nyuuzyou/wikis huggingface dataset:
Divisors of One More than Power of 10/Number of Zero Digits Even/Examples

== Examples of Divisors of One More than Power of 10: Number of Zero Digits Even ==
== Demonstration ==
 90909091
 x 11
 --------
 90909091
 909090910
-----------
 1000000001
== Also see ==
* Henry Ernest Dudeney: Modern Puzzles: $62$ -- Factorizing
== Sources ==
* : Solutions: $62$. -- Factorizing
* : Answers: $113$. Factorizing
Category:Divisors of One More than Power of 10
Divisors of Power of Prime

== Theorem ==
Let $p$ be a prime number.
Let $n \in \Z_{> 0}$ be a (strictly) positive integer.
Then the only divisors of $p^n$ are $1, p, p^2, \ldots, p^{n - 1}, p^n$.
== Proof ==
First it is necessary to establish that every element of the set $\set {1, p, p^2, \ldots, p^{n - 1}, p^n}$ is in fact a divisor of $p^n$.
For any $j \in \set {1, 2, \ldots, n}$:
:$p^n = p^j p^{n - j}$
and so each of $1, p, p^2, \ldots, p^{n - 1}, p^n$ is a divisor of $p^n$.
Let:
:$a \in \Z_{>0}: a \notin \set {1, p, p^2, \ldots, p^{n - 1}, p^n}$
Let $a = p^j$ where $j \in \Z: j > n$.
Then:
:$p^j = p^n p^{j - n}$
and so $p^n$ is a divisor of $p^j$.
Hence $p_j \nmid p^n$.
Now let:
:$a \notin \set {p^k: k \in \Z_{>0} }$
Then:
:$\exists q \in \Bbb P: q \divides a$
where:
:$\Bbb P$ is the set of all prime numbers
:$\divides$ denotes divisibility.
 $a \divides p^n$.
From Divisor Relation is Transitive it follows that $q \divides p^n$.
From Euclid's Lemma for Prime Divisors: General Result it follows that:
:$q \divides p$
As $p$ is a prime, by definition its only divisors are $1$ and $p$.
This contradicts the supposition that $q$ is a divisor of $p^n$.
Hence $a \nmid p^n$.
Category:Prime Numbers
Divisors of Product of Coprime Integers

== Theorem ==
Let $a, b, c \in \Z$ be integers.
Let the symbol $\divides$ denote the divisibility relation.
Let $a \divides b c$, where $b \perp c$.
Then $\tuple {r, s}$ satisfying:
:$a = r s$, where $r \divides b$ and $s \divides c$
is unique up to absolute value with:
:$\size r = \gcd \set {a, b}$
:$\size s = \gcd \set {a, c}$
=== Corollary ===
== Proof ==
By Divisor of Product, there exists $\tuple {r, s}$ satisfying:
:$r \divides b$
:$s \divides c$
:$r s = a$
We have:
:$r, s \divides a$
By definition of GCD:
:$\gcd \set {a, b} \divides r$
:$\gcd \set {a, c} \divides s$
By Absolute Value of Integer is not less than Divisors:
:$\gcd \set {a, b} \le \size r$
:$\gcd \set {a, c} \le \size s$
We also have:
This forces both inequalities to be equalities, i.e.:
:$\size r = \gcd \set {a, b}$
:$\size s = \gcd \set {a, c}$
Hence $\tuple {r, s}$ is unique up to absolute value.
Category:Coprime Integers
Divisors of Product of Coprime Integers/Corollary

== Corollary to Divisors of Product of Coprime Integers ==
Let $p$ be a prime.
Let $p \divides b c$, where $b \perp c$.
Then $p \divides b$ or $p \divides c$, but not both.
== Proof ==
From the main result, $p = r s$, where $r \divides b$ and $s \divides c$.
But as $p$ is prime, either:
:$r = 1$ and $s = p$, or:
:$r = p$ and $s = 1$.
So $p \divides b$ or $p \divides c$.
But $p$ can not divide both, as $b \perp c$.
Category:Coprime Integers
Category:Prime Numbers
Divisors of Repunit with Composite Index

