nyuuzyou/wikis huggingface dataset:
Number does not divide Number iff Cube does not divide Cube

== Theorem ==
Let $a, b \in \Z$ be integers.
Then:
:$a \nmid b \iff a^3 \nmid b^3$
where $a \nmid b$ denotes that $a$ is not a divisor of $b$.
== Proof ==
Let $a \nmid b$.
:
:$a^3 \divides b^3$
where $\divides$ denotes divisibility.
Then by Number divides Number iff Cube divides Cube:
:$a \divides b$
From Proof by Contradiction it follows that $a^2 \divides b^2$ is false.
Thus $a^3 \nmid b^3$.
Let $a^3 \nmid b^3$.
:$a \divides b$
Then by Number divides Number iff Cube divides Cube:
:$a^3 \divides b^3$
From Proof by Contradiction it follows that $a \divides b$ is false.
Thus $a \nmid b$.
== Sources ==
* : Book $\text{VIII}$. Propositions
Category:Divisors
Category:Cube Numbers
Number does not divide Number iff Square does not divide Square

== Theorem ==
Let $a, b \in \Z$ be integers.
Then:
:$a \nmid b \iff a^2 \nmid b^2$
where $a \nmid b$ denotes that $a$ is not a divisor of $b$.
== Proof ==
Let $a \nmid b$.
:
:$a^2 \divides b^2$
where $\divides$ denotes divisibility.
Then by Number divides Number iff Square divides Square:
:$a \divides b$
From Proof by Contradiction it follows that $a^2 \divides b^2$ is false.
Thus $a^2 \nmid b^2$.
Let $a^2 \nmid b^2$.
:
:$a \divides b$
Then by Number divides Number iff Square divides Square:
:$a^2 \divides b^2$
From Proof by Contradiction it follows that $a \divides b$ is false.
Thus $a \nmid b$.
== Sources ==
* : Book $\text{VIII}$. Propositions
Category:Divisors
Category:Square Numbers
Number greater than Integer iff Ceiling greater than Integer

== Theorem ==
Let $x \in \R$ be a real number.
Let $\ceiling x$ be the ceiling of $x$.
Let $n \in \Z$ be an integer.
Then:
:$\ceiling x > n \iff x > n$
== Proof ==
=== Necessary Condition ===
Let $x > n$.
By Number is between Ceiling and One Less:
:$\ceiling x \ge x$
Hence:
:$\ceiling x > n$
=== Sufficient Condition ===
Let $\ceiling x > n$.
We have that:
:$\forall m, n \in \Z: m < n \iff m \le n - 1$
and so:
:$(1): \quad \ceiling x - 1 \ge n$
Then:
Hence the result:
:$\ceiling x > n \iff x > n$
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $3 \ \text{(d)}$
Category:Ceiling Function
Number is Divisor iff Modulo is Zero

== Theorem ==
Let $x, y \in \R$ be real numbers.
Let $x \bmod y$ denote the modulo operation:
:$x \bmod y := \begin {cases}
x - y \floor {\dfrac x y} & : y \ne 0 \\
x & : y = 0
\end {cases}$
where $\floor {\dfrac x y}$ denotes the floor of $\dfrac x y$.
Then $x \bmod y = 0$ $x$ is an integer multiple of $y$.
== Proof ==
=== Sufficient Condition ===
Let $x \bmod y = 0$.
From Number minus Modulo is Integer Multiple:
:$x - \paren {x \bmod y}$
is an integer multiple of $y$.
As $x \bmod y = 0$ it follows that $x - 0 = x$ is an integer multiple of $y$.
=== Necessary Condition ===
If $y = 0$ it follows from Zero Divides Zero that $x = 0$.
Hence the result by definition of the modulo operation:
Otherwise, let $y \ne 0$.
Let $x$ be an integer multiple of $y$.
That is:
:$\exists n \in \Z: x = n y$
Hence:
:$\dfrac x y = n$
From Quotient of Modulo Operation with Modulus:
:$\dfrac x y - \floor {\dfrac x y} = \dfrac {x \bmod y} y$
From Real Number is Integer iff equals Floor it follows that:
:$\dfrac x y = \floor {\dfrac x y}$
Thus:
:$\dfrac {x \bmod y} y = 0$
and so multiplying both sides by $y$ the result follows.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory
Category:Modulo Arithmetic
Number is Sum of Five Cubes

