nyuuzyou/wikis huggingface dataset:
Divisibility by 12

== Theorem ==
Let $N \in \N$ be expressed as:
:$N = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$
Then $N$ is divisible by $12$ $a_0 - 2 a_1 + 4 \paren {\ds \sum_{r \mathop = 2}^n a_r}$ is divisible by $12$.
== Proof ==
We first prove that $100 \times 10^n = 4 \pmod {12}$, where $n \in \N$.
Proof by induction:
For all $n \in \N$, let $P \paren n$ be the proposition:
:$100 \times 10^n = 4 \pmod {12}$
=== Basis for the Induction ===
$\map P 0$ says $100 = 4 \pmod {12}$, which is true as $100 = 8 \times 12 + 4$.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$100 \times 10^k = 4 \pmod {12}$
Then we need to show:
:$100 \times 10^{k + 1} = 4 \pmod {12}$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$100 \times 10^n = 4 \pmod {12}$
And then:
Therefore:
:$N = 0 \pmod {12} \iff a_0 - 2 a_1 + 4 \paren {\ds \sum_{r \mathop = 2}^n a_r} = 0 \pmod {12}$
The result follows from Congruent to Zero iff Modulo is Divisor
== Sources ==
* : Chapter $2$. Equivalence Relations: Exercise $9$
Category:Divisibility Tests
Category:12
Divisibility by 19

== Theorem ==
Let $n$ be an integer expressed in the form:
:$n = 100 a + b$
Then $n$ is divisible by $19$ $a + 4 b$ is divisible by $19$.
== Proof ==
Let $a, b, c \in \Z$.
== Sources ==
* : $19$
* : $19$
Category:19
Category:Divisibility Tests
Divisibility by 2

== Theorem ==
An integer $N$ expressed in decimal notation is divisible by $2$ the of $N$ is divisible by $2$.
That is:
:$N = [a_n \ldots a_2 a_1 a_0]_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $2$ 
:
:$a_0$ is divisible by $2$.
== Proof ==
Let $N$ be divisible by $2$.
Then:
== Sources ==
* : $2$
* : $2$
* 
* 
Category:Divisibility Tests
Category:2
Divisibility by 37

== Theorem ==
Let $n$ be an integer which has at least $3$ digits when expressed in decimal notation.
Let the digits of $n$ be divided into groups of $3$, counting from the right, and those groups added.
Then the result is equal to a multiple of $37$ $n$ is divisible by $37$.
== Proof ==
Write $n = \ds \sum_{i \mathop = 0}^k a_i 10^{3 i}$, where $0 \le a_i < 1000$.
This divides the digits of $n$ into groups of $3$.
Then the statement is equivalent to:
:$37 \divides n \iff 37 \divides \ds \sum_{i \mathop = 0}^k a_i$
Note that $1000 = 37 \times 27 + 1 \equiv 1 \pmod {37}$.
Hence:
which proves our statement.
Category:Divisibility Tests
Divisibility by 4

== Theorem ==
An integer $N$ expressed in decimal notation is divisible by $4$ the $2$ s of $N$ form a $2$-digit integer divisible by $4$.
That is:
:$N = \sqbrk {a_n \ldots a_2 a_1 a_0}_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $4$ 
:
:$10 a_1 + a_0$ is divisible by $4$.
== Proof ==
Let $N$ be divisible by $4$.
Then:
== Sources ==
* : $4$
* : $4$
* 
* 
Category:Divisibility Tests
Category:4
Divisibility by 5

== Theorem ==
An integer $N$ expressed in decimal notation is divisible by $5$ the units digit of $N$ is divisible by $5$.
That is:
:$N = \sqbrk {a_n \ldots a_2 a_1 a_0}_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $5$ 
:
:$a_0$ is divisible by $5$.
== Proof ==
Let $N$ be divisible by $5$.
Then:
== Sources ==
* 
* 
Category:Divisibility Tests
Category:5
Divisibility by 6

