avewright/proofwiki-math huggingface dataset:
Theorem: Bott-Milnor-Kervaire 1,2,4,8 Theorem

Let $A$ be a division algebra with real scalars.
Then the dimension of $A$ is either:
:$1$: the real numbers $\R$
:$2$: the complex numbers $\C$
:$4$: the quaternions $\Bbb H$
or:
:$8$: the octonions $\Bbb O$.

Proof:
{{ProofWanted}}
{{Namedfor|Raoul Bott|name2 = John Willard Milnor|name3 = Michel André Kervaire|cat = Bott|cat2 = Milnor|cat3 = Kervaire}}
Theorem: Montel's Theorem

Let $U \subseteq \C$ be an open subset of the complex numbers.
Let $\map \HH U$ be the space of holomorphic mappings on $U$.
Then a family of mappings $\FF \subseteq \map \HH U$ is normal {{iff}} $\FF$ is locally bounded.

Proof:
=== Normal implies locally bounded ===
By the Arzelà-Ascoli Theorem, every normal family is locally bounded.
Theorem: Vitali's Convergence Theorem

Let $U$ be an  open,  connected subset of $\C$.
Let $S \subseteq U$ contain a limit point $\sigma$.
Let $\sequence {f_n}_{n \mathop \in \N}$ be a  normal family of holomorphic mappings $f_n : U \to \C$.
Let $\sequence {f_n}_{n \mathop \in \N}$  converge to some holomorphic mapping $f : U \to \C$ at $\sigma$.
Then $f_n$ converges uniformly to $f$ on all compact subsets of $U$.

Proof:
{{AimForCont}} there exists some compact subset $K$ of $U$ such that $f_n$ does not converge uniformly to $f$ on $K$.
Consider $K^* := K \cup \set \sigma$.
From Subsets Inherit Uniform Convergence, $f_n$ does not converge uniformly to $f$ on $K^*$.
From Uniformly Convergent iff Difference Under Supremum Norm Vanishes, the above is equivalent to:
:$\exists \epsilon > 0 : \forall N \in \N : \exists n \ge N : \norm {f_n - f}_{K^*} \ge \epsilon$
where $\norm {\cdot}_{K^*}$ denotes the supremum norm over $K^*$.
From Finite Union of Compact Sets is Compact, $K^{*}$ is compact.
Since $\sequence {f_n}$ is a normal family, there is some subsequence $\sequence {f_{n_r} }$ of $\sequence {f_n}$ and some mapping $g \in \map {\HH} U$ such that:
:$\sequence {f_{n_r} }$ converges uniformly to $g$ on $K^*$.
Further:
{{begin-eqn}}
{{eqn | l = \map f \sigma
      | r = \lim_{n \mathop \to \infty} \map {f_n} \sigma
      | c = Definition of $f$ at $\sigma$
}}
{{eqn | r = \lim_{r \mathop \to \infty} \map {f_{n_r} } \sigma
      | c = Limit of Subsequence equals Limit of Sequence
}}
{{eqn | r = \map g \sigma
      | c = Definition of $g$ at $\sigma$
}}
{{end-eqn}}
From the Identity Theorem, $f$ and $g$ agree on $U$.
From Uniformly Convergent iff Difference Under Supremum Norm Vanishes:
:$\exists N \in \N: r \ge N \implies \norm {f_{n_r} - f}_{K^*} < \epsilon$
This contradicts the result that:
:$\forall N \in \N: \exists n \ge N: \norm {f_n - f}_{K^*} \ge \epsilon$
Hence the result, by Proof by Contradiction.
{{qed}}
{{Namedfor|Giuseppe Vitali|cat = Vitali}}
Theorem: Identity Theorem

Let $U$ be an open connected subset of the complex plane $\C$.
Let $f$ and $g$ be complex functions whose domain is $U$.
Let $S = \set {z \in U: \map f z = \map g z}$.
Let $f$ and $g$ be analytic on $U$.
Let $S$ have a limit point in $U$.
Then:
:$\forall z \in U: \map f z = \map g z$

