nyuuzyou/wikis huggingface dataset:
Current in Electric Circuit/L, R in Series/Constant EMF at t = 0/Corollary 2

== Theorem ==
Consider the electric circuit $K$ consisting of:
:a resistance $R$
:an inductance $L$
in series with a source of electromotive force $E$ which is a function of time $t$.
:
Let the electric current flowing in $K$ at time $t = 0$ be $0$.
Let a constant EMF $E_0$ be imposed upon $K$ at time $t = 0$.
The electric current $I$ in $K$ is given by the equation:
:$I = \dfrac {E_0} R \paren {1 - e^{-R t / L} }$
== Proof ==
From Electric Current in Electric Circuit: L, R in Series: Constant EMF at $t = 0$:
:$I = \dfrac {E_0} R + \paren {I_0 - \dfrac {E_0} R} e^{-R t / L}$
Setting $I_0 = 0$:
Hence the result.
== Sources ==
* : $\S 2.13$: Simple Electric Circuits: Example $1$
Category:Electronics
Current in Electric Circuit/L, R in Series/Constant EMF at t = 0/Corollary 3

== Theorem ==
Consider the electric circuit $K$ consisting of:
:a resistance $R$
:an inductance $L$
in series with a source of electromotive force $E$ which is a function of time $t$.
:
Let the electric current flowing in $K$ at time $t = 0$ be $I_0$.
Let EMF imposed upon $K$ be zero.
The electric current $I$ in $K$ is given by the equation:
:$I = I_0 e^{-R t / L}$
== Proof ==
From Electric Current in Electric Circuit: L, R in Series: Constant EMF at $t = 0$:
:$I = \dfrac {E_0} R + \paren {I_0 - \dfrac {E_0} R} e^{-R t / L}$
Setting $E_0 = 0$:
Hence the result.
== Sources ==
* : $\S 2.13$: Simple Electric Circuits: Example $1$
Category:Electronics
Current in Electric Circuit/L, R in Series/Exponentially Decaying EMF at t = 0

== Theorem ==
Consider the electric circuit $K$ consisting of:
:a resistance $R$
:an inductance $L$
in series with a source of electromotive force $E$ which is a function of time $t$.
:
Let the electric current flowing in $K$ at time $t = 0$ be $I_0$.
Let an EMF $E$ be imposed upon $K$ at time $t = 0$ defined by the equation:
:$E = E_0 e^{-k t}$
The electric current $I$ in $K$ is given by the equation:
:$I = \dfrac {E_0} {R - k L} e^{-k t} + \paren {I_0 - \dfrac {E_0} {R - k L} } e^{-R t / L}$
== Proof ==
From Electric Current in Electric Circuit: L, R in Series:
:$L \dfrac {\d I} {\d t} + R I = E_0 e^{-k t}$
defines the behaviour of $I$.
This can be written as:
:$(1): \quad \dfrac {\d I} {\d t} + \dfrac R L I = \dfrac {E_0} L e^{-k t}$
$(1)$ is a linear first order ODE in the form:
:$\dfrac {\d I} {\d t} + \map P t I = \map Q t$
where:
:$\map P t = \dfrac R L$
:$\map Q t = \dfrac {E_0} L e^{-k t}$
Thus:
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
:$\map {\dfrac \d {\d t} } {e^{R t / L} I} = e^{R t / L} \dfrac {E_0} L e^{-k t}$
and the general solution becomes:
:$\ds e^{R t / L} I = \int e^{R t / L} \dfrac {E_0} L e^{-k t} \rd t$
and so:
When $t = 0$, we have $I = I_0$.
So:
So:
:$I e^{R t/ L} = \dfrac {E_0} {R - k L} e^{\paren {R / L - k} t} + I_0 - \dfrac {E_0} {R - k L}$
Multiplying by $e^{\frac {R t} L}$ and tidying up, we get:
== Sources ==
* : $\S 2.13$: Simple Electric Circuits: Problem $1 \ \text{(a)}$
Category:Electronics
Current in Electric Circuit/L, R in Series/Maximum Current implies Decreasing EMF

