nyuuzyou/wikis huggingface dataset:
Multiplicative Group of Reduced Residues/Examples/Modulo 5

== Example of Multiplicative Group of Reduced Residues ==
Consider the reduced residue system $\Z'_5$ modulo $5$ under modulo multiplication:
:$\Z'_5 = \set {\eqclass 1 5, \eqclass 2 5, \eqclass 3 5, \eqclass 4 5}$
$\struct {\Z'_5, \times_5}$ is the multiplicative group of reduced residues modulo $5$.
=== Cayley Table ===
== Also see ==
* Reduced Residues Modulo 5 under Multiplication form Cyclic Group
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 7$: Isomorphic Groups: Example $1 \ (\gamma)$
Category:Cyclic Group of Order 4
Category:Examples of Multiplicative Groups of Reduced Residues
Category:Multiplicative Group of Reduced Residues Modulo 5
Multiplicative Group of Reduced Residues/Examples/Modulo 7

== Example of Multiplicative Group of Reduced Residues ==
Consider the reduced residue system $\Z'_7$ modulo $7$ under modulo multiplication:
:$\Z'_7 = \set {\eqclass 1 7, \eqclass 2 7, \eqclass 3 7, \eqclass 4 7, \eqclass 5 7, \eqclass 6 7}$
$\struct {\Z'_7, \times_7}$ is the multiplicative group of reduced residues modulo 7.
=== Cayley Table ===
Category:Groups of Order 6
Category:Examples of Multiplicative Groups of Reduced Residues
Category:Multiplicative Group of Reduced Residues Modulo 7
Multiplicative Group of Reduced Residues/Examples/Modulo 8

== Example of Multiplicative Group of Reduced Residues ==
Consider the reduced residue system $\Z'_8$ modulo $8$ under modulo multiplication:
:$\Z'_8 = \set {\eqclass 1 8, \eqclass 3 8, \eqclass 5 8, \eqclass 7 8}$
$\struct {\Z'_8, \times_8}$ is the multiplicative group of reduced residues modulo 8.
=== Cayley Table ===
== Also see ==
* Multiplicative Group of Reduced Residues Modulo 8 is Klein Four-Group
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 7$: Isomorphic Groups: Example $2 \ \text{(c)}$
Category:Examples of Multiplicative Groups of Reduced Residues
Category:Multiplicative Group of Reduced Residues Modulo 8
Category:Klein Four-Group
Multiplicative Group of Reduced Residues Modulo 5/Cayley Table

== Cayley Table for Multiplicative Group of Reduced Residues Modulo 5 ==
The multiplicative group of reduced residues modulo $5$:
:$\struct {\Z'_5, \times_5} = \set {\eqclass 1 5, \eqclass 2 5, \eqclass 3 5, \eqclass 4 5}$
can be described completely by showing its Cayley table:
$\quad \begin {array} {r|rrrr}
\times_5 & \eqclass 1 5 & \eqclass 2 5 & \eqclass 3 5 & \eqclass 4 5 \\
\hline
\eqclass 1 5 & \eqclass 1 5 & \eqclass 2 5 & \eqclass 3 5 & \eqclass 4 5 \\
\eqclass 2 5 & \eqclass 2 5 & \eqclass 4 5 & \eqclass 1 5 & \eqclass 3 5 \\
\eqclass 3 5 & \eqclass 3 5 & \eqclass 1 5 & \eqclass 4 5 & \eqclass 2 5 \\
\eqclass 4 5 & \eqclass 4 5 & \eqclass 3 5 & \eqclass 2 5 & \eqclass 1 5 \\
\end {array}$
By arranging the rows and columns into a different order, its cyclic nature becomes clear:
$\quad \begin {array} {r|rrrr}
\times_5 & \eqclass 1 5 & \eqclass 2 5 & \eqclass 4 5 & \eqclass 3 5 \\
\hline
\eqclass 1 5 & \eqclass 1 5 & \eqclass 2 5 & \eqclass 4 5 & \eqclass 3 5 \\
\eqclass 2 5 & \eqclass 2 5 & \eqclass 4 5 & \eqclass 3 5 & \eqclass 1 5 \\
\eqclass 4 5 & \eqclass 4 5 & \eqclass 3 5 & \eqclass 1 5 & \eqclass 2 5 \\
\eqclass 3 5 & \eqclass 3 5 & \eqclass 1 5 & \eqclass 2 5 & \eqclass 4 5 \\
\end {array}$
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.5$
* : Chapter $2$: Maps and relations on sets: Exercise $5$
Category:Examples of Cayley Tables
Category:Multiplicative Group of Reduced Residues Modulo 5
Multiplicative Group of Reduced Residues Modulo 7/Cayley Table

