nyuuzyou/wikis huggingface dataset:
Equivalence of Definitions of Algebraically Closed Field

== Theorem ==
Let $K$ be a field. 
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
== Proof ==
=== Definition $(1)$ implies Definition $(2)$ ===
Let $K$ be algebraically closed by definition 1.
Let $f$ be an irreducible polynomial over $K$.
By Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal, the ideal $\gen f$ generated by $f$ is maximal.
So by Maximal Ideal iff Quotient Ring is Field:
:$L = K \sqbrk X / \gen f$ is a field
where $L$ is a field extension over $K \sqbrk X$.
Now:
:$L = \set {g + \gen f: g \in K \sqbrk X}$
From Division Theorem for Polynomial Forms over Field:
:$\forall g \in K \sqbrk X: \exists q, r \in K \sqbrk X: g = q f + r, \deg r < \deg f =: n$
Therefore:
:$L = \set {r + \gen f: r \in K \sqbrk X, \deg r < n}$
By Basis for Quotient of Polynomial Ring, this has basis:
:$1 + \gen f, \ldots, X^{n - 1} + \gen f$ span $L$.
Thus $L$ is finite.
By Finite Field Extension is Algebraic, $L$ is algebraic.
Also $K \subseteq L$.
So by hypothesis $K = L$.
This implies:
:$\index L K = 1$
where $\index L K$ is the degree of $L$ over $K$.
Hence:
:$n = \deg f = 1$
Thus $K$ is algebraically closed by definition 2.
=== Definition $(2)$ implies Definition $(3)$ ===
Let $K$ be algebraically closed by definition 2.
Let $f$ be a polynomial in $K \sqbrk X$ of strictly positive degree.
From Polynomial Forms over Field form Principal Ideal Domain, $K \sqbrk X$ is a principal ideal domain.
From Principal Ideal Domain is Unique Factorization Domain, $K \sqbrk X$ is a unique factorization domain.
So $f$ can be factorized $f = u g_1 \cdots g_r$ such that:
: $u$ is a unit
and:
: $g_i$ are irreducible for $i = 1, \ldots, r$.
By hypothesis, $g_1$ has degree $1$.
Therefore by the Polynomial Factor Theorem $g_1$, and hence $f$, has a root in $K$.
Thus $K$ is algebraically closed by definition 3.
=== Definition $(3)$ implies Definition $(1)$ ===
Let $K$ be algebraically closed by definition 3.
Let $L / K$ be an algebraic field extension of $K$.
Let $\alpha \in L$.
By hypothesis, the minimal polynomial $\mu_\alpha$ of $\alpha$ over $K$ has a root $\beta$ in $K$.
Therefore by the Polynomial Factor Theorem:
:$\mu_\alpha = \paren {X - \beta} g$
for some $g \in K \sqbrk X$.
Since $\mu_\alpha$ is irreducible and monic it follows that:
:$\mu_\alpha = X - \beta$
Also:
:$\map {\mu_\alpha} \alpha = \alpha - \beta = 0$
so $\alpha = \beta$.
Therefore $\alpha \in K$.
Therefore $L = K$ as required.
Thus $K$ is algebraically closed by definition 1.
Category:Field Extensions
Equivalence of Definitions of Almost-Everywhere Equality Relation on Lebesgue Space

== Theorem ==
=== Definition 1 ===
 
=== Definition 2 ===
 
== Proof ==
Let $f, g \in \map {\LL^p} {X, \Sigma, \mu}$. 
By P-Seminorm of Function Zero iff A.E. Zero, we have: 
:$\norm {f - g}_p = 0$ $f - g = 0$ $\mu$-almost everywhere.
From Pointwise Addition preserves A.E. Equality, we have: 
:$\norm {f - g}_p = 0$ $f = g$ $\mu$-almost everywhere.
That is: 
:$\norm {f - g}_p = 0$ $\map \mu {\set {x \in X : \map f x \ne \map g x} } = 0$.
Category:Almost-Everywhere Equality Relation
Category:Lebesgue Spaces
Equivalence of Definitions of Almost Perfect Number