== Theorem ==
Consider the electric circuit $K$ consisting of:
:a resistance $R$
:an inductance $L$
in series with a source of electromotive force $E$ which is a function of time $t$.
:
Let the current $I$ be at a maximum.
Then the EMF $E$ is decreasing.
== Proof ==
From Electric Current in Electric Circuit: L, R in Series:
:$L \dfrac {\d I} {\d t} + R I = E$
defines the behaviour of $I$.
Taking the derivative:
:$L \dfrac {\d^2 I} {\d t^2} + R \dfrac {\d I} {\d t} = \dfrac {\d E} {\d t}$
From Second Derivative of Real Function at Minimum:
:$L \dfrac {\d^2 I} {\d t^2} \le 0$
while from Derivative at Maximum or Minimum:
:$R \dfrac {\d I} {\d t} = 0$
and so:
:$\dfrac {\d E} {\d t} \le 0$
The result follows from Decreasing Function has Negative Derivative.
== Sources ==
* : $\S 2.13$: Simple Electric Circuits: Problem $2 \ \text{(b)}$
Category:Electronics
Current in Electric Circuit/L, R in Series/Minimum Current implies Increasing EMF

== Theorem ==
Consider the electric circuit $K$ consisting of:
:a resistance $R$
:an inductance $L$
in series with a source of electromotive force $E$ which is a function of time $t$.
:
Let the current $I$ be at a minimum.
Then the EMF $E$ is increasing.
== Proof ==
From Electric Current in Electric Circuit: L, R in Series:
:$L \dfrac {\d I} {\d t} + R I = E$
defines the behaviour of $I$.
Taking the derivative:
:$L \dfrac {\d^2 I} {\d t^2} + R \dfrac {\d I} {\d t} = \dfrac {\d E} {\d t}$
From Second Derivative of Real Function at Minimum:
:$L \dfrac {\d^2 I} {\d t^2} \ge 0$
while from Derivative at Maximum or Minimum:
:$R \dfrac {\d I} {\d t} = 0$
and so:
:$\dfrac {\d E} {\d t} \ge 0$
The result follows from Increasing Function has Positive Derivative.
== Sources ==
* : $\S 2.13$: Simple Electric Circuits: Problem $2 \ \text{(b)}$
Category:Electronics
Current in Electric Circuit/L, R in Series/Sinusoidal EMF

== Theorem ==
Consider the electric circuit $K$ consisting of:
:a resistance $R$
:an inductance $L$
in series with a source of electromotive force $E$ which is a function of time $t$.
:
Let the electric current flowing in $K$ at time $t = 0$ be $I_0$.
Let an EMF $E$ be imposed upon $K$ at time $t = 0$ defined by the equation:
:$E = E_0 \sin \omega t$
The electric current $I$ in $K$ is given by the equation:
:$I = \dfrac {E_0} {\sqrt {R^2 - L^2 \omega^2} } \map \sin {\omega t - \alpha} + \paren {I_0 - \dfrac {E_0 L \omega} {R^2 + L^2 \omega^2} } e^{-R t / L}$
where $\tan \alpha = \dfrac {L \omega} R$.
== Proof ==
From Electric Current in Electric Circuit: L, R in Series:
:$L \dfrac {\d I} {\d t} + R I = E_0 \sin \omega t$
defines the behaviour of $I$.
This can be written as:
:$(1): \quad \dfrac {\d I} {\d t} + \dfrac R L I = \dfrac {E_0} L \sin \omega t$
$(1)$ is a linear first order ODE in the form:
:$\dfrac {\d I} {\d t} + \map P t I = \map Q t$
where:
:$\map P t = \dfrac R L$
:$\map Q t = \dfrac {E_0} L e^{-k t}$
Thus:
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
:$\map {\dfrac \d {\d t} } {e^{R t / L} I} = e^{R t / L} \dfrac {E_0} L \sin \omega t$
and so the general solution becomes:
When $t = 0, I = I_0$.
The result follows by application of Multiple of Sine plus Multiple of Cosine: Sine Form.
== Sources ==
* : $\S 2.13$: Simple Electric Circuits: Problem $1 \ \text{(b)}$
Category:Electronics
Curry's Paradox