Proof:
{{ProofWanted}}
Category:Complex Analysis
Category:Named Theorems
3mddkpoi5myvk6aq2apc2myvxlfh6vu
Theorem: Exponential Sequence is Uniformly Convergent on Compact Sets

Let $\EE = \sequence {E_n}$ denote the sequence of complex functions $E_n: \C \to \C$ defined as:
:$\map {E_n} z = \paren {1 + \dfrac z n}^n$
Let $K$ be a compact subset of $\C$.
Then $\EE$ is uniformly convergent on $K$.

Proof:
=== $\EE$ is Uniformly Bounded on an open space containing $K$ ===
First, from Equivalence of Definitions of Complex Exponential Function  we see that $\EE$ is pointwise convergent to $\exp$.
{{refactor|extract this result from that page|level = medium}}
From Combination Theorem for Continuous Complex Functions, $E_n$ is continuous for each $n \in \N$.
From Compact Subspace of Metric Space is Bounded, $K$ is bounded by some real number $M \in \R$.
Let:
:$M' = M + 1$
:$U = \set {z \in \C: \cmod z < M'}$
:$z \in U$
:$n \in \N$
From the Triangle Inequality for Complex Numbers:
:$\cmod {1 + \dfrac z n} \le 1 + \dfrac {\cmod z} n$
Thus, from Power Function is Strictly Increasing over Positive Reals: Natural Exponent:
:$\cmod {1 + \dfrac z n}^n \le \paren {1 + \dfrac {\cmod z} n}^n$
Call this result $(1)$.
Also:
{{begin-eqn}}
{{eqn | l = \cmod z
      | o = <
      | r = M'
}}
{{eqn | ll= \leadsto
      | l = 1 + \frac {\cmod z} n
      | o = <
      | r = 1 + \frac {M'} n
      | c = Divide both sides by $n$, add $1$ to both sides
}}
{{eqn | n = 2
      | ll= \leadsto
      | l = \paren {1 + \frac {\cmod z} n}^n
      | o = <
      | r = \paren {1 + \frac {M'} n}^n
      | c = Power Function is Strictly Increasing over Positive Reals: Natural Exponent
}}
{{end-eqn}}
Thus it is sufficient to show that $\paren {1 + \dfrac {M'} n}^n$ is bounded.
From Exponential Sequence is Eventually Increasing:
:$\exists N \in \N: n \ge N \implies \paren {1 + \dfrac {M'} n}^n \le \paren {1 + \dfrac {M'} {n + 1} }^{n + 1}$
So, for $n \ge N$:
{{begin-eqn}}
{{eqn | l = \paren {1 + \frac {M'} n}^n
      | o = \le
      | r = \lim_{n \mathop \to \infty} \paren {1 + \frac {M'} n}^n
      | c = Limit of Bounded Convergent Sequence is Bounded
}}
{{eqn | r = \exp M'
}}
{{end-eqn}}
Now, for each $n \in \set {1, 2, \ldots, N - 1}$:
:$\exists M_n \in \R: \paren {1 + \dfrac {M'} n}^n \le M_n$
by Continuous Function on Compact Subspace of Euclidean Space is Bounded.
So:
:$\forall n \in \N: \paren {1 + \dfrac {M'} n}^n \le \max \set {M_1, M_2, \ldots, M_{N - 1}, \exp M'}$
Call this result $(3)$.
Hence, for $z \in U$; that is, for $\cmod z < {M'}$:
{{begin-eqn}}
{{eqn | q = \forall n \in \N
      | l = \cmod {\paren {1 + \frac z n}^n}
      | r = \cmod {1 + \dfrac z n}^n
      | c = Complex Modulus of Product of Complex Numbers
}}
{{eqn | o = \le
      | r = \paren {1 + \dfrac {\cmod z} n}^n
      | c = From $(1)$
}}
{{eqn | o = \le
      | r = \paren {1 + \frac {M'} n}^n
      | c = From $(2)$
}}
{{eqn | o = \le
      | r = \max \set {M_1, M_2, \ldots, M_{N - 1}, \exp M'}
      | c = From $(3)$
}}
{{end-eqn}}
That is, $\EE$ is uniformly bounded on $U$.
Theorem: Uniformly Convergent Sequence Evaluated on Convergent Sequence