== Cayley Table for Multiplicative Group of Reduced Residues Modulo 7 ==
The multiplicative group of reduced residues modulo $7$:
:$\Z'_7 = \set {\eqclass 1 7, \eqclass 2 7, \eqclass 3 7, \eqclass 4 7}, \eqclass 5 7, \eqclass 6 7$
can be described completely by showing its Cayley table:
$\quad \begin {array} {r|rrrr}
\times_5 & \eqclass 1 7 & \eqclass 2 7 & \eqclass 3 7 & \eqclass 4 7 & \eqclass 5 7 & \eqclass 6 7 \\
\hline
\eqclass 1 7 & \eqclass 1 7 & \eqclass 2 7 & \eqclass 3 7 & \eqclass 4 7 & \eqclass 5 7 & \eqclass 6 7 \\
\eqclass 2 7 & \eqclass 2 7 & \eqclass 4 7 & \eqclass 6 7 & \eqclass 1 7 & \eqclass 3 7 & \eqclass 5 7 \\
\eqclass 3 7 & \eqclass 3 7 & \eqclass 6 7 & \eqclass 2 7 & \eqclass 5 7 & \eqclass 1 7 & \eqclass 4 7 \\
\eqclass 4 7 & \eqclass 4 7 & \eqclass 1 7 & \eqclass 5 7 & \eqclass 2 7 & \eqclass 6 7 & \eqclass 3 7 \\
\eqclass 5 7 & \eqclass 5 7 & \eqclass 3 7 & \eqclass 1 7 & \eqclass 6 7 & \eqclass 4 7 & \eqclass 2 7 \\
\eqclass 6 7 & \eqclass 6 7 & \eqclass 5 7 & \eqclass 4 7 & \eqclass 3 7 & \eqclass 2 7 & \eqclass 1 7 \\
\end {array}$
Category:Examples of Cayley Tables
Category:Multiplicative Group of Reduced Residues Modulo 7
Multiplicative Group of Reduced Residues Modulo 7 is Cyclic

== Theorem ==
Let $\struct {\Z'_7, \times_7}$ denote the multiplicative group of reduced residues modulo $7$.
Then $\struct {\Z'_7, \times_7}$ is cyclic.
== Proof ==
From Reduced Residue System under Multiplication forms Abelian Group‎ it is noted that $\struct {\Z'_7, \times_7}$ is a group.
It remains to be shown that $\struct {\Z'_7, \times_7}$ is cyclic.
It will be demonstrated that:
:$\gen {\eqclass 3 7} = \struct {\Z'_7, \times_7}$
That is, that $\eqclass 3 7$ is a generator of $\struct {\Z'_7, \times_7}$.
We note that $\eqclass 1 7$ is the identity element of $\struct {\Z'_7, \times_7}$.
Thus successive powers of $\eqclass 3 7$ are taken, until $n \in \Z$ is found such that $\eqclass 3 7^n = \eqclass 1 7$:
All elements of $\struct {\Z'_7, \times_7}$ are seen to be in $\gen {\eqclass 3 7}$.
Hence the result by definition of cyclic group.
== Sources ==
* : Chapter $6$: An Introduction to Groups: Exercise $16$
Category:Multiplicative Groups of Reduced Residues
Category:Multiplicative Group of Reduced Residues Modulo 7
Category:Examples of Cyclic Groups
Multiplicative Group of Reduced Residues Modulo 8/Cayley Table

