hcju/proofwikiqa huggingface dataset:
Let us introduce the variable $z = \dfrac {dy}{dx}$. Then the initial ODE can be written: :$\begin{cases} \dfrac {dy}{dx} = z & : y \left({x_0}\right) = y_0 \\ & \\ \dfrac {dz}{dx} = - P \left({x}\right) z - Q \left({x}\right) y + R \left({x}\right) & : z \left({x_0}\right) = y_0' \end{cases}$ The converse is also true. The result follows from Existence of Solution to System of First Order ODEs. {{qed}} Category:Differential Equations 6vhz0eqftyu4brtbvbujanu266sw2gu
Let the cartesian metric be $\delta _{ij} = \langle e_i, e_j \rangle$. Also, let $\delta_{ij}^\prime$ be the metric of the coordinate system of $\delta _{ij}$ rotated by a rotation matrix $A$. Then, $\delta_{ij}^\prime = \langle A e_i, A e_j \rangle$. $\langle A x, y \rangle = \langle x, A^T y \rangle$, see Factor Matrix in the Inner Product, so $\delta_{ij}^\prime = \langle A e_i, A e_j \rangle = \langle e_i, A^T A e_j \rangle$. For rotation matrices, we have $A^T = A^{-1}$, so $\delta_{ij}^\prime$ reduces to: :$\delta_{ij}^\prime = \langle e_i, A^T A e_j \rangle = \langle e_i, A^{-1} A e_j \rangle = \langle e_i, I e_j \rangle = \langle e_i, e_j \rangle$ where $I$ is the unit matrix. Thus: : $\delta_{ij}^\prime = \delta_{ij}$ The result follows. {{Qed}} {{MissingLinks}} === Notes === This proof holds for any finite-dimensional Euclidian space. Category:Metric Spaces ij9y4gbjuqcnvd8nwr5rll8neaxvlqd
By the process of Polynomial Long Division, we can express $\map {P_n} x$ as: :$(1): \quad \map {P_n} x = \paren {x - a} \map {Q_{n - 1} } x + R$ where: :$\map {Q_{n - 1} } x$ is a polynomial in $x$ of degree $n - 1$ :$R$ is a polynomial in $x$ of degree no greater than $0$; that is, a constant. It follows that, by setting $x = a$ in $(1)$, we get $\map {P_n} a = R$. Hence the result. {{qed}}
:400px Let $O$ be the center of the circle, and define the vectors: :$\mathbf u = \overrightarrow{OC}$ :$\mathbf v = \overrightarrow{OB}$ :$\mathbf w = \overrightarrow{OA}$ Thus: :$\overrightarrow{AC} = \mathbf u - \mathbf w$ :$\overrightarrow{BC} = \mathbf u - \mathbf v$ Then: {{begin-eqn}} {{eqn | o =       | r = \overrightarrow{AC} \cdot \overrightarrow{BC}       | c = where $\cdot$ is the dot product }} {{eqn | r = \paren {\mathbf u - \mathbf w} \cdot \paren {\mathbf u - \mathbf v}       | c = from above }} {{eqn | r = \paren {\mathbf u + \mathbf v} \cdot \paren {\mathbf u - \mathbf v}       | c = Since $A$ is directly opposite $B$ in the circle, $\mathbf w = - \mathbf v$ }} {{eqn | r = \mathbf u \cdot \mathbf u - \mathbf u \cdot \mathbf v + \mathbf v \cdot \mathbf u - \mathbf v \cdot \mathbf v       | c = Dot Product Distributes over Addition }} {{eqn | r = \mathbf u \cdot \mathbf u - \mathbf u \cdot \mathbf v + \mathbf u \cdot \mathbf v - \mathbf v \cdot \mathbf v       | c = Dot Product Operator is Commutative }} {{eqn | r = \norm {\mathbf u}^2 - \mathbf u \cdot \mathbf v + \mathbf u \cdot \mathbf v - \norm {\mathbf v}^2       | c = Dot Product of Vector with Itself }} {{eqn | r = \norm {\mathbf u}^2 - \norm {\mathbf v}^2       | c = simplification }} {{eqn | r = \norm {\mathbf u}^2 - \norm {\mathbf u}^2       | c = as $\norm {\mathbf u} = \norm {\mathbf v}$ }} {{eqn | r = 0       | c =  }} {{end-eqn}} From Non-Zero Vectors Orthogonal iff Perpendicular, it follows that $AC$ and $BC$ are perpendicular to each other. {{Qed}}
