning of February, they competed on the 2021 Russian Junior Championships in Krasnoyarsk, placing third in the rhythm dance, second in the free dance and second overall. They claimed to be happy with their performances after such a difficult period, defining their result as a "silver medal with a golden shine."

They were scheduled to participate in the Russian Cup Final in Moscow but withdrew for medical reasons. On the 17 and 18 of April, they performed in Team Tutberidze’s show Champions on Ice in Krasnodar and Sochi.

2021–2022 season
Moving to the international senior level, Shanaeva/Naryzhnyy won the bronze medal at the Budapest Trophy. They went on to make their senior Grand Prix debut at the 2021 Skate Canada International, where they finished in ninth place.

In December, Shanaeva/Naryzhnyy competed on the 2022 Russian Championship in Saint Petersburg, placing eighth in the rhythm dance and fifth in the free dance and finishing fifth overall.

Competitive highlights
GP: Grand Prix; CS: Challenger Series; JGP: Junior Grand Prix

Detailed results
''Small medals awarded only at ISU Championships. ISU personal bests highlighted in bold.''

title: Elizaveta Shanaeva


text:
Wikipedia:Sockpuppet investigations/ScriptKKiddie/Archive

Suspected sockpuppets

More spam about "ScriptKKiddie"/this Praxidicae (talk) 13:36, 15 August 2020 (UTC)

Comments by other users

Clerk, CheckUser, and/or patrolling admin comments

 * . Sir Sputnik (talk) 19:39, 15 August 2020 (UTC)

Suspected sockpuppets

Created a draft Shobhit Sharma, which were spammed before by the blocked socks. Also note the user's link to social media accounts here. All bear the name "ScriptKKiddie", name of the sockmaster. Fylindfotberserk (talk) 19:21, 25 March 2021 (UTC)

Comments by other users

Clerk, CheckUser, and/or patrolling admin comments

 * Evidence provided by is completely conclusive, clearly the same user. Blocked and tagged. Thanks! ~  mazca  talk 14:15, 26 March 2021 (UTC)


text:
Michael Pereira (field hockey)

Michael Estevao Pereira (26 December 1932 – 20 February 2019) was a Kenyan field hockey player. He competed in the men's tournament at the 1956 Summer Olympics.

title: Michael Pereira (field hockey)


text:
User:White0078906316

= OURSLIKE = OursLike is an Indian e-commerce company based in Maharajganj, U.P, India. It was founded by  Krishna Murari Verma in 2017. The company initially focused on book sales, before expanding into other product categories ,fashion, groceries, and lifestyle products.

Owner of OursLike pursuing a ICSI ( The Institute of Company Secretaries of India) and he is passionate about exceeding your expectation in every field including what are your requirement at that particular website.

OursLike offers a range of product and services through website. He serves consumers through its retail website and focuses on selection, price and convenience. It designs its website to enable many products to be sold by them.

OursLike allows customers to access its website directly and through its mobile website and application.

Business Structure
Bussines structure of OursLike is According to a report in November 2019, OursLike was operating through a complex business structure which included several firms,  some registered in Local town.

OursLike give a trust if purchasing of any  things, it give a cinfidentent, and most important things is it give a CIA(Confidential Integrity Availability) to any product, it also providea online Transaction including home delivery.

Website Information
Type of service: Private

Type of website: E-Commerce

Owner: Krishna Murari Verma

URL: http://www.ourslike.com

Available Languages: Hindi, English

Founded: 2017

Advertising: No

Commercial: Yes

Registration: Required

Current status: Active

title: User:White0078906316


text:
Topic talk:Department of Sociology

Links for Wikibooks not yet created
These links don't seem to be needed on the front page, so saving the list here:

To be developed:
 * Contemporary Social Issues
 * Delinquency and Crime
 * Demography
 * Environmental Sociology (or Human Ecology)
 * Gender Stratification
 * Global Inequality
 * History of Sociology
 * Medical Sociology
 * Organizational Sociology (or Industrial Sociology)
 * Race & Ethnicity
 * Social Class
 * Social Policy
 * Social Movements
 * Urban Society

title: Topic talk:Department of Sociology


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BASIS Independent Silicon Valley

BASIS Independent Silicon Valley (BISV) is a private school operated by Spring Education Group in San Jose, California. BISV is a TK-Grade 12 STEM-focused school.

History
BASIS Independent Silicon Valley was founded in 2014 inside a former IBM office in San Jose. Based on the success of BASIS charter schools opened in Arizona, Texas, and Washington D.C., BASIS.ed decided to implement its educational model into a private school, opening two new private schools in San Jose and Brooklyn. Today, Spring Education Group has taken over the ownership and management of the seven private schools. The mission of BASIS Independent Schools is to raise the standards of students' learning to the highest international levels. At BISV, each student takes at least four AP courses before their senior year, where they do independent research and take capstone classes. The school offers a wide range of extracurricular activities including a competitive speech and debate club, volunteering clubs, and active sports teams.

In fall of 2022, BASIS Independent Silicon Valley opened a lower school campus in Sunnyvale, California. The lower school serves Transitional Kindergarten (TK) - Grade 5.

title: BASIS Independent Silicon Valley