nyuuzyou/wikis huggingface dataset:
Biconditional iff Disjunction implies Conjunction/Formulation 1/Proof

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, in all cases the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccc||ccccccc|} \hline
p & \iff & q & (p & \lor & q) & \implies & (p & \land & q) \\
\hline
F & T & F & F & F & F & T & F & F & F \\
F & F & T & F & T & T & F & F & F & T \\
T & F & F & T & T & F & F & T & F & F \\
T & T & T & T & T & T & T & T & T & T \\
\hline
\end{array}$
Category:Biconditional iff Disjunction implies Conjunction
Category:Truth Table Proofs
Biconditional in terms of NAND

== Theorem ==
:$p \iff q \dashv \vdash \paren {\paren {p \uparrow p} \uparrow \paren {q \uparrow q} } \uparrow \paren {p \uparrow q}$
where $\iff$ denotes logical biconditional and $\uparrow$ denotes logical NAND.
== Proof ==
== Sources ==
* : Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(5)$
Category:Logical NAND
Category:Biconditional
Biconditional is Associative/Formulation 1

== Theorem ==
The biconditional operator is associative:
:$p \iff \paren {q \iff r} \dashv \vdash \paren {p \iff q} \iff r$
== Proof ==
Proof of associativity by natural deduction is just too tedious to be considered.
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective match for all boolean interpretations.
$\begin{array}{|ccccc||ccccc|} \hline
p & \iff & (q & \iff & r) & (p & \iff & q) & \iff & r \\
\hline
\F & \F & \F & \T & \F & \F & \T & \F & \F & \F \\
\F & \T & \F & \F & \T & \F & \T & \F & \T & \T \\
\F & \T & \T & \F & \F & \F & \F & \T & \T & \F \\
\F & \F & \T & \T & \T & \F & \F & \T & \F & \T \\
\T & \T & \F & \T & \F & \T & \F & \F & \T & \F \\
\T & \F & \F & \F & \T & \T & \F & \F & \F & \T \\
\T & \F & \T & \F & \F & \T & \T & \T & \F & \F \\
\T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
== Sources ==
* : $\S 2.3.3$
Category:Biconditional is Associative
Biconditional is Associative/Formulation 2

== Theorem ==
:$\vdash \paren {p \iff \paren {q \iff r} } \iff \paren {\paren {p \iff q} \iff r}$
== Proof ==
== Sources ==
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T94}$
Category:Biconditional is Associative
Biconditional is Commutative/Formulation 1

== Theorems ==
The biconditional operator is commutative:
:$p \iff q \dashv \vdash q \iff p$
== Proof 1 ==
== Proof 2 ==
== Proof by Truth Table ==
== Sources ==
* : $\S 2.3.3$
Category:Biconditional is Commutative
Biconditional is Commutative/Formulation 1/Proof by Truth Table

== Theorems ==
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccc||ccc|} \hline
p & \iff & q & q & \iff & p \\
\hline
\F & \T & \F & \F & \T & \F \\
\F & \F & \T & \T & \F & \F \\
\T & \F & \F & \F & \F & \T \\
\T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
Category:Truth Table Proofs
Category:Biconditional is Commutative
Biconditional is Commutative/Formulation 2

== Theorems ==
The biconditional operator is commutative:
:$\vdash \paren {p \iff q} \iff \paren {q \iff p}$
== Proof ==
== Sources ==
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T92}$
Category:Biconditional is Commutative
Biconditional is Reflexive

== Theorem ==
The biconditional operator, considered as a relation, is reflexive:
: $\vdash p \iff p$
This can otherwise be stated as that equivalence destroys copies of itself.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 4.7$: The Derivation of Formulae: $D \, 23$
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T91}$
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems
* : $\S 2.3.3$
Category:Biconditional
Category:Biconditional is Reflexive
Biconditional is Reflexive/Proof 2

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective match for both boolean interpretations.
$\begin{array}{|ccc|} \hline
p & \iff & p \\
\hline
F & T & F \\
T & T & T \\
\hline
\end{array}$
Category:Biconditional is Reflexive
Category:Truth Table Proofs
Biconditional is Self-Inverse