== Theorem ==
Let $n \in \Z$ be an integer.
Then $n$ can be expressed as the sum of $5$ cubes (either positive or negative) in an infinite number of ways.
== Proof ==
We have for any $m, n \in \Z$:
By definition of modulo arithmetic:
:$\exists k \in \Z: \paren {6 m + n}^3 = n + 6 k$
We also have:
Thus $n = \paren {6 m + n}^3 + k^3 + k^3 - \paren {k + 1}^3 - \paren {k - 1}^3$ is an expression of $n$ as a sum of $5$ cubes.
In the equation above, $m$ is arbitrary and $k$ depends on both $m$ and $n$.
As there is an infinite number of choices for $m$, there is an infinite number of such expressions.
== Remark ==
From the result above we can also write each term of the sum explicitly:
However, because of the complexity of this expression it is recommended to first calculate $k$.
== Sources ==
* : No. $227$
* : $5$
* : $5$
Category:Cube Numbers
Category:Sums of Cubes
Category:5
Number is between Ceiling and One Less

== Theorem ==
:$\ceiling x - 1 < x \le \ceiling x$
where $\ceiling x$ denotes the ceiling of $x$.
== Proof ==
By definition of ceiling of $x$:
:$\forall x \in \R: \ceiling x = \map \inf {\set {m \in \Z: m \ge x} }$
By definition of infimum:
:$\ceiling x \ge x$
Also by definition of infimum:
:$\ceiling x - 1 \not \ge x$
as $\ceiling x$ is the smallest integer with that property.
That is:
:$x > \ceiling x - 1$
Hence the result.
Category:Ceiling Function
Number less than Integer iff Floor less than Integer

== Theorem ==
Let $x \in \R$ be a real number.
Let $\floor x$ denote the floor of $x$.
Let $n \in \Z$ be an integer.
Then:
:$\floor x < n \iff x < n$
== Proof ==
=== Necessary Condition ===
Let $x < n$.
By definition of the floor of $x$:
:$\floor x \le x$
Hence:
:$\floor x < n$
=== Sufficient Condition ===
Let $\floor x < n$.
We have that:
:$\forall m, n \in \Z: m < n \iff m + 1 \le n$
and so:
:$(1): \quad \floor x + 1 \le n$
Then:
Hence the result:
:$\floor x < n \iff x < n$
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $3 \ \text{(a)}$
Category:Floor Function
Number minus Modulo is Integer Multiple

== Theorem ==
Let $x, y \in \R$ be real numbers.
Let $x \bmod y$ denote the modulo operation:
:$x \bmod y := \begin{cases}
x - y \floor {\dfrac x y} & : y \ne 0 \\
x & : y = 0
\end{cases}$
where $\floor {\dfrac x y}$ denotes the floor of $\dfrac x y$.
Let $y < 0$.
Then:
:$x - \paren {x \bmod y}$
is an integer multiple of $y$.
== Proof ==
When $y = 0$ we have:
:$x \bmod y := x$
Thus:
:$x - \paren {x \bmod y} = 0$
From Zero is Integer Multiple of Zero it follows that:
:$x - \paren {x \bmod y}$
is an integer multiple of $y$.
Let $y \ne 0$.
Then:
From Floor Function is Integer, $\floor {\dfrac x y}$ is an integer.
Thus:
:$\exists n \in \Z: x - \paren {x \bmod y} = n y$
where in the case $n = \floor {\dfrac x y}$
Hence the result by definition of integer multiple.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: $(2) \ \text{c)}$
Category:Modulo Arithmetic
Number multiplied by Cube Number making Cube is itself Cube