== Paradox ==
Let $P$ be an arbitrary proposition.
Let a proposition $C$ be defined:
:$C \implies P$
 that $\neg C$.
Then $C \implies P$ is vacuously true.
By definition, $C$ is true, a contradiction.
This contradiction shows $C$ to be true.
By definition, $C \implies P$.
By Modus Ponendo Ponens, we conclude $P$, where $P$ is arbitrary.
But this means that any proof system expressing the above is inconsistent.
== Also known as ==
Curry's Paradox can also be seen referred to as Löb's Paradox after .
== Also see ==
* Löb's Theorem
== Sources ==
* 
* 
Category:Antinomies
Curve is Involute of Evolute

== Theorem ==
Let $C$ be a curve defined by a real function which is twice differentiable.
Let the curvature of $C$ be non-constant.
Let $E$ be the evolute $C$.
Then the involute of $E$ is $C$.
== Proof ==
From Length of Arc of Evolute equals Difference in Radii of Curvature:
:the length of arc of the evolute $E$ of $C$ between any two points $Q_1$ and $Q_2$ of $C$ is equal to the difference between the radii of curvature at the corresponding points $P_1$ and $P_2$ of $C$.
Thus $C$ exhibits precisely the property of the involute of $E$.
== Sources ==
* : Chapter $\text {B}.23$: Evolutes and Involutes. The Evolute of a Cycloid
* 
* 
Category:Evolutes
Category:Involutes
Curved Mirror producing Parallel Rays is Paraboloid

== Theorem ==
Let $M$ be a curved mirror embedded in a real cartesian $3$- space.
Let there be a source of light at the origin.
Let $M$ reflect the light in a beam parallel to the $x$-axis.
Then $M$ is the solid of revolution produced by rotating about the $x$-axis the parabola whose equation is:
:$y^2 = 2 c x + c^2$
== Proof ==
The mirror will have the shape of a surface of revolution generated by revolving a curve $APB$ in the cartesian plane around the $x$-axis.
Let $P = \tuple {x, y}$ be an arbitrary point on $APB$.
:
From the Law of Reflection:
:$\alpha = \beta$
By the geometry of the situation:
:$\phi = \beta$
:$\theta = \alpha + \phi = 2 \beta$
By definition of tangent:
:$\tan \theta = \dfrac y x$
and so:
Using the Quadratic Formula:
:$\dfrac {\d y} {\d x} = \dfrac {-x \pm \sqrt {x^2 + y^2} \rd x} y$
which can be expressed as:
:$x \rd x + y \rd y = \pm \sqrt {x^2 + y^2} \rd x$
Using Differential of Sum of Squares:
:$\pm \dfrac {\map \d {x^2 + y^2} } {2 \sqrt {x^2 + y^2} } = \d x$
and so:
:$\pm \sqrt {x^2 + y^2} = x + c$
Hence the result.
== Sources ==
* : $\S 2.9$: Integrating Factors: Example $2$
Category:Optics
Cusps of Hypocycloid from Irrational Ratio of Circle Radii

== Theorem ==
Consider the hypocycloid $H$ generated by a epicycle $C_1$ of radius $b$ rolling within a deferent $C_2$ of (larger) radius $a$.
Let $k = \dfrac a b$ be an irrational number.
Then $H$ has an infinite number of cusps, which are evenly and densely distributed around the circumference of $C_2$.
== Proof ==
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid
Category:Hypocycloids
Cut-Off Subtraction is Primitive Recursive

== Theorem ==
The cut-off subtraction function, defined as:
:$\forall \tuple {n, m} \in \N^2: n \mathop {\dot -} m = \begin{cases}
0 & : n < m \\
n - m & : n \ge m
\end{cases}$
is primitive recursive‎.
== Proof ==
We see that:
:$n \mathop {\dot -} \paren {m + 1} = \begin{cases}
0 & : n \mathop {\dot -} m = 0 \\
\paren {n \mathop {\dot -} m} - 1 & : n \mathop {\dot -} m > 0
\end{cases}$
Hence we can define cut-off subtraction as:
:$n \mathop {\dot -} m = \begin{cases}
n & : m = 0 \\
\operatorname{pred} \paren {n \mathop {\dot -} \paren {m - 1} } & : m > 0
\end{cases}$
This is a definition by primitive recursion from the primitive recursive function $\operatorname{pred}$.
Hence the result.
== Also see ==
* Natural Number Subtraction is not Closed
== Motivation ==
The usual subtraction operation is not defined on $\N^2$ for all pairs $\tuple {n, m}$.
If $m > n$, then $n - m$, although well-defined for the integers $\Z$, has no definition in the set of natural numbers $\N$ which go no lower than $0$.
Hence the need to define this hybrid operation.
Category:Primitive Recursive Functions
Cut-Vertex divides Graph into Two or More Components

