avewright/proofwiki-math huggingface dataset:
Theorem: Zero and One are the only Consecutive Perfect Squares

If $n$ is a perfect square other than $0$, then $n+1$ is not a perfect square.

Proof:
Suppose that $k, l \in \Z$ are such that their squares are consecutive.
That is:
:$l^2 - k^2 = 1$
Then we can factor the {{LHS}} as:
:$l^2 - k^2 = \paren {l + k} \paren {l - k}$
By Invertible Integers under Multiplication, it follows that:
:$l + k = \pm 1 = l - k$
Therefore, it must be that:
:$\paren {l + k} - \paren {l - k} = 0$
That is, $2 k = 0$, from which we conclude $k = 0$.
So if $n$ and $n + 1$ are squares, then necessarily $n = 0$.
The result follows.
{{qed}}
Theorem: ProofWiki:Jokes

If I cannot open these cans of food, I will die.

Proof:
Suppose not.
{{qed}}
Theorem: Finite Union of Bounded Subsets

Let $M = \struct {A, d}$ be a metric space.
Then the union of any finite number of bounded subsets of $M$ is itself bounded.

Proof:
It is sufficient to prove this for two subsets, as the general result follows by induction.
Suppose $S_1$ and $S_2$ are bounded subsets of $M = \struct {A, d}$.
Let $a_1, a_2 \in A$.
Let $K_1, K_2 \in \R$ such that:
:$(1): \quad \forall x \in S_1: \map d {x, a_1} \le K_1$
:$(2): \quad \forall x \in S_2: \map d {x, a_2} \le K_2$
{{WLOG}}, let $a = a_1$ and $K = \max \set {K_1, K_2 + \map d {a_1, a_2} }$.
Then $\forall x \in S_1 \cup S_2$:
Either:
:$x \in S_1$ and so $\map d {x, a} \le K_1 \le K$
or:
:$x \in S_2$ and so $\map d {x, a} = \map d {x, a_1} \le \map d {x, a_2} + \map d {a_2, a_1} \le K_2 + \map d {a_2, a_1} \le K$
Hence the result.
{{qed}}
Theorem: Euler Triangle Formula

Let $d$ be the distance between the incenter and the circumcenter of a triangle.
Then:
:$d^2 = R \paren {R - 2 \rho}$
where:
:$R$ is the circumradius
:$\rho$ is the inradius.

Proof:
{{WLOG}}, it will be demonstrated that $BP = IP$.
Let $CP$ be the bisector of $\angle ACB$.
:$\angle ACP$ and $\angle ICB$ both subtend $AP$.
Hence indirectly by the Inscribed Angle Theorem:
:$\angle ACP = \angle ICB$
From Angles on Equal Arcs are Equal:
:$\angle ACP = \angle ABP$
and so:
:$\angle ABP = \angle ICB$
By the construction of the incircle, $IB$ is the bisector of $B$.
Then:
:$\angle IBA = \angle IBC$
and so:
{{begin-eqn}}
{{eqn | l = \angle IBP
      | r = \angle IBA + \angle ABP
      | c = 
}}
{{eqn | r = \angle IBC + \angle ICB
      | c = 
}}
{{end-eqn}}
By Sum of Angles of Triangle equals Two Right Angles:
:$\angle CIB + \angle IBC + \angle ICB = \angle CIB + \angle PIB$
as $\angle CIB$ and $\angle PIB$ are supplementary.
Thus:
{{begin-eqn}}
{{eqn | l = \angle PIB
      | r = \angle IBC + \angle ICB
      | c = 
}}
{{eqn | r = \angle IBP
      | c = 
}}
{{end-eqn}}
and from Triangle with Two Equal Angles is Isosceles:
:$IP = BP$
The proof that $IP = AP$ follows the same lines.
Theorem: Euler Triangle Formula

Let $d$ be the distance between the incenter and the circumcenter of a triangle.
Then:
:$d^2 = R \paren {R - 2 \rho}$
where:
:$R$ is the circumradius
:$\rho$ is the inradius.