== Cayley Table for Multiplicative Group of Reduced Residues Modulo 8 ==
The multiplicative group of reduced residues modulo $8$:
:$\Z'_8 = \set {\eqclass 1 8, \eqclass 3 8, \eqclass 5 8, \eqclass 7 8}$
can be described completely by showing its Cayley table:
$\quad \begin {array} {r|rrrr}
\times_8 & \eqclass 1 8 & \eqclass 3 8 & \eqclass 5 8 & \eqclass 7 8 \\
\hline
\eqclass 1 8 & \eqclass 1 8 & \eqclass 3 8 & \eqclass 5 8 & \eqclass 7 8 \\
\eqclass 3 8 & \eqclass 3 8 & \eqclass 1 8 & \eqclass 7 8 & \eqclass 5 8 \\
\eqclass 5 8 & \eqclass 5 8 & \eqclass 7 8 & \eqclass 1 8 & \eqclass 3 8 \\
\eqclass 7 8 & \eqclass 7 8 & \eqclass 5 8 & \eqclass 3 8 & \eqclass 1 8 \\
\end{array}$
Category:Examples of Cayley Tables
Category:Multiplicative Group of Reduced Residues Modulo 8
Multiplicative Group of Reduced Residues Modulo 8 is Klein Four-Group

== Theorem ==
Let $K_4$ denote the Klein $4$-group.
Let $R_4$ be the multiplicative group of reduced residues Modulo $8$.
Then $K_4$ and $R_4$ are isomorphic algebraic structures.
== Proof ==
Establish the mapping $\phi: K_4 \to R_4$ as follows:
From Isomorphism by Cayley Table, the two Cayley tables can be compared by eye to ascertain that $\phi$ is an isomorphism:
=== Cayley Table of Klein $4$-Group ===
</onlyinclude>
=== Multiplicative Group of Reduced Residues Modulo $8$ ===
The Cayley table for $S$ is as follows:
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 7$: Isomorphic Groups: Example $2$
Category:Examples of Group Isomorphisms/Order 4
Category:Klein Four-Group
Category:Multiplicative Group of Reduced Residues Modulo 8
Multiplicative Identity for Quaternions

== Theorem ==
In the set of quaternions $\mathbb H$, the element:
:$\mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k$
serves as the identity element for quaternion multiplication.
This element is written $\mathbf 1$.
== Proof ==
Let $\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$.
From the definition of quaternion multiplication:
Similarly for $\mathbf 1 \mathbf x$.
Category:Quaternions
Multiplicative Identity is Unique

== Theorem ==
Let $\struct {F, +, \times}$ be a field.
Then the multiplicative identity $1_F$ of $F$ is unique.
== Proof ==
From the definition of multiplicative identity, $1_F$ is the identity element of the multiplicative group $\struct {F^*, \times}$.
The result follows from Identity of Group is Unique.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $1 \ \text{(ii)}$
Category:Field Theory
Category:Multiplication
Multiplicative Inverse in Field is Unique

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose zero is $0_F$.
Let $a \in F$ such that $a \ne 0_F$.
Then the multiplicative inverse $a^{-1}$ of $a$ is unique.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $1 \ \text{(iv)}$
Category:Field Theory
Category:Multiplicative Inverse in Field is Unique
Multiplicative Inverse in Monoid of Integers Modulo m

== Theorem ==
Let $\struct {\Z_m, \times_m}$ be the multiplicative monoid of integers modulo $m$.
Then:
:$\eqclass k m \in \Z_m$ has an inverse in $\struct {\Z_m, \times_m}$
:
:$k \perp m$
== Proof ==
First, suppose $k \perp m$.
That is:
:$\gcd \set {k, m} = 1$
By Bézout's Identity:
:$\exists u, v \in \Z: u k + v m = 1$
Thus:
:$\eqclass {u k + v m} m = \eqclass {u k} m = \eqclass u m \eqclass k m = \eqclass 1 m$
Thus:
:$\eqclass u m$ is an inverse of $\eqclass k m$
Suppose that:
:$\exists u \in \Z: \eqclass u m \eqclass k m = \eqclass {u k} m = 1$
Then:
:$u k \equiv 1 \pmod m$
and:
:$\exists v \in \Z: u k + v m = 1$
Thus from Bézout's Identity, $k \perp m$.
== Sources ==
* : $\S 1$: Some examples of groups: Example $1.11$
* : Chapter $2$: Examples of Group Structure: $\S 34$
Category:Modulo Arithmetic
Multiplicative Inverse in Nicely Normed Star-Algebra