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
By definition of abundance:
:$\map A n = \map {\sigma_1} n - 2 n$
By definition of divisor sum function:
:$\map {\sigma_1} n$ is the sum of all the divisors of $n$.
Thus $\map {\sigma_1} - n$ is the sum of the aliquot parts of $n$.
Hence the result.
Category:Almost Perfect Numbers
Equivalence of Definitions of Amicable Pair

== Theorem ==
Let $m \in \Z_{>0}$ and $n \in \Z_{>0}$ be (strictly) positive integers.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition 1 is equivalent to Definition 2 ===
Let $\map s n$ denote the aliquot sum of (strictly) positive integer $n$.
The sum of all the divisors of a (strictly) positive integer $n$ is $\map {\sigma_1} n$, where $\sigma_1$ is the divisor sum function.
The aliquot sum of $n$ is the sum of the divisors of $n$ with $n$ excluded.
Thus:
:$\map s n = \map {\sigma_1} n - n$
Suppose:
:$\map s n = m$
and:
:$\map s m = n$
Then:
Similarly:
Thus:
:$\map s n = \map s m = m + n$
The argument reverses.
=== Definition 1 is equivalent to Definition 3 ===
From the definition of definition 1 of an amicable pair:
From the definition of a sociable chain:
Here it is seen that setting $r = 2$ gives that:
:$\map s {a_0} = a_1$
:$\map s {a_1} = a_0$
and the equivalence follows.
Category:Amicable Pairs
Equivalence of Definitions of Amicable Triplet

== Theorem ==
Let $m_1, m_2, m_3 \in \Z_{>0}$ be (strictly) positive integers.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
For $n \in \Z_{>0}$, let $\map s n$ denote the aliquot sum of (strictly) positive integer $n$.
The sum of all the divisors of a (strictly) positive integer $n$ is $\map {\sigma_1} n$, where $\sigma_1$ is the divisor sum function.
The aliquot sum of $n$ is the sum of the divisors of $n$ with $n$ excluded.
Thus:
:$\map s n = \map {\sigma_1} n - n$
Suppose:
Then:
Similarly:
and:
Thus:
:$\map {\sigma_1} {m_1} = \map {\sigma_1} {m_2} = \map {\sigma_1} {m_3} = m_1 + m_2 + m_3$
The argument reverses.
Category:Amicable Triplets
Equivalence of Definitions of Analytic Basis

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $\BB$ be an analytic basis for $\tau$ by definition 1.
Let $U \in \tau$.
By definition 1 of an analytic basis, we can choose $\AA \subseteq \BB$ such that:
:$\ds U = \bigcup \AA$
By the definition of union:
:$\forall x \in U: \exists B \in \AA: x \in B$
By Union is Smallest Superset:
:$\forall B \in \AA: B \subseteq U$
Since $\AA \subseteq \BB$, the result follows.
Thus $\BB$ is an analytic basis for $\tau$ by definition 2.
=== Definition 2 implies Definition 1 ===
Let $\BB$ be an analytic basis for $\tau$ by definition 2.
Let $U \in \tau$.
Let $\AA = \set {B \in \BB: B \subseteq U}$.
Then $\AA \subseteq \BB$.
Let $x \in U$ be arbitrary.
Since $\BB$ is an analytic basis for $\tau$ by definition 2, there is some $B_x \in \BB$ such that:
:$x \in B_x \subseteq U$
Hence, by construction of $\AA$, $B_x \in \AA$.
Thus:
:$\ds x \in \bigcup \AA$
and it follows that:
:$\ds U \subseteq \bigcup \AA$
By Union is Smallest Superset applied to $\AA$ and $U$:
:$\ds \bigcup \AA \subseteq U$
By definition of set equality and definition 1 of an analytic basis, the result follows.
Thus $\BB$ is an analytic basis for $\tau$ by definition 1.
Category:Topological Bases
Equivalence of Definitions of Antisymmetric Relation

