avewright/proofwiki-math huggingface dataset:
Theorem: Inner Product Space is Uniformly Convex

Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space.
Then $V$ is uniformly convex.

Proof:
Let $\norm \cdot$ be the inner product norm for $\struct {V, \innerprod \cdot \cdot}$.
Let $\epsilon > 0$. 
Let $x, y \in V$ be such that: 
:$\norm x = \norm y = 1$
and:
:$\norm {x - y} > \epsilon$
Then from the Parallelogram Law (Inner Product Space), we have:
:$\norm {x + y}^2 + \norm {x - y}^2 = 2 \paren {\norm x^2 + \norm y^2} = 4$
So that: 
:$\norm {x + y}^2 = 4 - \norm {x - y}^2$
Since:
:$\norm {x - y} > \epsilon$
we have:
:$\norm {x + y}^2 < 4 - \epsilon^2$
From positive homogeneity, we have: 
:$\ds \norm {\frac {x + y} 2}^2 < 1 - \frac {\epsilon^2} 4$
We now pick $\delta$ such that: 
:$\ds \paren {1 - \delta}^2 = 1 - \frac {\epsilon^2} 4$
Note that since:
:$\norm {x - y} \le \norm x + \norm y$
we have: 
:$\norm {x - y} \le 2$ 
So if there exists $x, y \in V$ with $\norm x = \norm y = 1$ such that: 
:$\norm {x - y} > \epsilon$
we have $0 < \epsilon < 2$. 
Hence, if $\epsilon \ge 2$, the demand is vacuously true, and we can pick any positive real $\delta > 0$.
Now take $0 < \epsilon < 2$.
We then have: 
:$\ds 0 < 4 - \epsilon^2 < 4$
so that: 
:$\ds 0 < 1 - \frac {\epsilon^2} 4 < 1$
So, we may pick: 
:$\ds \delta = 1 - \sqrt {1 - \frac {\epsilon^2} 4} > 0$
Then, for each $x, y \in V$ such that $\norm x = \norm y = 1$ and $\norm {x - y} > \epsilon$, we have: 
:$\ds \norm {\frac {x + y} 2}^2 < \paren {1 - \delta}^2$
Since $0 < 1 - \delta < 1$, we have: 
:$\ds \norm {\frac {x + y} 2} < 1 - \delta$
Since $\epsilon > 0$ was arbitrary, for each $\epsilon > 0$ we have found $\delta > 0$ such that:
:whenever $x, y \in V$ have $\norm x = \norm y = 1$ and $\norm {x - y} > \epsilon$, we have: 
::$\ds \norm {\frac {x + y} 2} < 1 - \delta$
as required.
So $V$ is uniformly convex.
{{qed}}
Theorem: Schauder Basis is Linearly Independent

Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$.
Let $\set {e_n : n \in \N}$ be a Schauder basis for $X$. 
Then $\set {e_n : n \in \N}$ is linearly independent.

Proof:
Suppose that: 
:$\ds \sum_{k \mathop = 1}^n \alpha_{i_k} e_{i_k} = 0$
for some $n \in \N$, $i_1, \ldots, i_n \in \N$ and $\alpha_{i_1}, \ldots, \alpha_{i_n} \in \Bbb F$.
Define a sequence $\sequence {\alpha_j}_{j \mathop \in \N}$ in $\Bbb F$ by: 
:$\ds \alpha_j = \begin{cases}\alpha_{i_k} & \text { if there exists } k \text { such that } j = i_k \\ 0 & \text { otherwise}\end{cases}$
Then, we have: 
:$\ds \sum_{j \mathop = 1}^\infty \alpha_j e_j = \sum_{k \mathop = 1}^n \alpha_{i_k} e_{i_k} = 0$
From the definition of Schauder basis, we then have: 
:$\alpha_j = 0$ for each $j \in \N$
and in particular: 
:$\alpha_{i_k} = 0$ for each $k$. 
Since the coefficients $\alpha_{i_1}, \ldots, \alpha_{i_n}$ and $n \in \N$ were arbitrary, we have that:
:$\set {e_n : n \in \N}$ is linearly independent.
{{qed}}
Theorem: Inverse of Linear Transformation is Linear Transformation

Let $K$ be a field. 
Let $V$ and $U$ be vector spaces over $K$.
Let $A : V \to U$ be an invertible (in the sense of a mapping) linear transformation with inverse mapping $A^{-1} : U \to V$.
Then $A^{-1}$ is a linear transformation.