Let $X = \struct {A, d_X}$ and $Y = \struct {B, d_Y}$ be metric spaces.
Let $K$ be a subspace of $X$.
Let $f: X \to Y$ be a mapping.
Let $\FF = \sequence {f_n}$ be a sequence of continuous mappings $f_n: X \to Y$ that  converges to $f$ uniformly on $K$.
Let $\sequence {a_n}$ be a convergent sequence in $K$ with limit $a \in K$.
Then the sequence $\sequence {\map {f_n} {a_n} }$ is convergent such that:
:$\ds \lim_{n \mathop \to \infty} \map {f_n} {a_n} = \map f a$

Proof:
We want to show that:
:$\map {d_Y} {\map {f_n} {a_n} , \map f a} \to 0$ as $n \to \infty$
Let $\epsilon \in \R_{>0}$ be fixed.
From {{Metric-space-axiom|2}}:
:$\map {d_Y} {\map {f_n} {a_n} , \map f a} \le \map {d_Y} {\map {f_n} {a_n} , \map f {a_n} } + \map {d_Y} {\map f {a_n} , \map f a}$
From the Uniform Limit Theorem, it follows that $f$ is continuous.
Since $a_n \to a$, Sequential Continuity is Equivalent to Continuity in Metric Space tells us that:
:$\exists M \in \N : \forall n \ge M : \map {d_Y} {\map f {a_n} , \map f a} < \dfrac \epsilon 2$
Since $\sequence {f_n}$ is uniformly convergent on $K$, it follows that:
:$\exists N \in \N : \forall n \ge N : \map {d_Y} {\map {f_n} {a_n} , \map f {a_n} } < \dfrac \epsilon 2$
So if $n \ge \max \set {M, N}$, it follows that:
{{begin-eqn}}
{{eqn | o = \le
      | l = \map {d_Y} {\map {f_n} {a_n} , \map f a}  
      | r = \map {d_Y} {\map {f_n} {a_n} , \map f {a_n} } + \map {d_Y} {\map f {a_n} , \map f a} 
}}
{{eqn | o = < 
      | r = \dfrac \epsilon 2 + \dfrac \epsilon 2 
}}
{{eqn | r = \epsilon
}}
{{end-eqn}}
which shows that $\map {d_Y} {\map {f_n} {a_n} , \map f a} \to 0$ as $n \to \infty$. 
{{qed}}
Category:Metric Spaces
Category:Uniform Convergence
eggmm3va6k9v1tv4zov1rdi4rb9n7yd
Theorem: Meet-Continuous iff Meet of Suprema equals Supremum of Meet of Ideals

Let $\mathscr S = \struct {S, \vee, \wedge, \preceq}$ be an up-complete lattice.
Then
:$\mathscr S$ is meet-continuous {{iff}}
:for every ideals $I, J$ in $\mathscr S$: $\paren {\sup I} \wedge \paren {\sup J} = \sup \set {i \wedge j: i \in I, j \in J}$