== Theorem ==
An integer $N$ expressed in decimal notation is divisible by $6$ 
:$N$ is even
and:
:$N$ is divisible by $3$.
== Proof ==
We have that:
:$6 \divides N \iff 2 \divides N \land 3 \divides N$
The result follows by definition of even integer.
== Sources ==
* 
* 
Category:Divisibility Tests
Category:6
Divisibility by 7

== Theorem ==
An integer $X$ with $n$ digits ($X_0$ in the ones place, $X_1$ in the tens place, and so on) is divisible by $7$ :
:$\ds \sum_{i \mathop = 0}^{n - 1} \paren {3^i X_i}$ is divisible by $7$.
== Proof ==
From Difference of Two Powers, $10^i - 3^i$ always produces a number divisible by $7$.
Thus the first addend is always divisible by $7$. 
So $X$ will be divisible by $7$ the second addend is divisible by $7$.
Hence the result.
== Sources ==
* : $7$
* : $7$
Category:Divisibility by 7
Category:Divisibility Tests
Category:7
Divisibility by 8

== Theorem ==
An integer $N$ expressed in decimal notation is divisible by $8$ the $3$ s of $N$ form a $3$-digit integer divisible by $8$.
That is:
:$N = \sqbrk {a_n \ldots a_2 a_1 a_0}_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $8$ 
:
:$100 a_2 + 10 a_1 + a_0$ is divisible by $8$.
== Proof ==
Let $N$ be divisible by $8$.
Then:
== Sources ==
* : $8$
* : $8$
* 
* 
Category:Divisibility Tests
Category:8
Divisibility by 9

== Theorem ==
A number expressed in decimal notation is divisible by $9$ the sum of its digits is divisible by $9$.
That is:
:$N = \sqbrk {a_0 a_1 a_2 \ldots a_n}_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $9$ 
:
:$a_0 + a_1 + \cdots + a_n$ is divisible by $9$.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $2$: Some Properties of $\Z$: Exercise $2.8$
* : $\S 2.3$: Congruences: Exercise $9$
* : $9$
* : $9$
* 
* 
Category:Divisibility by 9
Category:Divisibility Tests
Category:9
Divisibility by 9/Corollary

== Corollary to Divisibility by 9 ==
A number expressed in decimal notation is divisible by $3$ the sum of its digits is divisible by $3$.
That is:
:$N = \sqbrk {a_0 a_1 a_2 \ldots a_n}_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $3$ 
:
:$a_0 + a_1 + \ldots + a_n$ is divisible by $3$.
== Proof ==
From Divisibility by 9 we have that:
:$N = \sqbrk {a_0 a_1 a_2 \ldots a_n}_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $3^2$ 
:
:$a_0 + a_1 + \ldots + a_n$ is divisible by $3^2$.
So:
== Sources ==
* : $\S 2.6$. Algebra of congruences: Example $41$
* : $3$
* : $3$
* 
* 
Category:Divisibility Tests
Category:3
Divisibility by Power of 10

== Theorem ==
Let $r \in \Z_{\ge 1}$ be a strictly positive integer.
An integer $N$ expressed in decimal notation is divisible by $10^r$ the last $r$ digits of $N$ are all $0$.
That is:
:$N = \sqbrk {a_n \ldots a_2 a_1 a_0}_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $10^r$ 
:
:$a_0 + a_1 10 + a_2 10^2 + \cdots + a_r 10^r = 0$
== Proof ==
Let $N$ be divisible by $10^r$.
Then:
Hence the result.
== Sources ==
* : $10$
* : $10$
Category:Divisibility Tests
Category:10
Divisibility by Power of 2