Proof:
We aim to show that: 
:$\map {A^{-1} } {\alpha x + \beta y} = \alpha A^{-1} x + \beta A^{-1} y$
for all $x, y \in U$ and $\alpha, \beta \in K$.
Since $A$ is a linear transformation, we have: 
:$\map A {\alpha u + \beta v} = \alpha A u + \beta A v$
for each $u, v \in V$. 
Note that we have $A^{-1} x \in V$ and $A^{-1} y \in V$, so that: 
:$\map A {\alpha A^{-1} x + \beta A^{-1} y} = \alpha x + \beta y$
for each $x, y \in U$ and $\alpha, \beta \in K$.
Composing with $A^{-1}$, we then have: 
:$\map {\paren {A^{-1} \circ A} } {\alpha A^{-1} x + \beta A^{-1} y} = \map {A^{-1} } {\alpha x + \beta y}$
From the definition of an inverse mapping, we therefore have: 
:$\ds \map {A^{-1} } {\alpha x + \beta y} = \alpha A^{-1} x + \beta A^{-1} y$
for all $x, y \in U$ and $\alpha, \beta \in K$.
So:
:$A^{-1}$ is a linear transformation.
{{qed}}
Category:Linear Transformations
e8pf49hdokqslogpcr1xopxe733nh6y
Theorem: Inverse of Linear Operator is Linear Operator

Let $X$ be a vector space.
Let $A : X \to X$ be an invertible (in the sense of a mapping) linear transformation with inverse mapping $A^{-1} : X \to X$.
Then $A^{-1}$ is a linear operator.

Proof:
Applying Inverse of Linear Transformation is Linear Transformation in the case $U = V = X$ we have: 
:$A^{-1}$ is a linear transformation.
Since $A^{-1}$ is a linear transformation $X \to X$, we have: 
:$A^{-1}$ is a linear operator.
{{qed}}
Category:Linear Operators
tu3wscntho696ilz8bjsxkfo5tx6zme
Theorem: Largest Parallelogram Contained in Triangle

Let $T$ be a triangle.
Let $P$ be a parallelogram contained within $T$.
Let $P$ have the largest area possible for the conditions given.
Then:
:$(1): \quad$ One side of $P$ is coincident with part of one side of $T$, and hence two vertices lie on that side of $T$
:$(2): \quad$ The other two vertices of $P$ bisect the other two sides of $T$
:$(3): \quad$ The area of $P$ is equal to half the area of $T$.

Proof:
We will first find the maximum area of $P$ when $(1)$ is satisfied, that is, when $P$ is inscribed in $T$.
Theorem: Relative Algebraic Closure with Algebraically Closed Extension is Algebraic Closure

Let $L$ be an algebraically closed field.
Let $L/K$ be a field extension.
Then the relative algebraic closure of $K$ contained in $L$ is an algebraic closure of $K$.

Proof:
Let $K'$ denote the relative algebraic closure of $K$ contained in $L$.
By definition, $K' = \{\alpha \in L \mid \alpha \text{ is algebraic over } K\}$.
First we show that $K'$ is a field extension of $K$.
By Field is Algebraic over itself, $K \subseteq K'$.
Since $K$ is a field, it and (therefore) $K'$ are nonzero.
Let $\alpha, \beta \in K'$ be arbitrary.
By definition of generated field extension, $\alpha, \beta \in K(\alpha, \beta) \subseteq L$.
By the field axioms, $\alpha - \beta, \alpha\beta^{-1} \in K(\alpha, \beta) \subseteq L$.
By Finitely Generated Algebraic Extension is Finite and Finite Field Extension is Algebraic, we have that $K(\alpha, \beta)/K$ is algebraic.
Therefore both $\alpha - \beta$ and $\alpha\beta^{-1}$ are algebraic over $K$.
So, by definition, $\alpha - \beta, \alpha\beta^{-1} \in K'$.
Using the Subfield Test/Three-Step, we conclude that $K'$ is a subfield of $L$.
So $K'$ is a field containing $K$, ergo $K'/K$ is a field extension.
Furthermore, $K'$ is algebraically closed.
Indeed, let $p \in K'[X]$, where $K'[X]$ is the ring of polynomials with coefficients in $K'$.
Since $K'[X] \subseteq L[X]$ and, by hypothesis, $L$ is algebraically closed, $p$ has a root in $L$.
Let $\alpha \in L$ be a root of $p$.
Consider the field extension $K'(\alpha)/K'$.
By Finitely Generated Algebraic Extension is Finite and Finite Field Extension is Algebraic, $K'(\alpha)/K'$ is algebraic.
By definition of $K'$, the extension $K'/K$ is algebraic.
Therefore, by Transitivity of Algebraic Extensions, $K'(\alpha)/K$ is algebraic.
Thus $\alpha \in K'(\alpha)$ is algebraic over $K$.
By definition again, we conclude $\alpha \in K'$.
Thus $p$ has a root in $K'$.
We have that $K'/K$ is an algebraically closed algebraic field extension.
Therefore, by definition, $K'$ is an algebraic closure of $K$.
Theorem: Modulus of Eigenvalue of Bounded Linear Operator is Bounded Above by Operator Norm

Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $T : X \to X$ be a bounded linear operator.
Let $\lambda$ be a eigenvalue of $T$. 
Then: 
:$\cmod \lambda \le \norm T$
where $\norm T$ denotes the norm of $T$.

Proof:
Since $\lambda$ is an eigenvalue of $T$, there exists $x \ne 0$ such that: 
:$T x = \lambda x$ 
Then we have: 
{{begin-eqn}}
{{eqn	| l = \norm {\map T {\frac x {\norm x} } }
		| r = \norm {\frac 1 {\norm x} \map T x}
		| c = {{Defof|Linear Transformation on Vector Space}}
}}
{{eqn	| r = \frac {\norm {\map T x} } {\norm x}
		| c = using {{NormAxiomVector|2}}
}}
{{eqn	| r = \frac {\cmod {\lambda x} } {\norm x}
}}
{{eqn	| r = \frac {\cmod \lambda \norm x} {\norm x}
		| c = using {{NormAxiomVector|2}}
}}
{{eqn	| r = \cmod \lambda
}}
{{end-eqn}}
while: 
:$\ds \norm {\frac x {\norm x} } = 1$
So we have: 
:$\cmod \lambda \in \set {\norm {T x} : \norm x = 1}$
So, from the definition of supremum and of the norm of a bounded linear transformation, we have:
:$\cmod \lambda \le \norm T$
{{qed}}
Theorem: Divisibility by 37

Let $n$ be an integer which has at least $3$ digits when expressed in decimal notation.
Let the digits of $n$ be divided into groups of $3$, counting from the right, and those groups added.
Then the result is equal to a multiple of $37$ {{iff}} $n$ is divisible by $37$.

Proof:
Write $n = \ds \sum_{i \mathop = 0}^k a_i 10^{3 i}$, where $0 \le a_i < 1000$.
This divides the digits of $n$ into groups of $3$.
Then the statement is equivalent to:
:$37 \divides n \iff 37 \divides \ds \sum_{i \mathop = 0}^k a_i$
Note that $1000 = 37 \times 27 + 1 \equiv 1 \pmod {37}$.
Hence:
{{begin-eqn}}
{{eqn | l = n
      | r = \sum_{i \mathop = 0}^k a_i 10^{3 i}
}}
{{eqn | r = \sum_{i \mathop = 0}^k a_i 1000^i
}}
{{eqn | o = \equiv
      | r = \sum_{i \mathop = 0}^k a_i 1^i
      | rr= \pmod {37}
      | c = Congruence of Powers
}}
{{eqn | o = \equiv
      | r = \sum_{i \mathop = 0}^k a_i
      | rr= \pmod {37}
}}
{{end-eqn}}
which proves our statement.
{{qed}}
Category:Divisibility Tests
glizwxt3apemxacs9ixsohipounxvar
Theorem: Inscribed Squares in Right-Angled Triangle/Side Lengths/Side Lies on Hypotenuse