== Theorem ==
Let $G$ be a graph.
Let $v$ be a cut-vertex of $G$.
Then the vertex deletion $G - v$ contains $2$ or more components.
== Proof ==
By definition of cut-vertex, $G - v$ contains at least $2$ components.
That it can contain more components than $2$ is best proved by illustration:
:
== Sources ==
* : $\S 2.4$: Cut-Vertices and Bridges
Category:Graph Theory
Cut Associated with 1 is Identity for Multiplication of Cuts

== Theorem ==
Let $\alpha$ be a cut.
Let $1^*$ denote the rational cut (rational) number $1$.
Then:
:$\alpha 1^* = \alpha$
where $\alpha 1^*$ denote the product of $\alpha$ and $1^*$.
== Proof ==
By definition, we have that:
:$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha < 0^*, \beta < 0^*
\end {cases}$
where:
:$\size \alpha$ denotes the absolute value of $\alpha$
:$\size \alpha \, \size \beta$ is defined as in Multiplication of Positive Cuts
:$0^*$ denotes the rational cut associated with the (rational) number $0$.
:$\ge$ denotes the ordering on cuts.
We have that:
:$0 < \dfrac 1 2 < 1$
and so:
:$\dfrac 1 2 \notin 0^*$
but:
:$\dfrac 1 2 \in 1^*$
Thus by definition of the strict ordering of cuts:
:$1^* > 0^*$
Thus by definition of absolute value of $\alpha$:
:$\size {1^*} = 1^*$
Thus:
:$\alpha 1^* := \begin {cases}
\size \alpha \, 1^* & : \alpha \ge 0^* \\
-\paren {\size \alpha \, 1^*} & : \alpha < 0^*
\end {cases}$
Let $\alpha > 0^*$.
Let $r \in \alpha 1^*$.
Then by definition of Multiplication of Positive Cuts, either:
:$r < 0$
or
:$\exists p \in \beta, q \in 1^*: r = p q$
where $p \ge 0$ and $q \ge 0$.
Because $0 \le q < 1$ it follows that $p q < p$.
Thus if $r > 0$ it follows that $r < p$.
That is:
:$r \in \alpha$
If $r < 0$ then $r \in \alpha$ because $0^* < \alpha$.
Thus in either case $r \in \alpha$.
That is:
:$\alpha 1^* \le \alpha$
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.26$. Theorem $\text {(f)}$
Category:Cuts
Category:Multiplication
Cycle Decomposition/Examples/Permutation in S7

== Example of Cycle Decomposition ==
Consider the permutation given in two-row notation as:
:$\rho = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 5 & 6 & 1 & 7 & 3 & 2 & 4 \end{pmatrix}$
The cycle decomposition for $\rho$ is:
:$\begin{pmatrix} 1 & 5 & 3 \end{pmatrix} \begin{pmatrix} 2 & 6 \end{pmatrix} \begin{pmatrix} 4 & 7 \end{pmatrix}$
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 80$
Category:Examples of Cycle Decompositions
Cycle Decomposition/Examples/Permutation in S9

== Example of Cycle Decomposition ==
Consider the permutation given in two-row notation as:
:$\rho = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 1 & 4 & 6 & 2 & 8 & 9 & 7 & 5 & 3 \end{pmatrix}$
The cycle decomposition for $\rho$ is:
:$\begin{pmatrix} 1 \end{pmatrix} \begin{pmatrix} 2 & 4 \end{pmatrix} \begin{pmatrix} 3 & 6 & 9 \end{pmatrix} \begin{pmatrix} 5 & 8 \end{pmatrix} \begin{pmatrix} 7 \end{pmatrix}$
or, omitting the $1$-cycles:
:$\begin{pmatrix} 2 & 4 \end{pmatrix} \begin{pmatrix} 3 & 6 & 9 \end{pmatrix} \begin{pmatrix} 5 & 8 \end{pmatrix}$
== Sources ==
* : Chapter $9$: Permutations: Example $9.6$
Category:Examples of Cycle Decompositions
Cycle Decomposition/Examples/Permutation in S9/Mistake