== Theorem ==
== Proof ==
Let $a, b \in \Z$ be integers such that $a^3 b$ is a cube number.
By Square of Cube Number is Cube, $a^3 \cdot a^3$ is a cube number.
By Multiples of Ratios of Numbers:
:$a^3 \cdot a^3 : a^3 b = a^3 : b$
From Between two Cubes exist two Mean Proportionals, there exist two mean proportionals between $a^3 \cdot a^3$ and $a^3 b$.
By Geometric Sequences in Proportion have Same Number of Elements, there exist two mean proportionals between $a^3$ and $b$.
The result follows from If First of Four Numbers in Geometric Sequence is Cube then Fourth is Cube.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Euclidean Number Theory
Number neither whose Half is Odd nor Power of Two is both Even-Times Even and Even-Times Odd

== Theorem ==
Let $a \in \Z$ be an integer such that:
:$(1): \quad a$ is not a power of $2$
:$(2): \quad \dfrac a 2$ is an even integer.
Then $a$ is both even-times even and even-times odd.
== Proof ==
As $\dfrac a 2$ is an even integer it follows that:
:$\dfrac a 2 = 2 r$
for some $r \in \Z$.
That is:
:$a = 2^2 r$
and so $a$ is even-times even by definition.
 $a$ is not even-times odd.
Then $a$ does not have an odd divisor.
Thus in its prime decomposition there are no odd primes.
Thus $a$ is in the form:
:$a = 2^k$
for some $k \in \Z_{>0}$.
That is, $a$ is a power of $2$.
This contradicts condition $(1)$.
From this contradiction it is deduced that $a$ is even-times odd.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Euclidean Number Theory
Number not greater than Integer iff Ceiling not greater than Integer

== Theorem ==
Let $x \in \R$ be a real number.
Let $\ceiling x$ be the ceiling of $x$.
Let $n \in \Z$ be an integer.
Then:
:$\ceiling x \le n \iff x \le n$
== Proof ==
=== Necessary Condition ===
Let $\ceiling x \le n$.
By Number is between Ceiling and One Less:
:$x \le \ceiling x$
Hence:
:$x \le n$
=== Sufficient Condition ===
Let $x \le n$.
 $\ceiling x > n$.
We have that:
:$\forall m, n \in \Z: m < n \iff m \le n - 1$
Hence:
:$\ceiling x - 1 \ge n$
and so by hypothesis:
:$\ceiling x - 1 \ge x$
This contradicts the result Number is between Ceiling and One Less:
:$\ceiling x - 1 < x$
Thus by Proof by Contradiction:
:$\ceiling x \le n$
Hence the result:
:$\ceiling x \le n \iff x \le n$
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $3 \ \text{(c)}$
Category:Ceiling Function
Number not less than Integer iff Floor not less than Integer

== Theorem ==
Let $x \in \R$ be a real number.
Let $\floor x$ denote the floor of $x$.
Let $n \in \Z$ be an integer.
Then:
:$x \ge n \iff \floor x \ge n$
== Proof ==
=== Necessary Condition ===
Let $\floor x \ge n$.
By definition of the floor of $x$:
:$x \ge \floor x$
Hence:
:$x \ge n$
=== Sufficient Condition ===
Let $x \ge n$.
 $\floor x < n$.
We have that:
:$\forall m, n \in \Z: m < n \iff m + 1 \le n$
Hence:
:$\floor x + 1 \le n$
and so by hypothesis:
:$\floor x + 1 \le x$
This contradicts the definition of the floor of $x$:
:$\floor x + 1 > x$
Thus by Proof by Contradiction:
:$\floor x \ge n$
Hence the result:
:$\floor x \ge n \iff x \ge n$
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $3 \ \text{(b)}$
Category:Floor Function
Number of Abelian Groups