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $\RR$ be a relation which fulfils the condition:
:$\tuple {x, y} \in \RR \land \tuple {y, x} \in \RR \implies x = y$
Let $\tuple {x, y} \in \RR$ such that $x \ne y$.
 that $\tuple {y, x} \in \RR$.
Then $\tuple {x, y} \in \RR \land \tuple {y, x} \in \RR$.
This implies that $x = y$.
From this contradiction it is concluded that $\tuple {y, x} \notin \RR$.
It follows that the condition:
:$\tuple {x, y} \in \RR \land x \ne y \implies \tuple {y, x} \notin \RR$
holds for $\RR$.
=== Definition 2 implies Definition 1 ===
Let $\RR$ be a relation which fulfils the condition:
:$\tuple {x, y} \in \RR \land x \ne y \implies \tuple {y, x} \notin \RR$
Let $\tuple {x, y} \in \RR$ such that $\tuple {y, x} \in \RR$ also.
 that $x \ne y$.
This implies that $\tuple {y, x} \notin \RR$.
From this contradiction it is concluded that $\tuple {y, x} \notin \RR$.
It follows that the condition:
:$\tuple {x, y} \in \RR \land \tuple {y, x} \in \RR \implies x = y$
holds for $\RR$.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $5$ Properties of Relations: Exercise $1 \ \text{(a)}$
Category:Antisymmetric Relations
Equivalence of Definitions of Antisymmetric Relation (Class Theory)

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $\RR$ be a relation which fulfils the condition:
:$\tuple {x, y} \in \RR \land \tuple {y, x} \in \RR \implies x = y$
Let $\tuple {x, y} \in \RR$ such that $x \ne y$.
 that $\tuple {y, x} \in \RR$.
Then $\tuple {x, y} \in \RR \land \tuple {y, x} \in \RR$.
Then this implies that $x = y$.
From this contradiction it is concluded that $\tuple {y, x} \notin \RR$.
It follows that the condition:
:$\tuple {x, y} \in \RR \land x \ne y \implies \tuple {y, x} \notin \RR$
holds for $\RR$.
=== Definition 2 implies Definition 1 ===
Let $\RR$ be a relation which fulfils the condition:
:$\tuple {x, y} \in \RR \land x \ne y \implies \tuple {y, x} \notin \RR$
Let $\tuple {x, y} \in \RR$ such that $\tuple {y, x} \in \RR$ also.
 that $x \ne y$.
Then this implies that $\tuple {y, x} \notin \RR$.
From this contradiction it is concluded that $\tuple {y, x} \notin \RR$.
It follows that the condition:
:$\tuple {x, y} \in \RR \land \tuple {y, x} \in \RR \implies x = y$
holds for $\RR$.
Category:Antisymmetric Relations
Equivalence of Definitions of Artinian Module

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 iff Definition 2 ===
Let $D$ be the set of all submodules of $M$.
We shall show that:
:descending chain condition
:minimal condition
with respect to $\struct {D, \supseteq}$ are equivalent.
This is nothing but:
:ascending chain condition 
:maximal condition
with respect to $\struct {D, \subseteq}$ are equivalent.
The latter follows from Increasing Sequence in Ordered Set Terminates iff Maximal Element.
Category:Artinian Modules
Equivalence of Definitions of Associate in Integral Domain

== Theorem ==
Let $\struct {D, +, \circ}$ be an integral domain.
Let $x, y \in D$.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== $(1)$ is Equivalent to $(2)$ ===
=== $(1)$ is Equivalent to $(3)$ ===
Category:Integral Domains
Category:Associates
Category:Equivalence of Definitions of Associate in Integral Domain
Equivalence of Definitions of Associate in Integral Domain/Definition 1 Equivalent to Definition 2

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
We are to show that:
:$x \divides y \text{ and } y \divides x \iff \ideal x = \ideal y$
Thus:
== Sources ==
* : $\S 62.3$ Factorization in an integral domain: $\text{(iii)}$
Category:Equivalence of Definitions of Associate in Integral Domain
Equivalence of Definitions of Associate in Integral Domain/Definition 1 Equivalent to Definition 3

== Theorem ==
=== Definition 1 ===
=== Definition 3 ===
== Proof ==
By the definition of divisor:
:$x \divides y$ and $y \divides x$
Let $x \divides y$ and $y \divides x$.
Then $\exists s, t \in D$ such that:
:$(1): \quad y = t \circ x$
and:
:$(2): \quad x = s \circ y$
If either $x = 0_D$ or $y = 0_D$, then so must be the other (as an integral domain has no zero divisors by definition).
So $x = 1_D \circ y$ and $y = 1_D \circ x$, and the result holds.
Otherwise:
So:
:$s \circ t = 1_D$
and both $s \in U_D$ and $t \in U_D$.
The result follows.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 26$. Divisibility: Theorem $50$
* : $\S 62.1$ Factorization in an integral domain: $\text{(iv)}$
Category:Equivalence of Definitions of Associate in Integral Domain
Equivalence of Definitions of Associated Prime of Module

