nyuuzyou/wikis huggingface dataset:
Multiple of Repdigit Base minus 1/Generalization

== Theorem ==
Let $b \in \Z_{>1}$ be an integer greater than $1$.
Let $n$ be a repdigit number of $k$ instances of the digit $b - 1$ for some integer $k$ such that $k \ge 1$.
Let $m \in \N$ be an integer such that $1 \le m \le b^k$.
Then $m \times n$, when expressed in base $b$, is the concatenation of $m - 1$ with $b^k - m$, that is:
:$m n = \sqbrk {\paren {m - 1} \paren {b^k - m} }_b$
If $b^k - m$ has less than $k$ digits, leading zeros are added to it until it has $k$ digits.
== Proof ==
which is exactly the representation $\sqbrk {\paren {m - 1} \paren {b^k - m} }_b$ with leading zeroes for $b^k - m$, as defined.
== Example ==
In base $10$, $972 \times 999 = 971 \, 028$.
Hence $971 = 972 - 1$ and $028 = 10^3 - 972$.
Category:Repdigit Numbers
Category:Multiple of Repdigit Base minus 1
Multiple of Ring Product

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $x, y \in \struct {R, +, \circ}$.
Then:
:$\forall n \in \Z_{> 0}: \paren {n \cdot x} \circ y = n \cdot \paren {x \circ y} = x \circ \paren {n \cdot y}$
where $n \cdot x$ denotes the $n$th multiple of $x$.
== Proof ==
By definition:
:$\ds n \cdot x := \sum_{j \mathop = 1}^n x$
Thus:
Category:Ring Theory
Multiple of Row Added to Row of Determinant

== Theorem ==
Let $\mathbf A = \begin {bmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
 \vdots & \vdots & \ddots & \vdots \\
a_{r 1} & a_{r 2} & \cdots & a_{r n} \\
 \vdots & \vdots & \ddots & \vdots \\
a_{s 1} & a_{s 2} & \cdots & a_{s n} \\
 \vdots & \vdots & \ddots & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n} \\
\end {bmatrix}$ be a square matrix of order $n$.
Let $\map \det {\mathbf A}$ denote the determinant of $\mathbf A$.
Let $\mathbf B = \begin{bmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
 \vdots & \vdots & \ddots & \vdots \\
a_{r 1} + k a_{s 1} & a_{r 2} + k a_{s 2} & \cdots & a_{r n} + k a_{s n} \\
 \vdots & \vdots & \ddots & \vdots \\
a_{s 1} & a_{s 2} & \cdots & a_{s n} \\
 \vdots & \vdots & \ddots & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n} \\
\end{bmatrix}$.
Then:
:$\map \det {\mathbf B} = \map \det {\mathbf A}$.
That is, the value of a determinant remains unchanged if a constant multiple of any row is added to any other row.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Multiple of Column Added to Column of Determinant
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercises -- Second Set
* 
* 
* 
Category:Determinants
Category:Multiple of Row Added to Row of Determinant
Multiple of Row Added to Row of Determinant/Proof 1

== Theorem ==
== Proof ==
Let $e$ be the elementary row operation that adds $k$ times row $r$ to row $s$.
Let $\mathbf B = \map e {\mathbf A}$.
Let $\mathbf E$ be the elementary row matrix corresponding to $e$.
From Elementary Row Operations as Matrix Multiplications:
:$\mathbf B = \mathbf E \mathbf A$
From Determinant of Elementary Row Matrix: Scale Row and Add:
:$\map \det {\mathbf E} = 1$
Then:
Hence the result.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Proposition $1.11$
Category:Multiple of Row Added to Row of Determinant
Multiple of Row Added to Row of Determinant/Proof 2