== Theorem ==
Let $n \in \Z_{\ge 1}$ be a (strictly) positive integer.
Let:
:$n = \ds \prod_{i \mathop = 1}^s p_i^{m_i}$
where the $p_i$ are distinct primes.
Let $\map {\nu_a} n$ denote the number of abelian groups of order $n$.
Then:
:$\map {\nu_a} n = \ds \prod_{i \mathop = 1}^s \map {\nu_a} {p_i^{m_i} }$
where:
:$\map {\nu_a} {p_i^{m_i} }$ is the number of integer partitions of $m_i$.
== Proof ==
== Sources ==
* : $1$: Introduction to Finite Group Theory: $1.5$
Category:Abelian Groups
Number of Arrangements of n Objects of m Types

== Theorem ==
Let $S$ be a collection of $n$ objects.
Let these $n$ objects be of $m$ different types, as follows:
Let there be:
:$k_1$ objects of type $1$
:$k_2$ objects of type $2$
:$\cdots$
:$k_m$ objects of type $m$
such that:
:for each $j \in \set {1, 2, \ldots, m}$, all objects of type $j$ are indistinguishable from each other
:$k_1 + k_2 + \cdots + k_m = n$
Then the total number $N$ of different arrangements of $S$ is given by the multinomial coefficient:
:$N = \dbinom n {k_1, k_2, \ldots, k_m} = \dfrac {n!} {k_1! \, k_2! \cdots k_m!}$
== Proof ==
Let $N$ be the number of different arrangements of $S$.
First suppose that all $n$ objects are distinct one from another.
Then from Number of Permutations of All Elements:
:$N = n!$
Now suppose that $k_j$ elements of $S$ are indistinguishable from each other.
From Number of Permutations of All Elements, there are $k_j!$ different arrangements of those $k_j$ elements.
Hence the $n!$ arrangements of $S$ can each be grouped into $\dfrac {n!} {k_j!}$ partitions, each with $k_j$ elements, such that all the arrangements in each partition are indistinguishable from each other.
This applies for all $m$ types of objects.
The result follows.
== Examples ==
Category:Number of Arrangements of n Objects of m Types
Category:Combinatorics
Number of Arrangements of n Objects of m Types/Examples

== Examples of Use of Number of Arrangements of $n$ Objects of $m$ Types ==
=== $2$ Types ===
=== $3$ Types ===
=== Set of $3 p$ Objects of $3$ Equal Sized Subsets ===
=== Letters in added ===
=== $6$ people in $3$ pairs ===
=== $10$ people in $3$ groups of sizes $5$, $3$ and $2$ ===
Category:Number of Arrangements of n Objects of m Types
Number of Arrangements of n Objects of m Types/Examples/10 people in 3 groups sizes 5, 3, 2

== Example of Use of Number of Arrangements of $n$ Objects of $m$ Types ==
Let $N$ be the number of ways $10$ people can be partitioned into $3$ sets: one with $5$, one with $3$ and one with $2$ people.
Then:
:$N = 2520$
== Proof ==
Here we have an instance of Number of Arrangements of $n$ Objects into $3$ Types, such that:
:$n = 10$
:$p = 5$
:$q = 3$
:$r = 2$
Hence we have:
Hence the result.
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: Exercises $\text I$: $7$
Category:Number of Arrangements of n Objects of m Types
Number of Arrangements of n Objects of m Types/Examples/2 Types

== Example of Use of Number of Arrangements of $n$ Objects of $m$ Types ==
Let $S$ be a collection of $n$ objects, consisting of:
:$p$ objects of one type
:$q$ objects of another type.
The total number $N$ of different arrangements of $S$ is given by:
:$N = \dfrac {n!} {p! \, q!}$
== Proof ==
An arbitrary arrangement can be made into $p!$ arrangements if the $p$ objects of type $1$ are replaced by $p$ objects that are all different.
Similarly, an arbitrary arrangement can be made into $q!$ arrangements if the $q$ objects of type $2$ are replaced by $q$ objects that are all different.
The total number of arrangements, if all objects are different, would be $n!$
Hence:
:$n! = N \times p! \times q!$
Hence the result.
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: The number of ways of arranging $q$ things in line if $p$ are alike of one kind and $q$ are alike of another kind
Category:Number of Arrangements of n Objects of m Types
Number of Arrangements of n Objects of m Types/Examples/3 Types

