nyuuzyou/wikis huggingface dataset:
Integer Addition is Closed

== Theorem ==
The set of integers is closed under addition:
:$\forall a, b \in \Z: a + b \in \Z$
== Proof ==
Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the formal definition of integers.
That is, $\eqclass {\tuple {a, b} } \boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$.
$\boxminus$ is the congruence relation defined on $\N \times \N$ by:
:$\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$
In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, as suggested.
Integer addition is defined as:
:$\forall a, b, c, d \in \N: \eqclass {a, b} {} + \eqclass {c, d} {} = \eqclass {a + c, b + d} {}$
We have that:
:$\forall a, b, c, d \in \N: \eqclass {a, b} {} \in \Z, \eqclass {c, d} {} \in \Z$
Also:
:$\forall a, b, c, d \in \N: \eqclass {a, b} {} + \eqclass {c, d} {} = \eqclass {a + c, b + d} {}$
But:
:$a + c \in \N, b + d \in \N$
So:
:$\forall a, b, c, d \in \N: \eqclass {a + c, b + d} {} \in \Z$
Therefore integer addition is closed.
== Sources ==
* : Chapter $1$: Integral Domains: $\S 2$. Operations: Example $1$
* : $1$: Complex Numbers: The Real Number System: $2$
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers
Category:Integer Addition
Category:Algebraic Closure
Integer Addition is Commutative

== Theorem ==
The operation of addition on the set of integers $\Z$ is commutative:
:$\forall x, y \in \Z: x + y = y + x$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Introduction $\S 5$: The system of integers
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: $\mathbf Z. \, 2$
* : $\S 1$
Category:Integer Addition
Category:Examples of Commutative Operations
Category:Integer Addition is Commutative
Category:Commutative Law of Addition
Integer Addition is Well-Defined

== Theorem ==
Let $\struct {\N, +}$ be the semigroup of natural numbers under addition.
Let $\struct {\N \times \N, \oplus}$ be the (external) direct product of $\struct {\N, +}$ with itself, where $\oplus$ is the operation on $\N \times \N$ induced by $+$ on $\N$.
Let $\boxtimes$ be the cross-relation defined on $\N \times \N$ by:
:$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$
Let $\eqclass {x, y} {}$ denote the equivalence class of $\tuple {x, y}$ under $\boxtimes$.
Let:
:$\paren {\N \times \N} / \boxtimes$
be the quotient set induced by $\boxtimes$.
The operation $\oplus$ on these equivalence classes is well-defined, in the sense that:
for all $\eqclass {a_1, b_1} {}, \eqclass {a_2, b_2} {}, \eqclass {c_1, d_1} {}, \eqclass {c_2, d_2} {} \in \paren {\N \times \N} / \boxtimes$
== Proof 1 ==
== Proof 2 ==
Category:Integer Addition
Category:Integer Addition is Well-Defined
Integer Addition is Well-Defined/Proof 1

== Theorem ==
== Proof ==
Let $\eqclass {a_1, b_1} {}, \eqclass {a_2, b_2} {}, \eqclass {c_1, d_1} {}, \eqclass {c_2, d_2} {}$ be $\boxtimes$-equivalence classes such that $\eqclass {a_1, b_1} {} = \eqclass {a_2, b_2} {}$ and $\eqclass {c_1, d_1} {} = \eqclass {c_2, d_2} {}$.
Then:
Then we have:
== Sources ==
* : Introduction $\S 5$: The system of integers
* : $\S 2.5$: Theorem $2.22$
Category:Integer Addition is Well-Defined
Integer Addition is Well-Defined/Proof 2