:$l = \dfrac {a b c} {a b + c^2}$

Proof:
:400px
In the figure above, let $BC = a$ and $AC = b$.
$CD$ is drawn such that $AB \perp CD$.
Since $CD$ is the height of $\triangle ABC$:
:$CD = \dfrac {a b} c$
Note that $FH \parallel AB$.
Therefore $\triangle CFH \sim \triangle CAB$ by Equiangular Triangles are Similar.
Thus:
{{begin-eqn}}
{{eqn | l = \frac {CG} {CD}
      | r = \frac {FH} {AB}
      | c = {{Defof|Similar Triangles}}
}}
{{eqn | l = \frac {\frac {a b} c - l} {\frac {a b} c}
      | r = \frac l c
}}
{{eqn | l = \frac {a b - c l} {a b}
      | r = \frac l c
}}
{{eqn | l = a b c - c^2 l
      | r = a b l
}}
{{eqn | l = a b c 
      | r = a b l + c^2 l
}}
{{eqn | l = l
      | r = \frac {a b c} {a b + c^2}
}}
{{end-eqn}}
{{qed}}
Category:Inscribed Squares in Right-Angled Triangle
16ouz2qiiwpp8cl4cg73wp1pbztezzy
Theorem: Inscribed Squares in Right-Angled Triangle/Side Lengths/Shared Right Angle

:$l = \dfrac {a b} {a + b}$

Proof:
:200px
In the figure above, let $BC = a$ and $AC = b$.
Note that $DE \parallel CF$.
Therefore $\triangle BDE \sim \triangle BCA$ by Equiangular Triangles are Similar.
Thus:
{{begin-eqn}}
{{eqn | l = \frac {BD} {DE}
      | r = \frac {BC} {CA}
      | c = {{Defof|Similar Triangles}}
}}
{{eqn | l = \frac {a - l} l
      | r = \frac a b
}}
{{eqn | l = b \paren {a - l}
      | r = a l
}}
{{eqn | l = b a
      | r = a l + b l
}}
{{eqn | l = l
      | r = \frac {a b} {a + b}
}}
{{end-eqn}}
{{qed}}
Theorem: Inscribed Squares in Right-Angled Triangle

For any right-angled triangle, two squares can be inscribed inside it.
One square would share a vertex with the right-angled vertex of the right-angled triangle:
:200px
The other square would have a side lying on the hypotenuse of the right-angled triangle:
::400px

Proof:
By definition of inscribed polygon, all four vertices of the inscribed square lies on the sides of the right-angled triangle.
By Pigeonhole Principle, at least two of the vertices must lie on the same side of the right-angled triangle.
The case where this side is the hypotenuse would be the second case above.
For the case where this side is not the hypotenuse, if none of the vertices coincide with the right angle of the right-angled triangle, the angle formed by the three vertices not on the hypotenuse would be obtuse, which is not possible since a square contains no obtuse angles.
Therefore a vertex must coincide with the right angle of the right-angled triangle, which is the first case above.
{{qed}}
Category:Inscribed Squares in Right-Angled Triangle
Category:Squares
Category:Right Triangles
2sm7qs8acbj47wv0mrpbibo44lgbun9
Theorem: Inverse of Linear Isometry is Linear Isometry

Let $\struct {X, \norm \cdot_X}$ and $\struct {Y, \norm \cdot_Y}$ be normed vector spaces.
Let $T : X \to Y$ be an invertible (in the sense of a mapping) linear isometry with inverse $T^{-1} : Y \to X$. 
Then $T^{-1}$ is a linear isometry.

Proof:
From Inverse of Linear Transformation is Linear Transformation, we have: 
:$T^{-1}$ is a linear transformation.
Since $T$ is a linear isometry, we have: 
:$\norm {T x}_Y = \norm x_X$
for each $x \in X$.
Note that for each $y \in Y$, we have $T^{-1} y \in X$. 
We then have: 
{{begin-eqn}}
{{eqn	| l = \norm {T^{-1} y}_X
		| r = \norm {\map T {T^{-1} y} }_Y
}}
{{eqn	| r = \norm {\map {\paren {T \circ T^{-1} } } y}_Y
}}
{{eqn	| r = \norm y_Y
		| c = {{Defof|Inverse Mapping}}
}}
{{end-eqn}}
So $T$ is a linear transformation with: 
:$\norm {T^{-1} y}_X = \norm y_Y$ 
for each $y \in Y$.
So $T^{-1}$ is a linear isometry.
{{qed}}
Category:Linear Isometries
9tt9nsd4wbahi5tu6amk93267fnvvgg
Theorem: Greatest Area of Quadrilateral with Sides in Arithmetic Sequence