== Theorem ==
Let $r \in \Z_{\ge 1}$ be a strictly positive integer.
An integer $N$ expressed in decimal notation is divisible by $2^r$ the last $r$ digits of $N$ form an integer divisible by $2^r$.
That is:
:$N = [a_n \ldots a_2 a_1 a_0]_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $2^r$ 
:
:$a_0 + a_1 10 + a_2 10^2 + \cdots + a_r 10^r$ is divisible by $2^r$.
== Proof ==
First note that:
:$10^r = 2^r 5^r$
and so:
:$2^r \divides 10^r$
where $\divides$ denotes divisibility.
Thus:
:$\forall s \in \Z: s \ge r: 2^r \divides 10^s$
but:
:$\forall s \in \Z: s < r: 2^r \nmid 10^s$
Thus let $N$ be divisible by $2^r$.
Then:
Hence the result.
== Sources ==
* : $2$
* : $2$
Category:Divisibility Tests
Category:2
Divisibility of Common Difference of Arithmetic Sequence of Primes

== Theorem ==
If $n$ terms of an arithmetic sequence are primes, then the common difference must be divisible by all primes less than $n$.
== Proof ==
 suppose the arithmetic sequence is increasing.
We also disregard the trivial case of zero common difference (do note that the theorem also holds in this case).
The proof proceeds by induction:
=== Basis for the Induction ===
As the smallest prime is $2$, the first nontrivial case is $n = 3$.
 there exists an arithmetic sequence of $3$ primes where the common difference $d$ is not divisible by $2$.
By Prime not Divisor implies Coprime, $d$ and $2$ are coprime.
By Exist Term in Arithmetic Sequence Divisible by Number, one of these primes must be divisible by $2$.
By definition of a prime number, that prime must be $2$.
As the smallest prime is $2$, it must be the first term of the sequence.
But then the third term of the sequence, $2 + 2 d$, would be divisible by $2$, and hence cannot be prime.
This contradiction shows that any arithmetic sequence of $3$ primes must have common difference divisible by $2$.
=== Induction Hypothesis ===
This is the induction hypothesis:
:For some $k \ge 3$, if $k$ terms of an arithmetic sequence are odd primes, then the common difference must be divisible by all primes less than $k$.
Now we need to show:
:If $k + 1$ terms of an arithmetic sequence are odd primes, then the common difference must be divisible by all primes less than $k + 1$.
=== Induction Step ===
This is the induction step:
We apply the induction hypothesis to the first $k$ terms of the $\paren {k + 1}$-term arithmetic sequence of primes.
If $k$ is not prime, then the set of primes less than $k$ is the same as the set of primes less than $k + 1$, and the conclusion follows trivially.
Now suppose $k$ is prime.
From the induction hypothesis we know that the common difference $d$ is divisible by all primes less than $k$.
We just need to show that $d$ is divisible by $k$ as well.
 this is not the case.
By Prime not Divisor implies Coprime, $d$ and $k$ are coprime.
By Exist Term in Arithmetic Sequence Divisible by Number, one of these primes must be divisible by $k$.
By definition of a prime number, that prime must be $k$.
If $k$ is the first term of the sequence, then the $\paren {k + 1}^{th}$ term of the sequence, $k + k d$, would be divisible by $k$, and hence cannot be prime.
Therefore $k$ is not the first term of the sequence.
Consider $k - d$, the term before $k$.
As $k - d$ is a prime less than $k$, it must be a divisor of $d$.
But then $k = k - d + d$ would be a multiple of $k - d$ greater than $k - d$, so it cannot be prime.
This contradiction shows that any arithmetic sequence of $k + 1$ primes must have common difference divisible by $k$.
The result follows by induction.
== Note ==
 attributes this result to an M. Cantor.
== Sources ==
* : No. $72$
Divisibility of Elements in Geometric Sequence of Integers

== Theorem ==
Let $Q_n = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of integers.
Let $j \ne k$.
Then:
:$\paren {\exists j \in \set {0, 1, \ldots, n - 1}: a_j \divides a_{j + 1} } \iff \paren {\forall j, k \in \set {0, 1, \ldots, n}, j < k: a_j \divides a_k}$
where $\divides$ denotes integer divisibility.
That is:
:One term of a geometric sequence of integers is the divisor of the next term
:All terms are divisors of all later terms.
== Proof ==
Let $a_j \divides a_{j + 1}$ for some $j \in \set {0, 1, \ldots, n - 1}$.
Then by definition of integer divisibility:
:$\exists r \in \Z: r a_j = a_{j + 1}$
Thus the common ratio of $Q_n$ is $r$.
So by definition of geometric sequence:
:$\forall j, k \in \set {0, 1, \ldots, n}, j < k: r^{k - j} a_j = a_k$
and so $a_j \divides a_k$.
The converse is trivial.
Category:Geometric Sequences of Integers
Divisibility of Elements of Geometric Sequence from One where First Element is Prime