== Theorem ==
:$\paren {p \iff q} \iff q \dashv \vdash p$
where $\iff$ denotes the biconditional operator.
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective on the match those for $p$ on the for all boolean interpretations:
:$\begin{array}{|ccccc||c|} \hline
(p & \iff & q) & \iff & q & p \\
\hline
\F & \T & \F & \F & \F & \F \\
\F & \F & \T & \F & \T & \F \\
\T & \F & \F & \T & \F & \T \\
\T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
Category:Biconditional
Category:Truth Table Proofs
Biconditional is Transitive/Formulation 1

== Theorem ==
The biconditional operator is transitive:
:$p \iff q, q \iff r \vdash p \iff r$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems
* : $\S 1.13$: Tableau Problems (TAB1): EQUIV
Category:Biconditional is Transitive
Category:Truth Table Proofs
Biconditional is Transitive/Formulation 1/Proof 2

== Theorem ==
:$p \iff q, q \iff r \vdash p \iff r$
== Proof ==
We apply the Method of Truth Tables.
As can be seen for all boolean interpretations by inspection, where the truth values under the main connective on the is $T$, that under the one on the is also $T$:
$\begin{array}{|ccccccc||ccc|} \hline
(p & \iff & q) & \land & (q & \iff & r) & p & \iff & r \\
\hline
F & T & F & T & F & T & F & F & T & F \\
F & T & F & F & F & F & T & F & F & T \\
F & F & T & F & T & F & F & F & T & F \\
F & F & T & F & T & T & T & F & F & T \\
T & F & F & F & F & T & F & T & F & F \\
T & F & F & F & F & F & T & T & T & T \\
T & T & T & F & T & F & F & T & F & F \\
T & T & T & T & T & T & T & T & T & T \\
\hline
\end{array}$
Hence the result.
Category:Biconditional is Transitive
Category:Truth Table Proofs
Biconditional is Transitive/Formulation 2

== Theorem ==
The biconditional operator is transitive:
:$\vdash \paren {\paren {p \iff q} \land \paren {q \iff r} } \implies \paren {p \iff r}$
== Proof ==
== Sources ==
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T93}$
Category:Biconditional is Transitive
Biconditional of Proposition and its Negation/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations.
$\begin{array}{|c|cccc|} \hline
\neg & (p & \iff & \neg & p)\\
\hline
\T & \F & \F & \T & \F \\
\T & \T & \F & \F & \T \\
\hline
\end{array}$
Category:Biconditional of Proposition and its Negation
Biconditional with Contradiction

== Theorem ==
A biconditional with a contradiction:
:$p \iff \bot \dashv \vdash \neg p$
== Proof by Natural Deduction ==
== Proof by Truth Table ==
== Sources ==
* : $\S 2.3.3$
Category:Contradiction
Category:Biconditional
Category:Biconditional with Contradiction
Biconditional with Contradiction/Proof 2

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations.
$\begin{array}{|c|ccc||cc|} \hline
p & p & \iff & \bot & \neg & p \\
\hline
F & F & T & F & T & F \\
T & T & F & F & F & T \\
\hline
\end{array}$
Category:Biconditional with Contradiction
Biconditional with Tautology

== Theorem ==
A biconditional with a tautology:
:$p \iff \top \dashv \vdash p$
== Proof by Natural Deduction ==
== Proof by Truth Table ==
== Sources ==
* : $\S 2.3.3$
Category:Tautology
Category:Biconditional
Category:Biconditional with Tautology
Biconditional with Tautology/Proof 2

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations.
$\begin{array}{|c|ccc||c|ccc|} \hline
p & p & \iff & \top & p \\
\hline
F & F & F & T & F \\
T & T & T & T & T \\
\hline
\end{array}$
Category:Biconditional with Tautology
Bieberbach Conjecture

== Theorem ==
Let $f$ be a holomorphic complex function defined as:
:$\forall z \in \C: \map f z = z + a_2 z^2 + a_3 z^3 + \cdots$
where the $a_n$ are complex.
Let $f$ be injective for $\size z < 1$.
Then:
:$\forall n \ge 2: \size {a_n} \le n$
== Proof ==
== Also known as ==
== Historical Note ==
== Sources ==
* 
Category:Bieberbach Conjecture
Category:Holomorphic Functions
Bienaymé-Chebyshev Inequality