Let $Q$ be a quadrilateral whose sides $a$, $b$, $c$ and $d$ are in arithmetic sequence.
Let $\AA$ be the area of $Q$.
Let $Q$ be such that $\AA$ is the greatest area possible for one with sides $a$, $b$, $c$ and $d$.
Then:
:$\AA = \sqrt {a b c d}$

Proof:
We are given that $\AA$ is the greatest possible for a quadrilateral whose sides are $a$, $b$, $c$ and $d$.
From Area of Quadrilateral with Given Sides is Greatest when Quadrilateral is Cyclic, $Q$ is cyclic.
Hence $\AA$ can be found using Brahmagupta's Formula.
Let $s$ denote the semiperimeter of $Q$:
:$s = \dfrac {a + b + c + d} 2$
We are given that $a$, $b$, $c$ and $d$ are in arithmetic sequence.
{{WLOG}}, that means there exists $k$ such that:
{{begin-eqn}}
{{eqn | l = b
      | r = a + k
}}
{{eqn | l = c
      | r = a + 2 k
}}
{{eqn | l = d
      | r = a + 3 k
}}
{{end-eqn}}
where $k$ is the common difference.
Then:
{{begin-eqn}}
{{eqn | l = s
      | r = \dfrac {a + b + c + d} 2
      | c = {{Defof|Semiperimeter}}
}}
{{eqn | r = \dfrac {a + \paren {a + k} + \paren {a + 2 k} + \paren {a + 3 k} } 2
      | c = 
}}
{{eqn | r = \dfrac {4 a + 6 k} 2
      | c = 
}}
{{eqn | n = 1
      | r = 2 a + 3 k
      | c = 
}}
{{end-eqn}}
and so:
{{begin-eqn}}
{{eqn | l = \AA
      | r = \sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} }
      | c = Brahmagupta's Formula
}}
{{eqn | r = \sqrt {\paren {a + 3 k} \times \paren {a + 2 k} \times \paren {a + k} \times a}
      | c = substituting $s = 2 a + 3 k$ from $(1)$ and simplifying
}}
{{eqn | r = \sqrt {a b c d}
      | c = from above
}}
{{end-eqn}}
{{qed}}
Category:Cyclic Quadrilaterals
4a0jhwdmgckjonkd3fw9ji505wn0zkw
Theorem: Thomsen Graph is Toroidal

The Thomsen graph, that is, the complete bipartite graph $K_{3, 3}$, is a toroidal graph.

Proof:
{{ProofWanted|More background work necessary}}
Theorem: Seven Touching Cylinders

It is possible to arrange $7$ identical cylinders so that each one touches each of the others.
The cylinders must be such that their heights must be at least $\dfrac {7 \sqrt 3} 2$ of the diameters of their bases.

Proof:
:600px
It remains to be proved that the heights of the cylinders must be at least $\dfrac {7 \sqrt 3} 2$ of the diameters of their bases.
{{ProofWanted|Prove the above}}
Theorem: Inscribed Squares in Right-Angled Triangle/Compass and Straightedge Construction/Shared Right Angle

:200px
Let $\triangle ABC$ be a right-angled triangle, where $\angle C = 90^\circ$.
Construct the angle bisector of $\angle C$.
Let the point of intersection of this angle bisector and side $AB$ be $E$.
Construct perpedicular lines from $E$ to sides $AC$ and $BC$, and name their intersections $F$ and $D$ respectively.
Then $CDEF$ is an inscribed square.

Proof:
Note that $\angle C = 90^\circ$ and $ED \perp BC$, $EF \perp AC$ by construction.
Therefore $CDEF$ is a rectangle.
By definition of an angle bisector, $\angle ECF = 45^\circ$.
Since $\angle CFE = 90^\circ$, by Sum of Angles of Triangle equals Two Right Angles:
:$\angle CEF + \angle CFE + \angle ECF = 180^\circ$
:$\therefore \angle CEF = 180^\circ - 90^\circ - 45^\circ = 45^\circ = \angle ECF$
By Triangle with Two Equal Angles is Isosceles, $CF = FE$.
Since $CDEF$ is a rectangle with two adjacent equal sides, it must also be a square.
{{qed}}
Category:Inscribed Squares in Right-Angled Triangle
2n9aewz2hvgt2y5cg44wtcyg83yv6o9
Theorem: Inscribed Squares in Right-Angled Triangle/Compass and Straightedge Construction/Side Lies on Hypotenuse