Proof:
We have {{hypothesis}}:
:$AI$, $BI$ and $CI$ bisect their respective angles.
Let the half-angles be:
:$\alpha = \dfrac 1 2 \angle CAB$
:$\beta = \dfrac 1 2 \angle ABC$
:$\gamma = \dfrac 1 2 \angle BCA$
:300px
{{begin-eqn}}
{{eqn | l = AP
      | r = BP
      | c = Equal Angles in Equal Circles
}}
{{eqn | l = \angle BAP
      | r = \gamma
      | c = Angles on Equal Arcs are Equal
}}
{{eqn | l = \angle API 
      | r = \angle ABC = 2 \beta
      | c = Angles on Equal Arcs are Equal
}}
{{eqn | ll = \leadsto
      | l = \angle IAP
      | r = \alpha + \gamma
      | c = 
}}
{{eqn | l = \angle AIP
      | r = \alpha + \gamma
      | c = Sum of Angles of Triangle equals Two Right Angles
}}
{{end-eqn}}
By Triangle with Two Equal Angles is Isosceles
:$\triangle AIP$ is isosceles
$\leadsto$:
:$AP = BP = IP$
Theorem: Euler Triangle Formula

Let $d$ be the distance between the incenter and the circumcenter of a triangle.
Then:
:$d^2 = R \paren {R - 2 \rho}$
where:
:$R$ is the circumradius
:$\rho$ is the inradius.

Proof:
=== Lemma 1 ===
{{:Euler Triangle Formula/Lemma 1}}{{qed|lemma}}
=== Lemma 2 ===
{{:Euler Triangle Formula/Lemma 2}}{{qed|lemma}}
By {{Lemma|Euler Triangle Formula|1}}:
:$GI \cdot IJ = IP \cdot CI$
substituting:
:$IP \cdot CI = \paren {R + d} \paren {R - d}$
By {{Lemma|Euler Triangle Formula|2}}:
:$IP = PB$
and so:
:$GI \cdot IJ = PB \cdot CI$
Now using the Extension of Law of Sines in $\triangle CPB$:
:$\dfrac {PB} {\map \sin {\angle PCB} } = 2 R$
and so:
:$GI \cdot IJ = 2 R \map \sin {\angle PCB} \cdot CI$
By the $4$th of Euclid's common notions:
:$\angle PCB = \angle ICF$
and so:
:$(1): \quad GI \cdot IJ = 2 R \map \sin {\angle ICF} \cdot CI$
We have that:
:$IF = \rho$
and by Radius at Right Angle to Tangent:
:$\angle IFC$ is a right angle.
By the definition of sine:
:$\map \sin {\angle ICF} = \dfrac {\rho} {CI}$
and so:
:$\map \sin {\angle ICF} \cdot CI = \rho$
Substituting in $(1)$:
{{begin-eqn}}
{{eqn | l = GI \cdot IJ
      | r = 2 R \rho
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {R + d} \paren {R - d}
      | r = 2 R \rho
      | c = 
}}
{{eqn | ll= \leadsto
      | l = R^2 - d^2
      | r = 2 R \rho
      | c = Difference of Two Squares
}}
{{eqn | ll= \leadsto
      | l = d^2
      | r = R^2 - 2 R \rho
      | c = 
}}
{{eqn | r = R \paren {R - 2 \rho}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Euler Triangle Formula

Let $d$ be the distance between the incenter and the circumcenter of a triangle.
Then:
:$d^2 = R \paren {R - 2 \rho}$
where:
:$R$ is the circumradius
:$\rho$ is the inradius.

Proof:
=== Lemma 1 ===
{{:Euler Triangle Formula/Lemma 1}}{{qed|lemma}}
=== Lemma 2 ===
{{:Euler Triangle Formula/Lemma 2}}{{qed|lemma}}
{{begin-eqn}}
{{eqn | ll= \leadsto
      | l = IP \cdot CI
      | r = \paren {R + d} \cdot \paren {R - d}
      | c = {{Lemma|Euler Triangle Formula|1}}
}}
{{eqn | l = IP
      | r = PB
      | c = {{Lemma|Euler Triangle Formula|2}}
}}
{{eqn | l = CI \cdot PB
      | r = \paren {R + d} \cdot \paren {R - d}
      | c = {{EuclidCommonNotionLink|1}}
}}
{{end-eqn}}
400px
Draw diameter $POQ$.
{{begin-eqn}}
{{eqn | l = \angle PBQ
      | r = 90 \degrees
      | c = Thales' Theorem
}}
{{eqn | l = \angle PCB
      | r = \angle BQP
      | c = Equal Angles in Equal Circles
}}
{{eqn | l = \triangle IFC
      | o = \sim
      | r = \triangle PBQ
      | c = Equiangular Right Triangles are Similar
}}
{{eqn | ll= \leadsto
      | l = \dfrac {2R} {PB}
      | r = \dfrac {CI} \rho
      | c =
}}
{{eqn | l = CI \cdot PB
      | r = 2 R \rho
      | c = rearranging
}}
{{eqn | n = 1
      | l = \paren {R + d} \paren {R - d}
      | r = 2 R \rho
      | c = substitution from above
}}
{{eqn | l = R^2 - d^2
      | r = 2 R \rho
      | c = Difference of Two Squares
}}
{{eqn | l = d^2
      | r = R \paren {R - 2 \rho}
      | c = rearranging
}}
{{end-eqn}}
{{qed}}
Theorem: Open Ball of Point Inside Open Ball/Metric Space