== Theorem ==
Let $R_n$ be a repunit number with $n$ digits.
Let $n$ be composite such that $n = r s$ where $1 < r < n$ and $1 < s < n$.
Then $R_r$ and $R_s$ are both divisors of $R_n$.
== Proof ==
Let $n = r s$.
Then:
Similarly:
:$R_n = R_s \paren {\ds \sum_{j \mathop = 0}^{r - 1} 10^{s j} }$
Thus, for example:
The pattern is clear.
Category:Repunits
Dixon's Hypergeometric Theorem

== Theorem ==
Let $x, y, n \in \C$.
Let $n \notin \Z_{\lt 0}$.
Let $\map \Re {x + y + \dfrac n 2 + 1} > 0$.
Then:
:$\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {\dfrac n 2 + 1} \map \Gamma {x + y + \dfrac n 2 + 1} } { \map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {y + \dfrac n 2 + 1} } $
where:
:$\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1}$ is the generalized hypergeometric function of $1$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { n^{\overline k} \paren {-x}^{\overline k} \paren {-y}^{\overline k} } { \paren {x + n + 1}^{\overline k} \paren {y + n + 1}^{\overline k} } \dfrac {1^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
:$\map \Gamma {n + 1} = n!$ is the Gamma function.
=== Corollary 1 ===
== Proof 1 ==
== Proof 2 ==
== Also known as ==
Dixon's Hypergeometric Theorem is usually known as Dixon's Theorem, but there are a number of such and similarly named theorems.
Some sources refer to it as Dixon's Summation Theorem.
== Examples ==
== Also see ==
* Dixon's Identity
* Dougall's Hypergeometric Theorem
* Gauss's Hypergeometric Theorem
* Kummer's Hypergeometric Theorem
* Properties of Generalized Hypergeometric Function
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
* 
Category:Dixon's Hypergeometric Theorem
Category:Hypergeometric Functions
Category:Gamma Function
Dixon's Hypergeometric Theorem/Corollary 1

== Corollary to Dixon's Hypergeometric Theorem ==
Let $n \in \C$ be a complex number.
Let $\map \Re n < \dfrac 2 3$.
Then:
:$\ds 1 + \paren {\dfrac n {1!} }^3 + \paren {\dfrac {n \paren {n + 1} } {2!} }^3 + \paren {\dfrac {n \paren {n + 1} \paren {n + 2} } {3!} }^3 + \cdots = \dfrac {6 \map \sin {\dfrac {\pi n} 2} \map \sin {\pi n} \map {\Gamma^3} {\dfrac n 2 + 1} } {\pi^2 n^2 \paren {1 + 2 \map \cos {\pi n} } \map \Gamma {\dfrac {3 n} 2 + 1} } $
== Proof ==
From Dixon's Hypergeometric Theorem:
:$\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {\dfrac n 2 + 1} \map \Gamma {x + y + \dfrac n 2 + 1} } { \map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {y + \dfrac n 2 + 1} } $
where:
:$\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1}$ is the generalized hypergeometric function of $1$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { n^{\overline k} \paren {-x}^{\overline k} \paren {-y}^{\overline k} } { \paren {x + n + 1}^{\overline k} \paren {y + n+ 1}^{\overline k} } \dfrac {1^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
:$\map \Gamma {n + 1} = n!$ is the Gamma function.
Set $x = y = -n$:
and:
Note the $\map \sin {\dfrac {3\pi n} 2}$ in the denominator which is why $\map \Re n < \dfrac 2 3$.
== Also see ==
* Morley's Formula
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
Category:Dixon's Hypergeometric Theorem‎
Dixon's Hypergeometric Theorem/Examples