Proof:
=== Sufficient Condition ===
Let $\mathscr S$ be meet-continuous.
Define $\II$, the set of all ideals in $\mathscr S$
Define a mapping $f: \II \to S$ such that
:$\forall I \in \II: \map f I = \sup I$
By Meet-Continuous iff Ideal Supremum is Meet Preserving:
:$f$ preserves meet.
Let $I, J \in \II$.
By definition of mapping preserves meet:
:$f$ preserves the infimum of $\set {I, J}$
Thus
{{begin-eqn}}
{{eqn | l = \paren {\sup I} \wedge \paren {\sup J}
      | r = \map f I \wedge \map f J
      | c = Definition of $f$
}}
{{eqn | r = \inf \set {\map f I, \map f J}
      | c = {{Defof|Meet (Order Theory)}}
}}
{{eqn | r = \map \inf {\map {f^\to} {\set {I, J} } }
      | c = {{Defof|Image of Subset under Mapping}}
}}
{{eqn | r = \map f {\inf \set {I, J} }
      | c = {{Defof|Mapping Preserves Infimum|subdef = Subset}}
}}
{{eqn | r = \map f {I \wedge J}
      | c = {{Defof|Meet (Order Theory)}}
}}
{{eqn | r = \map \sup {I \wedge J}
      | c = Definition of $f$
}}
{{eqn | r = \sup \set {i \wedge j: i \in I, j \in J}
      | c = Meet in Set of Ideals
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Area between Radii and Whorls of Archimedean Spiral

Let $S$ be the Archimedean spiral defined by the equation:
:$r = a \theta$
Let $\theta = \theta_1$ and $\theta = \theta_2$ be the two rays from the pole at angles $\theta_1$ and $\theta_b$ to the polar axis respectively.
Let $R$ be the figure enclosed by:
:$\theta_1$ and $\theta_2$
:the $n$th turn of $S$ and the $n+1$th turn of $S$
The area $\AA$ of $R$ is given by:
:$\AA = a^2 \pi \paren {\theta_2 - \theta_1} \paren {\theta_2 + \theta_1 + 2 \pi \paren {2 n + 1} }$
{{Proofread|My algebra may have failed me}}

Proof:
The straight line boundaries of $R$ are given as $\theta_1$ and $\theta_2$.
The corners of $R$ are located where:
:$\theta = \theta_1 + 2 n \pi$
:$\theta = \theta_2 + 2 n \pi$
:$\theta = \theta_1 + 2 \paren {n + 1} \pi$
:$\theta = \theta_2 + 2 \paren {n + 1} \pi$
500px
{{begin-eqn}}
{{eqn | l = \AA
      | r = \int_{\theta_1 + 2 \paren {n + 1} \pi}^{\theta_2 + 2 \paren {n + 1} \pi} \frac {\paren {a \theta}^2 \rd \theta} 2 - \int_{\theta_1 + 2 n \pi}^{\theta_2 + 2 n \pi} \frac {\paren {a \theta}^2 \rd \theta} 2
      | c = Area between Radii and Curve in Polar Coordinates
}}
{{eqn | r = \intlimits {\frac {a^2 \theta^3} 6} {\theta_1 + 2 \paren {n + 1} \pi} {\theta_2 + 2 \paren {n + 1} \pi} - \intlimits {\frac {a^2 \theta^3} 6} {\theta_1 + 2 n \pi} {\theta_2 + 2 n \pi}
      | c = Primitive of Power
}}
{{eqn | r = \frac {a^2} 6 \paren {\paren {\theta_2 + 2 \paren {n + 1} \pi}^3 - \paren {\theta_1 + 2 \paren {n + 1} \pi}^3 - \paren {\theta_2 + 2 n \pi}^3 + \paren {\theta_1 + 2 n \pi}^3}
      | c = 
}}
{{eqn | r = \frac {a^2} 6 \paren {\theta_2^3 + 3 \paren {2 \paren {n + 1} \pi} \theta_2^2 + 3 \paren {2 \paren {n + 1} \pi}^2 \theta_2 + \paren {2 \paren {n + 1} \pi}^3}
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \frac {a^2} 6 \paren {\theta_1^3 + 3 \paren {2 \paren {n + 1} \pi} \theta_1^2 + 3 \paren {2 \paren {n + 1} \pi}^2 \theta_1 + \paren {2 \paren {n + 1} \pi}^3}
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \frac {a^2} 6 \paren {\theta_2^3 + 3 \paren {2 n \pi} \theta_2^2 + 3 \paren {2 n \pi}^2 \theta_2 + \paren {2 n \pi}^3}
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \frac {a^2} 6 \paren {\theta_1^3 + 3 \paren {2 n \pi} \theta_1^2 + 3 \paren {2 n \pi}^2 \theta_1 + \paren {2 n \pi}^3}
      | c = 
}}
{{eqn | r = a^2 \pi \paren {\paren {n + 1} \theta_2^2 + 2 \pi \paren {n + 1}^2 \theta_2 - \paren {n \theta_2^2 + 2 \pi n^2 \theta_2} }
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = a^2 \pi \paren {\paren {n + 1} \theta_1^2 + 2 \pi \paren {n + 1}^2 \theta_1 - \paren {n \theta_1^2 + 2 \pi n^2 \theta_1} }
      | c = 
}}
{{eqn | r = a^2 \pi \paren {\theta_2^2 + 2 \pi \paren {2 n + 1} \theta_2}
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = a^2 \pi \paren {\theta_1^2 + 2 \pi \paren {2 n + 1} \theta_1}
      | c = 
}}
{{eqn | r = a^2 \pi \paren {\theta_2 - \theta_1} \paren {\theta_2 + \theta_1 + 2 \pi \paren {2 n + 1} }
      | c = 
}}
{{end-eqn}}
{{proofread}}
Theorem: Uniformly Convergent Sequence of Bounded Functions is Uniformly Bounded