== Theorem ==
Let $A = \struct {A_F, \oplus}$ be a nicely normed $*$-algebra whose conjugation is denoted $*$.
Let $a \in A$.
Then the multiplicative inverse of $a$ is given by:
:$a^{-1} = \dfrac {a^*} {\norm a^2}$
where:
:$a^*$ is the conjugate of $a$
:$\norm a$ is the norm of $a$.
== Proof ==
For the result to hold, we need to show that $a \oplus \dfrac {a^*} {\norm a^2} = 1 = \dfrac {a^*} {\norm a^2} \oplus a$.
Note that this construction works whether $\oplus$ is associative or not.
== Sources ==
* 2002: : The Octonions: 2.2 The Cayley-Dickson Construction
Category:Nicely Normed Star-Algebras
Category:Multiplicative Inverses
Multiplicative Inverse in Ring of Integers Modulo m

== Theorem ==
Let $\struct {\Z_m, +_m, \times_m}$ be the ring of integers modulo $m$.
Then $\eqclass k m \in \Z_m$ has an inverse in $\struct {\Z_m, \times_m}$ $k \perp m$.
== Proof 1 ==
== Proof 2 ==
Category:Ring of Integers Modulo m
Category:Multiplicative Inverse in Ring of Integers Modulo m
Category:Multiplicative Inverses
Multiplicative Inverse in Ring of Integers Modulo m/Proof 1

== Theorem ==
== Proof ==
First, suppose $k \perp m$.
That is:
:$\gcd \set {k, m} = 1$
Then, by Bézout's Identity:
:$\exists u, v \in \Z: u k + v m = 1$
Thus:
:$\eqclass {u k + v m} m = \eqclass {u k} m = \eqclass u m \eqclass k m = \eqclass 1 m$
Thus $\eqclass u m$ is an inverse of $\eqclass k m$.
Suppose that:
:$\exists u \in \Z: \eqclass u m \eqclass k m = \eqclass {u k} m = 1$.
Then:
:$u k \equiv 1 \pmod m$
and:
:$\exists v \in \Z: u k + v m = 1$
Thus from Bézout's Identity:
:$k \perp m$
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 1$. Rings and Fields: Example $4$
Category:Multiplicative Inverse in Ring of Integers Modulo m
Multiplicative Inverse in Ring of Integers Modulo m/Proof 2

== Theorem ==
== Proof ==
From Ring of Integers Modulo m is Ring, $\left({\Z_m, +_m, \times_m}\right)$ is a [[Definition:Commutative and Unitary Ring|commutative ring with unity $\left[\!\left[{1}\right]\!\right]_m$]].
Thus by definition $\left({\Z_m, \times_m}\right)$ is a commutative monoid.
The result follows from Multiplicative Inverse in Monoid of Integers Modulo m.
Category:Multiplicative Inverse in Ring of Integers Modulo m
Multiplicative Inverse of Quaternion

== Theorem ==
Let $\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$ be a quaternion such that $\mathbf x \ne \mathbf 0$.
Then $\mathbf x$ has an inverse $\mathbf x^{-1}$ under the operation of quaternion multiplication:
:$\mathbf x^{-1} = \lambda \overline {\mathbf x}$
where:
:$\lambda = \dfrac 1 {a^2 + b^2 + c^2 + d^2} \mathbf 1$
== Proof ==
From Multiplicative Identity for Quaternions‎, we need to show that $\lambda \overline{\mathbf x} \mathbf x = \mathbf 1 = \mathbf x \lambda \overline{\mathbf x}$.
From Product of Quaternion with Conjugate we have that:
:$\overline{\mathbf x} \mathbf x = \paren {a^2 + b^2 + c^2 + d^2} \mathbf 1$
Using the definition of $\lambda$ from above:
Unless all of $a, b, c, d = 0$ we have that $a^2 + b^2 + c^2 + d^2 \ne 0$.
Therefore $\lambda$ is defined for all quaternions not equal to $\mathbf 0$.
It follows from Left Inverse for All is Right Inverse that $\mathbf x \lambda \overline{\mathbf x} = \mathbf 1$.
Hence the result.
== Sources ==
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Ring Example $9$
* : Chapter $\text {B}.26$: Extensions of the Complex Number System. Algebras, Quaternions, and Lagrange's Four Squares Theorem
Category:Quaternions
Multiplicative Magic Square/Examples/Order 3/Smallest