== Theorem ==
Let $X$ be a random variable. 
Let $\expect X = \mu$ for some $\mu \in \R$. 
Let $\var X = \sigma^2$ for some $\sigma^2 \in \R_{> 0}$.
Then, for all $k > 0$: 
:$\map \Pr {\size {X - \mu} \ge k \sigma} \le \dfrac 1 {k^2}$
== Proof 1 ==
 
== Proof 2 ==
== Also known as ==
== Historical Note ==
== Sources ==
* 
* 
* 
Category:Bienaymé-Chebyshev Inequality
Category:Probability Theory
Category:Inequalities
Bienaymé-Chebyshev Inequality/Also known as

== Bienaymé-Chebyshev Inequality: Also known as ==
The Bienaymé-Chebyshev Inequality is also known as Chebyshev's Inequality.
However, some sources use this name to mean the Chebyshev's Sum Inequality, which is a completely different result.
Hence, to avoid confusion, the name Chebyshev's Inequality is not used on .
Variants on spellings can be seen: Bienaymé-Tchebyshev Inequality, Tchebyshev's Inequality, and so on.
== Sources ==
* 
* 
* 
* 
* 
Category:Bienaymé-Chebyshev Inequality
Bienaymé-Chebyshev Inequality/Historical Note

== Historical Note on Bienaymé-Chebyshev Inequality ==
The result now known as the Bienaymé-Chebyshev Inequality was first formulated, without proof, by in $1853$.
The proof was provided by his friend and colleague in $1867$.
's student provided another proof in his $1884$ Ph.D. thesis.
== Sources ==
* 
* 
Category:Historical Notes
Bienaymé-Chebyshev Inequality/Proof 1

== Theorem ==
== Proof ==
Let $f$ be the function:
:$\map f x = \begin{cases} k^2 \sigma^2 & : \size {x - \mu} \ge k \sigma \\ 0 & : \text{otherwise} \end{cases}$
By construction:
:$\forall x \in \Dom f: \map f x \le \size {x - \mu}^2 = \paren {x - \mu}^2$
Hence from Expectation Preserves Inequality:
:$\expect {\map f X} \le \expect {\paren {X - \mu}^2}$
By definition of variance:
:$\expect {\paren {X - \mu}^2} = \var X = \sigma^2$
By definition of expectation of discrete random variable, we can show that:
Putting this together, we have:
By dividing both sides by $k^2 \sigma^2$, we get: 
:$\map \Pr {\size {X - \mu} \ge k \sigma} \le \dfrac 1 {k^2}$
Category:Bienaymé-Chebyshev Inequality
Big-O Estimate for Real Function/Examples

== Examples of Big-$\OO$ Estimates for Real Functions ==
=== Example: $10 \times$ Function at $+\infty$ ===
=== Example: Sine Function at $+\infty$ ===
=== Example: $x = \map \OO {x^2}$ at $+\infty$ ===
=== Runtime of Arbitrary Algorithm ===
Category:Examples of Big-O Notation
Big-O Estimate for Real Function/Examples/10x Function at Infinity

== Example of Big-$\OO$ Estimate for Real Function ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = 10 x$
Then:
:$\map f x = \map \OO x$
as $x \to \infty$.
== Proof ==
Let us consider the real function $g: \R \to \R$ defined as:
:$\forall x \in \R: \map g x = x$
Then we have that:
:$\forall x \in \R_{>0}: \size {\map f x } < 11 \cdot \size x$
Hence the result.
== Sources ==
* : $\text I$: The Series of Primes: $1.6$ Some notations
Category:Examples of Big-O Notation
Big-O Estimate for Real Function/Examples/Runtime of Arbitrary Algorithm

== Example of Big-$\OO$ Estimate for Real Function ==
Let $n$ be the size of a set of inputs to an algorithm.
Let the number of steps needed to carry out the algorithm be $3 n^4$.
Then the complexity of the algorithm is $\map \OO {n^4}$.
== Proof ==
Let $f: \N_{>0} \to \R$ be defined as:
:$\forall n \in \N_{>0}: \map f n = 3 n^4$
Let us consider the real function $g: \R \to \R$ defined as:
:$\forall x \in \R: \map g x = 4 x^4$
Then we have that:
:$\forall n \in \N_{>0}: \size {\map f n} < 4 n^4$
Hence the result.
== Sources ==
* 
Category:Examples of Big-O Notation
Big-O Estimate for Real Function/Examples/Sine Function at Infinity