Let $X = \left({A, d}\right)$ and $Y = \left({B, \rho}\right)$ be metric spaces.
Let $\left \langle{f_i}\right \rangle_{i \in I}$ be a uniformly convergent sequence of mappings $f_i: X \to Y$.
$\forall i \in I$, let $f_i$ be bounded.
Then $\left \langle{f_i}\right \rangle$ is uniformly bounded.

Proof:
{{ProofWanted}}
Category:Metric Spaces
7d5j0zwent5gjtl0jwe89zyrpjqlp4t
Theorem: Product of Uniformly Convergent Sequences of Bounded Functions is Uniformly Convergent

Let $X = \struct {A, d}$ and $Y = \struct {B, \rho}$ be metric spaces.
Let $\sequence {f_n}$ and $\sequence {g_n}$ be sequences of mappings from $X$ to $Y$.
Let $\sequence {f_n}$ and $\sequence {g_n}$ be uniformly convergent on some subspace $S$ of $X$. 
$\forall n \in \N$, let $f_n$ and $g_n$ be bounded.
Then the sequence $\sequence {f_n g_n}$ is uniformly convergent on $S$.
{{explain|for the product to be defined, we probably want $Y$ to have some more structure}}

Proof:
{{ProofWanted}}
Category:Metric Spaces
i5xrjd237orij2ge9yvscxqz4r2g3zo
Theorem: Area between Radii and Curve in Polar Coordinates

Let $C$ be a curve expressed in polar coordinates $\polar {r, \theta}$ as:
:$r = \map g \theta$
where $g$ is a real function.
Let $\theta = \theta_a$ and $\theta = \theta_b$ be the two rays from the pole at angles $\theta_a$ and $\theta_b$ to the polar axis respectively.
Then the area $\AA$ between $\theta_a$, $\theta_b$ and $C$ is given by:
:$\ds \AA = \int \limits_{\theta \mathop = \theta_a}^{\theta \mathop = \theta_b} \frac {\paren {\map g \theta}^2 \rd \theta} 2$
as long as $\paren {\map g \theta}^2$ is integrable.