== Example of Use of Number of Arrangements of $n$ Objects of $m$ Types ==
Let $S$ be a collection of $\paren {p + q + r}$ objects.
Let $S$ need to be partitioned into $3$ subsets of size $p$, $q$ and $r$ such that $p \ne q$, $q \ne r$ and $r \ne p$.
The total number $N$ of ways this can be done is:
:$N = \dfrac {\paren {p + q + r}!} {p! \, q! \, r!}$
== Proof ==
The number of ways of choosing $p$ from $\paren {p + q + r}$ is $\dbinom {p + q + r} p$.
The number of ways of choosing $q$ from the remaining $\paren {q + r}$ is $\dbinom {q + r} q$.
We are left with the remaining $r$.
Hence $N$ can be given by:
Hence the result.
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: The number of ways of dividing $\paren {p + q + r}$ things into three unequal groups, the first to contain $p$, the second $q$ and the third $r$
Category:Number of Arrangements of n Objects of m Types
Number of Arrangements of n Objects of m Types/Examples/3p Objects into 3 Equal Sized Subsets

== Example of Use of Number of Arrangements of $n$ Objects of $m$ Types ==
Let $S$ be a collection of $3 p$ objects.
Let $S$ need to be partitioned into $3$ subsets of size $p$.
The total number $N$ of ways this can be done is:
:$N = \dfrac {\paren {3 p}!} {\paren {p!}^3 \times 3!}$
== Proof ==
From Number of Arrangements of $n$ Objects of $3$ Types, we can partition $S$ into $3$ subsets in $\dfrac {\paren {3 p}!} {\paren {p!}^3}$ different ways.
However, each of these subsets is of the same size $p$.
Hence they cannot be counted as different.
So we have counted $3!$ ways which are apparently all different, but are in fact the same.
Hence we need to divide our total by $3!$
Hence $N$ can be given by:
:$N = \dfrac {\paren {3 p}!} {\paren {p!}^3 \times 3!}$
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: The number of ways of dividing $3 p$ things into three equal groups each containing $p$ things
Category:Number of Arrangements of n Objects of m Types
Number of Arrangements of n Objects of m Types/Examples/6 people in 3 pairs

== Example of Use of Number of Arrangements of $n$ Objects of $m$ Types ==
Let $N$ be the number of ways $6$ people can be partitioned into $3$ (unordered) pairs.
Then:
:$N = 15$
== Proof ==
Here we have an instance of Number of Arrangements of $3 p$ Objects into $3$ Equal Sized Subsets, such that $p = 2$.
Hence we have:
:$N = \dfrac {\paren {3 \times 2}!} {\paren {2!}^3 \times 3!} = \dfrac {720} {8 \times 6} = \dfrac {720} {48} = 15$
Hence the result.
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: Exercises $\text I$: $3$
Category:Number of Arrangements of n Objects of m Types
Number of Arrangements of n Objects of m Types/Examples/Letters in added

== Example of Use of Number of Arrangements of $n$ Objects of $m$ Types ==
Let $N$ be the number of different arrangements of the letters in the word $\texttt{added}$.
Then:
:$N = 20$
== Proof ==
There are $3$ types of letter in $\texttt{added}$, that is:
:$\texttt a$: $1$ instance
:$\texttt d$: $3$ instances
:$\texttt e$: $1$ instance
Hence we have an instance of Number of Arrangements of $n$ Objects of $m$ Types, such that:
:$n = 5$
:$k_1 = 1$
:$k_2 = 3$
:$k_3 = 1$
Hence we have:
:$N = \dfrac {5!} {1! \times 3! \times 1!} = \dfrac {120} 6 = 20$
Hence the result.
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: Exercises $\text I$: $2$
Category:Number of Arrangements of n Objects of m Types
Number of Atoms in Observable Universe