== Theorem ==
== Proof ==
By Determinant as Sum of Determinants:
:$\map \det {\mathbf B} = \begin{vmatrix}
a_{11} & a_{12} & \ldots & a_{1n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{r1} + k a_{s1} & a_{r2} + k a_{s2} & \cdots & a_{rn} + k a_{sn} \\
\vdots & \vdots & \ddots & \vdots \\
a_{s1} & a_{s2} & \cdots & a_{sn} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn} \\
\end{vmatrix} = \begin{vmatrix}
a_{11} & a_{12} & \ldots & a_{1n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{r1} & a_{r2} & \cdots & a_{rn} \\
\vdots & \vdots & \ddots & \vdots \\
a_{s1} & a_{s2} & \cdots & a_{sn} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn} \\
\end{vmatrix} + \begin{vmatrix}
a_{11} & a_{12} & \ldots & a_{1n} \\
\vdots & \vdots & \ddots & \vdots \\
k a_{s1} & k a_{s2} & \cdots & k a_{sn} \\
\vdots & \vdots & \ddots & \vdots \\
a_{s1} & a_{s2} & \cdots & a_{sn} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn} \\
\end{vmatrix}$
Thus by Determinant with Row Multiplied by Constant:
:$\map \det {\mathbf B} = \begin{vmatrix}
a_{11} & a_{12} & \ldots & a_{1n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{r1} + k a_{s1} & a_{r2} + k a_{s2} & \cdots & a_{rn} + k a_{sn} \\
\vdots & \vdots & \ddots & \vdots \\
a_{s1} & a_{s2} & \cdots & a_{sn} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn} \\
\end{vmatrix} = \begin{vmatrix}
a_{11} & a_{12} & \ldots & a_{1n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{r1} & a_{r2} & \cdots & a_{rn} \\
\vdots & \vdots & \ddots & \vdots \\
a_{s1} & a_{s2} & \cdots & a_{sn} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn} \\
\end{vmatrix} + k \begin{vmatrix}
a_{11} & a_{12} & \ldots & a_{1n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{s1} & a_{s2} & \cdots & a_{sn} \\
\vdots & \vdots & \ddots & \vdots \\
a_{s1} & a_{s2} & \cdots & a_{sn} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn} \\
\end{vmatrix}$
But by Square Matrix with Duplicate Rows has Zero Determinant:
:$\begin{vmatrix}
a_{11} & a_{12} & \ldots & a_{1n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{s1} & a_{s2} & \cdots & a_{sn} \\
\vdots & \vdots & \ddots & \vdots \\
a_{s1} & a_{s2} & \cdots & a_{sn} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn} \\
\end{vmatrix} = 0$
The result follows.
Category:Multiple of Row Added to Row of Determinant
Multiple of Semiperfect Number is Semiperfect

== Theorem ==
Let $n \in \Z_{>0}$ be a semiperfect number.
Let $k \in \Z_{>0}$ be a (strictly) positive integer.
Then $k n$ is also a semiperfect number.
== Proof ==
Let $P$ be a subset of the divisors of $n$ such that the sum of the elements of $P$ equals $n$.
Let $\sigma = \ds \sum_{p \mathop \in P} p$ be the sum of the elements of $P$.
Let:
:$Q = \set {k p: p \in P}$
be the set of elements of $P$ multiplied by $k$.
We have by definition that:
:$\forall p \in P: p \divides P$
where $p \divides P$ denotes divisibility.
Then we have:
:$\forall p \in P: k p \divides k P$
Thus $Q$ constitutes a subset of the divisors of $k n$.
But:
demonstrating that $Q$ constitutes a subset of the divisors of $k n$ equal to $k n$.
Hence the result, by definition of semiperfect number.
Category:Semiperfect Numbers
Multiple of Sine plus Multiple of Cosine/Cosine Form

== Theorem ==
:$p \sin x + q \cos x = \sqrt {p^2 + q^2} \map \cos {x + \arctan \dfrac {-p} q}$
== Proof ==
Let it be assumed that $p \sin x + q \cos x$ can be expressed in the form $M \map \cos {x + \phi}$.
Then:
Then:
Hence the result.
Category:Sine Function
Category:Cosine Function
Multiple of Sine plus Multiple of Cosine/Sine Form

== Theorem ==
:$p \sin x + q \cos x = \sqrt {p^2 + q^2} \map \sin {x + \arctan \dfrac q p}$
== Proof ==
Let it be assumed that $p \sin x + q \cos x$ can be expressed in the form $M \map \sin {x + \phi}$.
Then:
Then:
Hence the result.
Category:Sine Function
Category:Cosine Function
Multiple of Supremum