== Source Work ==
:
:Chapter $9$: Permutations:
::Example $9.6$
== Mistake ==
:The cycle decomposition for the permutation
::$\pi = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 1 & 4 & 6 & 2 & 8 & 9 & 7 & 5 & 3 \end{pmatrix}$
:is $\begin{pmatrix} 1 \end{pmatrix} \begin{pmatrix} 2 & 4 \end{pmatrix} \begin{pmatrix} 3 & 6 & 9 \end{pmatrix} \begin{pmatrix} 5 & 8 \end{pmatrix}$. It is usual to omit the cycles of length $1$, those integers fixed by $\pi$, and so $\pi$ is abbreviated to $\begin{pmatrix} 2 & 4 \end{pmatrix} \begin{pmatrix} 3 & 6 & 9 \end{pmatrix} \begin{pmatrix} 5 & 8 \end{pmatrix}$.
== Correction ==
The cycle decomposition given above includes the $\begin{pmatrix} 1 \end{pmatrix}$, but for some reason omits the $\begin{pmatrix} 7 \end{pmatrix}$.
If you are going to include the one, then you also need to include the other.
== Sources ==
* : Chapter $9$: Permutations: Example $9.6$
Category:Mistakes/A Course in Group Theory
Cycle Decomposition/Examples/Product of Permutations in S9

== Example of Product of Cycle Decompositions ==
Consider the product permutations:
:$\rho = \begin{pmatrix} 1 & 2 & 4 \end{pmatrix} \begin{pmatrix} 3 & 5 & 7 & 9 \end{pmatrix} \begin{pmatrix} 1 & 3 & 9 \end{pmatrix} \begin{pmatrix} 2 & 3 & 4 & 5 & 6 & 8 \end{pmatrix}$
Then $\rho$ evaluates to:
:$\begin{pmatrix} 1 & 5 & 6 & 8 & 4 & 7 & 9 & 2 & 3 \end{pmatrix}$
== Proof ==
We start with $1$.
We start at the right hand end of the product and scan leftwards, looking for the first $1$.
We find it in $\begin{pmatrix} 1 & 3 & 9 \end{pmatrix}$, and see that it is followed by $3$.
We continue to scan leftwards, looking for the next occurrence of $3$, and see it in $\begin{pmatrix} 3 & 5 & 7 & 9 \end{pmatrix}$, where it is followed by $5$.
We continue to scan leftwards, looking for the next occurrence of $5$, and see that there are no more.
Thus we have that:
:$\map \rho 1 = 5$
We then start at the right hand end of the product and scan leftwards, looking for the first $5$.
We find it in $\begin{pmatrix} 2 & 3 & 4 & 5 & 6 & 8 \end{pmatrix}$, followed by $6$
There are no further occurrences of $6$.
Continuing this procedure, starting with $6$, we see that leads to $8$.
Starting with $8$, we see that this is at the end of its cycle, so by definition of cycle notation, we see $8$ is followed by $2$.
The next occurrence of $2$ to the left of this is found in $\begin{pmatrix} 1 & 2 & 4 \end{pmatrix}$, and is followed by $4$.
Similarly:
:$4$ leads to $5$ which leads to $7$
:$7$ leads to $9$
:$9$ leads to $1$ which leads to $2$
:$2$ leads to $3$ which leads to $9$ which leads back to $3$
:$3$ leads to $4$ which leads to $1$, and we are back to the start.
Thus we have the cycle:
:$\begin{pmatrix} 1 & 5 & 6 & 8 & 4 & 7 & 9 & 2 & 3 \end{pmatrix}$
which takes on the complete set of elements of $S_9$.
== Sources ==
* : Chapter $9$: Permutations: Example $9.7$
Category:Examples of Cycle Decompositions
Cycle Decomposition of Conjugate