== Example of Big-$\OO$ Estimate for Real Function ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \sin x$
Then:
:$\map f x = \map \OO 1$
as $x \to \infty$.
== Proof ==
Let us consider the real function $g: \R \to \R$ defined as:
:$\forall x \in \R: \map g x = 1$
Then we have that:
:$\forall x \in \R: \size {\map f x } < 2 \cdot \size 1$
Hence the result.
== Sources ==
* : $\text I$: The Series of Primes: $1.6$ Some notations
Category:Examples of Big-O Notation
Category:Sine Function
Big-O Estimate for Real Function/Examples/x is Big-O of x^2 at Infinity

== Example of Big-$\OO$ Estimate for Real Function ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = x$
Then:
:$\map f x = \map \OO {x^2}$
as $x \to \infty$.
== Proof ==
Let us consider the real function $g: \R \to \R$ defined as:
:$\forall x \in \R: \map g x = x^2$
Then we have that:
:$\forall x \in \R_{>1}: \size {\map f x } < x^2$
Hence the result.
== Sources ==
* : $\text I$: The Series of Primes: $1.6$ Some notations
Category:Examples of Big-O Notation
Big-O Notation for Sequences Coincides with General Definition

== Theorem ==
Let $\sequence {a_n}$ and $\sequence {b_n}$ be sequences of real or complex numbers.
Let $\N$ be given the discrete topology.
:$(1): \quad a_n = \map \OO {b_n}$, where $\OO$ denotes big-$\OO$ notation for sequences
:$(2): \quad a_n = \map \OO {b_n}$, where $\OO$ stands for the general definition of big-$\OO$ notation
== Proof ==
Category:Big-O Notation
Big Implies Saturated

== Theorem ==
Let $\MM$ be an $\LL$-structure.
Let $\kappa$ be a cardinal.
If $\MM$ is $\kappa$-big, then it is $\kappa$-saturated.
== Proof ==
The idea of the proof is:
:to go to some elementary equivalent structure where a type is realized
and:
:to interpret a new relation symbol as the singleton containing some realization of the type.
This lets us write sentences about the extension saying that:
:there must exist an element satisfying the relation
and:
:every element satisfying the relation satisfies all of the formulas in the type.
Then using bigness, these sentences carry over into $\MM$.
Let $p$ be a complete $n$-type over $A$, where $A$ is a subset of the universe of $\MM$ with cardinality strictly less than $\kappa$.
We are to show that $p$ is realized in $\MM$.
Let $\LL_A$ be the language obtained from $\LL$ by adding constant symbols for each element of $A$.
Viewing $\MM$ as an $\LL_A$-structure in the natural way, let $\map {\operatorname {Th}_A} \MM$ be the collection of $\LL_A$-sentences satisfied by $\MM$.
By definition of type, $p \cup \map {\operatorname {Th}_A} \MM$ is satisfiable by some $\LL_A$-structure $\NN$.
Since $\NN$ satisfies $p$, some $n$-ordered tuple $\bar b$ in it realizes $p$.
Now, let $\LL_A^*$ be the language obtained from $\LL_A$ by adding a new relation symbol $R$.
We can extend $\NN$ to be an $\LL_A^*$-structure by interpreting the symbol $R$ as the singleton $R^\NN = \set {\bar b}$.
Since $\NN \models \map {\operatorname {Th}_A} \MM$, we have that $\MM$ and $\NN$ are elementary equivalent as $\LL_A$-structures.
We have that $\MM$ is $\kappa$-big by assumption.
Hence there is some relation $R^\MM$ on the universe of $\MM$ such that:
:$\struct {\MM, R^\MM}$ is elementary equivalent to $\struct {\NN, R^\NN}$ as $\LL_A^*$-structures.
But $\struct {\NN, R^\NN}$ satisfies the $\LL_A^*$-sentences:
:$\exists \bar x \map R {\bar x}$
and
:$\forall \bar x \paren {\map R {\bar x} \to \map \phi {\bar x} }$
for each $\map \phi {\bar x}$ in $p$.
Hence $\struct {\MM, R^\MM}$ also satisfies these sentences, and so there must be some $\bar d \in R^\MM$ realizing $p$.
So $\MM$ realizes $p$.
Category:Model Theory for Predicate Logic
Biholomorphic Function from Open Unit Disk to Right Half-Plane