Let $O$ be the center of $ACB$. From the Inscribed Angle Theorem, $\angle AOB = 2 \angle ACB$. Then we have that $\angle AOB$ is a straight angle. Hence the result. {{qed}}
Let $O$ be the center of $ACB$. We have that $AO$, $BO$ and $CO$ are radii of the same circle. Thus by definition: :$AO = BO = CO$ Thus $\triangle AOC$ and $\triangle BOC$ are isosceles triangles. From External Angle of Triangle equals Sum of other Internal Angles: :$\angle COB = \angle ACO + \angle CAO$ Hence: {{begin-eqn}} {{eqn | l = \angle ACO       | r = \angle CAO       | c = Isosceles Triangle has Two Equal Angles }} {{eqn | ll= \leadsto       | l = \angle COB       | r = 2 \angle ACO       | c =  }} {{eqn | ll= \leadsto       | l = \frac 1 2 \angle COB       | r = \angle ACO       | c =  }} {{end-eqn}} Similarly, from External Angle of Triangle equals Sum of other Internal Angles: :$\angle COA = \angle BCO + \angle CBO$ Hence: {{begin-eqn}} {{eqn | l = \angle BCO       | r = \angle CBO       | c = Isosceles Triangle has Two Equal Angles }} {{eqn | ll= \leadsto       | l = \angle COA       | r = 2 \angle BCO       | c =  }} {{eqn | ll= \leadsto       | l = \frac 1 2 \angle COA       | r = \angle BCO       | c =  }} {{end-eqn}} Therefore: {{begin-eqn}} {{eqn | l = \angle ACO + \angle BCO       | r = \frac 1 2 \angle COB + \frac 1 2 \angle COA       | c =  }} {{eqn | ll= \leadsto       | l = \angle ACB       | r = \frac 1 2 \left({\angle COB + \angle COA}\right)       | c =  }} {{end-eqn}} But $\angle COB + \angle COA$ equals two right angles. Hence $\angle ACB$ is a right angle. {{qed}}
File:Curvature.png Let $X$ and $Y$ be two separate faces of a polyhedron separated by the edge $l$. Let $P$ be a point on $X$ and let $Q$ be a point on $Y$. The curvature inside an infinitesimal region $\delta a$ is given by the net angular displacement $\delta\theta$ a vector $v$ experiences as it is parallel transported along a closed path around $\delta a$. The curvature is then given by: :$R = \dfrac {\delta \theta} {\delta a}$ We must then prove that the vector $v$ experiences no net angular displacement as it is parallel transported from $P$ to $Q$ and back to $P$. The two open curves $r$ and $s$ make a closed curve. As the vector is parallel transported along the open curve $r$, it crosses the edge between the two faces $X$ and $Y$. In doing so, it gains a finite angular displacement $\delta\theta_1$. Then, when the vector is transported back along the open curve $s$, it gains another angular displacement $\delta\theta_2$. Notice that because it is not being transported the other way (from $Y$ to $X$), the new angular displacement will be: :$\delta\theta_2 = -\delta\theta_1$. The curvature inside the region $\delta a$ is therefore given by: {{begin-eqn}} {{eqn | l = R       | r = \frac {\delta \theta_1 + \delta \theta_2} {\delta a}       | c =  }} {{eqn | r = \frac {\delta \theta_1 - \delta \theta_1} {\delta a}       | c =  }} {{eqn | r = \frac 0 {\delta a}       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} The result follows. {{qed}} Category:Differential Geometry ceimnk356hrgednbl236ekkgzuqbxx2