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $M$ be a module over $A$.
Let $\mathfrak p$ be a prime ideal in $A$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Suppose that $x \in M$ satisfies:
:$\map {\operatorname {Ann}_A} x = \mathfrak p$
Define a submodule of $M$ by:
:$N := \set { a x : a \in A }$
Define a module homomorphism $\phi : A \to N$ by:
:$a \mapsto a x$
Then the kernel of $\phi$ is:
:$\map \ker \phi = \map {\operatorname {Ann}_A} x = \mathfrak p$
By First Isomorphism Theorem:
:$N \cong A / \mathfrak p$
=== Definition 2 implies Definition 1 ===
Let $N \subseteq M$ be a submodule.
Suppose that there is an isomorphism:
:$\psi : A / \mathfrak p \to N$
Let $a_0 \in A \setminus \mathfrak p$.
Let:
:$x := \map \psi {a_0 + \mathfrak p}$
We claim that:
:$\map {\operatorname {Ann}_A} x = \mathfrak p$
Indeed:
Category:Definitions/Module Theory
Category:Definitions/Commutative Algebra
Equivalence of Definitions of Asymmetric Relation

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $\RR$ be a relation which fulfills the condition:
:$\left({x, y}\right) \in \RR \implies \left({y, x}\right) \notin \RR$
Then:
Hence $\RR$ is asymmetric by definition 2.
=== Definition 2 implies Definition 1 ===
Let $\RR$ be a relation which fulfils the condition:
: $\RR \cap \RR^{-1} = \O$
Then:
Hence $\RR$ is asymmetric by definition 1.
Category:Asymmetric Relations
Equivalence of Definitions of Asymptotically Equal Sequences

== Theorem ==
Let $\sequence {a_n}$ and $\sequence {b_n}$ be sequences in $\R$.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== $(1)$ iff $(2)$ ===
=== Definition $(2)$ iff $(3)$ ===
Let $a_n - b_n = \map \oo {b_n}$.
Let $0 < \epsilon < 1/2$.
Then:
So:
:$\size {a_n - b_n} \le \dfrac {\epsilon \cdot \size {a_n} } {1 - \epsilon} \le 2 \epsilon \cdot \size {a_n}$
Thus:
:$a_n - b_n = \map \oo {a_n}$
The other implication follows by symmetry.
Category:Asymptotic Equality
Equivalence of Definitions of Balanced String

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
The proof proceeds by strong induction on the length of a string.
For all $n \in \N$, let $\map P n$ be the proposition:
:Every balanced string $S_n$ of length $n$ is a balanced string by definition $1$ it been generated by definition $2$.
=== Basis for the Induction ===
$\map P 0$ is the case:
:Every balanced string $S_0$ of length $0$ is a balanced string by definition $1$ it been generated by definition $2$.
By definition $2$ of a balanced string:
:$(1): \quad$ The null string $\epsilon$ is a balanced string.
This is the only string of length $0$.
$\epsilon$ is seen vacuously to be a balanced string by definition $1$.
Thus $\map P 0$ is seen to hold.
$\map P 1$ is the case:
:Every balanced string $S_1$ of length $n$ is a balanced string by definition $1$ it been generated by definition $2$.
Let $a$ be an arbitrary symbol.
We use $\map P 0$ and $\map P 1$ together to form the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P j$ is true, for all $j$ such that $0 \le j \le k$ and $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
This is the induction hypothesis:
:Every balanced string $S_j$ of length $j$ such that $j \le k$ is a balanced string by definition $1$ it been generated by definition $2$.
from which it is to be shown that:
:Every balanced string $S_{k + 1}$ of length $k + 1$ is a balanced string by definition $1$ it been generated by definition $2$.
=== Induction Step ===
This is the induction step:
Let $x$ be a balanced string of length $k - 1$ according to either definition $1$ or definition $2$.
By the induction hypothesis, $x$ is therefore a balanced string by both definition $1$ and definition $2$.
Thus we have shown that for all $k \in \N$, if $\sqbrk y$ reads the same backwards as it does forwards, then it has been created by one of the rules $(1)$, $(2)$ or $(3)$. 
So $\paren {\forall j: 0 \le j \le k: \map P j} \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.
Therefore:
:For all $n \in \N$: every balanced string $S_n$ of length $n$ is a balanced string by definition $1$ it been generated by definition $2$.
== Sources ==
* : Chapter $1$: Preliminaries: Exercises: $1.4$
Category:Balanced Strings
Equivalence of Definitions of Basis of Vector Space