== Theorem ==
Let $S \subseteq \R: S \ne \O$ be a non-empty subset of the set of real numbers $\R$.
Let $S$ be bounded above.
Let $z \in \R: z > 0$ be a positive real number.
Then:
:$\ds \map {\sup_{x \mathop \in S} } {z x} = z \map {\sup_{x \mathop \in S} } x$
== Proof ==
Let $B = \map \sup S$.
Then by definition, $B$ is the smallest number such that $x \in S \implies x \le B$.
Let $T = \set {z x: x \in S}$.
Because $z > 0$, it follows that:
:$\forall x \in S: z x \le z B$
So $T$ is bounded above by $z B$.
By the Continuum Property, $T$ has a supremum which we will call $C$.
We need to show that $C = z B$.
Since $z B$ is an upper bound for $T$, and $C$ is the smallest upper bound for $T$, it follows that $C \le z B$.
Now as $z > 0$ and is a real number:
:$\exists z^{-1} \in \R: z^{-1} > 0$
So we can reverse the roles of $S$ and $T$:
:$S = \set {z^{-1} y: y \in T}$
We know that $C$ is the smallest number such that:
:$\forall y \in T: y \le C$
So it follows that:
:$\forall y \in T: z^{-1} y \le z^{-1} C$
So $z^{-1} C$ is an upper bound for $S$.
But $B$ is the smallest upper bound for $S$.
So:
:$B \le z^{-1} C \implies z B \le C$
So we have shown that:
:$z B \le C$ and $C \le z B$
Hence the result.
== Sources ==
* : $\S 2$: Continuum Property: $2.11$: Manipulations with $\sup$ and $\inf$: Theorem $\S 2.12$
Category:Real Analysis
Multiple of Vector in Topological Vector Space Converges

== Theorem ==
Let $K$ be a topological field. 
Let $\struct {X, \tau}$ be a topological vector space. 
Let $x \in X$.
Let $\lambda \in K$. 
Let $\sequence {\lambda_n}_{n \mathop \in \N}$ be a sequence in $K$ such that $\lambda_n \to \lambda$. 
Then: 
:$\lambda_n x \to \lambda x$
== Proof ==
From the definition of a topological vector space, the map $f : K \times \struct {X, \tau} \to X$ defined by:
:$\map f {\lambda, y} = \lambda y$ 
is continuous. 
From Horizontal Section of Continuous Function is Continuous, we therefore have the map $c_x : X \to X$ defined by: 
:$\map {c_x} \lambda = \lambda x$ 
is continuous.
From Continuous Mapping is Sequentially Continuous, $c_x$ is sequentially continuous.
So we have: 
:$\lambda_n x \to \lambda x$
Category:Topological Vector Spaces
Multiples of Alternate Ratios of Equal Fractions

== Theorem ==
== Proof ==
Let the (natural) number $AB$ be an aliquant part of the (natural) number $C$, and another $DE$ be the same aliquant part of another, $F$.
We need to show that whatever aliquant part or aliquot part $AB$ is of $DE$, the same aliquant part or aliquot part is $C$ of $F$.
:
We have that whatever aliquant part $AB$ is of $C$, the same aliquant part also is $DE$ of $F$.
So as many aliquant parts of $C$ as there are in $AB$, the same aliquant parts also is $DE$ of $F$.
Let $AB$ be divided into the aliquant parts of $C$, that is, $AG, GB$, and $DE$ into the aliquant parts of $F$, that is, $DH, HE$.
Thus the multitude of $AG, GB$ will be equal to the multitude of $DH, HE$.
Now we have that whatever aliquot part $AG$ is of $C$, the same aliquot part also is $DH$ of $F$.
From , whatever aliquot part $AG$ is of $DH$, the same aliquot part or the same aliquant part is $C$ of $F$ also.
For the same reason, whatever part $GB$ is of $HE$, the same aliquot part or the same aliquant part is $C$ of $F$ also.
So from and , whatever aliquant part or aliquot part $AB$ is of $DE$, the same aliquant part also, or the same aliquot part, is $C$ of $F$.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Ratios
Multiples of Divisors obey Distributive Law

== Theorem ==
In modern algebraic language:
:$a = \dfrac m n b, c = \dfrac m n d \implies a + c = \dfrac m n \paren {b + d}$
== Euclid's Proof ==
== Modern Proof ==
Category:Algebra
Category:Divisors
Category:Multiples of Divisors obey Distributive Law
Multiples of Divisors obey Distributive Law/Proof 1