== Theorem ==
The number of atoms in the observable universe is approximately $10^{80}$.
== Proof ==
According to our current observations and understanding of the , the age of the universe is estimated to be $13.787$ billion years.
That means the light from the most distant galaxies has traveled $13.787$ billion years to reach us.
The speed of light is $299 \, 792 \, 458 \ \text {m s}^{-1}$
This implies that the radius of the observable universe (assuming no expansion) is:
However, contemplating expansion at an average rate of $8.8 \%$ per billion years, we obtain:
This in turn implies that the volume of the observable universe is:
Finally, assuming there is approximately $1$ atom per cubic metre brings us to our estimate of approximately $10^{80}$ atoms in the observable universe.
Category:Atoms
Number of Bijective Restrictions

== Theorem ==
Let $f: S \to T$ be a surjection.
Let $B$ be the set of all bijective restrictions of $f$.
Then the cardinality of $B$ is:
:$\ds \card {\prod_{i \mathop \in I} \family {S / \RR_f}_i}$
where $S / \RR_f$ denotes the quotient set of the induced equivalence of $f$ indexed by $I$.
== Proof ==
Category:Bijections
Category:Restrictions
Number of Binary Digits in Power of 10

== Theorem ==
Let $n$ be a positive integer.
Expressed in binary notation, the number of digits in the $n$th power of $10$:
:$10^n$
is equal to:
:$\ceiling {n \log_2 10}$
where $\ceiling x$ denotes the ceiling of $x$.
== Examples ==
== Proof ==
Let $10^n$ have $m$ digits when expressed in binary notation.
By the Basis Representation Theorem and its implications, a positive integer $x$ has $m$ digits :
:$2^{m - 1} \le x < 2^m$
Thus:
Because a power of $10$ cannot equal a power of $2$, it will always be the case that:
:$m - 1 < n \log_2 10 < m$
and so:
:$m - 1 < n \log_2 10 \le m$
Hence from Integer equals Ceiling iff Number between Integer and One Less:
:$m = \ceiling {n \log_2 10}$
== Sources ==
* : $3 \cdotp 321 \, 928 \ldots$
* : $3 \cdotp 32192 \, 8 \ldots$
Category:Powers of 10
Category:Logarithms
Category:Number of Binary Digits in Power of 10
Number of Binary Digits in Power of 10/Example/1000

== Theorem ==
When expressed in binary notation, the number of digits in $1000$ is $10$.
== Proof ==
Let $m$ be the number of digits in $1000$.
From Number of Binary Digits in Power of 10:
:$m = \ceiling {3 \log_2 10}$
From Logarithm Base 2 of 10:
:$\log_2 10 \approx 3 \cdotp 32192 \, 8 \ldots$
and so:
:$m \approx 9 \cdotp 96$
Hence the result.
The actual number is:
:$1000_{10} = 1 \, 111 \, 101 \, 100_2$
== Sources ==
* : $3 \cdotp 321 \, 928 \ldots$
* : $3 \cdotp 32192 \, 8 \ldots$
Category:1000
Number of Bits for Decimal Integer

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $n$ have $m$ digits when expressed in decimal notation.
Then $n$ may require as many as $\ceiling {\dfrac m {\log_{10} 2} }$ bits to represent it.
== Proof ==
Let $d$ be the number of bits that may be needed to represent $n$.
Let $n$ have $m$ digits.
Then:
:$n \le 10^m - 1$
and so:
== Examples ==
Category:Number Theory
Category:Number of Bits for Decimal Integer
Number of Bits for Decimal Integer/Examples/14 Digits