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\pi, \rho \in S_n$.
The cycle decomposition of the permutation $\pi \rho \pi^{-1}$ can be obtained from that of $\rho$ by replacing each $i$ in the cycle decomposition of $\rho$ with $\map \pi i$.
== Proof ==
Consider the effect of $\pi \rho \pi^{-1}$ on $\map \pi i$:
:$\map {\pi \rho \pi^{-1} } {\map \pi i} = \map \pi {\map \rho i}$
That is:
:$\pi \rho \pi^{-1}$ maps $\map \pi i$ to $\map \pi {\map \rho i}$
In the cycle decomposition of $\pi \rho \pi^{-1}$, $\map \pi i$ lies to the left of $\map \pi {\map \rho i}$, whereas in the cycle decomposition of $\rho$, $i$ lies to the left of $\map \rho i$.
== Sources ==
* : Appendix: Elementary set and number theory
* : Chapter $9$: Permutations: Proposition $9.20$
Category:Symmetric Groups
Category:Conjugacy
Category:Permutations
Cycle Graph is 2-Regular

== Theorem ==
Let $G$ be a cycle graph.
Then $G$ is regular.
== Proof ==
Let $G$ be a cycle graph.
By definition, a cycle graph is a graph which consists of a single cycle $C$.
By definition, a cycle is a circuit in which no vertex except the first (which is also the last) appears more than once.
By definition, a circuit is a closed trail with at least one edge.
By definition, a trail is a walk in which all edges are distinct.
Let $v$ be a vertex of $G$.
Then $v$ is a vertex of a closed walk.
Hence for every edge which is incident with $v$, there exists another edge which is also incident with $v$.
Suppose $v$ were odd.
Then at least one edge must appear at least twice in $C$.
But then $C$ would not be a trail.
Thus $v$ is even.
Suppose $v$ has degree zero.
Then there are no edges incident with $v$.
Hence $v$ is not on a closed walk.
Hence the degree of $v$ is greater than $0$.
Suppose $v$ has degree greater than $2$.
Then $v$ must be on a circuit in which $v$ appears more than once.
Hence that circuit is not a cycle
Hence $G$ is not a cycle graph.
Hence $v$ is of degree $2$.
The result follows.
Category:Cycle Graphs
Category:Regular Graphs
Cycle Graph is Bipartite iff Order is Even

== Theorem ==
Let $n \in \N$ be a natural number.
Let $C_n$ be the cycle graph of order $n$.
Then $C_n$ is a bipartite graph $n$ is even.
== Proof ==
Let $V$ be the set of vertices of $C_n$.
Let the elements of $V$ be denoted $v_1, v_2, \ldots, v_n$.
Then by definition of cycle graph, by appropriate selection of subscripts, $C_n$ consists of one cycle $C$ that can be expressed as:
:$C := \tuple {v_1 v_2 \ldots, v_n v_1}$
Let $v_k \in C_n$.
Then for $1 < k < n$, $v_k$ is adjacent to $v_{k - 1}$ and $v_{k + 1}$ and no other vertices of $C_n$.
We also have that:
:$v_1$ is adjacent to $v_2$ and $v_n$
and by the same coin:
:$v_n$ is adjacent to $v_{n - 1}$ and $v_1$.
=== Necessary Condition ===
Let $n$ be even.
Then we can partition $V$ into $A$ and $B$ such that:
:$A = \set {V_k: k = 2 r: r \in \N}$
:$B = \set {V_k: k = 2 r + 1: r \in \N}$
That is:
:$A$ contains the vertices with even subscripts
:$B$ contains the vertices with odd subscripts.
By construction, every edge of $C_n$ is incident to one vertices with even subscript and one vertices with odd subscript.
This also applies to the edge $v_1 v_n$, as $1$ is odd and $n$ is even.
Hence by definition $C_n$ is a bipartite graph.
=== Sufficient Condition ===
Let $C_n$ is a bipartite graph.
Let $V$ be partitioned into $A$ and $B$.
, let $v_1 \in A$.
Then:
:$v_2 \in B$
:$v_3 \in A$
and so on, such that:
:$v_k \in A \implies k = 2 r + 1$ for some $r \in \N$
:$v_k \in B \implies k = 2 r$ for some $r \in \N$
That is:
:$A$ contains all vertices with odd subscript
and:
:$B$ contains all vertices with even subscript.
 $n$ is odd.
For $1 < k < n$, each pair of adjacent vertices is joined by an edge which is incident to one vertices with even subscript and one vertices with odd subscript.
But the edge $v_1 v_n$ is incident to vertices which both have odd subscripts.
Hence $C_n$ is not bipartite.
It follows by Proof by Contradiction that $n$ is even.
== Examples ==
Category:Cycle Graphs
Category:Bipartite Graphs
Category:Cycle Graph is Bipartite iff Order is Even
Cycle Graph is Bipartite iff Order is Even/Examples/C 4