== Examples of Use of Dixon's Hypergeometric Theorem‎==
=== Example: $\map { {}_3 \operatorname F_2} {\dfrac 1 2, \dfrac 1 2, \dfrac 1 4; 1, \dfrac 5 4; 1}$ ===
=== Example: $\map { {}_3 \operatorname F_2} {\dfrac 1 2, \dfrac 1 4, \dfrac 1 4; \dfrac 5 4, \dfrac 5 4; 1}$ ===
=== Example: $\map { {}_3 \operatorname F_2} {\dfrac 1 2, \dfrac 1 2, \dfrac 1 2; 1, 1; 1}$ ===
Category:Examples of Use of Dixon's Hypergeometric Theorem‎
Dixon's Hypergeometric Theorem/Examples/3F2(0.5,0.25,0.25;1.25,1.25;1)

== Example of Use of Dixon's Hypergeometric Theorem ==
:$1 + \dfrac 1 {5^2} \paren {\dfrac 1 2} + \dfrac 1 {9^2} \paren {\dfrac {1 \times 3} {2 \times 4} } + \dfrac 1 {13^2} \paren {\dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} } + \cdots = \dfrac {\pi^{\frac 5 2} } {8 \sqrt 2 \paren {\map \Gamma {\dfrac 3 4} }^2}$
== Proof ==
From Dixon's Hypergeometric Theorem:
:$\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {\dfrac n 2 + 1} \map \Gamma {x + y + \dfrac n 2 + 1} } { \map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {y + \dfrac n 2 + 1} }$
where:
:$\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1}$ is the generalized hypergeometric function of $1$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { n^{\overline k} \paren {-x}^{\overline k} \paren {-y}^{\overline k} } { \paren {x + n + 1}^{\overline k} \paren {y + n+ 1}^{\overline k} } \dfrac {1^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
:$\map \Gamma {n + 1} = n!$ is the Gamma function.
We have:
and:
Recall the Euler's Reflection Formula:
:$\map \Gamma z \map \Gamma {1 - z} = \dfrac \pi {\map \sin {\pi z} }$
Therefore:
Substituting these results back into our equation above:
Therefore:
:$1 + \dfrac 1 {5^2} \paren {\dfrac 1 2} + \dfrac 1 {9^2} \paren {\dfrac {1 \times 3} {2 \times 4} } + \dfrac 1 {13^2} \paren {\dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} } + \cdots = \dfrac {\pi^{\frac 5 2} } {8 \sqrt 2 \paren {\map \Gamma {\dfrac 3 4} }^2 }$
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
Category:Examples of Use of Dixon's Hypergeometric Theorem‎
Dixon's Hypergeometric Theorem/Examples/3F2(0.5,0.5,0.25;1,1.25;1)

== Example of Use of Dixon's Hypergeometric Theorem ==
:$1 + \dfrac 1 5 \paren {\dfrac 1 2}^2 + \dfrac 1 9 \paren {\dfrac {1 \times 3} {2 \times 4} }^2 + \dfrac 1 {13} \paren {\dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} }^2 + \cdots = \dfrac {\pi^2} {4 \paren {\map \Gamma {\dfrac 3 4} }^4}$
== Proof ==
From Dixon's Hypergeometric Theorem:
:$\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {\dfrac n 2 + 1} \map \Gamma {x + y + \dfrac n 2 + 1} } {\map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {y + \dfrac n 2 + 1} }$
where:
:$\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1}$ is the generalized hypergeometric function of $1$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { n^{\overline k} \paren {-x}^{\overline k} \paren {-y}^{\overline k} } { \paren {x + n + 1}^{\overline k} \paren {y + n+ 1}^{\overline k} } \dfrac {1^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
:$\map \Gamma {n + 1} = n!$ is the Gamma function.
We have:
and:
Recall the Euler's Reflection Formula:
:$\map \Gamma z \map \Gamma {1 - z} = \dfrac \pi {\map \sin {\pi z} }$
Therefore:
Substituting this result back into our equation above:
Therefore:
:$1 + \dfrac 1 5 \paren {\dfrac 1 2}^2 + \dfrac 1 9 \paren {\dfrac {1 \times 3} {2 \times 4} }^2 + \dfrac 1 {13} \paren {\dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} }^2 + \cdots = \dfrac {\pi^2} {4 \paren {\map \Gamma {\dfrac 3 4} }^4}$
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
Category:Examples of Use of Dixon's Hypergeometric Theorem‎
Dixon's Hypergeometric Theorem/Examples/3F2(0.5,0.5,0.5;1,1;1)