== Theorem ==
== Proof ==
Let the (natural) number $AB$ be an aliquant part of the (natural) number $C$.
Let the (natural) number $DE$ be the same aliquant part of another (natural) number $F$ that $AB$ is of $C$.
We need to show that $AB + DE$ is the same aliquant part of $C + F$.
:
We have that whatever aliquant part $AB$ is of $C$, $DE$ is also the same aliquant part as $F$.
It follows that as many aliquant parts of $C$ as there are in $AB$, so many aliquant parts of $F$ are there also in $DE$.
Let $AB$ be divided into the aliquant parts of $C$, namely $AG, GB$, and $DE$ into the aliquant parts of $F$, namely $DH, HE$.
Thus the multitude of $AG, GB$ will be equal to the multitude of $DH, HE$.
We have that whatever aliquot part $AG$ is of $C$, the same aliquot part is $DH$ of $F$ also.
Therefore, from , whatever aliquot part $AG$ is of $C$, the same aliquot part also is $AG + DH$ of $C + F$ also.
For the same reason, whatever aliquot part $GB$ is of $C$, the same aliquot part also is $GB + HE$ of $C + F$.
Therefore whatever aliquant part $AB$ is of $C$, the same aliquant part also is $AB + DE$ of the $C + F$
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Multiples of Divisors obey Distributive Law
Multiples of Factorial Plus One are Coprime

== Theorem ==
Let $a \in \N$ be a natural number.
Let $\sequence {x_1, x_2, \dotsc, x_n}$ be a sequence of natural numbers such that:
:$\forall i: 1 \le i \le n: x_i = 1 + i \times a!$
where $a!$ denotes the factorial of $a$.
Let $a \ge n - 1$.
Then $\set {x_i}$ are pairwise coprime.
== Proof ==
Let $p$ be a prime number.
Suppose that $p \divides x_i$.
Then, by definition of $x_i$:
:$p \divides 1 + i \times a!$
Thus, by Consecutive Integers are Coprime:
:$p \nmid i \times a!$
Therefore:
:$p \nmid a!$
 $p \divides x_j$, where $j \ne i$.
Then $p \divides 1 + j \times a!$.
By Common Divisor Divides Difference:
:$p \divides \paren {i - j} \times a!$
By Euclid's Lemma, along with $p \nmid a!$ above:
:$p \divides i - j$
But as $0 < \size {i - j} < n$:
:$p \divides \paren {n - 1}!$
Additionally, since $a \ge n - 1$:
:$\paren {n - 1}! \divides a!$
By Divisor Relation is Transitive:
:$p \divides a!$
However, $p \nmid a!$ from above, a contradiction.
Thus, by Proof by Contradiction, there is no such $x_j$.
Hence the result.
Category:Coprime Integers
Category:Factorials
Multiples of Homogeneous Cartesian Coordinates represent Same Point

== Theorem ==
Let $\CC$ denote the Cartesian plane.
Let $P$ be an arbitrary point in $\CC$.
Let $P$ be expressed in homogeneous Cartesian coordinates as:
:$P = \tuple {X, Y, Z}$
Then $P$ can also be expressed as:
:$P = \tuple {\rho X, \rho Y, \rho Z}$
where $\rho \in \R$ is an arbitrary real number such that $\rho \ne 0$.
== Proof ==
By definition of homogeneous Cartesian coordinates, $P$ can be expressed in conventional Cartesian coordinates as:
:$P = \tuple {x, y}$
where:
for arbitrary $Z$.
We have that:
The result follows.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $9$. Parallel lines. Points at infinity
Category:Homogeneous Cartesian Coordinates
Multiples of Ratios of Numbers