== Theorem ==
Let $\mathbb D = \set {z : \cmod z < 1}$ be the open unit disk. 
Let $\mathbb H_r = \set {z : \map \Re z > 0}$ be the right half-plane.
Define $f : \mathbb D \to \C$ by:
:$\ds \map f z = \frac {1 - z} {1 + z}$
for each $z \in \mathbb D$.
Then $f$ is a biholomorphic function $\mathbb D \to \mathbb H_r$.
== Proof ==
From Möbius Transformation is Bijection, defining the extended map $f_\ast : \overline \C \to \overline \C$ by:
:$\map {f_\ast} z = \begin {cases} \dfrac {1 - z} {1 + z} & : z \ne -1 \\
\infty & : z = -1 \\
-1 & : z = \infty \end{cases}$
we obtain a bijection.
Hence $f : \mathbb D \to \C$ is an injection.
To show that $f$ is a surjection $\mathbb D \to \mathbb H_r$ it is enough to show that if $\map {f_\ast} z \in \mathbb H_r$, then $\cmod z < 1$.
We first demonstrate that this is the case.
Suppose that it is true that if $\map {f_\ast} z \in \mathbb H_r$ for $z \in \overline \C$, then $\cmod z < 1$.
For each $w \in \mathbb H_r$, there will exist $z \in \overline \C$ such that $\map {f_\ast} z = w$.
We will then have $z \in \mathbb D$, and hence $w = \map {f_\ast} z = \map f z$.
Hence we will have shown that $f$ is surjective.
Now, we have, for $z \not \in \set {-1, \infty}$:
If $z = \infty$, then $\map {f_\ast} z = -1 \not \in \mathbb H_r$. 
Similarly $\map {f_\ast} {-1} = \infty \not \in \mathbb H_r$. 
Hence $\map \Re {\map {f_\ast} z} > 0$ :
:$\ds \frac {1 - \cmod z^2} {\cmod {1 + z}^2} > 0$
which is the case $1 - \cmod z^2 > 0$, which is the case $\cmod z < 1$.
Hence, due to the prior discussion, $f$ is surjective.
Since we have $\map \Re {\map f z} > 0$ $\cmod z < 1$, we can consider $f$ as a mapping $\mathbb H_r \to \mathbb D$.
Denote this mapping as $f^{-1} : \mathbb H_r \to \mathbb D$. 
From Combination Theorem for Complex Derivatives: Quotient Rule, $f : \mathbb D \to \mathbb H_r$ is holomorphic.
We have, for $z \in \mathbb D$:
Hence $f^{-1}$ is the inverse of $f$. 
From Combination Theorem for Complex Derivatives: Quotient Rule again, $f^{-1}$ is holomorphic.
$f$ is therefore biholomorphic as required.
Category:Biholomorphic Functions
Category:Möbius Transformations
Bijection/Examples

== Examples of Bijections ==
=== Arbitrary Mapping on Sets ===
=== $\paren {-1}^x \floor {\dfrac x 2}$ from $\N$ to $\Z$ ===
=== $x^3$ Function on Real Numbers is Bijective ===
=== Negative Functions on Standard Number Systems are Bijective ===
=== $2 x + 1$ Function on Real Numbers is Bijective ===
=== $a x + b$ Function on Real Numbers is Bijective ===
=== $n + 1$ Mapping on Integers is Bijective ===
Category:Examples of Bijections
Bijection/Examples/-1^x of Floor of Half x from Natural Numbers to Integers

== Example of Bijection ==
Let $f: \N \to \Z$ be the mapping defined from the natural numbers to the integers as:
:$\forall x \in \N: f \paren x = \paren {-1}^x \floor {\dfrac x 2}$
Then $f$ is a bijection.
== Proof ==
Let $x_1, x_2 \in \N$.
Suppose $f \paren {x_1} = f \paren {x_2}$.
Then, for a start:
:$\paren {-1}^{x_1} = \paren {-1}^{x_2}$
and so $x_1$ and $x_2$ are either both even or both odd.
Suppose $x_1$ and $x_2$ are both even.
Then:
Suppose $x_1$ and $x_2$ are both odd.
Then:
Thus:
:$\forall x_1, x_2 \in \N: f \paren {x_1} = f \paren {x_2} \implies x_1 = x_2$
and so $f$ is an injection by definition.
Let $y \in \Z$ such that $y \ge 0$.
Consider the natural number $x = 2 y$.
Then:
Thus:
:$\forall y \in \Z_{\ge 0}: \exists x \in \N: f \paren x = y$
Let $y \in \Z$ such that $y < 0$.
Consider the natural number $x = 2 \paren {-y} + 1$.
Then:
Thus:
:$\forall y \in \Z_{<0}: \exists x \in \N: f \paren x = y$
Thus by definition $f$ is a surjection by definition.
Thus $f$ is both an injection and a surjection, and so by definition a bijection.
== Sources ==
* : Chapter $4$: Mappings: Exercise $5 \ \text{(b)}$
Category:Examples of Bijections
Bijection/Examples/2x+1 Function on Real Numbers