Let $\mathbf u, \mathbf v \in \R^n$. We will check the four defining properties of an inner product in turn. === Conjugate Symmetry === {{begin-eqn}} {{eqn | l = \mathbf u \cdot \mathbf v       | r = \mathbf v \cdot \mathbf u       | c = Dot Product Operator is Commutative }} {{eqn | r = \overline{\mathbf v \cdot \mathbf u}       | c = Complex Number equals Conjugate iff Wholly Real }} {{end-eqn}} Thus the dot product possesses conjugate symmetry. {{qed|lemma}} === Bilinearity === From Dot Product Operator is Bilinear, the dot product possesses bilinearity. {{qed|lemma}} === Non-Negative Definiteness === From Dot Product with Self is Non-Negative, the dot product possesses non-negative definiteness. {{qed|lemma}} === Positiveness === From Dot Product with Self is Zero iff Zero Vector, the dot product possesses positiveness. {{qed|lemma}} Hence the dot product satisfies the definition of an inner product. {{Qed}} Category:Dot Product bptif9dwnicrdc3izmxfl5ljtv2g5fq
{{begin-eqn}} {{eqn|l=\left \langle {A \mathbf u,\mathbf v} \right \rangle      |r=\left({A \mathbf u}\right)^T \mathbf v      |c=Definition of Dot Product }} {{eqn|r=\mathbf u^T A^T \mathbf v      |c=Transpose of Matrix Product }} {{eqn|r=\left \langle {\mathbf u, A^T \mathbf v} \right \rangle      |c=Definition of Dot Product }} {{end-eqn}} {{Qed}} Category:Vector Algebra rrqdb9x5aqek4w4jjl5s25b9pbnyhsd
From Intersection with Empty Set, we have that, for all sets, $S \cap \varnothing = \varnothing$. The result follows from the definition of disjoint sets. {{qed}} Category:Empty Set stzriqqx98lh9mo6u4remwzdsvx5fi0
For each $n$, let $\mathbf A_n$ be the formula: :$\exists x_1 \exists x_2 \ldots \exists x_n: \left({x_1 \ne x_2 \land x_1 \ne x_3 \land \ldots \land x_{n-1} \ne x_n }\right)$ Then $\mathbf A_i$ is true in a structure $\mathcal A$ {{iff}} $\mathcal A$ has at least $n$ elements. Take: : $\displaystyle \Gamma := A \cup \bigcup_{i \mathop = 1}^\infty A_i$ Since $F$ has models of arbitrarily large size, every finite subset of $\Gamma$ is satisfiable. From the Compactness Theorem, $\Gamma$ is satisfiable in some model $\mathcal M$. But since $\mathcal M \models A_i$ for each $i$, $\mathcal M$ must be infinite. So $A$ has an infinite model. {{qed}}
Let $P$ be the set of axioms of Peano arithmetic. Let $Q = P \cup \left\{{\neg x = 0, \neg x = s0, \neg x = ss0, \ldots}\right\}$ where $x$ is a variable of the language. Then each finite subset of $Q$ is satisfied by the standard model of arithmetic Hence $Q$ is satisfiable by the Compactness theorem. But any model satisfying $Q$ must assign $x$ to an element which cannot be obtained by iterating the successor operator on zero a finite number of times. {{qed}} {{MissingLinks}} {{disambiguate|Definition:Model (Logic)}} Category:Mathematical Logic m3vzgmfsr7waas32t1qlq8ivfn035pc
{{AimForCont}} $\pi$ is rational. Then $\pi = \dfrac p q$ where $p$ and $q$ are integers and $q \ne 0$.  Now let: :$\displaystyle A_n = \frac {q^n} {n!} \int_0^\pi \paren {x \paren {\pi - x} }^n \sin x \rd x $ Integration by Parts twice gives: {{begin-eqn}} {{eqn | l = A_n       | r = \frac {q^n} {n!} \int_0^\pi n \paren {x \paren {\pi - x} }^{n - 1} \paren {\pi - 2 x} \cos x \rd x       | c =  }} {{eqn | r = \frac {q^n} {n!} \int_0^\pi 2 n \paren {x \paren {\pi - x} }^{n - 1} \sin x - n \paren {n - 1} \paren {x \paren {\pi - x} }^{n - 2} \paren {\pi - 2 x}^2 \sin x \rd x       | c =  }} {{end-eqn}} Writing: :$\paren {\pi - 2 x}^2 = \pi^2 - 4 x \paren {\pi - x}$ gives the result: :$(1): \quad A_n = \paren {4 n - 2} q A_{n - 1} - \paren {q \pi}^2 A_{n - 2}$  We will deduce that $A_n$ is an integer for all $n$. First confirm by direct integration that $A_0$ and $A_1$ are integers: :$\displaystyle A_0 = \int_0^\pi \sin x \rd x = 2$ :$\displaystyle A_1 = q \int_0^\pi x \paren {\pi - x} \sin x \rd x = 4 q$ Suppose $A_{k - 2}$ and $A_{k - 1}$ are integers.  