nyuuzyou/wikis huggingface dataset:
Integer to Power of Multiple of Order

== Theorem ==
Let $a$ and $n$ be integers.
Let $a \perp n$, that is, let $a$ and $b$ be coprime.
Let $c \in \Z_{>0}$ be the multiplicative order of $a$ modulo $n$.
Then $a^k \equiv 1 \pmod n$ $k$ is a multiple of $c$.
=== Corollary ===
== Proof ==
First note from Integer has Multiplicative Order Modulo n iff Coprime to n that unless $a \perp n$ the multiplicative order of $a$ modulo $n$ does not exist.
=== Necessary Condition ===
Let $k$ be a multiple of $c$.
Then $k = c r$, say.
Then:
:$a^k = a^{c r} = \paren {a^c}^r \equiv 1^r \equiv 1 \pmod n$
=== Sufficient Condition ===
Let $a^k \equiv 1 \pmod n$.
Then:
:$1 \equiv a^k \equiv a^{q c + r} \equiv \paren {a^c}^q a^r \equiv 1^q a^r \equiv a^r \pmod n$
So $r = 0$ or else (from the Division Theorem) $0 < r < c$ and this contradicts $c$ being the smallest integer such that $a^c \equiv 1 \pmod n$.
Hence $k$ is a multiple of $c$.
Category:Integer to Power of Multiple of Order
Category:Number Theory
Integer to Power of Multiple of Order/Corollary

== Theorem ==
Let $a$ and $n$ be integers.
Let $a \perp n$, that is, let $a$ and $b$ be coprime.
Let $c \in \Z_{>0}$ be the multiplicative order of $a$ modulo $n$.
Then $\map \phi n$ is a multiple of $c$, where $\map \phi n$ is the Euler phi function of $n$.
== Proof ==
From Euler's Theorem (Number Theory):
:$a^{\map \phi n} \equiv 1 \pmod n$
Applying Integer to Power of Multiple of Order we see that $\map \phi n$ is a multiple of $c$.
Category:Integer to Power of Multiple of Order
Category:Number Theory
Integer to Power of p-1 over 2 Modulo p

== Theorem ==
Let $a \in \Z$.
Let $p$ be an odd prime.
Let $b = a^{\frac {\paren {p - 1} } 2}$.
Then one of the following cases holds:
:$b \bmod p = 0$
which happens exactly when $a \equiv 0 \pmod p$, or:
:$b \bmod p = 1$
or:
:$b \bmod p = p - 1$
where:
:$b \bmod p$ denotes the modulo operation
:$x \equiv y \pmod p$ denotes that $x$ is congruent modulo $p$ to $y$.
== Proof ==
By definition of congruence modulo $p$:
:$\forall x, y \in \R: x \equiv y \pmod p \iff x \bmod p = y \bmod p$ 
We have that:
:$b = a^{\frac{\paren {p - 1} } 2}$
and so:
:$b^2 = a^{p - 1}$
Let $a \equiv 0 \pmod p$.
Then by definition of congruence modulo $p$:
:$p \divides a$
and so:
:$p \divides a^{\frac{\paren {p - 1} } 2}$
where $\divides$ denotes divisibility.
Thus by definition of congruence modulo $p$:
:$b \equiv 0 \pmod p$
and so:
:$b \bmod p = 0$
Otherwise, from Fermat's Little Theorem:
:$b^2 \equiv 1 \pmod p$
That is:
:$b^2 - 1 \equiv 0 \pmod p$
From Difference of Two Squares:
:$b^2 - 1 = \paren {b + 1} \paren {b - 1}$
So either:
:$p \divides b + 1$
or:
:$p \divides b - 1$
 both $p \divides b + 1$ and $p \divides b - 1$.
Then by Modulo Subtraction is Well-Defined:
:$p \divides \paren {b + 1} - \paren {b - 1} = 2$
But $p$ is an odd prime.
So it cannot be the case that $p \divides 2$.
From this contradiction it follows that 
Note that $p$ cannot divide both $b + 1$ and $b - 1$.
So either:
or:
Hence the result.
== Also see ==
* Euler's Criterion
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $26$
Category:Modulo Arithmetic
Category:Number Theory
Integer to Rational Power is Irrational iff not Integer or Reciprocal

