nyuuzyou/wikis huggingface dataset:
Graph Isomorphism is Equivalence Relation/Proof 2

== Theorem ==
== Proof ==
In the following, let:
be arbitrary graphs.
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
The identity mapping on $\map V {G_1}$ is a bijection that maps each vertex to itself.
This is seen to preserve adjacency.
Thus graph isomorphism preserves reflexivity.
=== Symmetry ===
Let $G_1$ be isomorphic to $G_2$
Then there exists an isomorphism $\phi: \map V {G_1} \to \map V {G_2}$ from $\map V {G_1}$ to $\map V {G_2}$.
Let its inverse $\phi^{-1}: \map V {G_2} \to \map V {G_1}$ be defined by:
:$\map {\phi^{-1} } {v_2} = v_1 \iff \map \phi {v_1} = v_2$
By Inverse of Bijection is Bijection, $\phi^{-1}$ is a bijection.
Let $u_2, v_2 \in \map V {G_2}$ such that $\map {\phi^{-1} } {u_2} = u_1$ and $\map {\phi^{-1} } {v_2} = v_1$.
Then:
So $u_2$ and $v_2$ are adjacent $\map \phi {u_1}$ and $\map \phi {v_1}$ are adjacent.
Because $G_1$ is isomorphic to $G_2$, $\map \phi {u_1}$ and $\map \phi {v_1}$ are adjacent $u_1 = \map {\phi^{-1} } {u_2}$ and $v_1 = \map {\phi^{-1} } {v_2}$ are adjacent.
So $u_2$ and $v_2$ are adjacent $u_1 = \map {\phi^{-1} } {u_2}$ and $v_1 = \map {\phi^{-1} } {v_2}$ are adjacent.
That is, $G_2$ is isomorphic to $G_1$
Thus graph isomorphism preserves symmetry.
=== Transitivity ===
Let $G_1$ be isomorphic to $G_2$ and $G_2$ be isomorphic to $G_3$.
Then there exist isomorphisms:
:$\alpha: \map V {G_1} \to \map V {G_2}$ from $\map V {G_1}$ to $\map V {G_2}$
:$\beta: \map V {G_2} \to \map V {G_3}$ from $\map V {G_2}$ to $\map V {G_3}$.
Consider the composite mapping $\beta \circ \alpha$.
From Composite of Bijections is Bijection, $\beta \circ \alpha$ is a bijection from $\map V {G_1}$ to $\map V {G_3}$.
Let $u_1, v_1 \in \map V {G_1}$.
Let $\map \alpha {u_1} = u_2$ and $\map \alpha {v_1} = v_2$.
Let $\map \beta {u_2} = u_3$ and $\map \beta {v_2} = v_3$.
We have that $\alpha$ and $\beta$ are isomorphisms.
So:
:$u_1$ and $v_1$ are adjacent $\map \alpha {u_1} = u_2$ and $\map \alpha {v_1} = v_2$ are adjacent
:$u_2$ and $v_2$ are adjacent $\map \beta {u_2} = u_3$ and $\map \beta {v_2} = v_3$ are adjacent.
Thus:
:$u_1$ and $v_1$ are adjacent $\map \beta {\map \alpha {u_1} } = u_3$ and $\map \beta {\map \alpha {v_1} } = v_3$ are adjacent.
That is, $\map {\paren {\beta \circ \alpha} } {u_1} = u_3$ and $\map {\paren {\beta \circ \alpha} } {v_1} = v_3$ are adjacent.
Thus it has been demonstrated that $\beta \circ \alpha$ is an isomorphism.
Thus graph isomorphism has been shown to preserve transitivity.
Graph isomorphism has been shown to be reflexive, symmetric and transitive.
Hence by definition it is an equivalence relation.
== Sources ==
* : Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.2$: Isomorphic Graphs: Theorem $2.3$
Category:Graph Isomorphism is Equivalence Relation
Graph Paper/Examples/Logarithmic

== Example of Logarithmic Graph Paper ==
This is an example of graph paper that can be used for presenting graphs of functions of the form $y = a x^n$:
:
== Also see ==
* Graph of $y = a x^n$ on Logarithmic Scales
== Sources ==
* 
* 
Category:Examples of Logarithmic Graph Paper
Category:Examples of Graph Paper
Graph Parametrization of Continuous Mapping is Embedding