Then by $(1)$ and the assumption that $q$ and $q \pi$ are integers, $A_k$ is also an integer.  So $A_n$ is an integer for all $n$ by Second Principle of Mathematical Induction. For $x \in \closedint 0 \pi$, we have: :$0 \le \sin x \le 1$ and: :$0 \le x \paren {\pi - x} \le \pi^2 / 4$ hence: :$0 < A_n < \pi \dfrac {\paren {q \pi^2 / 4}^n} {n!}$ From Power over Factorial: :$\displaystyle \lim_{n \mathop \to \infty} \frac {\paren {q \pi^2 / 4}^n} {n!} = 0$ It follows from the Squeeze Theorem that: :$\displaystyle \lim_{n \mathop \to \infty} A_n = 0$ Hence for sufficiently large $n$, $A_n$ is strictly between $0$ and $1$. This contradicts the supposition that $A_n$ is an integer. It follows that $\pi$ is irrational. {{qed}}
From Rational Points on Graph of Sine Function, the only rational point on the graph of the sine function in the real Cartesian plane $\R^2$: :$f := \left\{ {\left({x, y}\right) \in \R^2: y = \sin x}\right\}$ is the point $\left({0, 0}\right)$. But $\left({\pi, 0}\right)$ is also on $f$. Hence $\pi$ cannot be rational. {{qed}}
The equivalence: :$\displaystyle \sum_{p^k \mathop \le x} \ln p \equiv \sum_{1 \mathop \le n \mathop \le x} \map \Lambda n$ follows directly from the definition of the von Mangoldt function. {{qed|lemma}} Let $N = \floor x$. It can be seen directly that all the above summations are exactly the same whether performed on $N$ or $x$. Hence we need only to prove the equivalence for integral arguments. First we expand the von Mangoldt function: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 1}^N \map \Lambda n       | r = \map \Lambda 1 + \map \Lambda 2 + \cdots + \map \Lambda N       | c =  }} {{eqn | r = 0 + \map \ln 2 + \map \ln 3 + \map \ln 2 + \map \ln 5 + 0 + \map \ln 7 + \map \ln 2 + \map \ln 3 + 0 + \cdots       | c =  }} {{end-eqn}} Notice this sum will have: :as many $\map \ln 2$ terms as there are powers of $2$ less than or equal to $N$ :as many $\map \ln 3$ terms as there are powers of $3$ less than or equal to $N$ and in general, if $p$ is a prime less than $N$, $\ln p$ will occur in this sum $\floor {\log_p N}$ times. Hence: :$\displaystyle \sum_{1 \mathop \le n \mathop \le x} \map \Lambda n \equiv \sum_{p \mathop \le x} \floor {\log_p x} \ln p$ {{qed}} Category:Number Theory 0szraoupncb6ksxnvlrktcrf22sgp7d
=== Closure of Hadamard Product === {{:Closure of Hadamard Product}} === Associativity of Hadamard Product === {{:Associativity of Hadamard Product}} === Commutativity of Hadamard Product === {{:Commutativity of Hadamard Product}} Category:Hadamard Product 9o8hi3biogmwh2i4ezufn49x1df57f7
This follows as $\map {\MM_R} {m, n}$ is a direct instance of the module given in the module of all mappings, where $\map {\MM_R} {m, n}$ is the $R$-module $R^{\closedint 1 m \times \closedint 1 n}$. The $S$ of that example is the set $\closedint 1 m \times \closedint 1 n$, while the $G$ of that example is the $R$-module $R$. {{finish}}