== Example of Number of Bits for Decimal Integer ==
A positive integer $n \in \Z_{>0}$ which has $14$ digits when expressed in decimal notation may require $47$ bits to represent in binary.
== Proof ==
Let $n$ have $m$ digits.
Let $d$ be the number of bits that may be needed to represent $n$.
From Number of Bits for Decimal Integer:
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $19$
Category:Number of Bits for Decimal Integer
Number of Boolean Interpretations for Finite Set of Variables

== Theorem ==
Let $\PP_0$ be the vocabulary of language of propositional logic.
Let $S \subseteq \PP_0$ be a finite set of $n$ letters from $\PP_0$.
Then there are $2^n$ different partial boolean interpretations for $S$.
== Proof ==
A partial boolean interpretation for $S$ is a mapping from $S$ to the set of truth values $\set {T, F}$.
By Cardinality of Set of All Mappings, the total number of mappings from $S$ to $T$ is:
:$\card {T^S} = \card T^{\card S}$
The result follows directly.
== Sources ==
* : $\S 1.5$: Semantics of Propositional Logic
Category:Boolean Interpretations
Number of Characters on Finite Abelian Group

== Theorem ==
Let $G$ be a finite abelian group.
Then the number of characters $G \to \C^\times$ is $\order G$.
== Proof ==
=== Lemma ===
Let $H \le G$ be a subgroup.
Let $\chi: H \to \C^\times$ be a character on $G$.
Let $a \in G \divides H$ where $\divides$ denotes divisibility.
Then:
:$\chi$ extends to $\index G H$ distinct characters on $G$
where $\index G H$ denotes the index of $H$ in $G$.
=== Proof ===
The proof proceeds by stroing induction on $\index G H$.
If $\index G H = 1$, then $H = G$ by Langrange's theorem.
Hence (trivially) the result.
Suppose that $\index G H > 1$, and the result holds for all subgroups of smaller index in $G$.
Let $a \notin H$ and let $n$ be the indicator of $a$ in $H$.
Let $K = \gen {H, a}$ denote the subgroup generated by $H$ and $a$.
From Subgroup Generated by Subgroup and Element, each element of $K$ has a unique representation in the form $x a^k$ with:
:$x \in H$ and $0 \le k \le n-1$
and:
:$\order K = n \order H$
Let $\tilde \chi$ be a character extending $\chi$.
Then:
:$(1): \quad \map {\tilde \chi} {x a^n} = \map {\tilde \chi} x \map {\tilde \chi} a^n$
and:
:$(2): \quad \map {\tilde \chi} {x a^n} = \map {\tilde \chi} x \map {\tilde \chi} {a^n}$
Since $x, a^n \in H$, $\tilde \chi$ and $\chi$ agree on these values, so equating $(1)$ and $(2)$ we obtain:
:$\map {\tilde \chi} a^n = \map {\tilde \chi} {a^n}$
That is:
:$\map {\tilde \chi} a$ is an $n$th root of $\map {\tilde \chi} {a^n}$.
So by nth roots of a complex number are distinct there are $n$ distinct possibilities.
We choose one of these possibilities and define:
:$\map {\tilde \chi} {x a^k} = \map \chi x \map {\tilde \chi} a^k$
for $x a^k \in K$.
We check that $\tilde \chi$ is multiplicative:
For $x, y \in H$:
By Lagrange's theorem:
:$n = \dfrac {\order K} {\order H} = \index K H$
Also by Lagrange's Theorem:
:$\index G K < \index G H$
since $H \subsetneq K$.
So by the induction hypothesis, each $\tilde \chi$ extends to $\index G K$ characters on $G$.
Therefore by the Tower Law for Subgroups, $\chi$ extends to $\index K H \index G K = \index G H$ characters on $G$.
Since a character is a homomorphism, it must preserve the identity.
Therefore there is only one character $\chi: \set e \to \C^\times : e \mapsto 1$ on the trivial subgroup.
Moreover, by Langrange's theorem, the trivial subgroup has index $\index G {\set e} = \order G$.
So by the lemma there are $\order G$ distinct characters on $G$.
Category:Analytic Number Theory