== Theorem ==
Let $m$ be a rational number.
Let $n$ be a positive integer.
Then $n^m$ is an irrational number $n^{\size m}$ is not an integer.
== Proof ==
=== Necessary Condition ===
Let $n^{\size m} \notin \Z$.
We have that:
:$n^m = n^{u/v}$ for some $u \in\Z$ and $v \in \Z_{\ne 0}$
Then it follows from the definition of a rational power and the existence of a real $v$th root of $n^u$ that:
:$n^m = n^{u/v} = \paren {n^u}^{1/v} \in \R$
 $n^m \in \Q$.
We have that:
:$\size m = 0 \implies n^{\size m} \in \Z$
It follows that:
:$\size m > 0$
Then:
:$\size m > 0$
:$m \in \Q$
:$n^{\size m} > 0$
:$n^m \in \Q$
imply that
:$n^{\size m} = n^{p/q} = \dfrac r s$
for some $\tuple {p, q, r, s} \in \Z_{>0}^4$ where $r$ and $s$ have no common prime factors.
Raising both sides of $n^{p/q} = \dfrac r s$ to the power of $q$ yields:
:$n^p = \dfrac {r^q} {s^q}$
We have that:
:$\dfrac {r^q} {s^q} = n^p \in \Z$
Hence $r^q$ is divisible by $s^q$.
By the Fundamental Theorem of Arithmetic, $r^q$ and $s^q$ have the same prime factors as $r$ and $s$ respectively.
But we have that:
:$s^q \divides r^q$
and:
:$s \ne 1$
imply that $r$ and $s$ have a common prime factor.
But this would contradict the fact that $r$ and $s$ have no common prime factors.
Therefore $s = 1$.
Then:
:$n^{\size m} = \dfrac r s$
and:
:$s = 1$
imply that:
:$r = n^{\size m} \notin \Z$
From this contradiction it follows that:
:$n^m \notin \Q$
and since $n^m \in \R$:
:$n^m \in \R \setminus \Q$
That is, $n^m$ is an irrational number.
=== Sufficient Condition ===
Now let $n^m$ be an irrational number.
That is:
:$n^m \in \R \setminus \Q$
Then because $n^{-m}$ is the reciprocal of $n^m$:
:$n^{-m} \in \R \setminus \Q$
So:
:$n^{\size m} \in \R \setminus \Q$
and so since $\Z \subseteq \Q$:
:$n^{\size m} \notin \Z$
Category:Number Theory
Category:Irrational Numbers
Integer which is Multiplied by 9 when moving Last Digit to First

== Theorem ==
Let $N$ be the positive integer:
:$N = 10 \, 112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 808 \, 988 \, 764 \, 044 \, 943 \, 820 \, 224 \, 719$
$N$ is the smallest positive integer $N$ such that if you move the last digit to the front, the result is the positive integer $9 N$.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Integer which is Multiplied by Last Digit when moving Last Digit to First: the example given here is a special case
== Sources ==
* : $10,112,359,550,561,797,752,808,988,764,044,943,820,224,719$
Category:Integer which is Multiplied by 9 when moving Last Digit to First
Category:Recreational Mathematics
Integer which is Multiplied by 9 when moving Last Digit to First/Corollary

== Corollary to Integer which is Multiplied by 9 when moving Last Digit to First ==
Let $N$ be the positive integer:
:$N = 10 \, 112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 808 \, 988 \, 764 \, 044 \, 943 \, 820 \, 224 \, 719$
We have that $N$ is the smallest positive integer $N$ such that if you move the last digit to the front, the result is the positive integer $9 N$.
The positive integers formed by concatenating the decimal representation of $N$ with itself any number of times have the same property:
:$\sqbrk {NN}, \sqbrk {NNN}, \sqbrk {NNNN}, \ldots$
== Proof ==
Let $N$ be written in the form:
:$\sqbrk {abc \ldots xyz}$
where $a \ne 0$.
Let us define the recurring decimal:
:$F = \sqbrk {0 \cdotp \dot abc \ldots xy \dot z}$
Hence :
:$9 F = \sqbrk {0 \cdotp \dot zabc \ldots x \dot y}$
Therefore:
:$90 F = \sqbrk {z \cdotp abc \ldots xy \dot z}$
Hence by subtraction:
:$90 F - F = 89 F = z$
so:
:$F = \dfrac z {89}$
Because $a \ne 0$ :
:$F > 0 \cdotp 1$
Hence:
:$z = 9$
So:
=== Decimal Expansion ===
Hence the value for $N$.
We can take the recurring part an arbitrary number of times, and they all solve the problem.
Category:Integer which is Multiplied by 9 when moving Last Digit to First
Integer which is Multiplied by 9 when moving Last Digit to First/Decimal Expansion