Let $\struct {S_1, \RR_1}$, $\struct {S_2, \RR_2}$ and $\struct {S_3, \RR_3}$ be relational structures. Let $S \cong T$ denote the relation that $S$ is (relation) isomorphic to $T$. Checking in turn each of the criteria for equivalence: === Reflexive === The fact that relation isomorphism is reflexive follows immediately from Identity Mapping is Relation Isomorphism. {{qed|lemma}} === Symmetric === Suppose $S_1 \cong S_2$. Let $\phi: S_1 \to S_2$ be a relation isomorphism from $S_1$ to $S_2$. Then by Inverse of Relation Isomorphism is Relation Isomorphism, $\phi^{-1}: S_2 \to S_1$ is also a relation isomorphism. So $S_2 \cong S_1$ and so relation isomorphism is symmetric. {{qed|lemma}} === Transitive === Suppose $S_1 \cong S_2$ and $S_2 \cong S_3$. Let: :$\alpha: S_1 \to S_2$ be a relation isomorphism from $S_1$ to $S_2$ :$\beta: S_2 \to S_3$ be a relation isomorphism from $S_2$ to $S_3$. From Composite of Relation Isomorphisms is Relation Isomorphism, the composite mapping $\beta \circ \alpha$ is also a relation isomorphism. That is, $S_1 \cong S_3$. So relation isomorphism is transitive. {{qed}} Category:Graph Theory Category:Equivalence Relations Category:Relation Isomorphisms h5so7uycqxoozwxf9fozzvzznvq7dk2
Let $\phi: \map V {G_1} \to \map V {G_2}$ be an isomorphism. Let $u \in \map V {G_1}$ be an arbitrary vertex of $G_1$ such that $\map \phi u = v \in \map V {G_2}$. Let $\map {\deg_{G_1} } u = n$. We need to show that $\map {\deg_{G_2} } v = n$. As $\map {\deg_{G_1} } u = n$, there exist $u_1, u_2, \ldots, u_n \in \map V {G_1}$ which are adjacent to $u$. Every other vertex of $G_1$ is ''not'' adjacent to $u$. Let $\map \phi {u_i} = v_i$ for $1, 2, \ldots, n$. Because $\phi$ is an isomorphism, each of the vertices $v_1, v_2, \ldots, v_n \in \map V {G_2}$ are adjacent to $v$. Similarly, every other vertex of $G_2$ is ''not'' adjacent to $v$. Thus $\map {\deg_{G_2} } v = n$. This applies to all vertices $u \in  \map V {G_1}$. Hence the result. {{qed}}
{{AimForCont}} no subset $S_i$ of $S$ has as many as $\ceiling {\dfrac n k}$ elements. Then the maximum number of elements of any $S_i$ would be $\ceiling {\dfrac n k} - 1$. So the total number of elements of $S$ would be no more than $k \paren {\ceiling {\dfrac n k} - 1} = k \ceiling {\dfrac n k} - k$. There are two cases: :$n$ is divisible by $k$ :$n$ is not divisible by $k$. Suppose $k \divides n$. Then $\ceiling {\dfrac n k} = \dfrac n k$ is an integer and: :$k \ceiling {\dfrac n k} - k = n - k$ Thus: :$\displaystyle \card S = \sum_{i \mathop = 1}^k \card {S_i} \le n - k < n$ This contradicts the fact that $\card S = n$. Hence our assumption that no subset $S_i$ of $S$ has as many as $\ceiling {\dfrac n k}$ elements was false. Next, suppose that $k \nmid n$. Then: :$\card S = k \ceiling {\dfrac n k} - k < \dfrac {k \paren {n + k} } k - k = n$ and again this contradicts the fact that $\card S = n$. In the same way, our assumption that no subset $S_i$ of $S$ has as many as $\ceiling {\dfrac n k}$ elements was false. Hence, by Proof by Contradiction, there has to be at least $\ceiling {\dfrac n k}$ elements in at least one $S_i \subseteq S$. {{qed}}
We have that Graph Connectedness is Equivalence Relation. The result follows directly from the definition of component. {{qed}}
Let $u, v, w$ be arbitrary vertices of a graph $G$. Checking in turn each of the criteria for equivalence: === Reflexive === By definition, all vertices are connected to themselves. Hence $\to$ is reflexive. === Symmetric === Suppose $u \to v$. By definition there exists a walk $W$ between $u$ and $v$. Note that there is nothing in the definition of connectedness to insist that the walk has to be in a particular direction. So by definition, if $u \to v$ then $v \to u$. Hence $\to$ is symmetric. === Transitive === Suppose $u \to v$ and $v \to w$. Let: : $W_1 = \tuple {u, u_1, u_2, \ldots, u_{k - 1}, u_k, v}$ be a walk from $u$ to $v$ : $W_2 = \tuple {v, v_1, v_2, \ldots, v_{k - 1}, v_k, w}$ be a walk from $v$ to $w$. Then $W = \tuple {u, u_1, u_2, \ldots, u_{k - 1}, u_k, v, v_1, v_2, \ldots, v_{k - 1}, v_k, w}$ is a walk from $u$ to $w$. Hence $u \to w$, and so $\to$ is transitive. Thus '''is connected to''' is an equivalence relation. {{qed}}