== Theorem ==
== Proof ==
Consider the formal definition of the integers: $x = \eqclass {a, b} {}$ is an equivalence class of ordered pairs of natural numbers.
Consider the mapping $\phi: \N_{>0} \to \Z_{>0}$ defined as:
:$\forall u \in \N_{>0}: \map \phi u = u'$
where $u' \in \Z$ be the (strictly) positive integer $\eqclass {b + u, b} {}$.
Let $v' = \eqclass {c + v, c} {}$.
Then:
== Sources ==
* : Introduction $\S 5$: The system of integers
Category:Integer Addition is Well-Defined
Integer Arbitrarily Close to Rational in Valuation Ring of P-adic Norm

== Theorem ==
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime number $p$.
Let $x \in \Q$ such that $\norm x_p \le 1$.
Then for all $i \in \N$ there exists $\alpha \in \Z$ such that:
:$\norm {x - \alpha}_p \le p^{-i}$
== Proof ==
Let $i \in \N$.
Let $x = \dfrac a b: a, b \in \Z \text{ and } b \ne 0$.
 we can assume that $\dfrac a b$ is in canonical form.
By Valuation Ring of P-adic Norm on Rationals:
:$\dfrac a b \in \Z_{\paren p} = \set {\dfrac c d \in \Q : p \nmid d}$
So $p \nmid b$.
Since $p \nmid b$, by Prime not Divisor implies Coprime then $p^i \perp b$.
By Integer Combination of Coprime Integers, there exists $m, l \in \Z: m b + l p^i = 1$.
Then:
As $am \in \Z$, the result follows with $\alpha = am$.
== Also see ==
* Unique Integer Close to Rational in Valuation Ring of P-adic Norm
== Sources ==
* : $\S 1.4$ The field of $p$-adic numbers $\Q_p$: Lemma $1.29$
Category:P-adic Number Theory
Integer Combination of Coprime Integers

== Theorem ==
Let $a, b \in \Z$ be integers, not both zero.
Then:
:$a$ and $b$ are coprime
:
:there exists an integer combination of them equal to $1$:
::$\forall a, b \in \Z: a \perp b \iff \exists m, n \in \Z: m a + n b = 1$
In such an integer combination $m a + n b = 1$, the integers $m$ and $n$ are also coprime.
=== General Result ===
== Proof ==
The proof can conveniently be divided into two parts:
=== Sufficient Condition ===
=== Necessary Condition ===
== Also known as ==
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Theorem $2 \text{-} 4$
* : Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$: Lemma $1$
Category:Integer Combination of Coprime Integers
Category:Coprime Integers
Category:Integer Combinations
Integer Combination of Coprime Integers/Also known as

== Integer Combination of Coprime Integers: Also known as ==
This result is sometimes known as Bézout's Identity, although that result is usually used for the more general result for not necessarily coprime integers.
Some sources refer to this result as the Euclidean Algorithm, but the latter as generally understood is the procedure that can be used to establish the values of $m$ and $n$, and for any pair of integers, not necessarily coprime.
Category:Integer Combination of Coprime Integers
Integer Combination of Coprime Integers/General Result