== Lemma for Integer which is Multiplied by 9 when moving Last Digit to First ==
:$\dfrac 9 {89} = 0 \cdotp \dot 10112 \, 35955 \, 05617 \, 97752 \, 80898 \, 87640 \, 44943 \, 82022 \, 471 \dot 9$
== Proof ==
From Reciprocal of 89 we have:
:$\dfrac 1 {89} = 0 \cdotp \dot 01123 \, 59550 \, 56179 \, 77528 \, 08988 \, 76404 \, 49438 \, 20224 \, 719 \dot 1$
Hence:
 0.01123 59550 56179 77528 08988 76404 49438 20224 7191
x 9
 ------------------------------------------------------
 0.10112 35955 05617 97752 80898 87640 44943 82022 4719
 11235 844 5 51786 6427 88776 53 44 83371 2246 18
The lowest line is the carry digit.
As there is no overall carry digit on the left of the product, it follows that:
:$\dfrac 9 {89} = 0 \cdotp \dot 10112 \, 35955 \, 05617 \, 97752 \, 80898 \, 87640 \, 44943 \, 82022 \, 471 \dot 9$
Category:Integer which is Multiplied by 9 when moving Last Digit to First
Integer which is Multiplied by 9 when moving Last Digit to First/Mistake

== Source Work ==
:
:The Dictionary
::$10,112,359,550,561,797,752,808,988,764,044,943,820,224,719$
== Mistake ==
:The unique number which, when the last digit, $9$, is moved to the front, is the number multiplied by $9$.
== Correction ==
The number is not actually unique.
Let $N$ be the number given.
Then the positive integers formed by concatenating the decimal representation of $N$ with itself any number of times have the same property:
:$\sqbrk {NN}, \sqbrk {NNN}, \sqbrk {NNNN}, \ldots$
== Sources ==
* : $10,112,359,550,561,797,752,808,988,764,044,943,820,224,719$
Category:Mistakes/Curious and Interesting Numbers
Integer which is Multiplied by 9 when moving Last Digit to First/Proof 1

== Theorem ==
== Proof ==
First we demonstrate that $N$ has this property:
A way to find $N$ is given below.
Suppose $N = \sqbrk {a_k a_{k - 1} \dots a_1}$ satisfies the property above.
Consider the rational number represented by $q = 0.\dot {a_k} a_{k - 1} \dots \dot {a_1}$.
Since $a_k \ne 0$, we require $q \ge 0.1$.
By the property we have:
:$9 q = 0 . \dot {a_1} a_k \dots \dot {a_2}$
Then:
:$90 q = a_1 . \dot {a_k} a_{k - 1} \dots \dot {a_1}$
Notice that the recurring part of $90 q$ is the same as the recurring part of $q$.
Subtracting, we get $89 q = a_1$.
We have:
:$a_1 \ge 89 \times 0.1 = 8.9$
so:
:$a_1 = 9$
which gives:
:$q = \dfrac 9 {89}$
=== Decimal Expansion ===
which gives us:
:$N = 10 \, 112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 808 \, 988 \, 764 \, 044 \, 943 \, 820 \, 224 \, 719$
as the smallest positive integer satisfying the property.
Category:Integer which is Multiplied by 9 when moving Last Digit to First
Integer which is Multiplied by 9 when moving Last Digit to First/Proof 2