Let $G - e$ denote the edge deletion of $e$ from $G$. === Necessary Condition === We use a Proof by Contraposition. To that end, suppose $e = u v$ such that $e$ lies on a circuit, say: :$C = \tuple {u, v, w, \ldots, x, u}$ Thus $G - e$ contains a $u-v$ path, that is, $\tuple {v, x, \ldots, w, u}$. So $u$ is connected to $v$ in $G - e$. Now we need to show that $G - e$ is connected. Let $u_1, v_1 \in V$. Since $G$ is connected, there is a $u_1-v_1$ path $P$ in $G$. If $e \notin P$, then $P$ is also a path in $G - e$ and $u_1$ is connected to $v_1$ in $G - e$. On the other hand, suppose $e \notin P$. Then $P$ can be expressed as $\tuple {u_1, \ldots, u, v, \ldots, v_1}$ or $\tuple {u_1, \ldots, v, u, \ldots, v_1}$. In the first case, $u_1$ is connected to $u$ and $v$ is connected to $v_1$ in $G - e$. In the second case, $u_1$ is connected to $v$ and $u$ is connected to $v_1$ in $G - e$. But we have already established that $u$ is connected to $v$ in $G - e$. From Graph Connectedness is Equivalence Relation it follows that $u_1$ is connected to $v_1$. So, if $e$ lies on a circuit, then $G - e$ is connected. Therefore $e$ is not a bridge. From Rule of Transposition it follows that if $e$ is a bridge, it does not lie on a circuit. {{qed|lemma}} === Sufficient Condition === We use a Proof by Contraposition. To that end, suppose $e = u v$ is not a bridge. Then $G - e$ is connected, and thus there is a $u-v$ path $P$ in $G - e$. But $P$ together with $e = u v$ produces a circuit containing $e$. Thus $e$ lies on a circuit of $G$. From Rule of Transposition it follows that if $e$ does not lie on a circuit, it is a bridge. {{qed}}
{{refactor|this can be presented in a neater way}} Let $G$ be a graph. Suppose all the vertices are even, that is, there are no odd vertices. Then $G$ is Eulerian, and the result holds. Similarly, by the same result, if $G$ is Eulerian, it is by definition traversable. So the question of graphs all of whose vertices are even is settled. === Sufficient Condition === Suppose $G$ is a connected graph for which exactly two vertices $u, v$ are odd. Let $G'$ be the graph formed by adding the edge $u v$. Note that if there is already an edge $u v$ in $G$, that will mean there is now more than one edge $u v$ in $G'$. Thus $G'$ will then be a multigraph or loop-multigraph, and $uv$ a multiple edge. Then $G'$ will have all its vertices even, and thus by the above result be Eulerian and by definition traversable. Such an Eulerian circuit that traverses $G'$ can be written, for example, $P' = \tuple {v, w, x, \ldots, t, u, v}$. Let us then remove the final edge $u v$ from $P'$ to get the path $P = \tuple {v, w, x, \ldots, t, u}$. It follows that the path $P = \tuple {v, w, x, \ldots, t, u}$ is a path in $G$ which includes all edges in $G$. Hence $P$ traverses $G$ and so $G$ is traversable. We note that $u$ and $v$ are the odd vertices of $G$. === Necessary Condition === Suppose a graph $G$ is traversable. Then it has a