== Theorem ==
Let $a_1, a_2, \ldots, a_n$ be integers.
Then $\gcd \set {a_1, a_2, \ldots, a_n} = 1$ there exists an integer combination of them equal to $1$:
:$\exists m_1, m_2, \ldots, m_n \in \Z: \ds \sum_{k \mathop = 1}^n m_k a_k = 1$
== Proof ==
First let $\exists m_1, m_2, \ldots, m_n \in \Z: \ds \sum_{k \mathop = 1}^n m_k a_k = 1$.
Let $\gcd \set {a_1, a_2, \ldots, a_n} = d$.
Then $\ds \sum_{k \mathop = 1}^n m_k a_k$ has $d$ as a divisor.
That means $d$ is a divisor of $1$.
Thus $\gcd \set {a_1, a_2, \ldots, a_n} = 1$.
It remains to be shown that if $\gcd \set {a_1, a_2, \ldots, a_n} = 1$, then $\exists m_1, m_2, \ldots, m_n \in \Z: \ds \sum_{k \mathop = 1}^n m_k a_k = 1$.
The proof proceeds by induction.
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
:$\gcd \set {a_1, a_2, \ldots, a_n} = 1 \iff \exists m_1, m_2, \ldots, m_n \in \Z: \ds \sum_{k \mathop = 1}^n m_k a_k = 1$
=== Basis for the Induction ===
$\map P 2$ is the case:
:$\gcd \set {a_1, a_2} = 1 \iff \exists m_1, m_2 \in \Z: m_1 a_1 + m_2 a_2 = 1$
This is demonstrated in Integer Combination of Coprime Integers.
Thus $\map P 2$ 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 $r \ge 2$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$\gcd \set {a_1, a_2, \ldots, a_r} = 1 \iff \exists m_1, m_2, \ldots, m_r \in \Z: \ds \sum_{k \mathop = 1}^r m_k a_k = 1$
from which it is to be shown that:
:$\gcd \set {a_1, a_2, \ldots, a_r, a_{r + 1} } = 1 \iff \exists m_1, m_2, \ldots, m_r, m_{r + 1} \in \Z: \ds \sum_{k \mathop = 1}^{r + 1} m_k a_k = 1$
=== Induction Step ===
This is the induction step:
Let $\gcd \set {a_1, a_2, \ldots, a_r, a_{r + 1} } = 1$.
Let $\gcd \set {a_1, a_2, \ldots, a_r} = d$.
Dividing through by $d$, we have:
:$\gcd \set {\dfrac {a_1} d, \dfrac {a_2} d, \ldots, \dfrac {a_d} d} = 1$
By the induction hypothesis:
:$\exists m_1, m_2, \ldots, m_r \in \Z: \ds \sum_{k \mathop = 1}^r m_k a_k = 1$
whence multiplying through by $d$:
:$\exists m'_1, m'_2, \ldots, m'_r \in \Z: \ds \sum_{k \mathop = 1}^r m'_k a_k = d$
where $m'_k = m_k d$ throughout.
Then :
:$\gcd \set {d, a_{r + 1} } = 1$
From the basis for the induction:
:$\exists p, q \in \Z: p d + 1 a_{r + 1} = 1$
We take:
:$m_1 = m'_1 p, m_2 = m'_2 p, \ldots, m_r = m'_r p$
and:
:$m_{r + 1} = q$
and thus we have:
:$\gcd \set {a_1, a_2, \ldots, a_r, a_{r + 1} } = 1 \iff \exists m_1, m_2, \ldots, m_r, m_{r + 1} \in \Z: \ds \sum_{k \mathop = 1}^{r + 1} m_k a_k = 1$
== Sources ==
* : Appendix: Elementary set and number theory
Category:Integer Combination of Coprime Integers
Integer Combination of Coprime Integers/Necessary Condition

== Theorem ==
Let $a, b \in \Z$ be integers, not both zero.
Let $a$ and $b$ be such that there exists an integer combination of them equal to $1$.
Then $a$ and $b$ are coprime:
:$\forall a, b \in \Z: \exists m, n \in \Z: m a + n b = 1 \implies a \perp b$
In such an integer combination $m a + n b = 1$, the integers $m$ and $n$ are also coprime.
== Proof 1 ==
== Proof 2 ==
== Also known as ==
Category:Integer Combination of Coprime Integers
Integer Combination of Coprime Integers/Necessary Condition/Proof 2

== Theorem ==
== Proof ==
Let $a, b \in \Z$ be such that $\exists m, n \in \Z: m a + n b = 1$.
Let $d$ be a divisor of both $a$ and $b$.
Then:
:$d \divides m a + n b$
and so:
:$d \divides 1$
Thus:
:$d = \pm 1$
and so:
:$\gcd \set {a, b} = 1$
Thus, by definition, $a$ and $b$ are coprime.
== Also known as ==
== Sources ==
* : Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$: Lemma $1$
Category:Integer Combination of Coprime Integers
Integer Combination of Coprime Integers/Sufficient Condition