== Example of Use of Cycle Graph is Bipartite iff Order is Even ==
The cycle graph of order $4$ is the complete bipartite graph $K_{2, 2}$.
== Proof ==
From Cycle Graph is Bipartite iff Order is Even, $C_4$ is a bipartite graph, consisting of the cycle $v_1 v_2 v_3 v_4 v_1$.
We have that $C_4$ is partitioned into two sets $A$ and $B$ such that:
:$A = \set {v_1, v_3}$
:$B = \set {v_2, v_4}$
and it is seen by inspection that every element of $A$ is adjacent to every element of $B$.
The result follows by definition of complete bipartite graph.
Category:Cycle Graph is Bipartite iff Order is Even
Category:Examples of Complete Bipartite Graphs
Category:Examples of Bipartite Graphs
Cycle Graph is Eulerian

== Theorem ==
Let $G$ be a cycle graph.
Then $G$ is Eulerian.
== Proof ==
From Cycle Graph is Connected, $G$ is a connected graph.
From Cycle Graph is $2$-Regular, $G$ is $2$-regular.
It follows directly from Characteristics of Eulerian Graph that $G$ is Eulerian.
Category:Cycle Graphs
Category:Eulerian Graphs
Cycle Graph is Hamiltonian

== Theorem ==
Let $n \in \Z$ be an integer such that $n \ge 3$.
Let $C_n$ denote the cycle graph of order $n$.
Then $C_n$ is Hamiltonian.
== Proof ==
Recall the definition of a Hamiltonian graph:
:$G$ is Hamiltonian it is an undirected graph that contains a cycle that contains every vertex of $G$ (but not necessarily every edge of $G$).
A cycle graph is a graph which consists of a single cycle.
So all the vertices of $C_n$ are on the same cycle.
The result follows.
Category:Cycle Graphs
Category:Hamiltonian Graphs
Cycle Graph of Order 1 is Loop-Graph

== Theorem ==
Let $C_1$ denote the cycle graph of order $1$.
Then $C_1$ is a loop-graph.
== Proof ==
By definition, the vertex set of $C_1$ is singleton, $\set v$, say.
The only vertex of $C_1$ that an edge can be incident to is $v$.
Hence there exists an edge which is incident to $v$ at both ends.
That is, $C_1$ has a loop.
Hence the result by definition of loop-graph.
Category:Examples of Cycle Graphs
Category:Examples of Loop-Graphs
Cycle Graph of Order 2 is Multigraph

== Theorem ==
Let $C_2$ denote the cycle graph of order $2$.
Then $C_2$ is a multigraph.
== Proof ==
By definition, the vertex set of $C_2$ is doubleton, $\set {v_1, v_2}$, say.
By definition of cycle graph, there is a circuit $v_1 v_2 v_1$.
That is:
:there exists an edge which is incident to $v_1$ and $v_2$
:there exists an edge which is incident to $v_2$ and $v_1$
That is, there are $2$ edges which are both incident to $v_1$ and $v_2$
Hence the result by definition of multigraph.
Category:Examples of Cycle Graphs
Category:Examples of Multigraphs
Cycle Graph of Order 3 is Complete Graph

== Theorem ==
Let $C_3$ denote the cycle graph of order $2$.
Then $C_3$ is the complete graph of of order $3$.
== Proof ==
Let the vertex set of $C_3$ is $\set {v_1, v_2, v_3}$.
By definition of cycle graph, $C_3$ consists of the cycle $v_1 v_2 v_3 v_1$.
It is seen by inspection that:
:$v_1$ is adjacent to $v_2$ and $v_3$
:$v_2$ is adjacent to $v_1$ and $v_3$
:$v_3$ is adjacent to $v_1$ and $v_2$.
Hence the result by definition of complete graph.
Category:Examples of Cycle Graphs
Category:Examples of Complete Graphs