== Theorem ==
== Proof ==
Let the number $A$ by multiplying the two numbers $B, C$ to make $D, E$.
We need to show that:
:$B : C = D : E$
:
We have that:
:$A \times B = D$
Therefore $B$ measures $D$ according to the units in $A$.
But the unit $F$ also measures $A$ according to the units in it.
Therefore $F$ measures $A$ the same number of times that $B$ measures $D$.
So from :
:$F : A = B : D$
For the same reason:
:$F : A = C : E$
Therefore also:
:$B : D = C : E$
So from :
:$B : C = D : E$
== Examples ==
== Sources ==
* : Book $\text{VII}$. Propositions
* 
* 
Category:Multiples of Ratios of Numbers
Category:Ratios
Multiples of Ratios of Numbers/Examples/Arbitrary Example 1

== Examples of Multiples of Ratios of Numbers ==
We have:
:$12 : 15$ is the same ratio as $4 : 5$, and also the same ratio as $24 : 30$.
== Proof ==
We have:
== Sources ==
* 
* 
Category:Examples of Multiples of Ratios of Numbers
Multiples of Reciprocals with Maximum Period form Magic Square

== Theorem ==
Let $p$ be a prime number whose reciprocal has a decimal expansion which has the maximum period, that is, $p - 1$.
Let the first $p - 1$ digits of $\dfrac 1 p$ be expressed as an integer $n$ (with one or more leading zeroes as it is presented).
Then the first $p - 1$ multiples of $n$, when listed in order of size, arrange themselves as a magic square such that:
:the sum of the elements of each row
:the sum of the elements in each column
are the same.
== Proof ==
== Examples ==
== Sources ==
* : $052,631,578,947,368,421$
* : $052,631,578,947,368,421$
Category:Reciprocals
Category:Long Period Primes
Category:Multiples of Reciprocals with Maximum Period form Magic Square
Multiples of Reciprocals with Maximum Period form Magic Square/Examples/19

== Example of Multiples of Reciprocals with Maximum Period form Magic Square ==
We have that:
:$\dfrac 1 {19} = 0 \cdotp \dot 05263 \, 15789 \, 47368 \, 42 \dot 1$
Hence:
:$\begin{array}{|r|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline \frac 1 {19} & 0 & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 8 & 9 & 4 & 7 & 3 & 6 & 8 & 4 & 2 & 1 \\
\hline \frac 2 {19} & 1 & 0 & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 8 & 9 & 4 & 7 & 3 & 6 & 8 & 4 & 2 \\
\hline \frac 3 {19} & 1 & 5 & 7 & 8 & 9 & 4 & 7 & 3 & 6 & 8 & 4 & 2 & 1 & 0 & 5 & 2 & 6 & 3 \\
\hline \frac 4 {19} & 2 & 1 & 0 & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 8 & 9 & 4 & 7 & 3 & 6 & 8 & 4 \\
\hline \frac 5 {19} & 2 & 6 & 3 & 1 & 5 & 7 & 8 & 9 & 4 & 7 & 3 & 6 & 8 & 4 & 2 & 1 & 0 & 5 \\
\hline \frac 6 {19} & 3 & 1 & 5 & 7 & 8 & 9 & 4 & 7 & 3 & 6 & 8 & 4 & 2 & 1 & 0 & 5 & 2 & 6 \\
\hline \frac 7 {19} & 3 & 6 & 8 & 4 & 2 & 1 & 0 & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 8 & 9 & 4 & 7 \\
\hline \frac 8 {19} & 4 & 2 & 1 & 0 & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 8 & 9 & 4 & 7 & 3 & 6 & 8 \\
\hline \frac 9 {19} & 4 & 7 & 3 & 6 & 8 & 4 & 2 & 1 & 0 & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 8 & 9 \\
\hline \frac {10} {19} & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 8 & 9 & 4 & 7 & 3 & 6 & 8 & 4 & 2 & 1 & 0 \\
\hline \frac {11} {19} & 5 & 7 & 8 & 9 & 4 & 7 & 3 & 6 & 8 & 4 & 2 & 1 & 0 & 5 & 2 & 6 & 3 & 1 \\
\hline \frac {12} {19} & 6 & 3 & 1 & 5 & 7 & 8 & 9 & 4 & 7 & 3 & 6 & 8 & 4 & 2 & 1 & 0 & 5 & 2 \\
\hline \frac {13} {19} & 6 & 8 & 4 & 2 & 1 & 0 & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 8 & 9 & 4 & 7 & 3 \\
\hline \frac {14} {19} & 7 & 3 & 6 & 8 & 4 & 2 & 1 & 0 & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 8 & 9 & 4 \\
\hline \frac {15} {19} & 7 & 8 & 9 & 4 & 7 & 3 & 6 & 8 & 4 & 2 & 1 & 0 & 5 & 2 & 6 & 3 & 1 & 5 \\
\hline \frac {16} {19} & 8 & 4 & 2 & 1 & 0 & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 8 & 9 & 4 & 7 & 3 & 6 \\
\hline \frac {17} {19} & 8 & 9 & 4 & 7 & 3 & 6 & 8 & 4 & 2 & 1 & 0 & 5 & 2 & 6 & 3 & 1 & 5 & 7 \\
\hline \frac {18} {19} & 9 & 4 & 7 & 3 & 6 & 8 & 4 & 2 & 1 & 0 & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 8 \\
\hline
\end{array}$
As can be seen, each of the rows and columns add to $81$.
As a bonus, so do the numbers along each diagonal.
== Sources ==
* 
* : $052,631,578,947,368,421$
* : $052,631,578,947,368,421$
Category:Multiples of Reciprocals with Maximum Period form Magic Square
Category:19
Multiples of Terms in Equal Ratios