Proof:
:600px
{{MissingLinks}}
{{tidy|In particular the terms are to be defined.}}
{{improve|We haven't covered NSA on {{ProofWiki}}, and for a result as trivial as this one it seems like overkill.}}
Consider the area of the brown triangle.
This would be:
:$a_\triangle = \dfrac 1 2 r^2 \map \sin {\delta \theta}$
We will be using nonstandard analysis, so let $\delta \theta = \varepsilon > 0$, an infinitesimal. 
Thus:
:$a_\triangle = \dfrac 1 2 r^2 \sin \varepsilon$
Using the Power Series Expansion for Sine Function: 
{{begin-eqn}}
{{eqn | l = A_\triangle
      | r = \frac 1 2 r^2 \paren {\varepsilon - \frac {\varepsilon^3} {3!} + \frac {\varepsilon^5} {5!} - \map \OO {\varepsilon^7} }
      | c = Power Series Expansion for Sine Function
}}
{{eqn | r = \frac 1 2 r^2 \varepsilon
      | c = $\varepsilon ^2 = 0$
}}
{{eqn | l = \AA
      | r = \int_{\theta_a}^{\theta_b} \frac {\paren {\map g \theta}^2} 2 \rd \theta
      | c = Summing all areas of the triangles
}}
{{end-eqn}}
{{qed}}
Theorem: Exponential Sequence is Eventually Increasing

Let $\sequence {E_n}$ be the sequence of real functions $E_n: \R \to \R$ defined as:
:$\map {E_n} x = \paren {1 + \dfrac x n}^n$
Then, for sufficiently large $n \in \N$, $\sequence {\map {E_n} x}$ is increasing {{WRT}} $n$.
That is:
:$\forall x \in \R: \forall n \in \N: n \ge \ceiling {\size x} \implies \map {E_n} x \le \map {E_{n + 1} } x$
where $\ceiling x$ denotes the ceiling of $x$.

Proof:
Fix $x \in \R$.
Then:
{{begin-eqn}}
{{eqn | l = n
      | o = \ge
      | r = \ceiling {\size x}
}}
{{eqn | ll= \leadsto
      | l = n
      | o = >
      | r = -x
      | c = Real Number is between Ceiling Functions and Negative of Absolute Value
}}
{{eqn | ll= \leadsto
      | l = 1
      | o = >
      | r = \frac {-x} n
      | c = Divide both sides by $n$
}}
{{eqn | ll= \leadsto
      | l = 1 + \frac x n
      | o = >
      | r = 0
}}
{{end-eqn}}
So we may apply the AM-GM inequality, with:
:$x_1 := 1$
:$x_j := 1 + \dfrac x n$ for each $j \in \set {2, 3, \dotsc, n + 1}$
to obtain that:
:$\dfrac {1 + n \paren {1 + \dfrac x n} } {n + 1} > \paren {\paren {1 + \dfrac x n}^n}^{1 / \paren {n + 1} }$
After simplification:
:$1 + \dfrac x {n + 1} > \paren {1 + \dfrac x n}^{n / \paren {n + 1} }$
From Power Function is Strictly Increasing over Positive Reals: Natural Exponent:
:$\paren {1 + \dfrac x {n + 1} }^{n + 1} > \paren {1 + \dfrac x n}^n$
where we have raised both sides to the power of $n + 1$.
Hence the result.
{{qed}}
Category:Exponential Function
046sbaptmmllbj8z7yarfya6odvrlm3
Theorem: Euler's Number: Limit of Sequence implies Base of Logarithm

Let $e$ be Euler's number defined by:
:$\ds e := \lim_{n \mathop \to \infty} \paren {1 + \frac 1 n}^n$
Then $e$ is the unique solution to the equation $\map \ln x = 1$.
That is:
:$\map \ln x = 1 \iff x = e$