== Theorem ==
== Proof ==
From Integer which is Multiplied by Last Digit when moving Last Digit to First, $N$ is equal to the recurring part of the fraction:
:$q = \dfrac {a_1} {10 a_1 - 1}$
where $a_1 = 9$.
Thus:
:$q = \dfrac 9 {10 \times 9 - 1} = \dfrac 9 {89}$
Hence:
=== Decimal Expansion ===
The result follows.
Category:Integer which is Multiplied by 9 when moving Last Digit to First
Integer which is Multiplied by Last Digit when moving Last Digit to First

== Theorem ==
Let $N$ be a positive integer expressed in decimal notation in the form:
:$N = \sqbrk {a_k a_{k - 1} a_{k - 2} \ldots a_2 a_1}_{10}$
Let $N$ be such that when you multiply it by $a_1$, you get:
:$a_1 N = \sqbrk {a_1 a_k a_{k - 1} \ldots a_3 a_2}_{10}$
Then at least one such $N$ is equal to the recurring part of the fraction:
:$q = \dfrac {a_1} {10 a_1 - 1}$
== Proof ==
Let us consider:
:$q = 0 \cdotp \dot a_k a_{k - 1} a_{k - 2} \ldots a_2 \dot a_1$
Let:
:$a_1 q = 0 \cdotp \dot a_1 a_k a_{k - 1} \ldots a_3 \dot a_2$
Then:
Category:Recreational Mathematics
Integer which is Sum of 3 Fourth Powers in 2 Ways and Products of Those Roots

== Theorem ==
The positive integer $256 \, 103 \, 393$ can be expressed as the sum of $3$ fourth powers in $2$ different ways:
Also note that:
== Proof ==
We have that:
Then we have:
== Historical Note ==
== Sources ==
* : $256,103,393$
Category:Fourth Powers
Category:256,103,393
Integer whose Digits when Grouped in 3s add to Multiple of 999 is Divisible by 999

== 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 $999$ $n$ is divisible by $999$.
== Proof ==
The mistake is either and conversely or equal to $999$, since $999 \, 999$ is an easy counterexample.
Here we will show that the result is equal to a multiple of $999$ $n$ is divisible by $999$.
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:
:$999 \divides n \iff 999 \divides \ds \sum_{i \mathop = 0}^k a_i$
This statement is true since:
== Examples ==
== Sources ==
* : $999$
* : $142,857$
* : $999$
* : $142,857$
Category:Recreational Mathematics
Category:Divisibility Tests
Category:Integer whose Digits when Grouped in 3s add to Multiple of 999 is Divisible by 999
Integer whose Digits when Grouped in 3s add to Multiple of 999 is Divisible by 999/Mistake

== Source Work ==
:
:The Dictionary
::$142,857$
:
:The Dictionary
::$142,857$
== Mistake ==
:Now any number whose digits when grouped in $3$s from the units end add up to $999$ is a multiple of $999$, and conversely, so $142857$ must be a multiple of $999$.
== Correction ==
The and conversely is incorrect.
The simplest counterexample is:
The digits of $999 \, 999$, when grouped in $3$s from the units end, do not add up to $999$:
:$999 + 999 = 1998$
However, they do add to a multiple of $999$.
So perhaps the result could be worded:
:Now any number whose digits when grouped in $3$s from the units end add up to a multiple of $999$ is a multiple of $999$, and conversely, so $142857$ must be a multiple of $999$.
(emphasis added)
This mistake is repeated in a slightly different form on the page $999$.
== Also see ==
* Integer whose Digits when Grouped in 3s add to Multiple of 999 is Divisible by 999
== Sources ==
* : $142,857$
* : $142,857$
Category:Mistakes/Curious and Interesting Numbers
Category:Integer whose Digits when Grouped in 3s add to Multiple of 999 is Divisible by 999
Integers Coprime to Zero

== Theorem ==
The only integers which are coprime to zero are $1$ and $-1$.
That is:
:$n \in \Z: n \perp 0 \iff n \in \set {1, -1}$
In particular, note that two integers which are coprime to each other cannot both be $0$.
== Proof ==
Let $n \in \Z$ such that $n \perp 0$.
First we note that from Coprime Integers cannot Both be Zero, it cannot be the case that $n = 0$.
From the definition of coprime, we have:
:$m \perp n \iff \gcd \set {m, n} = 1$
From the definition of greatest common divisor:
:$\gcd \set {n, 0} = \size n$
where $\size n$ is the absolute value of $n$.
Let $n \in \set {1, -1}$.
Then:
:$\gcd \set {n, 0} = \size n = 1$
and so $n \perp 0$.
== Sources ==
* : Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $13$
Category:Coprime Integers
Integers Differing by 2 with Same Divisor Sum