== Theorem ==
Let $Q_n = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of length $n$ consisting of integers only.
Let $a_0 = 1$.
Let $a_1$ be a prime number.
Then the only divisors of $a_n$ are $a_j$ for $j \in \set {1, 2, \ldots, n}$.
== Proof ==
From Form of Geometric Sequence of Integers from One, the elements of $Q_n$ are given by:
:$Q_n = \tuple {1, a, a^2, \ldots, a^n}$
From Divisors of Power of Prime, each of $a_j$ for $j \in \set {1, 2, \ldots, n}$ are the only divisors of $a_n$.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Geometric Sequences of Integers
Divisibility of Elements of Pythagorean Triple by 7/Mistake

== Source Work ==
:
:The Dictionary
::$7$
== Mistake ==
:If $a$, $b$ are the shorter sides of a Pythagorean triangle, then $7$ divides one of $a$, $b$, $a - b$ or $a - b$.
== Correction ==
An obvious misprint for:
:If $a$, $b$ are the shorter sides of a Pythagorean triangle, then $7$ divides one of $a$, $b$, $a - b$ or $a + b$.
It can also be argued that it ought to say:
:... then $7$ divides at least one of ...
Interestingly and oddly, this is correct in the first edition of of $1986$.
== Sources ==
* : $7$
Category:Mistakes/Curious and Interesting Numbers
Divisibility of Fibonacci Number

== Theorem ==
Let $F_k$ denote the $k$th Fibonacci number.
Then:
:$\forall m, n \in \Z_{> 2} : m \divides n \iff F_m \divides F_n$
where $\divides$ denotes divisibility.
=== Corollary ===
== Proof ==
From the initial definition of Fibonacci numbers:
:$F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$
Let $n = k m - r$ where $0 \le r < m$
We have:
:$m \divides n \iff r = 0$
The proof proceeds by induction on $k$.
For all $k \in \N_{>0}$, let $\map P k$ be the proposition:
:$r = 0 \iff F_m \divides F_{k m - r}$
=== Basis for the Induction ===
$\map P 1$ is the case:
:$r = 0 \iff F_m \divides F_{m - r}$
which holds because $F_{m - r} < F_m$ unless $r = 0$.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k > 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$r = 0 \iff F_m \divides F_{k m - r}$
Then we need to show:
:$r = 0 \iff F_m \divides F_{k m + m - r}$
=== Induction Step ===
This is our induction step:
Let $F_{k m - r} = a F_m + b$ where $0 \le b < F_m$.
We have:
:$b = 0 \iff F_m \divides F_{k m - r} \iff r = 0$
by the induction hypothesis.
We have that $F_{m - 1}$ and $F_m$ are coprime by Consecutive Fibonacci Numbers are Coprime.
Let $F_m \divides b F_{m - 1}$.
Then there exists an integer $k$ such that $k F_m \divides b F_{m - 1}$, by the definition of divisibility.
Then:
:$\dfrac k b = \dfrac {F_{m - 1} } {F_m}$
We have that $F_{m - 1}$ and $F_m$ are coprime.
Thus by Coprime Numbers form Fraction in Lowest Terms:
:$\dfrac {F_{m - 1} } {F_m}$ is in canonical form.
Then by Ratios of Fractions in Lowest Terms
:$F_m \divides b$
Because $0 \le b < F_m$, the only case is when $b = 0$.
Therefore:
:$F_m \divides b F_{m - 1} \iff b = 0$
Therefore:
:$F_m \divides F_{k m + m - r} \iff F_m \divides b F_{m - 1} \iff b = 0 \iff r = 0$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall m, n > 2 : m \divides n \iff F_m \divides F_n$
== Sources ==
* : $5$
* : $5$
Category:Fibonacci Numbers
Category:Divisors
Category:Proofs by Induction
Divisibility of Fibonacci Number/Corollary