== Theorem ==
Let $U \subseteq \R^n$ be an open subset of $n$-dimensional Euclidean space.
Let $f : U \to \R^k$ be a continuous mapping.
Let $\map \Gamma f$ be the graph of $f$.
Let $\gamma_f : U \to \R^n \times \R^k$ be the graph parametrization of $\map \Gamma f$.
Then $\gamma_f$ is an embedding of $\R^n$ to $\R^{n + k}$.
== Proof ==
== Sources ==
* : $\S 3$: New Spaces From Old: Subspaces. Topological Embeddings
Graph containing Closed Walk of Odd Length also contains Odd Cycle

== Theorem ==
Let $G$ be a graph.
Let $G$ have a closed walk of odd length.
Then $G$ has an odd cycle.
== Proof ==
Let $G = \struct {V, E}$ be a graph with closed walk whose length is odd.
From Closed Walk of Odd Length contains Odd Circuit, such a walk contains a circuit whose length is odd.
Let $C_1 = \tuple {v_1, \ldots, v_{2 n + 1} = v_1}$ be such a circuit.
 $G$ has no odd cycles.
Then $C_1$ is not a cycle.
Hence, there exist a vertex $v_i$ where $2 \le i \le 2 n - 1$ and an integer $k$ such that $i + 1 \le k \le 2 n$ and $v_i = v_k$.
If $k - i$ is odd, then we have an odd circuit $\tuple {v_i, \ldots, v_k = v_i}$ smaller in length than $C_1$.
If $k - i$ is even, then $\tuple {v_1, \ldots, v_i, v_{k + 1}, \ldots, v_{2 n + 1} }$ is a circuit whose length is odd smaller in length than $C_2$.
This new odd length circuit is named $C_2$, and the same argument is applied as to $C_1$.
Thus at each step a circuit whose length is odd is reduced.
At the $n$th step for some $n \in \N$, either:
:$(1): \quad C_n$ is a cycle, which contradiction the supposition that $G$ has no odd cycles
or:
:$(2): \quad C_n$ is a circuit whose length is $3$.
But from Circuit of Length 3 is Cycle, $C_n$ is a cycle, which by definition has odd length.
From this contradiction it follows that $G$ has at least one odd cycle.
Category:Graph Theory
Graph is 0-Regular iff Edgeless

== Theorem ==
Let $G$ be a graph.
Then $G$ is $0$-regular graph $G$ is edgeless.
== Proof ==
=== Sufficient Condition ===
Let $G$ be an edgeless graph.
By definition of edgeless graph, $G$ has no edges.
That means every vertex of $G$ is incident with no edges.
That is, every vertex of $G$ is of degree $0$.
Hence the result, by definition of $0$-regular graph.
=== Necessary Condition ===
Let $G$ be a $0$-regular graph.
Then every vertex is incident with no edges.
So as there are no vertices with incident edges, there can be no edges.
Hence $G$ is an edgeless graph.
Category:Edgeless Graphs
Category:Regular Graphs
Graph is Bipartite iff No Odd Cycles