== Example of Use of Dixon's Hypergeometric Theorem ==
:$1 + \paren {\dfrac 1 2}^3 + \paren {\dfrac {1 \times 3} {2 \times 4} }^3 + \paren {\dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} }^3 + \cdots = \dfrac \pi {\paren {\map \Gamma {\dfrac 3 4} }^4 }$
== Proof ==
From Dixon's Hypergeometric Theorem:
:$\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {\dfrac n 2 + 1} \map \Gamma {x + y + \dfrac n 2 + 1} } { \map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {y + \dfrac n 2 + 1} } $
where:
:$\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1}$ is the generalized hypergeometric function of $1$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { n^{\overline k} \paren {-x}^{\overline k} \paren {-y}^{\overline k} } { \paren {x + n + 1}^{\overline k} \paren {y + n+ 1}^{\overline k} } \dfrac {1^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
:$\map \Gamma {n + 1} = n!$ is the Gamma function.
We have:
and:
Recall from the Euler's Reflection Formula: $\map \Gamma z \map \Gamma {1 - z} = \dfrac \pi {\map \sin {\pi z} }$
Therefore:
Substituting this result back into our equation above:
Therefore:
:$1 + \paren {\dfrac 1 2}^3 + \paren {\dfrac {1 \times 3} {2 \times 4} }^3 + \paren {\dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} }^3 + \cdots = \dfrac \pi {\paren {\map \Gamma {\dfrac 3 4} }^4}$
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
Category:Examples of Use of Dixon's Hypergeometric Theorem‎
Dixon's Hypergeometric Theorem/Proof 1

== Theorem ==
== Proof ==
From Dougall's Hypergeometric Theorem, we have:
:$\ds \map { {}_5 \operatorname F_4} { { {\dfrac n 2 + 1, n, -x, -y, -z} \atop {\dfrac n 2, x + n + 1, y + n + 1, z + n + 1} } \, \middle \vert \, 1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {z + n + 1} \map \Gamma {x + y + z + n + 1} } { \map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {y + z + n + 1} \map \Gamma {x + z + n + 1} } $
where:
:$\ds \map { {}_5 \operatorname F_4} { { {\dfrac n 2 + 1, n, -x, -y, -z} \atop {\dfrac n 2, x + n + 1, y + n + 1, z + n + 1} } \, \middle \vert \, 1}$ is the generalized hypergeometric function of $1$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { \paren {\dfrac n 2 + 1}^{\overline k} n^{\overline k} \paren {-x}^{\overline k} \paren {-y}^{\overline k} \paren {-z}^{\overline k} } { \paren {\dfrac n 2}^{\overline k} \paren {x + n + 1}^{\overline k} \paren {y + n + 1}^{\overline k} \paren {z + n + 1}^{\overline k} } \dfrac {1^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
:$\map \Gamma {n + 1} = n!$ is the Gamma function.
Setting $z = -\dfrac n 2$, we obtain:
Therefore:
:$\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {\dfrac n 2 + 1} \map \Gamma {x + y + \dfrac n 2 + 1} } {\map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {y + \dfrac n 2 + 1} }$
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
Category:Dixon's Hypergeometric Theorem
Dixon's Hypergeometric Theorem/Proof 2

== Theorem ==
== Proof ==
From Gauss's Hypergeometric Theorem, we have:
:$\map F {a, b; c; 1} = \dfrac {\map \Gamma c \map \Gamma {c - a - b} } {\map \Gamma {c - a} \map \Gamma {c - b} }$
Therefore:
Therefore:
We now have:
Substituting $a = n$, $b = -x$ and $c = -y$, we obtain:
== Sources ==
* Chapter $\text {3}$. The hypergeometric series
Category:Dixon's Hypergeometric Theorem
Dixon's Identity