== Corollary to Divisibility of Fibonacci Number ==
Let $F_k$ denote the $k$th Fibonacci number.
Then:
:$\forall m, n \in \Z_{> 0}: F_m \divides F_{m n}$
where $\divides$ denotes divisibility.
== Proof ==
When $m = 1$ or $n = 1$ the result is trivially true.
Otherwise, by definition of divisibility:
:$m \divides m n$
and the result follows from Divisibility of Fibonacci Number.
== Sources ==
* : $11$
* : $\S 1.2.8$: Fibonacci Numbers: $(6)$
* : $11$
Category:Fibonacci Numbers
Category:Divisors
Divisibility of Numerator of Sum of Sequence of Reciprocals

== Theorem ==
Let $p$ be a prime number such that $p > 3$.
Consider the sum of the finite sequence of reciprocals as follows:
:$S = 1 + \dfrac 1 2 + \dfrac 1 3 + \cdots + \dfrac 1 {p - 1}$
Let $S$ be expressed as a fraction in canonical form, that is:
:$S = \dfrac a b$
where $a$ and $b$ are coprime.
Then:
:$p^2 \divides a$
where $\divides$ denotes divisibility.
== Proof ==
=== Lemma ===
== Sources ==
* 
Category:Divisibility of Numerator of Sum of Sequence of Reciprocals
Category:Divisibility
Category:Reciprocals
Divisibility of Numerator of Sum of Sequence of Reciprocals/Lemma

== Lemma for Divisibility of Numerator of Sum of Sequence of Reciprocals ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
:$\ds \sum_{k \mathop = 1}^n \dfrac {\paren {-1}^{k - 1} } k \dbinom n k = \sum_{k \mathop = 1}^n \dfrac 1 k$
where $\dbinom n k$ denotes a binomial coefficient.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds \sum_{k \mathop = 1}^n \dfrac {\paren {-1}^{k - 1} } k \dbinom n k = \sum_{k \mathop = 1}^n \dfrac 1 k$
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$\ds \sum_{k \mathop = 1}^r \dfrac {\paren {-1}^{k - 1} } k \dbinom r k = \sum_{k \mathop = 1}^r \dfrac 1 k$
from which it is to be shown that:
:$\ds \sum_{k \mathop = 1}^{r + 1} \dfrac {\paren {-1}^{k - 1} } k \dbinom {r + 1} k = \sum_{k \mathop = 1}^{r + 1} \dfrac 1 k$
=== Induction Step ===
This is the induction step:
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{>0}: \sum_{k \mathop = 1}^n \dfrac {\paren {-1}^{k - 1} } k \dbinom n k = \sum_{k \mathop = 1}^n \dfrac 1 k$
== Sources ==
* 
Category:Divisibility of Numerator of Sum of Sequence of Reciprocals
Divisibility of Product of Consecutive Integers

== Theorem ==
The product of $n$ consecutive positive integers is divisible by the product of the first $n$ consecutive positive integers.
That is:
:$\ds \forall m, n \in \Z_{>0}: \exists r \in \Z: \prod_{k \mathop = 1}^n \paren {m + k} = r \prod_{k \mathop = 1}^n k$
== Proof ==
Hence the result, and note that for a bonus we have identified exactly what the divisor is: 
:$\dbinom {m + n} m$
== Examples ==
== Sources ==
* : $\text {3-1}$ Permutations and Combinations: Theorem $\text {3-3}$
* : $24$
* : $24$
Category:Divisibility of Product of Consecutive Integers
Category:Divisibility
Category:Factorials
Divisibility of Sum of 3 Fourth Powers