== Theorem ==
Let $\sigma_1: \Z_{>0} \to \Z_{>0}$ be the divisor sum function, defined on the strictly positive integers.
The equation:
:$\map {\sigma_1} n = \map {\sigma_1} {n + 2}$
is satisfied by integers in the sequence:
:$33, 54, 284, 366, 834, 848, 918, 1240, 1504, 2910, 2913, 3304, \ldots$
== Examples ==
Category:Divisor Sum Function
Category:Integers Differing by 2 with Same Divisor Sum
Integers Differing by 2 with Same Divisor Sum/Examples

== Examples of Integers Differing by 2 with Same Divisor Sum == 
Let $\sigma_1: \Z_{>0} \to \Z_{>0}$ denote the divisor sum function.
The following are solutions to the equation:
:$\map {\sigma_1} n = \map {\sigma_1} {n + 2}$
=== $\sigma_1$ of $33$ equals $\sigma_1$ of $35$ ===
=== $\sigma_1$ of $54$ equals $\sigma_1$ of $56$ ===
=== $\sigma_1$ of $284$ equals $\sigma_1$ of $286$ ===
=== $\sigma_1$ of $366$ equals $\sigma_1$ of $368$ ===
Category:Integers Differing by 2 with Same Divisor Sum
Integers Differing by 2 with Same Divisor Sum/Examples/284

== Example of Integers Differing by 2 with Same Divisor Sum ==
Let $\sigma_1: \Z_{>0} \to \Z_{>0}$ denote the divisor sum function.
Then:
:$\map {\sigma_1} {284} = \map {\sigma_1} {286} = 504$
== Proof ==
From :
:$\map {\sigma_1} {284} = 504$
From :
:$\map {\sigma_1} {286} = 504$
Hence the result.
Category:Integers Differing by 2 with Same Divisor Sum
Category:284
Category:286
Integers Differing by 2 with Same Divisor Sum/Examples/33

== Example of Integers Differing by 2 with Same Divisor Sum ==
Let $\sigma_1: \Z_{>0} \to \Z_{>0}$ denote the divisor sum function.
Then:
:$\map {\sigma_1} {33} = \map {\sigma_1} {35} = 48$
== Proof ==
From :
:$\map {\sigma_1} {33} = 48$
From :
:$\map {\sigma_1} {35} = 48$
Hence the result.
== Sources ==
* : $33$
Category:Integers Differing by 2 with Same Divisor Sum
Category:33
Category:35
Integers Differing by 2 with Same Divisor Sum/Examples/366

== Example of Integers Differing by 2 with Same Divisor Sum ==
Let $\sigma_1: \Z_{>0} \to \Z_{>0}$ denote the divisor sum function.
Then:
:$\map {\sigma_1} {366} = \map {\sigma_1} {368} = 744$
== Proof ==
From :
:$\map {\sigma_1} {366} = 744$
From :
:$\map {\sigma_1} {368} = 744$
Hence the result.
Category:Integers Differing by 2 with Same Divisor Sum
Category:366
Category:368
Integers Differing by 2 with Same Divisor Sum/Examples/54

== Example of Integers Differing by 2 with Same Divisor Sum ==
Let $\sigma_1: \Z_{>0} \to \Z_{>0}$ denote the divisor sum function.
Then:
:$\map {\sigma_1} {54} = \map {\sigma_1} {56} = 120$
== Proof ==
From :
:$\map {\sigma_1} {54} = 120$
From :
:$\map {\sigma_1} {56} = 120$
Hence the result.
Category:Integers Differing by 2 with Same Divisor Sum
Category:54
Category:56
Integers Divided by GCD are Coprime