== Theorem ==
Let $G$ be an undirected graph.
Then $G$ is bipartite it has no odd cycles.
== Proof ==
=== Sufficient Condition ===
Let $G = \struct {V, E}$ be bipartite.
So, let $V = A \cup B$ such that $A \cap B = \O$ and that all edges $e \in E$ are such that $e$ is of the form $\set {a, b}$ where $a \in A$ and $b \in B$.
(This is the definition of a bipartite graph.)
Suppose $G$ has (at least) one odd cycle $C$.
Let the length of $C$ be $n$.
Let $C = \tuple {v_1, v_2, \ldots, v_n, v_1}$.
, let $v_1 \in A$. It follows that $v_2 \in B$ and hence $v_3 \in A$, and so on.
Hence we see that $\forall k \in \set {1, 2, \ldots, n}$, we have:
:$v_k \in \begin{cases}
A : & k \text{ odd} \\
B : & k \text{ even}
\end{cases}$
But as $n$ is odd, $v_n \in A$.
But $v_1 \in A$, and $v_n v_1 \in C_n$.
So $v_n v_1 \in E$ which contradicts the assumption that $G$ is bipartite.
Hence if $G$ is bipartite, it has no odd cycles.
=== Necessary Condition ===
It is enough to consider $G$ as being connected, as otherwise we could consider each component separately.
Suppose $G$ has no odd cycles.
Choose any vertex $v \in G$.
Divide $G$ into two sets of vertices like this:
:Let $A$ be the set of vertices such that the shortest path from each element of $A$ to $v$ is of odd length
:Let $B$ be the set of vertices such that the shortest path from each element of $B$ to $v$ is of even length.
Then $v \in B$ and $A \cap B = \O$.
Suppose $a_1, a_2 \in A$ are adjacent.
Then there would be a closed walk of odd length $\tuple {v, \ldots, a_1, a_2, \ldots, v}$.
But from Graph containing Closed Walk of Odd Length also contains Odd Cycle, it follows that $G$ would then contain an odd cycle.
This contradicts our initial supposition that $G$ contains no odd cycles.
So no two vertices in $A$ can be adjacent.
By the same argument, neither can any two vertices in $B$ be adjacent.
Thus $A$ and $B$ satisfy the conditions for $G = \struct {A \cup B, E}$ to be bipartite.
Category:Bipartite Graphs
Graph of Adjoint of Densely-Defined Linear Operator

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\tuple {\map D T, T}$ be a densely-defined linear operator.
Let $\tuple {\map D {T^\ast}, T^\ast}$ be the adjoint of $\tuple {\map D T, T}$.
Let $\struct {\HH \times \HH, \innerprod \cdot \cdot_{\HH \times \HH} }$ be the Hilbert space direct sum of $\HH$ with itself.
Define $V : \HH \times \HH \to \HH \times \HH$ by:
:$\map V {x, y} = \tuple {-y, x}$
for each $\tuple {x, y} \in \HH \times \HH$.
Then:
:$\map G {T^\ast} = \paren {V \sqbrk {\map \GG T} }^\bot$
where $\bot$ denotes orthocomplementation in $\struct {\HH \times \HH, \innerprod \cdot \cdot_{\HH \times \HH} }$.
== Proof ==
Let $y, z \in \HH$. 
By the definition of the adjoint, we have $\tuple {y, z} \in \map \GG {T^\ast}$ :
:$\innerprod {T x} y = \innerprod x z$ for each $x \in \map D T$.
This is equivalent to:
:$\innerprod {-T x} y + \innerprod x z = 0$
That is:
:$\innerprod {\tuple {-T x, x} } {\tuple {y, z} }_{\HH \times \HH} = 0$ for each $x \in \map D T$.
Note that $\tuple {-T x, x} = \map V {x, T x} \in V \sqbrk {\map \GG T}$ and conversely any element of $V \sqbrk {\map \GG T}$ has this form.
So we have $\tuple {y, z} \in \map \GG {T^\ast}$ $\tuple {y, z} \in \paren {V \sqbrk {\map \GG T} }^\bot$. 
Hence we obtain:
:$\map G {T^\ast} = \paren {V \sqbrk {\map \GG T} }^\bot$
== Sources ==
* : $13.7$: Graphs
Category:Adjoints (Densely-Defined Linear Operators)
Graph of Continuous Mapping between Metric Spaces is Closed in Chebyshev Product