== Theorem ==
Let $n \in \Z_{\ge 0}$ be the sum of three $4$th powers.
Then:
:$n$ is divisible by $5$ all three addends are also divisible by $5$
:$n$ is divisible by $29$ all three addends are also divisible by $29$.
== Proof ==
Let $n = a^4 + b^4 + c^4$ for $a, b, c \in \Z$.
=== Necessary Condition ===
Let $a$, $b$ and $c$ all be divisible by either $5$ or $29$.
That is:
:$a, b, c \equiv 0 \pmod 5$
or:
:$a, b, c \equiv 0 \pmod {29}$
We have that for integer $p$:
Hence if all three addends are divisible by either $5$ or $29$, then so is the sum.
=== Sufficient Condition ===
Let $n$ be divisible by $5$.
Then:
:$a^4 + b^4 + c^4 \equiv 0 \mod 5$
From Fourth Power Modulo 5:
:$x \in \Z_5 \implies x^4 \in \set {0, 1}$
So:
:$a^4 + b^4 + c^4 \equiv 0 \mod 5 \implies a, b, c \equiv 0 \mod 5$
Let $n$ be divisible by $29$.
Then $a^4 + b^4 + c^4 \equiv 0 \mod {29}$.
But for $x \in \Z_{29}$, $x^4 \in \set {0, 1, 7, 16, 20, 23, 24, 25} = R$ necessarily.
So:
:$a^4 + b^4 + c^4 \equiv 0 \mod {29} \implies a, b, c \equiv 0 \mod {29}$
as $29$ and its non-zero multiples are not partitioned by any 3 elements in $R$
== Historical Note ==
== Sources ==
* : $29$
* : $29$
Category:Number Theory
Category:29
Category:5
Divisibility of n-1 Factorial by Composite n

== Theorem ==
Let $n \in \Z$ be composite.
Then:
:$n \divides \paren {n - 1}! \iff n \ne 4$
where:
:$\divides$ denotes divisibility
:$n!$ denotes the factorial of $n$.
== Proof ==
=== Necessary Condition ===
We have that $3! = 6$ and that $4$ does not divide $6$.
So in order for $n$ to divide $\paren {n - 1}!$ it is necessary that $n \ne 4$.
=== Sufficient Condition ===
Let $n \ne 4$ be composite.
Let $n = r s$ where:
:$r, s \in \Z_{> 1}$
:$r \ne s$
:$r, s < n$
This is always possible unless $n = p^2$ for some prime number $p$.
, let $r < s$.
Then by definition of factorial:
:$\paren {n - 1}! = 1 \times 2 \times \ldots \times r \times \ldots \times s \times \ldots \times \paren {n - 2} \times \paren {n - 1}$
and so:
:$n = r s \divides \paren {n - 1}!$
Let $n = p^2$.
As $n \ne 4$, we have that $p \ne 2$.
Hence $p > 2$.
Hence:
:$2 p < p^2$
and so:
:$2 p \divides \paren {n - 1}!$
By definition of factorial:
:$\paren {n - 1}! = 1 \times 2 \times \ldots \times p \times \ldots \times 2 p \times \ldots \times \paren {n - 2} \times \paren {n - 1}$
and so:
:$n = p^2 \divides \paren {n - 1}!$
Hence the result.
== Sources ==
* : $4$
* : $4$
Category:Factorials
Division Algebra has No Zero Divisors