Let $M = \struct {A, d}$ be a metric space.
Let $\map {B_\epsilon} x$ be an open $\epsilon$-ball in $M = \struct {A, d}$.
Let $y \in \map {B_\epsilon} x$.
Then:
:$\exists \delta \in \R: \map {B_\delta} y \subseteq \map {B_\epsilon} x$
That is, for every point in an open $\epsilon$-ball in a metric space, there exists an open $\delta$-ball of that point entirely contained within that open $\epsilon$-ball.

Proof:
Let $\delta = \epsilon - \map d {x, y}$.
From the definition of open ball, this is strictly positive, since $y \in \map {B_\epsilon} x$.
If $z \in \map {B_\delta} y$, then $\map d {y, z} < \delta$.
So:
:$\map d {x, z} \le \map d {x, y} + \map d {y, z} < \map d {x, y} + \delta = \epsilon$
Thus $z \in \map {B_\epsilon} x$.
So $\map {B_\delta} y \subseteq \map {B_\epsilon} x$.
{{qed}}
Theorem: Open Sets in Metric Space

Let $M = \struct {A, d}$ be a metric space.
Then $\O$ and $A$ are both open in $M$.

Proof:
We have the results:
:Empty Set is Open in Metric Space
:Metric Space is Open in Itself
{{qed}}
Theorem: Equivalence of Definitions of Continuity on Metric Spaces

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f: A_1 \to A_2$ be a mapping from $A_1$ to $A_2$.
{{TFAE|def = Continuous on Metric Space|view = continuity|context = Metric Space|contextview = metric spaces}}

Proof:
=== Definition by Points implies Definition by Open Sets ===
Suppose that $f$ is continuous at every point $x \in A_1$.
Let $U \subseteq M_2$ be open in $M_2$.
Let $x \in f^{-1} \sqbrk U$.
Since $U$ is open in $M_2$:
:$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} {\map f x; d_2} \subseteq U$
where $\map {B_\epsilon} {\map f x; d_2}$ denotes the open $\epsilon$-ball of $\map f x$ in $M_2$.
By the definition of continuity at a point:
:$\exists \delta \in \R_{>0}: f \sqbrk {\map {B_\delta} {x; d_1} } \subseteq \map {B_\epsilon} {\map f x; d_2}$
So:
:$f \sqbrk {\map {B_\delta} {x; d_1} } \subseteq U$
and so:
:$\map {B_\delta} {x; d_1} \subseteq f^{-1} \sqbrk U$
Thus $f^{-1} \sqbrk U$ is open in $M_1$.
{{qed|lemma}}
Theorem: Union of Open Sets of Metric Space is Open

Let $M = \struct {A, d}$ be a metric space.
The union of a set of open sets of $M$ is open in $M$.

Proof:
Let $I$ be any indexing set.
Let $U_i$ be open in $M$ for all $i \in I$.
Let $\ds x \in \bigcup_{i \mathop \in I} U_i$.
Then by definition of set union, $x \in U_k$ for some $k \in I$.
Since $U_k$ is open in $M$:
:$\ds \exists \epsilon > 0: \map {B_\epsilon} x \subseteq U_k$
where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$ in $M$.
By Set is Subset of Union:
:$\ds U_k \subseteq \bigcup_{i \mathop \in I} U_i$
Hence by Subset Relation is Transitive:
:$\ds \map {B_\epsilon} x \subseteq \bigcup_{i \mathop \in I} U_i$
and the result follows by definition of open set in metric space.
{{qed}}
Theorem: Lipschitz Equivalent Metric Spaces are Homeomorphic

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $M_1$ and $M_2$ be Lipschitz equivalent.
Then $M_1$ and $M_2$ are homeomorphic.