== Theorem ==
Let $a, b \in \Z$ be integers which are not both zero.
Let $d$ be a common divisor of $a$ and $b$, that is:
:$\dfrac a d, \dfrac b d \in \Z$
Then:
:$\gcd \set {a, b} = d$
:
:$\gcd \set {\dfrac a d, \dfrac b d} = 1$
that is:
:$\dfrac a {\gcd \set {a, b} } \perp \dfrac b {\gcd \set {a, b} }$
where:
:$\gcd$ denotes greatest common divisor
:$\perp$ denotes coprimality.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also presented as ==
It can be expressed so as not to include fractions:
:$\gcd \set {a, b} = d \iff \exists s, t \in \Z: a = d s \land b = d t \land \gcd \set {s, t} = 1$
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.3$ The Euclidean Algorithm: Problems $2.3$: $3$
Category:Greatest Common Divisor
Category:Coprime Integers
Category:Integers Divided by GCD are Coprime
Integers Divided by GCD are Coprime/Proof 1

== Theorem ==
== Proof ==
Let $d = \gcd \set {a, b}$.
By definition of divisor:
:$d \divides a \iff \exists s \in \Z: a = d s$
:$d \divides b \iff \exists t \in \Z: b = d t$
So:
== Sources ==
* : $\text {2-2}$ Divisibility: Example $\text {2-10}$
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Theorem $2 \text{-} 4$: Corollary $1$
Category:Integers Divided by GCD are Coprime
Integers Divided by GCD are Coprime/Proof 2

== Theorem ==
== Proof ==
Let $d = \gcd \set {a, b}$.
We have:
:$(1): d \divides a \iff \exists s \in \Z: a = d s$
:$(2): d \divides b \iff \exists t \in \Z: b = d t$
We have to prove:
:$\gcd \set {s, t} = 1$
 $\gcd \set {s, t} \ne 1$.
So:
:$(3): \exists k \in \N \setminus \set 1$ such that $k \divides s \land k \divides t$
So:
:$(4): \exists m, n \in \N: s = k m, t = k n$
Substituting from $(4)$ in $(1)$ and $(2)$:
:$a = d k m$, $b = d k n$
Therefore:
:$ d k \divides a \land d k \divides b$
From $(3)$ we have:
As $d k$ is a common divisor of $a$ and $b$ greater than $d$, this contradicts $d = \gcd \set {a, b}$. 
So our initial assumption that $\gcd \set {s, t} \ne 1$ is false. 
Therefore, from Proof by Contradiction, we have:
:$\gcd \set {s, t} = 1 \implies \gcd \set {\dfrac a d, \dfrac b d} = 1$
Category:Integers Divided by GCD are Coprime
Integers Modulo m under Addition form Abelian Group

== Theorem ==
Let $\Z_m$ is the set of integers modulo $m$
Let $+_m$ be the operation of addition modulo $m$.
Then the structure $\struct {\Z_m, +_m}$ is an abelian group.
== Proof ==
From Integers Modulo m under Addition form Cyclic Group, $\struct {\Z_m, +_m}$ is a cyclic group.
The result follows from Cyclic Group is Abelian.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.1$
* : Chapter $\text{II}$: Groups: A Little Number Theory
* : $\S 1$: Some examples of groups: Example $1.10$
* : Chapter $2$: Examples of Group Structure: $\S 33$
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups
* : $\S 34$. Examples of groups: $(3)$
* : Chapter $2$: Maps and relations on sets: Corollary $2.32$
* 
* 
Category:Additive Groups of Integers Modulo m
Category:Modulo Addition
Category:Examples of Abelian Groups
Integers Modulo m under Addition form Cyclic Group