== Theorem ==
Let $A = \struct {A_F, \oplus}$ be an algebra over a field $F$.
Then $A$ is a division algebra it has no zero divisors.
That is:
:$\forall a, b \in A_F: a \oplus b = \mathbf 0_A \implies a = \mathbf 0_A \lor b = \mathbf 0_A$
If the product of two elements of $A$ is zero, then at least one of those elements must itself be zero.
Some sources use this as the definition of a division algebra and from it deduce:
:$\forall a, b \in A_F, b \ne \mathbf 0_A: \exists_1 x \in A_F, y \in A_F: a = b \oplus x, a = y \oplus b$
== Proof ==
Let $A$ be a division algebra, in the sense that:
:$\forall a, b \in A_F, b \ne \mathbf 0_A: \exists_1 x \in A_F, y \in A_F: a = b \oplus x, a = y \oplus b$
Suppose that:
:$\exists a, b \in A_F \setminus \set {\mathbf 0_A}: \mathbf 0_A = b \oplus a$
Then by definition of the zero vector we also have that $\mathbf 0_A = b \oplus \mathbf 0_A$.
So there are two elements $x$ of $A_F$ such that $\mathbf 0_A = b \oplus x$, that is, $a$ and $\mathbf 0_A$.
Similarly, suppose that:
:$\exists a, b \in A_F \setminus \set {\mathbf 0_A}: \mathbf 0_A = a \oplus b$
Then by definition of the zero vector we also have that:
:$\mathbf 0_A = \mathbf 0_A \oplus b$
So there are two elements $y$ of $A_F$ such that:
:$\mathbf 0_A = y \oplus b$
that is:
:$a$ and $\mathbf 0_A$
So $A$ can not be a division algebra.
So it follows that if:
:$\forall a, b \in A_F, b \ne \mathbf 0_A: \exists_1 x \in A_F, y \in A_F: a = b \oplus x, a = y \oplus b$
then:
:$\forall a, b \in A_F: a \oplus b = \mathbf 0_A \implies a = \mathbf 0_A \lor b = \mathbf 0_A$
Now suppose that:
:$\forall a, b \in A_F: a \oplus b = \mathbf 0_A \implies a = \mathbf 0_A \lor b = \mathbf 0_A$
Suppose $\exists x_1, x_2 \in A_F, x_1 \ne x_2$ such that:
:$a = b \oplus x_1$
:$a = b \oplus x_2$
We have that $x_1 \ne x_2$ and so $x_1 - x_2 = z \ne \mathbf 0_A$.
Thus $b \oplus x_1 - b \oplus x_2 = \mathbf 0_A$.
As $\oplus$ is a bilinear mapping, it follows that:
:$b \oplus \paren {x_1 - x_2} = \mathbf 0_A$ and so $b \oplus z = \mathbf 0_A$
But $z \ne \mathbf 0_A$ and so it is not the case that:
:$\forall a, b \in A_F: a \oplus b = \mathbf 0_A \implies a = \mathbf 0_A \lor b = \mathbf 0_A$
Similarly it can be shown that if $\exists y_1, y_2 \in A_F, y_1 \ne y_2$ such that:
:$a = y_1 \oplus b$
:$a = y_2 \oplus b$
then it is not the case that:
:$\forall a, b \in A_F: a \oplus b = \mathbf 0_A \implies a = \mathbf 0_A \lor b = \mathbf 0_A$
Now suppose that:
:$\exists a, b \in A_F, b \ne \mathbf 0_A: \not \exists_1 x \in A_F: a = b \oplus x$
or
:$\exists a, b \in A_F, b \ne \mathbf 0_A: \not \exists_1 y \in A_F: a = y \oplus b$
Thus it is not the case that $\forall a, b \in A_F: a \oplus b = \mathbf 0_A \implies a = \mathbf 0_A \lor b = \mathbf 0_A$.
So it follows that if:
:$\forall a, b \in A_F: a \oplus b = \mathbf 0_A \implies a = \mathbf 0_A \lor b = \mathbf 0_A$
then:
:$\forall a, b \in A_F, b \ne \mathbf 0_A: \exists_1 x \in A_F, y \in A_F: a = b \oplus x, a = y \oplus b$
Category:Division Algebras
Division Laws for Groups

== Theorem ==
Let $G$ be a group.
Let $a, b, x \in G$.
Then:
:$(1): \quad a x = b \iff x = a^{-1} b$
:$(2): \quad x a = b \iff x = b a^{-1}$
== Proof ==
All derivations can be achieved using applications of the group axioms:
=== Proof of $(1)$ ===
and the converse:
=== Proof of $(2)$ ===
and the converse:
== Sources ==
* : $\S 35.2$: Elementary consequences of the group axioms
Category:Group Theory