Eulerian trail $P$. If $P$ is a circuit, then $G$ is Eulerian and therefore has all even vertices. Now, suppose $P = \tuple {v, w, x, \ldots, t, u}$ is ''not'' a circuit. Let $G'$ be the graph formed by adding the edge $uv$. Then the path $P' = \tuple {v, w, x, \ldots, t, u, v}$ is an Eulerian circuit and so $G$ is Eulerian. Hence all the vertices of $G'$ are even. So the degrees of vertices $u$ and $v$ in $G$ (and no other) are odd. Again, we note that $u$ and $v$ are the odd vertices of $G$. Hence the result. {{qed}}
From the definition, the complete graph $K_n$ is $n-1$-regular. That is, every vertex of $K_n$ is of degree $n-1$. Suppose $n$ is odd. Then $n-1$ is even, and so $K_n$ is Eulerian. Suppose $n$ is even. Then $n-1$ is odd. Hence for $n \ge 4$, $K_n$ has more than $2$ odd vertices and so can not be traversable, let alone Eulerian. If $n = 2$, then $K_n$ consists solely of two odd vertices (of degree $1$). Hence, by Characteristics of Traversable Graph (or trivially, by inspection), $K_2$ has an Eulerian trail, and so is traversable (although not Eulerian). {{qed}}
We prove this result by induction: For all $n \in \N_{>0}$, let $\map P N$ be the proposition: :if $G$ has $2 n$ odd vertices, it consists entirely of $n$ edge-disjoint trails, each starting and ending on an odd vertex. First note that from the corollary to the Handshake Lemma, no graph can have an odd number of odd vertices. === Basis for the Induction === $\map P 1$ is true, as this is the Characteristics of Traversable Graph. This is our basis for the induction. Note that, from the same result, the trail starts and ends on an odd vertex. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :If $G$ has $2 k$ odd vertices, it consists entirely of $k$ edge-disjoint trails, each starting and ending on an odd vertex. Then we need to show: :If $G$ has $2 \paren {k + 1}$ odd vertices, it consists entirely of $k + 1$ edge-disjoint trails, each starting and ending on an odd vertex. === Induction Step === This is our induction step: Let $G_{k + 1}$ be a graph with $2 \paren {k + 1}$ odd vertices. Let $u, v$ be any pair of odd vertices in $G_{k + 1}$. We construct the graph $G'_{k + 1}$ which consists of $G_{k + 1}$ with the edge $u v$ added. It can be seen that as $u, v$ are now even vertices, $G'_{k + 1}$ has $2 k$ odd vertices. By our induction hypothesis, we know that $G'_{k + 1}$ consists entirely of $k$ edge-disjoint trails. So, let us construct these trails: let them be $T_1, T_2, \ldots, T_k$. The edge $u v$ must be somewhere in one of the above trails. As the $k$ trails are all edge-disjoint, it will be in exactly one. Suppose it is in $T_j$. As $T_j$ must start and end on an odd vertex, $u v$ must be neither at the start nor the end of such a trail. Suppose $T_j = \tuple {t_1, t_2, \ldots, u, v, \ldots, t_r}$. Then it follows that $t_1$ and $t_r$ are odd vertices. Now we remove $u v$ from $G'_{k + 1}$ so as to return to $G_{k + 1}$. We see that $G_{k + 1}$ consists entirely of the edge-disjoint trails: :$T_1, T_2, \ldots, T_{j - 1}$ :$T_{j + 1}, T_{j + 2}, \ldots, T_k$ :The two trails: $\tuple {t_1, t_2, \ldots, u}$ and $\tuple {v, \ldots, t_r}$. That makes $k + 1$ edge-disjoint trails. Also note that as $u$ and $v$ are odd vertices, both of $\tuple {t_1, t_2, \ldots, u}$ and $\tuple {v, \ldots, t_r}$ start and end on odd vertices. So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. {{qed}}