== Theorem ==
For $n \in \Z_{\ge 0}$:
:$\ds \sum_{k \mathop \in \Z} \paren {-1}^k \binom {2 n} {n + k}^3 = \dfrac {\paren {3 n}!} {\paren {n!}^3}$
=== General Case ===
=== Gaussian Binomial Form ===
== Proof ==
Follows directly from Dixon's Identity/General Case:
:$\ds \sum_{k \mathop \in \Z} \paren {-1}^k \dbinom {l + m} {l + k} \dbinom {m + n} {m + k} \dbinom {n + l} {n + k} = \dfrac {\paren {l + m + n}!} {l! \, m! \, n!}$
setting $l = m = n$.
== Sources ==
* 
* : $\S 1.2.6$: Binomial Coefficients: Exercise $62$
Category:Binomial Coefficients
Dixon's Identity/Gaussian Binomial Form/Formulation 1

== Theorem ==
For $l, m, n \in \Z_{\ge 0}$:
:$\ds \sum_{k \mathop \in \Z} \paren {-1}^k \dbinom {m - r - s} k_q \dbinom {n + r - s} {n - k}_q \dbinom {r + k} {m + n}_q = \dbinom r m_q \dbinom s n_q$
where $\dbinom r m_q$ denotes a Gaussian binomial coefficient
== Proof ==
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $62$
Category:Gaussian Binomial Coefficients
Dixon's Identity/Gaussian Binomial Form/Formulation 2

== Theorem ==
For $l, m, n \in \Z_{\ge 0}$:
:$\ds \sum_{k \mathop \in \Z} \paren {-1}^k \dbinom {l + m} {l + k}_q \dbinom {m + n} {m + k}_q \dbinom {n + l} {n + k}_q = \dfrac {\paren {l + m + n}!_q} {l!_q \, m!_q \, n!_q}$
where:
:$\dbinom {l + m} {l + k}_q$ denotes a Gaussian binomial coefficient
:$l!_q$ is defined as $\ds \prod_{k \mathop = 1}^n \paren {1 + q + \cdots + q^{k - 1} }$
== Proof ==
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $62$
Category:Gaussian Binomial Coefficients
Dixon's Identity/General Case

== Theorem ==
For $l, m, n \in \Z_{\ge 0}$:
:$\ds \sum_{k \mathop \in \Z} \paren {-1}^k \dbinom {l + m} {l + k} \dbinom {m + n} {m + k} \dbinom {n + l} {n + k} = \dfrac {\paren {l + m + n}!} {l! \, m! \, n!}$
== Proof ==
From Sum over $k$ of $\dbinom {m - r + s} k$ by $\dbinom {n + r - s} {n - k}$ by $\dbinom {r + k} {m + n}$:
:$\ds \sum_{k \mathop \in \Z} \binom {m - r + s} k \binom {n + r - s} {n - k} \binom {r + k} {m + n} = \binom r m \binom s n$
Setting $\tuple {m, n, r, s, k} \gets \tuple {m + k, l - k, m + n, n + l, j}$ into the equation above, we obtain:
On the , we have:
Unless $j = l$ and $k = 0$ the sum on $k$ vanishes.
Therefore:
:$\ds \sum_{k \mathop \in \Z} \paren {-1}^k \dbinom {l + m} {l + k} \dbinom {m + n} {m + k} \dbinom {n + l} {n + k} = \dfrac {\paren {l + m + n}!} {l! \, m! \, n!}$
== Sources ==
* 
* : $\S 1.2.6$: Binomial Coefficients: Exercise $62$
Category:Binomial Coefficients
Dixon's Theorem (Group Theory)

== Theorem ==
Let $P_1$ and $P_2$ be distinct elements of the symmetric group on $n$ letters.
The probability that $\set {P_1, P_2}$ forms a generator of $S_n$ approaches $\dfrac 3 4$ as $n$ tends to infinity.
== Proof ==
== Historical Note ==
== Sources ==
* : $0,75$
Category:Symmetric Groups
Category:Generators of Groups
Dixon's Theorem (Group Theory)/Historical Note

== Historical Note on Dixon's Theorem ==
Dixon's Theorem started out as a conjecture made by , and published by him in his $1882$ work .
As a consequence, it was referred to as Netto's Conjecture.
It was finally proved by in $1967$.
Since then it has been called Dixon's Theorem.
== Sources ==
* : $0,75$
Category:Historical Notes
Do Quasiperfect Numbers Exist?