:400px
Let $\triangle ABC$ be a right-angled triangle, where $\angle C = 90^\circ$.
Construct a perpedicular line from $C$ to side $AB$, and name the intersection $D$.
Construct the angle bisector of $\angle ADC$.
Let the point of intersection of this angle bisector and side $AC$ be $E$.
Construct a line parallel to $AB$ that passes through $E$ to meet $CD$ at $F$.
Extend $AF$ to side $BC$ at $G$.
Construct a line parallel to $AB$ that passes through $G$ to meet side $AC$ at $H$.
Construct perpedicular lines from $H$ and $G$ to side $AB$, and name the intersections $I$ and $J$ respectively.
Then $GHIJ$ is an inscribed square.

Proof:
Note that $HG \perp HI$, $HG \perp GJ$ and $HG \parallel IJ$ by construction.
Therefore $GHIJ$ is a rectangle.
By definition of an angle bisector, $\angle FDE = 45^\circ$.
Since $\angle EFD = 90^\circ$ by construction, by Sum of Angles of Triangle equals Two Right Angles:
:$\angle DEF + \angle EFD + \angle FDE = 180^\circ$
:$\therefore \angle DEF = 180^\circ - 90^\circ - 45^\circ = 45^\circ = \angle FDE$
By Triangle with Two Equal Angles is Isosceles, $DF = FE$.
Since $EF \parallel HG$, $\triangle AEF \sim \triangle AHG$ by Equiangular Triangles are Similar.
Similarly, since $FD \parallel GJ$, $\triangle AFD \sim \triangle AGJ$ by Equiangular Triangles are Similar.
By definition of similar triangles:
:$\dfrac {EF} {HG} = \dfrac {AF} {AG} = \dfrac {FD} {GJ}$
As $EF = FD$, we must have $HG = GJ$.
Since $GHIJ$ is a rectangle with two adjacent equal sides, it must also be a square.
{{qed}}
Category:Inscribed Squares in Right-Angled Triangle
c7abds0q17esci7kquv5ny3pexgbu29
Theorem: Queen's Tour

Consider a chessboard $\CC$ of size $n \times n$ such that $n > 2$.
Then the shortest queen's tour on $\CC$ is of $2 n - 2$ moves.
For $n < 5$ it is necessary for the queen to move outside the boundary of the chessboard in order for this to happen.

Proof:
First it is shown that at least $2 n - 2$ moves are needed.
Let there be $R$ rows and $S$ columns which have none of the given moves on them.
The $R \times S$-square segment of this chessboard has a $2 R + 2 S - 4$ squares on its edge if $R$ and $S$ are both greater than $1$.
Each diagonal moves covers at most $2$ of these boundary square.
Thus in the set of moves there are at least:
:$R + S - 2$ diagonal moves
:$n - R$ horizontal moves
:$n - S$ vertical moves.
Thus there are at least $2 n - 2$ moves needed to cover the entire chessboard.
{{qed|lemma}}
It remains to demonstrate that no more than $2 n - 2$ moves are needed.
The proof proceeds by induction.
Let us raise the following hypothesis:
For all $n \in \Z_{\ge 3}$, let $\map P n$ be the proposition:
:A chessboard of size $n \times n$ has a queen's tour $T$ of $2 n - 2$ moves such that $T$ exits from the top right square:
:300px
Theorem: Re-entrant Queen's Tour

Consider a chessboard $\CC$ of size $n \times n$ such that $n > 3$.
Then there exists a re-entrant queen's tour on $\CC$ of $2 n - 2$ moves.
For $n < 6$ it is necessary for the queen to move outside the boundary of the chessboard in order for this to happen.