== Example of Order $3$ Multiplicative Magic Square ==
The order $3$ multiplicative magic square with the smallest magic constant is:
:$\begin {array} {|c|c|c|} \hline 18 & 1 & 12 \\ \hline 4 & 6 & 9 \\ \hline 3 & 36 & 2 \\ \hline \end{array}$
Its magic constant is $216$.
== Also see ==
* Smallest Multiplicative Magic Square is of Order 3
* Smallest Magic Constant of Order 3 Multiplicative Magic Square
== Historical Note ==
== Sources ==
* 
Category:Multiplicative Magic Squares
Multiplicative Magic Square/Examples/Order 3/Smallest/Historical Note

== Historical Note on Multiplicative Magic Square with Smallest Magic Constant ==
The order $3$ multiplicative magic square with the smallest magic constant is believed to have been discovered by in $1893$.
Some sources erroneously ascribe it to in $1913$, as work by has only recently come to light.
It is also sometimes erroneously ascribed to , who in $1917$ also rediscovered it.
== Sources ==
* : $216$
* : $216$
* 
Category:Historical Notes
Multiplicative Magic Square/Examples/Order 3/Smallest/Historical Note/Mistake

== Source Work ==
:
:The Dictionary
::$216$
:
:The Dictionary
::$216$
== Mistake ==
:$216$ is the magic constant in the smallest possible multiplicative magic square, as discovered by .
 only rediscovered a multiplicative magic square which had already been discovered by in $1893$, and in $1913$ by .
== Sources ==
* : $216$
* : $216$
Category:Mistakes/Curious and Interesting Numbers
Multiplicative Magic Square/Examples/Order 4/Smallest

== Example of Order $4$ Multiplicative Magic Square ==
The order $4$ multiplicative magic square with the smallest magic constant is:
:$\begin {array} {|c|c|c|c|} \hline 1 & 15 & 24 & 14 \\ \hline 12 & 28 & 3 & 5 \\ \hline 21 & 6 & 10 & 4 \\ \hline 20 & 2 & 7 & 18 \\ \hline \end{array}$
Its magic constant is $5040$.
== Also see ==
* Smallest Magic Constant of Order 4 Multiplicative Magic Square
== Historical Note ==
== Sources ==
* 
Category:Multiplicative Magic Squares
Multiplicative Magic Square from 9 Terms of Geometric Sequence from 1, 2, 4

== Problem ==
:With the first $9$ terms of the geometric sequence $1, 2, 4, \ldots$,
:form a product of $4096$ each way.
That is, form a multiplicative magic square whose magic constant is $4096$.
== Solution ==
:$\begin{array}{|c|c|c|} \hline 2 & 64 & 32 \\ \hline 256 & 16 & 1 \\ \hline 8 & 4 & 128 \\ \hline \end{array}$
== Proof ==
Take the order $3$ (additive) magic square:
Subtract $1$ from each element:
:$\begin{array}{|c|c|c|} \hline 1 & 6 & 5 \\ \hline 8 & 4 & 0 \\ \hline 3 & 2 & 7 \\ \hline \end{array}$
This square is still magic.
Now replace each element $n$ with $2^n$:
:$\begin{array}{|c|c|c|} \hline 2^1 & 2^6 & 2^5 \\ \hline 2^8 & 2^4 & 2^0 \\ \hline 2^3 & 2^2 & 2^7 \\ \hline \end{array}$
which is:
:$\begin{array}{|c|c|c|} \hline 2 & 64 & 32 \\ \hline 256 & 16 & 1 \\ \hline 8 & 4 & 128 \\ \hline \end{array}$
== Sources ==
* 
* : Rational Amusements for Winter Evenings: $152$
Category:Multiplicative Magic Squares
Multiplicative Order of Roots of Cyclotomic Polynomial Modulo Prime

== Theorem ==
Let $n\geq1$ be a natural number.
Let $p$ be a prime number
Let $n=p^\alpha q$ where $\alpha = \nu_p(n)$ is the valuation of $p$ in $n$.
Let $a\in\Z$ with $\Phi_n(a)\equiv0\pmod p$.
Then the order of $a$ modulo $p$ is $q$:
:$\operatorname{ord}_p(a) = q$.
== Proof ==
By Product of Cyclotomic Polynomials, $p\mid \Phi_n(a) \mid a^n-1$.
Thus $a$ is coprime to $p$.
By Fermat's Little Theorem, $1\equiv a^n\equiv a^q\pmod p$.
Thus $\operatorname{ord}_p(a) \leq q$.
Suppose $\operatorname{ord}_p(a) = k < q$.
By Product of Cyclotomic Polynomials, $p\mid \Phi_d(a)$ for some $d\mid k$.
Then $a$ is a double root of $x^q-1$ modulo $p$.
By Double Root of Polynomial is Root of Derivative, $q\equiv0\pmod p$.
This is a contradiction, thus $k=q$.
Category:Cyclotomic Polynomials
Multiplicative Ordering on Integers