== Open Question ==
It is not known whether any quasiperfect numbers actually exist.
== Progress ==
As of January $2014$, there are no known quasiperfect numbers.
It was proved by and that if $n$ is a quasiperfect number, then:
:$\map \omega n \ge 7$
and:
:$n > 10^{35}$
where $\map \omega n$ denotes the number of distinct prime factors of $n$.
== Also see ==
* Quasiperfect Number is Square of Odd Integer
== Sources ==
* 
* : $16$
* 
* : $16$
Category:Quasiperfect Numbers
Category:Open Questions
Domain Topology Contains Initial Topology iff Mappings are Continuous

== Theorem ==
Let $\struct{Y, \tau}$ be a topological space.
Let $\family {\struct{X_i, \tau_i}}_{i \mathop \in I}$ be a family of topological spaces.
Let $\family {f_i}_{i \mathop \in I}$ be a family of mappings $f_i : Y \to X_i$.
Let $\tau'$ be the initial topology on $Y$ with respect to $\family {f_i}_{i \mathop \in I}$.
Then:
:$\tau' \subseteq \tau$ $\forall i \in I : f_i: \struct{Y, \tau} \to \struct{X_i, \tau_i}$ is $\tuple{\tau, \tau_i}$-continuous.
== Proof ==
=== Necessary Condition ===
Let $\tau' \subseteq \tau$.
From Equivalence of Definitions of Initial Topology:
:for each $i \in I$, $f_i: \struct{Y, \tau'} \to \struct{X_i, \tau_i}$ is $\tuple{\tau', \tau_i}$-continuous
From Continuous Mapping on Finer Domain and Coarser Codomain Topologies is Continuous:
:for each $i \in I$, $f_i: \struct{Y, \tau} \to \struct{X_i, \tau_i}$ is $\tuple{\tau, \tau_i}$-continuous
=== Sufficient Condition ===
For all $i \in I$, let $f_i: \struct{Y, \tau} \to \struct{X_i, \tau_i}$ be $\tuple{\tau, \tau_i}$-continuous.
From Equivalence of Definitions of Initial Topology:
:$\tau'$ is the coarsest topology on $Y$ such that each $f_i: Y \to X_i$ is $\tuple{\tau', \tau_i}$-continuous.
Then:
:$\tau' \subseteq \tau$
Category:Initial Topology
Domain of Bijection is Codomain of Inverse

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a bijection.
Let $f^{-1}: T \to S$ be the inverse of $f$.
Then the codomain of $f^{-1}$ equals the domain of $f$.
== Proof ==
Follows directly from the definition of domain and codomain:
:$\Dom f = S$ and $\Cdm f = T$
:$\Dom {f^{-1} } = T$ and $\Cdm {f^{-1} } = S$
That is:
:$\Cdm {f^{-1} } = S = \Dom f$
== Also see ==
* Codomain of Bijection is Domain of Inverse
== Sources ==
* : Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $3$
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries
Category:Bijections
Domain of Composite Mapping