Proof:
Let $M_1$ and $M_2$ be Lipschitz equivalent.
Then, by definition, $\exists h, k \in \R_{>0}$ such that:
:$\forall x, y \in A_1: h \map {d_1} {x, y} \le \map {d_2} {\map f x, \map f y} \le k \map {d_1} {x, y}$
From the definition of open $\epsilon$-ball:
{{begin-eqn}}
{{eqn | l = y
      | o = \in
      | r = \map {B_{h \epsilon} } {\map f x; d_2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map {d_2} {\map f x, \map f y}
      | o = <
      | r = h \epsilon
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map {d_1} {x, y}
      | o = \le
      | r = \frac {\map {d_2} {\map f x, \map f y} } h < \epsilon
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | o = \in
      | r = \map {B_\epsilon} {x; d_1}
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = y
      | o = \in
      | r = \map {B_{\epsilon / k} } {x; d_1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map {d_1} {x, y}
      | o = <
      | r = \frac \epsilon k
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map {d_2} {\map f x, \map f y}
      | o = \le
      | r = k \map {d_1} {x, y} < \epsilon
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | o = \in
      | r = \map {B_\epsilon} {\map f x; d_2}
      | c = 
}}
{{end-eqn}}
Thus:
:$\map {B_{h \epsilon} } {\map f x; d_2} \subseteq \map {B_\epsilon} {x; d_1}$
:$\map {B_{\epsilon / k} } {x; d_1} \subseteq \map {B_\epsilon} {\map f x; d_2}$
{{qed|lemma}}
Now, suppose $U$ is $d_2$-open.
Let $x \in U$.
Then:
:$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} {\map f x; d_2} \subseteq U$.
Hence:
:$\map {B_{\epsilon / k} } {x; d_1} \subseteq U$
Thus $U$ is $d_1$-open.
Similarly, suppose $U$ is $d_1$-open.
Let $x \in U$.
Then:
:$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} {x; d_1} \subseteq U$
Hence:
:$\map {B_{h \epsilon} } {\map f x; d_2} \subseteq U$
Thus $U$ is $d_2$-open.
The result follows by definition of homeomorphic metric spaces.
{{qed}}
Theorem: P-Product Metrics on Real Vector Space are Topologically Equivalent

For $n \in \N$, let $\R^n$ be an Euclidean space.
Let $p \in \R_{\ge 1}$.
Let $d_p$ be the $p$-product metric on $\R^n$.
Let $d_\infty$ be the Chebyshev distance on $\R^n$.
Then $d_p$ and $d_\infty$ are topologically equivalent.

Proof:
Let $r, t \in \R_{\ge 1}$.
{{WLOG}}, assume that $r \le t$.
For all $x, y \in \R^n$, we are going to show that:
:$\map {d_r} {x, y} \ge \map {d_\infty} {x, y} \ge n^{-1} \map {d_r} {x, y}$
Then we can demonstrate Lipschitz equivalence between all of these metrics, from which topological equivalence follows.
Let $d_r$ be the metric defined as:
:$\ds \map {d_r} {x, y} = \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^r}^{1/r}$
Theorem: Derivative of Function of Constant Multiple

Let $f$ be a real function which is differentiable on $\R$.
Let $c \in \R$ be a constant.
Then:
:$\map {D_x} {\map f {c x} } = c \map {D_{c x} } {\map f {c x} }$

Proof:
First it is shown that $\map {D_x} {c x} = c$:
{{begin-eqn}}
{{eqn | l = \map {D_x} {c x}
      | r = c \map {D_x} x + x \map {D_x} c
      | c = Product Rule for Derivatives
}}
{{eqn | r = c + x \map {D_x} c
      | c = Derivative of Identity Function
}}
{{eqn | r = c + 0
      | c = Derivative of Constant
}}
{{eqn | r = c
}}
{{end-eqn}}
Next:
{{begin-eqn}}
{{eqn | l = \map {D_x} {\map f {c x} }
      | r = \map {D_x} {c x} \map {D_{c x} } {\map f {c x} }
      | c = Chain Rule for Derivatives
}}
{{eqn | r = c \map {D_{c x} } {\map f {c x} }
      | c = from above
}}
{{end-eqn}}
{{qed}}
Theorem: Cotangent Function is Periodic on Reals

The real cotangent function is periodic with period $\pi$.