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f: A_1 \to A_2$ be a continuous mapping.
Let $\AA = A_1 \times A_2$ be the cartesian product of $A_1$ and $A_2$.
Let $d_\infty: \AA \times \AA \to \R$ be the Chebyshev distance on $\AA$:
:$\ds \map {d_\infty} {x, y} = \max \set {\map {d_1} {x_1, y_1}, \map {d_2} {x_2, y_2} }$
where $x = \tuple {x_1, x_2}, y = \tuple {y_1, y_2} \in \AA$.
Let $\Gamma_f$ be the graph of $f$.
Then $\Gamma_f$ is a closed set of $\struct {\AA, d_\infty}$.
== Proof ==
Let $f: A_1 \to A_2$ be continuous.
Then:
:$\forall a \in A_1: \ds \lim_{n \mathop \to \infty} x_n = a \implies \lim_{n \mathop \to \infty} \map f {x_n} = \map f a$
This can be extended to ordered pairs and ordered tuples, because:
:$\ds \lim_{n \mathop \to \infty} x_n = a \iff \exists N \in \N: n > N \implies x_n \in V$
where $V$ is some neighborhood of $a$.
That is:
:$n > N \implies x_n \in \map {B_\delta} a$
for some $N \in \N$ and all $\delta \in \R_{>0}$.
So:
That is, each sequence of points $\sequence {a_n, b_n}$ such that:
:$\map f {a_n} = b_n$
converges to a point $\tuple {a, b}$ such that:
:$\map f a = b$
Hence:
:$\ds \lim_{n \mathop \to \infty} \tuple {a_n, b_n} = \tuple {a, \map f a} = \Gamma_f$
That is, $\Gamma_f$ contains all its limit points.
Hence the result by definition of closed set.
== Also see ==
* Mapping whose Graph is Closed in Chebyshev Product is not necessarily Continuous
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Exercise $3$
Category:Chebyshev Distance
Category:Continuous Mappings
Graph of Continuous Mapping is Homeomorphic to Domain

== Theorem ==
Let $T_1$ and $T_2$ be topological spaces.
Let $f$ be a homeomorphism from $T_1$ to $T_2$.
Let $G_f$ denote the graph of $f$.
Then $f$ is homeomorphic to $G_f$.
== Prooof ==
Let $\iota: G_f \to T_1 \times T_2$ be the inclusion mapping from $G_f$ into $T_1 \times T_2$.
Let $\theta: G_f \to T_1$ and $\psi: T_1 \to G_f$ be given by:
:$\theta = \pr_1 \circ \iota$
:$\forall x \in T_1: \map \psi x = \tuple {x, \map f x}$
$\theta$ and $\psi$ are seen to be inverses of each other.
The fact that $\theta$ is continuous follows from the fact that $\pr_1$ and $\iota$ are both continuous.
Then $\pr_1 \circ \iota \circ \psi$ is the identity mapping on $T_1$ and so is continuous.
Also, $\pr_2 \circ \iota \circ \psi$ is also continuous.
Hence by Continuous Mapping to Product Space, $\iota \circ \psi$ is continuous.
It follows by Continuity of Composite with Inclusion/Inclusion on Mapping that $\psi$ is continuous.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.6$: Homeomorphisms
Category:Homeomorphisms (Topological Spaces)
Graph of Continuous Mapping to Hausdorff Space is Closed in Product

== Theorem ==
Let $T_A = \struct {A, \tau_A}$ and $T_B = \struct {B, \tau_B}$ be topological spaces.
Let $T_B$ be a Hausdorff space.
Let $f: T_A \to T_B$ be a continuous mapping.
Then the graph of $f$ is a closed subset of $T_A \times T_B$ under the product topology.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $4$: The Hausdorff condition: Exercise $4.3: 3$
Category:Graph of Continuous Mapping to Hausdorff Space is Closed in Product
Category:Hausdorff Spaces
Category:Continuous Mappings
Category:Product Spaces
Graph of Continuous Mapping to Hausdorff Space is Closed in Product/Proof 1

== Theorem ==
== Proof ==
Let $G_f$ be the graph of $f$:
:$G_f = \set {\tuple {x, y} \in A \times B: \map f x = y}$
Let $I_B: T_B \to T_B$ be the identity mapping on $B$:
:$\forall y \in B: \map {I_B} y = y$
From Identity Mapping is Continuous, $I_B$ is continuous on $T_B$.
Let $f \times I_B: T_A \times T_B \to T_B \times T_B$ be the product map:
:$\map {f \times I_B} {x, y} = \tuple {\map f x, y}$
From Continuous Mapping to Product Space, $f \times I_B$ is continuous.
Let $\Delta_B$ be the diagonal relation on $B$.
From Hausdorff Space iff Diagonal Set on Product is Closed, $\Delta_B$ is a closed set of $T_B \times T_B$ under the product topology.
From Continuity Defined from Closed Sets:
is closed in $T_A \times T_B$ endowed with the product topology.
Category:Graph of Continuous Mapping to Hausdorff Space is Closed in Product
Graph of Continuous Mapping to Hausdorff Space is Closed in Product/Proof 2