== Theorem ==
Let $x, y, z \in \Z$ such that $z > 0$.
Then:
:$x < y \iff z x < z y$
:$x \le y \iff z x \le z y$
== Proof ==
Let $z > 0$.
Let $M_z: \Z \to \Z$ be the mapping defined as:
:$\forall x \in \Z: \map {M_z} x = z x$
It is sufficient to show that $M_z$ is an order embedding from $\struct {\Z, +, \le}$ to itself.
By Monomorphism from Total Ordering, it is sufficient to show that:
:$x < y \implies z x < z y$
Let $x < y$.
Then:
:$0 < y - x$
So $z \in \N$.
Hence by Natural Numbers are Non-Negative Integers:
:$y - x \in \N$
Thus by Ordering on Natural Numbers is Compatible with Multiplication:
:$z \paren {y - x} \in \N$
Therefore:
:$0 < z \paren {y - x} = z y - z x$
That is:
:$z x < z y$
== Sources ==
* : Introduction $\S 5$: The system of integers
* : Chapter $\text {IV}$: Rings and Fields: $20$. The Integers: Theorem $20.10: 1^\circ, 2^\circ$
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: $\mathbf Z. \, 8$
* : $1$ Divisibility: $1.1$ The Euclidean algorithm and unique factorization
Category:Integer Multiplication
Category:Orderings on Integers
Multiplicative Persistence/Examples/25

== Examples of Multiplicative Persistence ==
$25$ is the smallest positive integer which has a multiplicative persistence of $2$.
== Proof ==
Trivially:
All positive integers between $10$ and $19$ are seen to have a multiplicative persistence of $1$:
:$1 \times n = n$
where $n$ is a single digit.
Then for 2-digit positive integers starting with $2$:
:$2 \times n > 9 \implies n > 4$
by inspection.
Hence the result.
Category:Multiplicative Persistence
Category:25
Multiplicative Persistence/Examples/277,777,788,888,899

== Examples of Multiplicative Persistence ==
$277 \, 777 \, 788 \, 888 \, 899$ is the smallest positive integer which has a multiplicative persistence of $11$.
== Proof ==
We have:
The fact that it is the smallest can be demonstrated by brute-force methods.
== Sources ==
* : $277,777,788,888,899$
Category:Multiplicative Persistence
Category:277,777,788,888,899
Multiplicative Persistence/Examples/39

== Examples of Multiplicative Persistence ==
$39$ is the smallest positive integer which has a multiplicative persistence of $3$.
== Proof ==
We have:
The product of the $2$ digits of a positive integer smaller than $39$ is less than $25$, for example:
From Multiplicative Persistence of 25, all such positive integers have a multiplicative persistence of $2$.
The result follows.
Category:Multiplicative Persistence
Category:39
Multiplicative Persistence/Examples/77

== Examples of Multiplicative Persistence ==
$77$ is the smallest positive integer which has a multiplicative persistence of $4$.
== Proof ==
We have:
$39$ is the smallest positive integer with multiplicative persistence of $3$.
Hence if the product of digits of $n$ is less than $39$, its multiplicative persistence cannot exceed $3$.
Therefore we only need to check:
:$58, 59, 67, 68, 69, 76$
and we have:
so none of those numbers have a multiplicative persistence of $4$.
Category:Multiplicative Persistence
Category:77
Multiplicative Regular Representations of Units of Topological Ring are Homeomorphisms