Proof:
First we prove that $e$ is a solution to $\map \ln x = 1$:
{{begin-eqn}}
{{eqn | l = \map \ln e
      | r = \map \ln {\lim_{n \mathop \to \infty} \paren {1 + \frac 1 n}^n }
      | c = {{Defof|Euler's Number as Limit of Sequence}}
}}
{{eqn | r = \lim_{n \mathop \to \infty} \paren {\map \ln {1 + \frac 1 n}^n }
      | c = {{Defof|Limit at Infinity}}
}}
{{eqn | r = \lim_{n \mathop \to \infty} \dfrac {n \map \ln {\dfrac {n + 1} n } } 1
      | c = Logarithm of Power
}}
{{eqn | r = \lim_{n \mathop \to \infty} \frac {\map \ln {\dfrac {n + 1} n} } {\paren {\dfrac 1 n} }
      | c = multiplying numerator and denominator by $\dfrac 1 n$
}}
{{eqn | r = \lim_{n \mathop \to \infty} \frac{\paren {-\dfrac 1 {n^2} } \paren {\dfrac n {n + 1} } } {\paren {-\dfrac 1 {n^2} } }
      | c = L'Hôpital's Rule
}}
{{eqn | r = \lim_{n \mathop \to \infty} \paren {\frac n {n + 1} }
}}
{{eqn | r = 1
}}
{{end-eqn}}
From Logarithm is Strictly Increasing, $\ln$ is strictly monotone.
By Strictly Monotone Mapping with Totally Ordered Domain is Injective it follows that $\ln$ is an injection.
So the solution to $\map \ln x = 1$ is unique.
Hence the result.
{{qed}}
Category:Logarithms
Category:Euler's Number
1kf8ixi26wubh7fbtmjgdurqktexfqk
Theorem: Diophantine Equation y cubed equals x squared plus 2

The indeterminate Diophantine equation:
:$y^3 = x^2 + 2$
has only one solution in the Natural Numbers:
:$x = 5, y = 3$

Proof:
Assume that $x$ is even:
{{begin-eqn}}
{{eqn | l = \paren {2 k}^2 + 2
      | r = 4 k^2 + 2
}}
{{eqn | r = 2 \paren {2 k^2 + 1}
      | c = 
}}
{{end-eqn}}
Therefore, the {{RHS}} is $2 \paren {2 k^2 + 1} \equiv 2 \pmod 4$
If $y$ is odd, then the {{LHS}} will be odd:
{{begin-eqn}}
{{eqn | l = \paren {2 k + 1}^3
      | r = 8 k^3 + 6 k^2 + 6 k + 1
}}
{{eqn | r = 2 \paren {4 k^3 + 3 k^2 + 3 k} + 1
      | c = 
}}
{{end-eqn}}
and if $y$ is even, then the {{LHS}} will be $\equiv 0 \pmod 4$
{{begin-eqn}}
{{eqn | l = \paren {2 k }^3
      | r = 8 k^3
}}
{{eqn | r = 4 \paren {2 k^3}
      | c = 
}}
{{end-eqn}}
Therefore, '''$x$ and $y$ must both be odd''' since the {{LHS}} can never be $\equiv 2 \pmod 4$
Let us rewrite $x$ as $x = y + a$
We now demonstrate that $a$ must be even:
{{begin-eqn}}
{{eqn | l = x
      | r = y + a
      | c = 
}}
{{eqn | ll = \leadsto
      | l = 2 r + 1
      | r = \paren {2 s + 1 } + 2 k
      | c = $x$ and $y$ are both odd and $a$ is even
}}
{{eqn | r = \paren {2 \paren {s + k} + 1 } 
      | c = 
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | l = y^3
      | r = \paren {y + a}^2 + 2
      | c = substituting $x = y + a$
}}
{{eqn | r = y^2 + 2 a y + a^2 + 2
      | c = 
}}
{{eqn | ll = \leadsto
      | l = y^3 - y^2
      | r = 2 a y + a^2 + 2
      | c = subtracting $y^2$ from both sides
}}
{{eqn | ll = \leadsto
      | l = y^2 \paren {y - 1}
      | r = 2 \paren {a y + \dfrac {a^2} 2 + 1}
      | c = factoring both sides
}}
{{eqn | r = 2 \paren {2 k y + 2 k^2 + 1}
      | c = $a$ is even: substituting $a = 2 k$
}}
{{eqn | r = 2 \paren {2 k \paren {y + k} + 1}
      | c = 
}}
{{end-eqn}}
From the {{LHS}}, we know that $y$ is odd and therefore $y - 1$ is even and the only even term on the {{RHS}} is $2$.
Therefore, our only solution is: $y - 1 = 2 \leadsto y = 3$ and $x = 5$
{{qed}}
Theorem: Continuous Extension from Dense Subset

Let $X$ be a metric space.
Let $D$ be a dense subset of $X$.
Let $f: D \to \R$ be a uniformly continuous mapping.
Then there exists a unique continuous extension of $f$ to $X$.