== Theorem ==
Let $a, b, c, d$ be quantities.
Let $a : b = c : d$ where $a : b$ denotes the ratio between $a$ and $b$.
Then for any numbers $m$ and $n$:
:$m a : n b = m c : n d$
== Euclid's Proof ==
== Modern Proof ==
Category:Ratios
Category:Multiples of Terms in Equal Ratios
Multiples of Terms in Equal Ratios/Euclid's Proof

== Theorem ==
== Proof ==
Let a first magnitude $A$ have to a second magnitude $B$ the same ratio as a third $C$ to a fourth $D$.
Let equimultiples $E, F$ be taken of $A, C$, and let different equimultiples $G, H$ be taken of $B, D$.
We need to show that $E : G = F : H$.
:
Let equimultiples $K, L$ be taken of $E, F$ and other arbitrary equimultiples $M, N$ be taken of $G, H$.
We have that $E$ is the same multiple of $A$ that $L$ is of $C$.
So from Proposition $3$: Associative Law of Multiplication, $K$ is the same multiple of $A$ that $L$ is of $C$.
For the same reason, $M$ is the same multiple of $B$ that $N$ is of $D$.
We have that:
:$A$ is to $B$ as $C$ is to $D$
:of $A, C$ equimultiples $K, L$ have been taken
:of $B, D$ other equimultiples $M, N$ have been taken.
So from the definition of equality of ratios:
:if $K$ is in excess of $M$, $L$ is also in excess of $N$
:if $K$ is equal to $M$, $L$ is equal to $N$
:if $K$ is less than $M$, $L$ is less than $N$
But $K, L$ are equimultiples of $E, F$.
Therefore as $E$ is to $G$, so is $F$ to $H$.
== Sources ==
* : Book $\text{V}$. Propositions
Category:Ratios
Category:Multiples of Terms in Equal Ratios
Multiplication Function on Ring with Unity is Zero if Characteristic is Divisor

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let the characteristic of $R$ be $p$.
Let $g_a: \Z \to R$ be the mapping from the integers into $R$ defined as:
:$\forall n \in \Z:\forall a \in R: \map {g_a} n = n \cdot a$
where $\cdot$ denotes the multiple operation.
Then:
:$p \divides n \implies n \cdot a = 0_R$
where $p \divides n$ denotes that $p$ is a divisor of $n$.
== Proof ==
Let $p > 0$.
From Principal Ideal of Characteristic of Ring is Subset of Kernel of Multiple Function:
:$\ideal p \subseteq \map \ker {g_a}$
where:
:$\map \ker {g_a}$ is the kernel of $g_a$
:$\ideal p$ is the principal ideal of $\Z$ generated by $p$.
We have:
Then we have:
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Theorem $24.8 \ 1^\circ$
Category:Homomorphism from Integers into Ring with Unity
Multiplication Property of Characteristic Functions