== Theorem ==
Let $\struct{R, + , \circ, \tau}$ be a topological ring with unity $1_R$.
For all $y \in R$, let $\lambda_y$ and $\rho_y$ denote the left and right regular representations of $\struct{R, \circ}$ with respect to $y$.
Let $x \in R$ be a unit of $R$ with product inverse $x^{-1}$.
Then $\lambda_x, \, \rho_x: \struct{R, \tau} \to \struct{R, \tau}$ are homeomorphisms with inverse mappings $\lambda_{x^{-1} }, \, \rho_{x^{-1} }: \struct{R, \tau} \to \struct{R, \tau}$ respectively.
== Proof ==
Let $I_{_R} : R \to R$ be the identity mapping on $R$.
For all $y \in R$, let $y * I_{_R} : R \to R$ be the mapping defined by:
:$\forall z \in R: \map {\paren {y * I_{_R} } } z = y * \map {I_{_R}} z$
For all $y \in R$, let $I_{_R} * y : R \to R$ be the mapping defined by:
:$\forall z \in R: \map {\paren {I_{_R} * y}} z = \map {I_{_R}} z * y$
=== Lemma 1 ===
From Identity Mapping is Continuous, $I_{_R} : \struct{R, \tau} \to \struct{R, \tau}$ is continuous.
From Multiple Rule for Continuous Mappings into Topological Ring:
:$x * I_{_R}, \, I_{_R} * x, x^{-1} * I_{_R}, \,I_{_R} * x^{-1}$ are continuous.
=== Lemma 2 ===
By definition of homeomorphism, $\lambda_x, \,\rho_x : \struct{R, \tau} \to \struct{R, \tau}$ are homeomorphisms with inverse mappings $\lambda_{x^{-1} }, \,\rho_{x^{-1} } : \struct{R, \tau} \to \struct{R, \tau}$ respectively.
Category:Topological Rings
Category:Multiplicative Regular Representations of Units of Topological Ring are Homeomorphisms
Multiplicative Regular Representations of Units of Topological Ring are Homeomorphisms/Lemma 1

== Theorem ==
Let $\struct {R, + , \circ}$ be a ring.
For all $y \in R$, let $\lambda_y$ and $\rho_y$ denote the left and right regular representations of $\struct {R, \circ}$ with respect to $y$.
Let $I_{_R} : R \to R$ be the identity mapping on $R$.
For all $y \in R$, let $y * I_{_R} : R \to R$ be the mapping defined by:
:$\forall z \in R: \map {\paren{y * I_{_R}}} z = y * \map {I_{_R}} z$
For all $y \in R$, let $I_{_R} * y : R \to R$ be the mapping defined by:
:$\forall z \in R: \map {\paren{I_{_R} * y}} z = \map {I_{_R}} z * y$
Then:
:$\forall y \in R: \lambda_y = y * I_{_R} \text { and } \rho_y = I_{_R} * y$
== Proof ==
Let $y \in R$.
From Equality of Mappings:
:$\lambda_y = y * I_{_R}$
From Equality of Mappings:
:$\rho_y = I_{_R} * y$
Category:Multiplicative Regular Representations of Units of Topological Ring are Homeomorphisms
Multiplicative Regular Representations of Units of Topological Ring are Homeomorphisms/Lemma 2

== Theorem ==
Let $\struct{R, + , \circ}$ be a ring with unity $1_R$.
Let $I_R : R \to R$ be the identity mapping on $R$.
For all $y \in R$, let $y * I_R : R \to R$ be the mapping defined by:
:$\forall z \in R: \map {\paren {y * I_R} } z = y * \map {I_R} z$
For all $y \in R$, let $I_R * y : R \to R$ be the mapping defined by:
:$\forall z \in R: \map {\paren {I_R * y} } z = \map {I_R} z * y$
Let $x \in R$ be a unit of $R$ with product inverse $x^{-1}$.
Then:
:$x * I_R$ is a bijection and $x^{-1} * I_R$ is the inverse of $x * I_R$
:$I_R * x$ is a bijection and $I_R * x^{-1}$ is the inverse of $I_R * x$
== Proof ==
Consider the composite of $x * I_R$ with $x^{-1} * I_R$.
From Equality of Mappings, $\paren {x * I_R} \circ \paren {x^{-1} * I_R} = I_R$.
Consider the composite of $x^{-1} * I_R$ with $x * I_R$.
From Equality of Mappings, $\paren {x^{-1} * I_R} \circ \paren {x * I_R} = I_R$.
Hence $x * I_R$ has a left inverse and a right inverse.
From definition 2 of bijection, $x * I_R$ is a bijection and the inverse is $x^{-1} * I_R$.
Similarly, $I_R * x$ is a bijection and the inverse is $I_R * x^{-1}$.
Category:Multiplicative Regular Representations of Units of Topological Ring are Homeomorphisms