== Theorem ==
Let $K$ be a division ring.
Let $\struct {G, +_G, \circ}_K$ be an vector space over $K$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $\BB$ be a linearly independent subset of $G$ which is a generator for $G$.
Suppose $\BB \subseteq \BB'$ is a linearly independent subset of $G$.
We aim to show that $\BB = \BB'$, proving maximality. 
Suppose that $\BB \ne \BB'$. 
Let $x \in \BB' \setminus \BB$. 
Since $G$ is a generator for $G$, there exists $x_1, \ldots, x_n \in \BB \subseteq \BB'$ and $\alpha_1, \ldots, \alpha_n \in K$ such that: 
:$\ds x = \sum_{i \mathop = 1}^n \alpha_i x_i$ 
so that: 
:$\ds \sum_{i \mathop = 1}^n \alpha_i x_i - x = 0$ 
Since $K$ is a field, we have that $-1 \ne 0$. 
This shows that $\set {x_1, \ldots, x_n, x}$ and hence $\BB'$ is linearly dependent. 
This is a contradiction, so we must have $\BB = \BB'$. 
 
=== Definition 2 implies Definition 1 ===
Suppose $\BB$ is a maximal linearly independent subset of $G$.
Let: 
:$G = \map \span \BB \subseteq E$
Suppose that $E \ne F$. 
Suppose that $G \ne E$.
Let $x \in E \setminus G$.
Then from Vector not contained in Linear Span of Linearly Independent Set is Linearly Independent of Set, $\BB \cup \set x$ is linearly independent, contradicting the maximality of $\BB$. 
So there exists no such $x \in E \setminus G$, and we have $G = E$.
So $\BB$ is a linearly independent subset of $G$ which is a generator for $G$.
== Sources ==
* : Appendix $\text{A}$ Preliminaries: $\S 1.$ Linear Algebra: Proposition $1.1$
Category:Bases of Vector Spaces
Equivalence of Definitions of Bernoulli Numbers

== Theorem ==
=== Generating Function ===
=== Recurrence Relation ===
== Proof ==
From the generating function definition:
Equating coefficients:
For $n = 0$:
For $n > 0$:
Hence the result:
:$B_n = \begin{cases} 1 & : n = 0 \\
\ds - \sum_{k \mathop = 0}^{n-1} \binom n k \frac {B_k} {n - k + 1} & : n > 0
\end{cases}$
Category:Bernoulli Numbers
Equivalence of Definitions of Beta Function

== Theorem ==
For $\map \Re x, \map \Re y > 0$:
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition 1 is equivalent to Definition 2 ===
=== Definition 2 is equivalent to Definition 3 ===
By definition of Gamma function:
:$\ds \map \Gamma x \, \map \Gamma y = \int_0^\infty t^{x - 1} e^{-t} \rd t \int_0^\infty s^{y - 1} e^{-s} \rd s$
Substitute $t = u^2$ and $s = v^2$:
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.32$
Category:Beta Function
Equivalence of Definitions of Bijection

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
=== Definition 5 ===
== Proof ==
=== Definition 1 iff Definition 2 ===
=== Definition 1 iff Definition 3 ===
=== Definition 1 iff Definition 4 ===
=== Definition 3 iff Definition 4 ===
=== Definition 3 iff Definition 5 ===
Category:Bijections
Category:Equivalence of Definitions of Bijection
Equivalence of Definitions of Bijection/Definition 1 iff Definition 2