== Theorem ==
== Proof ==
Let $G_f$ be the graph of $f$:
:$G_f = \set {\tuple {x, y} \in A \times B: \map f x = y}$
Fix a point $\tuple {x_0, y_0}$ in $\paren {A \times B} \setminus G_f = \set {\tuple {x, y} \in A \times B : \map f x \ne y}$.
As $\map f {x_0} \ne y_0$ and $T_B$ is Hausdorff, there exists disjoint, non-empty, open sets $V_{x_0}$ and $V_{y_0}$ of $T_B$ containing $\map f {x_0}$ and $y_0$ respectively. 
From the definition of continuous mapping, $f^{-1} \sqbrk {V_{x_0} }$ is an open set of $T_A$. 
So we have that:
:$f^{-1} \sqbrk {V_{x_0} } \times V_{y_0}$ is open in $T_A \times T_B$
:$f^{-1} \sqbrk {V_{x_0} } \times V_{y_0}$ is a subset of $\paren {A \times B} \setminus G_f$ containing $\tuple {x_0, y_0}$
Hence it follows that $\paren {A \times B} \setminus G_f$ is open in $T_A \times T_B$.
Hence $G_f$ is closed in $T_A \times T_B$.
Category:Graph of Continuous Mapping to Hausdorff Space is Closed in Product
Graph of Continuous Real Function is Manifold

== Theorem ==
Let $U \subseteq \R^n$ be an open subset of $n$-dimensional Euclidean space.
Let $f : U \to \R^k$ be a continuous mapping.
Let $\map \Gamma f$ be the graph of $f$ equipped with the subspace topology.
Then $\map \Gamma f$ is a $n$-manifold.
== Proof ==
Let $\gamma_f : U \to \R^{n + k}$ be the graph parametrization of $\map \Gamma f$.
From Graph Parametrization of Continuous Mapping is Embedding, $\gamma_f$ is an embedding.
The image of $\gamma_f$ is $\map \Gamma f$.
Therefore, $U$ and $\map \Gamma f$ are homeomorphic.
In other words, for each point in $\map \Gamma f$, $\map \Gamma f$ is a open neighborhood that is homeomorphic to an open subset of $\R^n$.
So $\map \Gamma f$ is locally Euclidean of dimension $n$ by definition.
From Locally Euclidean Subspace of Euclidean Space is Manifold, $\map \Gamma f$ is a $n$-manifold.
== Sources ==
* : $\S 3$: New Spaces From Old: Subspaces. Topological Embeddings
Graph of Even Function is Symmetrical about y-Axis

== Theorem ==
Let $f$ be an even real function.
Let the graph $G$ of $f$ be presented in a Cartesian plane.
Then $G$ is bilaterally symmetric whose axis of symmetry is the $y$-axis.
== Proof ==
== Sources ==
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.3$ Functions of a Real Variable: $\text {(h)}$ Even and Odd Functions
* 
* 
Category:Even Functions
Category:Bilateral Symmetry
Graph of Icosahedron is Hamiltonian

== Theorem ==
The graph of the icosahedron is Hamiltonian.
== Proof ==
Proof by demonstration:
:
A Hamiltonian cycle is indicated in blue.
== Sources ==
* : $\S 3.2$: The Salesman's Problem: An Introduction to Hamiltonian Graphs: Problem $20$
Category:Hamiltonian Graphs
Graph of Linear Function of One Variable

== Theorem ==
Let $f: \R \to \R$ be a linear function of one variable expressed by the equation:
:$\map f x = a_0 + a_1 x$
The graph of $f$ in a cartesian plane consists of a straight line:
:whose $y$-intercept is $a_0$
:whose slope is $a_1$.
== Proof ==
Expressing $f$ in the form $y = \map f x$, we have:
:$y = a_0 + a_1 x$
Thus it is in the same form as the slope-intercept form of an equation of a straight line in the plane.
Hence the result.
== Sources ==
* 
* 
Category:Linear Functions
Category:Graphs of Mappings
Graph of Linear Transformation is Vector Subspace