== Theorem ==
Let $a, b \in \Z$ be integers, not both zero.
Let $a$ and $b$ be coprime to each other.
Then there exists an integer combination of them equal to $1$:
:$\forall a, b \in \Z: a \perp b \implies \exists m, n \in \Z: m a + n b = 1$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also known as ==
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 6$: Examples of Finite Groups: $\text{(iii)}$: $(1.34)$
* : $0$: Some Conventions and some Basic Facts
Category:Integer Combination of Coprime Integers
Integer Combination of Coprime Integers/Sufficient Condition/Proof 2

== Theorem ==
== Proof ==
Let $a \perp b$.
Thus they are not both $0$.
Let $S$ be defined as:
:$S = \set {a m + b n: m, n \in \Z}$
$S$ contains at least one strictly positive integer, because for example $a^2 + b^2 \in S$.
By Set of Integers Bounded Below has Smallest Element, let $d$ be the smallest element of $S$ which is strictly positive.
Let $d = a x + b y$.
It remains to be shown that $d = 1$.
By the Division Theorem:
:$a = d q + r$ where $0 \le r < d$
Then:
But we have that $0 \le r < d$.
We have defined $d$ as the smallest element of $S$ which is strictly positive
Hence it follows that $r$ cannot therefore be strictly positive itself.
Hence $r = 0$ and so $a = d q$.
That is:
:$d \divides a$
By a similar argument:
:$d \divides b$
and so $d$ is a common divisor of both $a$ and $b$.
But the GCD of $a$ and $b$ is $1$.
Thus it follows that, as $d \in S$:
:$\exists m, n \in \Z: m a + n b = 1$
== Also known as ==
== Sources ==
* : Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$: Lemma $1$
Category:Integer Combination of Coprime Integers
Integer Combination of Coprime Integers/Sufficient Condition/Proof 3

== Theorem ==
== Proof ==
Let $a \perp b$.
Thus they are not both $0$.
Let $S$ be defined as:
:$S = \set {\lambda a + \mu b: \lambda, \mu \in \Z}$
$S$ contains at least one strictly positive integer, because for example:
:$a \in S$ (setting $\lambda = 1$ and $\mu = 0$)
:$b \in S$ (setting $\lambda = 0$ and $\mu = 1$)
By Set of Integers Bounded Below has Smallest Element, let $d$ be the smallest element of $S$ which is strictly positive.
Let $d = \alpha a + \beta b$.
Let $c \in S$, such that $\lambda_0 a + \mu_0 b = c$ for some $\lambda_0, \mu_0 \in \Z$.
By the Division Algorithm:
:$\exists \gamma, \delta \in \Z: c = \gamma d + \delta$
where $0 \le \delta < d$
Then:
But we have that $0 \le \delta < d$.
We have defined $d$ as the smallest element of $S$ which is strictly positive
Hence it follows that $\delta$ cannot therefore be strictly positive itself.
Hence $\delta = 0$ and so $c = \gamma d$.
That is:
:$d \divides c$
and so the smallest element of $S$ which is strictly positive is a divisor of both $a$ and $b$.
But $a$ and $b$ are coprime.
Thus it follows that, as $d \divides 1$:
:$d = 1$
and so, by definition of $S$:
:$\exists m, n \in \Z: m a + n b = 1$
== Also known as ==
== Sources ==
* : Appendix: Elementary set and number theory
* : Appendix $1$: Some Results from the Elementary Theory of Numbers: Theorem $7$
Category:Integer Combination of Coprime Integers
Integer Coprime to Factors is Coprime to Whole

== Theorem ==
Let $a, b, c \in \Z$ be integers.
Let:
:$a \perp b$
:$a \perp c$
where $\perp$ denotes coprimality.
Then:
:$a \perp b c$
=== Integer Coprime to all Factors is Coprime to Whole ===
== Proof 1 ==
== Proof 2 ==
Category:Integer Coprime to Factors is Coprime to Whole
Category:Coprime Integers
Integer Coprime to Factors is Coprime to Whole/Proof 1