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
From Injection iff Left Inverse, $f$ is an injection $f$ has a left inverse mapping.
From Surjection iff Right Inverse, $f$ is a surjection $f$ has a right inverse mapping.
Putting these together, it follows that:
: $f$ is both an injection and a surjection
:
: $f$ has both a left inverse and a right inverse.
Category:Equivalence of Definitions of Bijection
Equivalence of Definitions of Bijection/Definition 1 iff Definition 4

== Theorem ==
=== Definition 1 ===
=== Definition 4 ===
== Proof ==
Let $f: S \to T$ be a bijection by definition 1.
Then by definition:
:$f$ is an injection
:$f$ is a surjection
By definition of injection:
:every element of $T$ is the image of at most $1$ element of $S$.
By definition of surjection:
:every element of $T$ is the image of at least $1$ element of $S$.
So:
:for each $y \in T$ there exists one and only one $x \in S$ such that $\tuple {x, y} \in f$.
Thus $f$ is a bijection by definition 4.
Let $f: S \to T$ be a bijection by definition 4.
Then by definition:
:for each $y \in T$ there exists one and only one $x \in S$ such that $\tuple {x, y} \in f$.
But:
:every element of $T$ is the image of at most $1$ element of $S$
defines an injection
and:
:every element of $T$ is the image of at least $1$ element of $S$
defines a surjection.
From Injection iff Left Inverse, $f$ is an injection $f$ has a left inverse mapping.
From Surjection iff Right Inverse, $f$ is a surjection $f$ has a right inverse mapping.
Putting these together, it follows that:
:$f$ is an injection
:$f$ is a surjection
Thus $f$ is a bijection by definition 1.
== Sources ==
* : $\S 22$: Injections; bijections; inverse of a bijection
Category:Equivalence of Definitions of Bijection
Equivalence of Definitions of Bijection/Definition 3 iff Definition 4

== Theorem ==
=== Definition 3 ===
=== Definition 4 ===
== Proof ==
==== Necessary Condition ====
Let $f^{-1}: T \to S$ be a mapping.
Then by definition:
:$\forall y \in T: \exists x \in S: \tuple {y, x} \in f^{-1}$
Thus for all $y \in T$ there exists at least one $x \in S$ such that $\tuple {y, x} \in f^{-1}$.
Also by definition of mapping:
:$\tuple {x_1, y} \in f^{-1} \land \tuple {x_2, y} \in f^{-1} \implies x_1 = x_2$
Thus for all $y \in T$ there exists at most one $x \in S$ such that $\tuple {y, x} \in f^{-1}$.
Hence it has been demonstrated that $y \in T$ there exists a unique $x \in S$ such that $\tuple {y, x} \in f^{-1}$.
==== Sufficient Condition ====
Let $f$ be such that for all $y \in T$ there exists a unique $x \in S$ such that $\tuple {y, x} \in f^{-1}$.
Then by definition $f^{-1}$ is a mapping.
Category:Equivalence of Definitions of Bijection
Equivalence of Definitions of Bijection/Definition 3 iff Definition 5

== Theorem ==
=== Definition 3 ===
=== Definition 5 ===
== Proof ==
==== Necessary Condition ====
Let $f: S \to T$ be a mapping such that $f^{-1}: T \to S$ is also a mapping.
Then as $f$ is a mapping:
:for all $x \in S$ there exists a unique $y \in T$ such that $\tuple {x, y} \in f$.
Similarly, as $f^{-1}$ is also a mapping:
:for all $y \in T$ there exists a unique $x \in S$ such that $\tuple {y, x} \in f^{-1}$.
==== Sufficient Condition ====
Let $f \subseteq S \times T$ be a relation such that:
:$(1): \quad$ for each $x \in S$ there exists one and only one $y \in T$ such that $\tuple {x, y} \in f$
:$(2): \quad$ for each $y \in T$ there exists one and only one $x \in S$ such that $\tuple {x, y} \in f$.
Then by definition:
: from $(1)$, $f$ is a mapping.
: from $(2)$, $f^{-1}$ is a mapping.
Category:Equivalence of Definitions of Bijection
Equivalence of Definitions of Binomial Coefficient

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Follows directly from Cardinality of Set of Subsets.
=== Definition 1 implies Definition 3 ===
This is proved in the Binomial Theorem.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 19$: Combinatorial Analysis
Category:Binomial Coefficients