== Theorem ==
Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$.
Let $X \times Y$ be the direct product of $X$ and $Y$.
Let $T : X \to Y$ be a linear transformation.
Define:
:$\map \GG T = \set {\tuple {x, T x} : x \in X} \subseteq X \times Y$ 
Then $\map \GG T$ is a vector subspace of $X \times Y$. 
== Proof ==
Since $\map T { {\mathbf 0}_X} = {\mathbf 0}_Y$, we have:
:$\tuple { {\mathbf 0}_X, {\mathbf 0}_Y} \in \map \GG T$ 
So $\map \GG T \ne \O$.
Hence from One-Step Vector Subspace Test, it is enough to show that $u + \lambda v \in \map \GG T$ for $u, v \in \map \GG T$ and $\lambda \in K$. 
Let $\alpha \in K$ and $\tuple {x, T x}, \tuple {y, T x} \in \map \GG T$.
Then we have:
:$\tuple {x, T x} + \alpha \tuple {y, T y} = \tuple {x + \alpha y, T x + \alpha T y}$ 
Since $T$ is linear, we have that $T x + \alpha T y = \map T {x + \alpha y}$. 
Hence:
:$\tuple {x, T x} + \alpha \tuple {y, T y} = \tuple {x + \alpha y, \map T {x + \alpha y} } \in \map \GG T$ 
Hence from One-Step Vector Subspace Test, $\map \GG T$ is a vector subspace of $X \times Y$.
Category:Direct Product of Vector Spaces
Graph of Nonlinear Additive Function is Dense in the Plane

== Theorem ==
Let $f: \R \to \R$ be an additive function which is not linear.
Then the graph of $f$ is dense in the real number plane.
== Proof ==
From Additive Function is Linear for Rational Factors:
:$\map f q = q \map f 1$
for all $q \in \Q$.
, let:
:$\map f q = q$
for all $q \in \Q$.
Since $f$ is not linear, let $\alpha \in \R \setminus \Q$ be such that:
:$\map f \alpha = \alpha + \delta$
for some $\delta \ne 0$.
Consider an arbitrary nonempty circle in the plane.
Let:
:its centre be $\tuple {x, y}$ where $x \ne y$ and $x, y \in \Q$
:its radius be $r > 0$.
We will show how to find a point of the graph of $f$ inside this circle.
As $x \ne y$ and $r$ can be arbitrarily small, this will prove the theorem.
Since $\delta \ne 0$, let $\beta = \dfrac {y - x} \delta$.
Since $x \ne y$:
:$\beta \ne 0$
As Rationals are Everywhere Dense in Topological Space of Reals,
there exists a rational number $b \ne 0$ such that:
:$\size {\beta - b} < \dfrac r {2 \size \delta}$
As Rationals are Everywhere Dense in Topological Space of Reals, there also exists a rational number $a$ such that:
:$\size {\alpha - a} < \dfrac r {2 \size b}$
Now put:
:$X = x + b \paren {\alpha - a}$
:$Y = \map f X$
Then:
:$\size {X - x} = \size {b \paren {\alpha - a} } < \frac r 2$
so $X$ is in the circle.
Then:
Therefore 
:$\size {Y - y} = \size {b \paren {\alpha - a} - \delta \paren {\beta - b} } \le \size {b \paren {\alpha - a} } + \size {\delta \paren {\beta - b} } \le r$
so $Y$ is in the circle as well.
Hence the point $\tuple {X, Y}$ is inside the circle.
== Also see == 
* Definition:Additive Function
* Definition:Cauchy Functional Equation
* Definition:Everywhere Dense
Category:Additive Functions
Graph of Odd Function is Rotationally Symmetric about Origin

== Theorem ==
Let $f$ be an odd real function.
Let the graph $G$ of $f$ be presented in a Cartesian plane.
Then $G$ is rotationally symmetric about the origin.
== Proof ==
== Sources ==
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.3$ Functions of a Real Variable: $\text {(h)}$ Even and Odd Functions
* 
* 
Category:Odd Functions
Category:Symmetries (Geometry)
Graph of Quadratic describes Parabola