== Theorem ==
== Proof ==
This proof follows the structure of 's proof, if not its detail.
Let $a, b, c \in \Z$ such that $c$ is coprime to each of $a$ and $b$.
Let $d = a b$.
 $c$ and $d$ are not coprime.
Then:
:$\exists e \in \Z_{>1}: e \divides c, e \divides d$
We have that $c \perp a$ and $e \divides c$
From :
:$e \perp a$
As $e \divides d$ it follows by definition of divisibility that:
:$\exists f \in \Z: e f = d$
We also have that:
:$a b = d$
So:
:$a b = e f$
But from :
:$e : a = b : f$
or in more contemporary language:
:$\dfrac a e = \dfrac b f$
But $a \perp e$.
From , $\dfrac a e$ is in canonical form.
From :
:$e \divides b$
But we already have that:
:$e \divides c$
So $e$ is a divisor of both $b$ and $c$.
But this is a contradiction of the assertion that $c$ and $b$ are coprime.
Hence the result.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Integer Coprime to Factors is Coprime to Whole
Integer Coprime to Factors is Coprime to Whole/Proof 2

== Theorem ==
== Proof ==
Let $a, b, c \in \Z$ such that $a$ is coprime to each of $b$ and $c$.
We have:
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $16 \ \text {(a)}$
Category:Integer Coprime to Factors is Coprime to Whole
Integer Coprime to Modulus iff Linear Congruence to 1 exists

== Theorem ==
Let $a, m \in \Z$.
The linear congruence:
:$a x \equiv 1 \pmod m$
has a solution $x$ $a$ and $m$ are coprime.
=== Corollary ===
== Proof ==
From Integer Combination of Coprime Integers:
:$a \perp m \iff \exists x, y \in \Z: a x + m y = 1$
That is, such an $x$ exists $a$ and $m$ are coprime.
== Sources ==
* : $\S 2.3$: Congruences: Proposition $2$
Category:Modulo Arithmetic
Integer Coprime to Modulus iff Linear Congruence to 1 exists/Corollary

== Theorem ==
Let $p$ be a prime number.
The linear congruence:
:$a x \equiv 1 \pmod p$
has a solution $x$ $a \not \equiv 0 \pmod p$.
== Proof ==
By definition of congruence:
:$a \not \equiv 0 \pmod p \iff p \nmid a$
where $p \nmid a$ denotes that $p$ is not a divisor of $a$.
From Prime not Divisor implies Coprime:
:$p \nmid a \iff p \perp a$
where $p \perp a$ denotes that $p$ and $a$ are coprime.
The result follows from Integer Coprime to Modulus iff Linear Congruence to 1 exists.
== Sources ==
* : $\S 2.3$: Congruences: Proposition $2$: Corollary
Category:Modulo Arithmetic
Integer Coprime to all Factors is Coprime to Whole

== Theorem ==
Let $a, b \in \Z$.
Let $\ds b = \prod_{j \mathop = 1}^r b_j$
Let $a$ be coprime to each of $b_1, \ldots, b_r$.
Then $a$ is coprime to $b$.
== Proof ==
From Integer Combination of Coprime Integers:
:$\forall j \in \set {1, 2, \ldots, r}: a x_j + b_j y_j = 1$
for some $x_j, y_j \in \Z$.
Thus:
:$\ds \prod_{j \mathop = 1}^r b_j y_j = \prod_{j \mathop = 1}^r \paren {1 - a x_j}$
But $\ds \prod_{j \mathop = 1}^r \paren {1 - a x_j}$ is of the form $1 - a z$.
Thus:
and so, by Integer Combination of Coprime Integers, $a$ and $b$ are coprime.
== Also see ==
* Integer Coprime to Factors is Coprime to Whole: the specific case where $r = 2$
== Sources ==
* : Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$: Lemma $2$
Category:Integer Coprime to Factors is Coprime to Whole
Integer Divided by Divisor is Integer