== Theorem ==
Let $X$ and $Y$ be independent random variables on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $\phi_X$ and $\phi_Y$ denote the characteristic functions of $X$ and $Y$ respectively.
Then:
:$\phi_{X + Y} = \phi_X \phi_Y$
== Proof ==
Let $i = \sqrt{-1}$.
Let $\expect X$ denote the expectation of $X$.
Hence: 
:$\phi_{X + Y} = \phi_X \phi_Y$
== Sources ==
* : $\S 11.1$
Category:Probability Theory
Multiplication by 2 over 3 in Egyptian Fractions

== Theorem ==
Let $\dfrac 1 n$ be an Egyptian fraction not equal to $\dfrac 2 3$.
In order to multiply $\dfrac 1 n$ by $\dfrac 2 3$ and have it that $\dfrac 1 n \times \dfrac 2 3$ is also expressed in Egyptian form, we have:
:$\dfrac 1 n \times \dfrac 2 3 = \dfrac 1 {2 n} + \dfrac 1 {6 n}$
== Proof ==
Note the case where we multiply $\dfrac 2 3$ by $\dfrac 2 3$ itself:
== Examples ==
== Sources ==
* : Egyptian Fractions
Category:Egyptian Fractions
Category:Multiplication by 2 over 3 in Egyptian Fractions
Multiplication by 2 over 3 in Egyptian Fractions/Examples

== Examples of Use of Multiplication by 2 over 3 in Egyptian Fractions ==
=== Example: $\dfrac 2 3 \times \dfrac 1 5$ ===
=== Example: $\dfrac 2 3 \times \dfrac 1 8$ ===
Category:Multiplication by 2 over 3 in Egyptian Fractions
Multiplication by 2 over 3 in Egyptian Fractions/Examples/1 over 5

== Example of Use of Multiplication by 2 over 3 in Egyptian Fractions ==
:$\dfrac 2 3 \times \dfrac 1 5 = \dfrac 1 {10} + \dfrac 1 {30}$
== Proof ==
== Sources ==
* : Egyptian Fractions
Category:Multiplication by 2 over 3 in Egyptian Fractions
Multiplication by 2 over 3 in Egyptian Fractions/Examples/1 over 8

== Example of Use of Multiplication by 2 over 3 in Egyptian Fractions ==
:$\dfrac 2 3 \times \dfrac 1 8 = \dfrac 1 {16} + \dfrac 1 {48}$
despite the fact that it is apparent that:
:$\dfrac 2 3 \times \dfrac 1 8 = \dfrac 1 {12}$
== Proof ==
== Sources ==
* : Egyptian Fractions
Category:Multiplication by 2 over 3 in Egyptian Fractions
Multiplication by Imaginary Unit is Equivalent to Rotation through Right Angle

== Theorem ==
Let $z \in \C$ be a complex number.
Let $z$ be interpreted as a vector in the complex plane.
Let $w \in \C$ be the complex number defined as $z$ multiplied by the imaginary unit $i$:
:$w = i z$
Then $w$ can be interpreted as the vector $z$ after being rotated through a right angle in an anticlockwise direction.
== Proof ==
:
Let $z$ be expressed in polar form as:
:$z = r \left({\cos \theta + i \sin \theta}\right)$
From Polar Form of Complex Number: $i$:
:$i = \cos \dfrac \pi 2 + i \sin \dfrac \pi 2$
and so:
: the modulus of $i$ is $1$
: the argument of $i$ is $\dfrac \pi 2$.
By Product of Complex Numbers in Polar Form:
: the modulus of $i z$ is $r$
: the argument of $i$ is $\theta + \dfrac \pi 2$.
That is, the result of multiplying $z$ by $i$ is the same as rotating $z$ through $\dfrac \pi 2$, which is a right angle.
== Sources ==
* : $\S 2$. Geometrical Representations
Category:Complex Multiplication
Category:Geometry of Complex Plane
Multiplication by Negative Real Number

== Theorem ==
:$\forall x, y \in \R: x \times \paren {-y} = -\paren {x \times y} = \paren {-x} \times y$
=== Corollary ===
== Proof ==
Similarly:
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $1 \ \text{(e)}$
Category:Real Multiplication
Multiplication by Negative Real Number/Corollary

== Corollary to Multiplication by Negative Real Number ==
: $\forall x \in \R: \paren {-1} \times x = -x$
== Proof ==
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $1 \ \text{(f)}$
Category:Real Multiplication