Proof:
{{ProofWanted}}
Category:Real Analysis
Category:Metric Spaces
jde7psdzy42exqmb2yjik6axp2bc7ye
Theorem: Sum of two Fourth Powers cannot be Fourth Power

$\forall a, b, c \in \Z_{>0}$, the equation $a^4 + b^4 = c^4$ has no solutions.

Proof:
This is a direct consequence of Fermat's Right Triangle Theorem.
{{qed}}
Theorem: Moore-Osgood Theorem

Let $X$ and $Y$ be metric spaces.
Let $S$ be a subspace of $X$.
Let $c$ be a limit point of $S$.
Let $\sequence {f_n}$ be a sequence of mappings $f_n : X \to Y$.
Suppose that:
:$(1): \quad \sequence {f_n}$ is uniformly convergent on $S$
:$(2): \quad \ds \forall n \in \N : \lim_{x \mathop \to c} \map {f_n} x$ exists
Then:
:$\ds \lim_{x \mathop \to c} \lim_{n \mathop \to \infty} \map {f_n} x = \lim_{n \mathop \to \infty} \lim_{x \mathop \to c} \map {f_n} x$
{{questionable|The {{LHS}} may diverge, since $\lim_{n \mathop \to \infty} \map {f_n} x $ is not assumed to converge for $x \not \in S$. Overall $S$ does not seem effectively used in the statement. Some condition is missing.}}

Proof:
{{ProofWanted}}
{{Namedfor|Eliakim Hastings Moore|name2 = William Fogg Osgood|cat = Moore, Eliakim|cat2 = Osgood}}
Category:Real Analysis
Category:Metric Spaces
qgaeu2tjdpwt5olrbi4l67r2ex0ufe6
Theorem: Cycloid is Best Shape for Arch

The optimal shape for an arch is a cycloid.

Proof:
{{ProofWanted|... and of course an elaboration of what "optimal" means would be useful.}}
Theorem: Ore Graph is Connected

Let $G = \struct {V, E}$ be an Ore graph.
Then $G$ is connected.

Proof:
Let $G$ be an Ore graph of order $n$.
{{AimForCont}} $G$ is not connected.
Then it has at least two components.
Call these components $C_1$ and $C_2$.
Thus, there exist non-adjacent vertices $u$ and $v$ such that $u$ is in $C_1$ and $v$ is in $C_2$.
Let $m_1$ and $m_2$ be the number of vertices in $C_1$ and $C_2$ respectively.
It is clear that:
:$m_1 + m_2 \le n$
By definition of Ore graph, $G$ is simple.
Thus it follows that:
:$\map {\deg_G} u \le m_1 - 1$ and $\map {\deg_G} v \le m_2 - 1$
Thus:
{{begin-eqn}}
{{eqn | l = \map {\deg_G} u + \map {\deg_G} v
      | o = \le
      | r = m_1 - 1 + m_2 - 1
      | c = 
}}
{{eqn | r = m_1 + m_2 - 2
      | c = 
}}
{{eqn | o = <
      | r = m_1 + m_2
      | c = 
}}
{{eqn | o = \le
      | r = n
      | c = 
}}
{{end-eqn}}
That is:
:$\map {\deg_G} u + \map {\deg_G} v < n$
But by definition of Ore graph:
:$\map {\deg_G} u + \map {\deg_G} v \ge n$
By Proof by Contradiction, it follows that our assumption that $G$ is not connected was false.
Hence the result.
{{Qed}}