== Theorem ==
Let $a, b \in \N$.
Then:
:$b \divides a \implies \dfrac 1 b \times a \in \N$
where $\divides$ denotes divisibilty.
== Proof ==
Let $b \divides a$.
By definition of divisibilty:
:$\exists c \in \N: c \times b = a$
Then also:
:$c \times 1 = c$
So by :
:$1 : b = c : a$
Hence the result.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Divisors
Integer Division/Examples/29 Divided by 8

== Example of Integer Division ==
:$29 \div 8 = 3 \rem 5$
== Proof ==
We have that:
:$29 = 3 \times 8 + 5$
Hence the result by definition of integer division.
== Sources ==
* : $\S 2.5$. Congruence of integers: Example $39$
Category:Examples of Integer Division
Integer Divisor Results

== Theorem ==
Let $m, n \in \Z$ be integers.
Let $m \divides n$ denote that $m$ is a divisor of $n$.
The following results all hold:
=== One Divides all Integers ===
=== Integer Divides Itself ===
=== Integer Divides its Negative ===
=== Integer Divides its Absolute Value ===
=== Integer Divides Zero ===
=== Divisors of Negative Values ===
Category:Divisors
Category:Integers
Category:Integer Divisor Results
Integer Divisor Results/Divisors of Negative Values

== Theorem ==
Let $m, n \in \Z$, i.e. let $m, n$ be integers.
:$m \divides n \iff -m \divides n \iff m \divides -n \iff -m \divides -n$
That is, $m$ divides $n$ $-m$ divides $n$ $m$ divides $-n$ $-m$ divides $-n$.
== Proof ==
Let $m \divides n$.
From Integer Divides its Negative, we have $-m \divides m$.
From Divisor Relation on Positive Integers is Partial Ordering it follows that $-m \divides n$.
From Integer Divides its Negative, we have $n \divides -n$.
From Divisor Relation on Positive Integers is Partial Ordering it follows that $m \divides -n$.
From above, we have $-m \divides n$.
From Integer Divides its Negative, we have $n \divides -n$.
From Divisor Relation on Positive Integers is Partial Ordering it follows that $-m \divides -n$.
The reverse implications follow similarly.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $1$
Category:Divisors
Integer Divisor Results/Integer Divides Itself

== Theorem ==
Let $n \in \Z$ be an integer.
Then:
:$n \divides n$
That is, $n$ divides itself.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $3$: The Integers: $\S 10$. Divisibility
* : $\text {2-2}$ Divisibility: Example $\text {2-2}$
* : Chapter $1$: Properties of the Natural Numbers: $\S 22$
* : $\text I$: The Series of Primes: $1.1$ Divisibility of integers
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Theorem $2 \text{-} 2 \ (1)$
* : Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$: $\mathbf D. \, 2$
* : Appendix $1$: Some Results from the Elementary Theory of Numbers
* : $1$ Divisibility: $1.1$ The Euclidean algorithm and unique factorization: $\text {(i)}$
Category:Divisors
Category:Integer Divisor Results/Integer Divides Itself
Integer Divisor Results/Integer Divides Zero

== Theorem ==
Let $n \in \Z$ be an integer.
Then:
:$n \divides 0$
That is, $n$ divides $0$.
== Proof ==
From Integers form Integral Domain, the concept divides is fully applicable to the integers.
Therefore this result follows directly from Element of Integral Domain Divides Zero.
== Sources ==
* : Chapter $3$: The Integers: $\S 10$. Divisibility
* : $\text {2-2}$ Divisibility: Example $\text {2-3}$
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Theorem $2 \text{-} 2 \ (1)$
* : Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$
* : $1$ Divisibility: $1.1$ The Euclidean algorithm and unique factorization: $\text {(i)}$
Category:Divisors
Category:Integer Divisor Results
Integer Divisor Results/Integer Divides its Absolute Value