== Theorem ==
Let $S_1, S_2, S_3$ be sets.
Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be mappings.
Let $f_2 \circ f_1: S_1 \to S_3$ be the composite mapping of $f_1$ and $f_2$.
Then:
:$\Dom {f_1} = \Dom {f_2 \circ f_1}$
where $\Dom {f_1}$ denotes the domain of $f_1$.
== Proof ==
By definition of composition of mappings:
:$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \exists y \in S_2: \tuple {x, y} \in f_1 \land \tuple {y, z} \in f_2}$
Let $x \in \Dom {f_2 \circ f_1}$.
Then:
:$\exists z \in S_3: \tuple {x, z} \in S_1 \times S_3$
and:
:$\exists y \in S_2: \tuple {x, y} \in f_1$
That is:
:$x \in \Dom {f_1}$
Thus by definition of subset:
:$\Dom {f_2 \circ f_1} \subseteq \Dom {f_1}$
Now suppose $x \in \Dom {f_1}$.
By the definition of mapping:
:$\exists y \in S_2: \tuple {x, y} \in f_1$
As $f_2$ is likewise a mapping:
:$\exists z \in S_3: \tuple {y, z} \in f_2$
and so by definition of composition of mappings:
:$\tuple {x, z} \in S_1 \times S_3$
and so:
:$x \in \Dom {f_2 \circ f_1}$
Thus by definition of subset:
:$\Dom {f_1} \subseteq \Dom {f_2 \circ f_1}$
By definition of set equality:
:$\Dom {f_1} = \Dom {f_2 \circ f_1}$
== Sources ==
* : $\S 1.14$: Composition of Functions
Category:Composite Mappings
Domain of Composite Relation

== Theorem ==
Let $\RR_2 \circ \RR_1$ be a composite relation.
Then the domain of $\RR_2 \circ \RR_1$ is the domain of $\RR_1$:
:$\Dom {\RR_2 \circ \RR_1} = \Dom {\RR_1}$
== Proof ==
Let $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$.
The domain of $\RR_1$ is $S_1$.
The composite of $\RR_1$ and $\RR_2$ is defined as:
:$\RR_2 \circ \RR_1 = \set {\tuple {x, z}: x \in S_1, z \in S_3: \exists y \in S_2: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2}$
From this definition:
:$\RR_2 \circ \RR_1 \subseteq S_1 \times S_3$.
Thus the domain of $\RR_2 \circ \RR_1$ is $S_1$.
Thus:
:$\Dom {\RR_2 \circ \RR_1} = S_1 = \Dom {\RR_1}$
Category:Composite Relations
Domain of Continuous Injection to Hausdorff Space is Hausdorff

== Theorem ==
Let $T_\alpha = \struct {S_\alpha, \tau_\alpha}$ and $T_\beta = \struct {S_\beta, \tau_\beta}$ be topological spaces.
Let $f: S_\alpha \to S_\beta$ be a continuous mapping which is an injection.
If $T_\beta$ is a $T_2$ (Hausdorff) space, then $T_\alpha$ is also a $T_2$ (Hausdorff) space.
== Proof ==
Let $x, y \in S_\alpha$ be distinct points.
We want to find disjoint open sets $U, V \in \tau_\alpha$ containing $x$ and $y$ respectively.
Since $f$ is injective the points $\map f x, \map f y \in S_\beta$ are distinct.
By assumption $T_\beta$ is Hausdorff.
Therefore we can choose disjoint open sets $U', V'$ of $T_\beta$ such that $\map f x \in U'$ and $\map f y \in V'$.
Since $f$ is continuous, the sets $U = f^{-1} \sqbrk {U'}$ and $V = f^{-1} \sqbrk {V'}$ are open.
Moreover we have $x \in U$ and $y \in V$.
Finally we show that the sets $U$ and $V$ are disjoint.
 $z \in U \cap V$.
Then by the definition of $U$, $V$ we have $\map f z \in U'$ and $\map f z \in V'$.
This is a contradiction, since $U' \cap V' = \O$.
== Sources ==
* : $4$: The Hausdorff condition: $4.2$: Separation axioms: Proposition $4.2.4 \ \text{(c)}$
* : $4$: The Hausdorff condition: Exercise $4.3: 2$
Category:Hausdorff Spaces
Category:Injections
Category:Continuous Mappings
Domain of Injection Not Larger than Codomain

== Theorem ==
Let $f: S \rightarrowtail T$ be an injection.
Then:
:$\left|{S}\right| \le \left|{T}\right|$
where $\left|{S}\right|$ denotes the cardinality of $S$.
== Proof ==
== Sources ==
* : Appendix $\text{A}.10$: Proposition $\text{A}.10.1: 1$
Category:Injections