== Theorem ==
The locus of the equation defining a quadratic:
:$y = a x^2 + b x + c$
describes a parabola.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
Consider the focus-directrix property of a parabola $P$.
Let the focus of $P$ be the point $\tuple {0, f}$ on a Cartesian plane.
Let the directrix of $P$ be the straight line $y = -d$.
Let $\tuple {x, y}$ be an arbitrary point on $P$.
Then by the focus-directrix property:
:$y + d = \sqrt {\paren {x - k}^2 + \tuple {y - f}^2}$
where:
:$y + d$ is the distance from $\tuple {x, y}$ to the straight line $y = -d$
:$\sqrt {\paren {x - k}^2 + \paren {y - f}^2}$ is the distance from $\tuple {x, y}$ to the point $\tuple {k, f}$ by the Distance Formula.
Hence:
This is in the form $y = a x^2 + b^2 + c$.
By setting $k$, $f$ and $d$ appropriately in terms of $a$, $b$ and $c$, the specific focus and directrix can be appropriately positioned.
Category:Parabolas
Category:Quadratic Equations
Category:Graph of Quadratic describes Parabola
Graph of Quadratic describes Parabola/Corollary 1

== Theorem ==
The locus of the equation of the square function:
:$y = x^2$
describes a parabola.
== Proof ==
This is a particular instance of Graph of Quadratic describes Parabola, where:
:$y = a x^2 + b x + c$
is the equation of a parabola.
The result follows by setting $a = 1, b = 0, c = 0$.
== Sources ==
* : Chapter $6$: Curves and Coordinates: Functions
Category:Graph of Quadratic describes Parabola
Graph of Quadratic describes Parabola/Corollary 2

== Theorem ==
The locus of the equation of the square root function on the non-negative reals:
:$\forall x \in \R_{\ge 0}: \map f x = \sqrt x$
describes half of a parabola.
== Proof ==
From Graph of Quadratic describes Parabola: Corollary 1, where:
:$y = x^2$
is the equation of a parabola.
Let $f: \R \to \R$ be the real function defined as:
:$\map f x = x^2$
From Square of Real Number is Non-Negative, the image of $f$ is $\R_{\ge 0}$.
Also we have from Positive Real Number has Two Square Roots:
:$\forall x \in \R: \paren {-x}^2 = x^2$
Thus it is necessary to apply a bijective restriction upon $f$.
Let $g: \R_{\ge 0} \to \R_{\ge 0}$ be the bijective restriction of $f$ to $\R_{\ge 0} \times \R_{\ge 0}$:
:$\forall x \in \R_{\ge 0}: \map g x = x^2$
From Inverse of Bijection is Bijection, $g^{-1}: \R_{\ge 0} \to \R_{\ge 0}$ is also a bijection.
By definition:
:$\forall x \in \R_{\ge 0}: \map {g^{-1} } x = +\sqrt x$
Then from Graph of Inverse Mapping, the graph of $g^{-1}$ is the same as the graph of $g$, reflected in the line $x = y$.
As the graph of $f$ is a parabola, the graph of $g$ is also a parabola, but because of the restriction to $\R_{\ge 0}$, just half of it.
Thus the graph of $g^{-1}$ is also half a parabola.
== Sources ==
* : Chapter $6$: Curves and Coordinates: Functions
Category:Graph of Quadratic describes Parabola
Graph of Real Bijection in Coordinate Plane intersects Horizontal Line at One Point

== Theorem ==
Let $f: \R \to \R$ be a real function which is bijective.
Let its graph be embedded in the Cartesian plane $\CC$:
:
Every horizontal line through a point $b$ in the codomain of $f$ intersects the graph of $f$ on exactly one point $P = \tuple {a, b}$ where $b = \map f a$.
== Proof ==
By definition, a bijection is a mapping which is both an injection and a surjection.
Let $\LL$ be a horizontal line through a point $b$ in the codomain of $f$.
From Graph of Real Surjection in Coordinate Plane intersects Every Horizontal Line:
:$\LL$ intersects the graph of $f$ on at least one point $P = \tuple {a, b}$ where $b = \map f a$.
From Graph of Real Injection in Coordinate Plane intersects Horizontal Line at most Once:
:$\LL$ intersects the graph of $f$ on at most one point $P = \tuple {a, b}$ where $b = \map f a$.
The result follows.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $6 \ \text {(c)}$
Category:Real Functions
Category:Bijections
Category:Graphs of Mappings