== Example of Bijection ==
Let $f: \R \to \R$ be the mapping defined on the set of real numbers as:
:$\forall x \in \R: \map f x = 2 x + 1$
Then $f$ is a bijection.
== Proof ==
Let $x_1$ and $x_2$ be real numbers.
Then:
Hence by definition $f$ is an injection.
Let $y \in \R$.
Let $x = \dfrac {y - 1} 2$.
We have that:
:$x \in \R$
:$\map f x = y$
Hence by definition $f$ is a surjection.
Thus $f$ is both an injection and a surjection, and so a bijection.
== Sources ==
* : Chapter $2$: Maps and relations on sets: Example $2.5$
Category:Examples of Bijections
Bijection/Examples/Arbitrary Mapping on Sets

== Example of Bijection ==
Let $A = \set {a_1, a_2, a_3, a_4}$.
Let $B = \set {b_1, b_2, b_3, b_4}$.
Let $f \subseteq {A \times B}$ be the mapping defined as:
:$f = \set {\tuple {a_1, b_3}, \tuple {a_2, b_2}, \tuple {a_3, b_4}, \tuple {a_4, b_1} }$
Then $f$ is a bijection.
== Sources ==
* : Appendix $\text{A}.4$: Functions
Category:Examples of Bijections
Bijection/Examples/Negative Functions

== Example of Bijection ==
Let $\mathbb S$ be one of the standard number systems $\Z$, $\Q$, $\R$, $\C$.
Let $h: \mathbb S \to \mathbb S$ be the negation function defined on $\mathbb S$:
:$\forall x \in \mathbb S: \map h x = -x$
Then $h$ is a bijection.
== Proof ==
=== Complex Numbers ===
By the definition of the negative of complex number, the complex negation function is defined on the complex numbers $\C$ as:
:$-z := -x - i y$
Let $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2 \in \C$ such that $\map h {z_1} = \map h {z_2}$.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Composition of Functions
Category:Examples of Bijections
Bijection/Examples/Real Cube Function

== Example of Bijection ==
Let $f: \R \to \R$ be the mapping defined on the set of real numbers as:
:$\forall x \in \R: \map f x = x^3$
Then $f$ is a bijection.
== Proof ==
A direct application of Integer Power Function is Bijective iff Index is Odd.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Composition of Functions
Category:Examples of Bijections
Bijection/Examples/ax+b on Real Numbers

== Example of Bijection ==
Let $a, b \in \R$ such that $a \ne 0$.
Let $f_{a, b}: \R \to \R$ be the mapping defined on the set of real numbers as:
:$\forall x \in \R: \map f x = a x + b$
Then $f$ is a bijection.
== Proof ==
Let $x_1$ and $x_2$ be real numbers.
Then:
Hence by definition $f$ is an injection.
Let $y \in \R$.
Let $x = \dfrac {y - b} a$.
We have that:
:$x \in \R$
:$\map f x = y$
Hence by definition $f$ is a surjection.
Thus $f$ is both an injection and a surjection, and so a bijection.
Category:Examples of Bijections
Bijection/Examples/n+1 Mapping on Integers

== Example of Bijection ==
Let $f: \Z \to \Z$ be the mapping defined on the set of integers as:
:$\forall n \in \Z: \map f n = n + 1$
Then $f$ is a bijection.
== Proof ==
Let $n_1$ and $n_2$ be integers.
Then:
Hence by definition $f$ is an injection.
Let $m \in \Z$.
Let $n = m - 1$.
We have that:
:$n \in \Z$
and:
:$\map f n = m$
Hence by definition $f$ is a surjection.
Thus $f$ is both an injection and a surjection, and so a bijection
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections
Category:Examples of Bijections