Proof:
First it is shown that at least $2 n - 2$ moves are needed.
Let there be $R$ rows and $S$ columns which have none of the given moves on them.
The $R \times S$-square segment of this chessboard has a $2 R + 2 S - 4$ squares on its edge if $R$ and $S$ are both greater than $1$.
Each diagonal moves covers at most $2$ of these boundary square.
Thus in the set of moves there are at least:
:$R + S - 2$ diagonal moves
:$n - R$ horizontal moves
:$n - S$ vertical moves.
Thus there are at least $2 n - 2$ moves needed to cover the entire chessboard.
{{qed|lemma}}
It remains to demonstrate that no more than $2 n - 2$ moves are needed.
{{ProofWanted}}
Theorem: Coefficients of Polynomial add to 0 iff 1 is a Root

Let $\map E x$ be the equation in $x$ represented as:
:$\ds \sum_{j \mathop = 0}^n a_j x^j = 0$
where the $a_j$s are constants.
Then $1$ is a root of $\map E x$ {{iff}}:
:$\ds \sum_{j \mathop = 0}^n a_j = 0$
That is, $1$ is a root of $\map E x$ {{iff}} all the coefficients of the polynomial in $x$ sum to zero.

Proof:
Letting $x = 1$ in $E$;
{{begin-eqn}}
{{eqn | l = x
      | r = 1
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \sum_{j \mathop = 0}^n a_j \times 1^j
      | r = 0
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \sum_{j \mathop = 0}^n a_j
      | r = 0
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Algebra
fdu6974gv04363kxpo6bt0omyy4c8xm
Theorem: Subspace of Complete Metric Space is Relatively Compact iff Every Sequence has Cauchy Subsequence

Let $\struct {X, d}$ be a complete metric space.
Let $\struct {H, d_H}$ be a metric subspace of $\struct {X, d}$. 
Then $\struct {H, d_H}$ is relatively compact {{iff}}:
:every sequence in $H$ has a Cauchy subsequence.

Proof:
=== Necessary Condition ===
Suppose that: 
:$\struct {H, d_H}$ is relatively compact
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $H$.
Then $\sequence {x_n}_{n \mathop \in \N}$ is a sequence in $\map \cl H$, where $\map \cl H$ is the closure of $H$ in $\struct {X, d}$.
Since $\struct {H, d_H}$ is relatively compact, we have: 
:$\map \cl H$ is compact.
From Compact Subspace of Metric Space is Sequentially Compact in Itself, we have: 
:$\map \cl H$ is sequentially compact.
So:
:$\sequence {x_n}_{n \mathop \in \N}$ has a subsequence $\sequence {x_{n_j} }_{j \mathop \in \N}$ that converges to a point in $\map \cl H$.
From Convergent Sequence in Metric Space is Cauchy Sequence, we have: 
:$\sequence {x_{n_j} }_{j \mathop \in \N}$ is a Cauchy sequence.
So:
:$\sequence {x_n}_{n \mathop \in \N}$ has a Cauchy subsequence $\sequence {x_{n_j} }_{j \mathop \in \N}$.
Since $\sequence {x_n}_{n \mathop \in \N}$ was an arbitrary sequence in $H$, we have: 
:every sequence in $H$ has a Cauchy subsequence.
{{qed|lemma}}
Theorem: Five Cube Theorem

Every integer can be expressed as a sum of five cube numbers.

Proof:
Let $r$ be an integer.
Then $r$ can be expressed in exactly one the following forms:
:$\paren {6 m + 0}$
:$\paren {6 m + 1}$
:$\paren {6 m + 2}$
:$\paren {6 m + 3}$
:$\paren {6 m + 4}$
:$\paren {6 m + 5}$
for some $m \in \Z$.
:$\paren {6 m + 0} = \paren {m + 1}^3 + \paren {m - 1}^3 + \paren {- m}^3 + \paren {- m}^3 + 0^3$
:$\paren {6 m + 1} = \paren {m + 1}^3 + \paren {m - 1}^3 + \paren {- m}^3 + \paren {- m}^3 + 1^3$
:$\paren {6 m + 2} = \paren {m}^3 + \paren {m - 2}^3 + \paren {1 - m}^3 + \paren {1 - m}^3 + 2^3$
:$\paren {6 m + 3} = \paren {m - 3}^3 + \paren {m - 5}^3 + \paren {- m + 4}^3 + \paren {- m + 4}^3 + 3^3$
:$\paren {6 m + 4} = \paren {m + 3}^3 + \paren {m + 1}^3 + \paren {- m - 2}^3 + \paren {- m - 2}^3 + \paren {-2}^3$
:$\paren {6 m + 5} = \paren {m + 2}^3 + \paren {m}^3 + \paren {- m - 1}^3 + \paren {- m - 1}^3 + \paren {-1}^3$
{{qed}}