== Theorem ==
Let $n \in \Z$ be an integer.
Then:
where:
:$\size n$ is the absolute value of $n$
:$\divides$ denotes divisibility.
== Proof ==
Let $n > 0$.
Then $\size n = n$ and Integer Divides Itself applies.
Let $n = 0$.
Then Integer Divides Itself holds again.
Let $n < 0$.
Then $\size n = -n$ and Integer Divides its Negative applies.
Category:Divisors
Category:Absolute Value Function
Integer Divisor Results/Integer Divides its Negative

== Theorem ==
Let $n \in \Z$ be an integer.
Then:
where $\divides$ denotes divisibility.
== Proof ==
From Integers form Integral Domain, the integers are an integral domain.
Hence we can apply Product of Ring Negatives:
:$\forall n \in \Z: \exists -1 \in \Z: n = \paren {-1} \times \paren {-n}$
and Product with Ring Negative:
:$\forall n \in \Z: \exists -1 \in \Z: -n = \paren {-1} \times \paren n$
== Sources ==
* : $\text {2-2}$ Divisibility: Example $\text {2-2}$
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor
Category:Divisors
Integer Divisor Results/One Divides all Integers

== Theorem ==
Let $n \in \Z$ be an integer.
Then:
where $\divides$ denotes divisibility.
== Proof ==
From Integers form Integral Domain, the concept divisibility is fully applicable to the integers.
Therefore this result follows directly from Unity Divides All Elements.
== Sources ==
* : $\text {2-2}$ Divisibility: Example $\text {2-2}$
* : Chapter $1$: Properties of the Natural Numbers: $\S 22$
* : $\text I$: The Series of Primes: $1.1$ Divisibility of integers
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Theorem $2 \text{-} 2 \ (1)$
* : Appendix $1$: Some Results from the Elementary Theory of Numbers
* : $1$
* : $1$ Divisibility: $1.1$ The Euclidean algorithm and unique factorization: $\text {(i)}$
Category:Divisors
Integer Divisor is Equivalent to Subset of Ideal

== Theorem ==
Let $\Z$ be the set of all integers.
Let $\Z_{>0}$ be the set of strictly positive integers.
Let $m \in \Z_{>0}$ and let $n \in \Z$.
Let $\ideal m$ be the principal ideal of $\Z$ generated by $m$.
Then:
:$m \divides n \iff \ideal n \subseteq \ideal m$
== Proof ==
The ring of integers is a principal ideal domain.
The result follows directly from Principal Ideals in Integral Domain.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Theorem $24.5$
Category:Integers
Category:Ideal Theory
Integer Heronian Triangle can be Scaled so Area equals Perimeter

== Theorem ==
Let $T_1$ be an integer Heronian triangle whose sides are $a$, $b$ and $c$.
Then there exists a rational number $k$ such that the Heronian triangle $T_2$ whose sides are $k a$, $k b$ and $k c$ such that the perimeter of $T$ is equal to the area of $T$.
== Proof ==
For a given triangle $T$:
:let $\map \AA T$ denote the area of $T$
:let $\map P T$ denote the perimeter of $T$.
We are given that $T_1$ is an integer Heronian triangle whose sides are $a$, $b$ and $c$.
Let $\map P {T_1} = k \map \AA {T_1}$.
Let $T_2$ have sides $k a$, $k b$ and $k c$.
Then we have that:
Category:Heronian Triangles
Integer Lattice/Examples/2-Dimensional

== Example of Integer Lattice ==
This is an instance of an integer lattice of $2$ dimensions:
:
Not only is $\set {\tuple {0, 1}, \tuple {1, 0} }$ a basis of this integer lattice, so is the set $\set {\tuple {0, -1}, \tuple {1, 1} }$.
== Sources ==
* 
* 
Category:Examples of Integer Lattices