== Theorem ==
Let $\Z_m$ be the set of integers modulo $m$.
Let $+_m$ be the operation of addition modulo $m$.
Let $\struct {\Z_m, +_m}$ denote the additive group of integers modulo $m$.
Then $\struct {\Z_m, +_m}$ is a cyclic group of order $m$, generated by the element $\eqclass 1 m \in \Z_m$.
== Proof ==
From the definition of integers modulo $m$, we have:
:$\Z_m = \dfrac \Z {\RR_m} = \set {\eqclass 0 m, \eqclass 1 m, \ldots, \eqclass {m - 1} m}$
It is established that Modulo Addition is Well-Defined:
:$\eqclass a m +_m \eqclass b m = \eqclass {a + b} m$
The group axioms are fulfilled:
:: Addition modulo $m$ is closed.
:: Addition modulo $m$ is associative.
:: The identity element of $\struct {\Z_m, +_m}$ is $\eqclass 0 m$.
:: The inverse of $\eqclass k m \in \Z_m$ is $-\eqclass k m = \eqclass {-k} m = \eqclass {n - k} m$.
:Commutativity: Addition modulo $m$ is commutative.
From Integers under Addition form Infinite Cyclic Group and Quotient Group of Cyclic Group, $\struct {\dfrac \Z {\RR_m}, +_m}$ is cyclic order $m$.
== Also see ==
* Integers Modulo m under Addition form Abelian Group
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 9$: Cyclic Groups: Example $2$
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Theorem $25.4$
* : Chapter $2$: Subgroups and Cosets: $\S 43$
* : $\S 3.4$: Cyclic groups: Theorem $1$
* : $\S 1.1$: Example $3$
* : Chapter $8$: The Homomorphism Theorem: Example $8.5$
* 
* 
Category:Additive Groups of Integers Modulo m
Category:Modulo Addition
Integers Modulo m under Max Operation form Ordered Semigroup

== Theorem ==
Let $\Z_m$ denote the set of integers modulo $m$:
:$\Z_m = \set {0, 1, \ldots, m - 1}$
Let $\vee_m$ be the operation on $\Z_m$ defined as:
:$\forall a, b \in \Z_m: a \vee_m b = \max \set {a, b}$
Then the ordered algebraic structure $\struct {\Z_m, \vee_m, \le}$ is an ordered semigroup.
== Proof ==
Taking the semigroup axioms in turn:
=== ===
We have that:
:$\forall a, b \in \Z_m: \max \set {a, b} \in \set {a, b}$
and so:
:$\max \set {a, b} \in \Z_m$
Thus $\struct {\Z_m, \vee_m}$ is closed.
=== ===
We have that Max Operation is Associative.
Thus $\vee_m$ is associative.
The semigroup axioms are thus seen to be fulfilled, and so $\struct {\Z_m, \vee_m}$ is a semigroup.
=== Ordered Semigroup ===
We now need to show that:
:$(1): \quad x \le y \implies x \vee_m z \le y \vee_m z$
and:
:$(2): \quad x \le y \implies z \vee_m x \le z \vee_m y$
Because Max Operation is Commutative, $(2)$ follows directly from $(1)$.
Hence we need demonstrate $(1)$ only.
Let $x, y, z \in S$.
Let $x \le y$.
;Case $1$. $z \le x$
;Case $2$. $x \le z \le y$
;Case $3$. $y \le z$
Thus $\le$ is compatible with $\vee_m$.
The result follows.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.2$
Category:Examples of Ordered Semigroups
Integers Modulo m under Multiplication form Commutative Monoid

== Theorem ==
Let $\struct {\Z_m, \times_m}$ denote the algebraic structure such that:
:$\Z_m$ is the set of integers modulo $m$
:$\times_m$ denotes the operation of multiplication modulo $m$.
Then $\struct {\Z_m, \times_m}$ is a commutative monoid.
== Proof ==
Multiplication modulo $m$ is closed.
Multiplication modulo $m$ is associative.
Multiplication modulo $m$ has an identity:
:$\forall k \in \Z: \eqclass k m \times_m \eqclass 1 m = \eqclass k m = \eqclass 1 m \times_m \eqclass k m$
This identity is unique.
Multiplication modulo $m$ is commutative.
Thus all the conditions are fulfilled for this to be a commutative monoid.
== Sources ==
* : $\S 1.1$
Category:Modulo Multiplication
Category:Examples of Commutative Monoids
Integers Representable as Product of both 3 and 4 Consecutive Integers

== Theorem ==
There are $3$ integers which can be expressed as both $x \paren {x + 1} \paren {x + 2} \paren {x + 3}$ for some $x$, and $y \paren {y + 1} \paren {y + 2}$ for some $y$:
:$24, 120, 175 \, 560$
== Proof ==
We have:
== Sources ==
* 
* 
* : $175,560$
Category:Number Theory