Proof:
{{begin-eqn}}
{{eqn | l = \map \cot {x + \pi}
      | r = \frac {\map \cos {x + \pi} } {\map \sin {x + \pi} }
      | c = {{Defof|Cotangent}}
}}
{{eqn | r = \frac {-\cos x} {-\sin x}
      | c = Cosine of Angle plus Straight Angle, Sine of Angle plus Straight Angle
}}
{{eqn | r = \cot x
      | c = 
}}
{{end-eqn}}
Also, from Derivative of Cotangent Function:
:$\map {D_x} {\cot x} = -\dfrac 1 {\sin^2 x}$
provided $\sin x \ne 0$.
From Shape of Sine Function, $\sin$ is strictly positive on the interval $\openint 0 \pi$.
From Derivative of Monotone Function, $\cot x$ is strictly decreasing on that interval, and hence cannot have a period of ''less'' than $\pi$.
Hence the result.
{{qed}}
Category:Cotangent Function
127zhl56tdepsxzwe8i3bi0589oyq82
Theorem: Shape of Cotangent Function

The nature of the cotangent function on the set of real numbers $\R$ is as follows:
:$\cot x$ is continuous and strictly decreasing on the interval $\openint 0 \pi$
:$\cot x \to +\infty$ as $x \to 0^+$
:$\cot x \to -\infty$ as $x \to \pi^-$
:$\cot x$ is not defined on $\forall n \in \Z: x = n \pi$, at which points it is discontinuous
:$\forall n \in \Z: \map \cot {n + \dfrac 1 2} \pi = 0$

Proof:
$\cot x$ is continuous and strictly decreasing on $\openint 0 \pi$:
Continuity follows from the Quotient Rule for Continuous Real Functions:
:$(1): \quad$ Both $\sin x$ and $\cos x$ are continuous on $\openint 0 \pi$ from Real Sine Function is Continuous and Cosine Function is Continuous
:$(2): \quad \sin x > 0$ on this interval.
The fact of $\cot x$ being strictly decreasing on this interval has been demonstrated in the discussion on Cotangent Function is Periodic on Reals.
$\cot x \to + \infty$ as $x \to 0^+$:
From Sine and Cosine are Periodic on Reals, we have that both $\sin x > 0$ and $\cos x > 0$ on $\openint 0 {\dfrac \pi 2}$.
We have that:
:$(1): \quad \cos x \to 1$ as $x \to 0^+$
:$(2): \quad \sin x \to 0$ as $x \to 0^+$
Thus it follows that $\cot x = \dfrac {\cos x} {\sin x} \to + \infty$ as $x \to 0^+$.
* $\tan x \to - \infty$ as $x \to \pi^-$:
From Sine and Cosine are Periodic on Reals, we have that $\sin x > 0$ and $\cos x < 0$ on $\openint {\dfrac \pi 2} \pi$.
We have that:
:$(1): \quad \cos x \to -1$ as $x \to \pi^-$
:$(2): \quad \sin x \to 0$ as $x \to \pi^-$
Thus it follows that $\cot x = \dfrac {\cos x} {\sin x} \to - \infty$ as $x \to \pi^-$.
$\cot x$ is not defined and discontinuous at $x = n \pi$:
From the discussion of Sine and Cosine are Periodic on Reals, it was established that $\forall n \in \Z: x = n \pi \implies \sin x = 0$.
As division by zero is not defined, it follows that at these points $\cot x$ is not defined either.
Now, from the above, we have:
:$(1): \quad \cot x \to + \infty$ as $x \to 0^+$
:$(2): \quad \cot x \to - \infty$ as $x \to \pi^-$
As $\map \cot {x + \pi} = \cot x$ from Cotangent Function is Periodic on Reals, it follows that $\cot x \to + \infty$ as $x \to \pi^+$.
Hence the left hand limit and right hand limit at $x = \pi$ are not the same.
From the periodic nature of $\cot x$, it follows that the same applies $\forall n \in \Z: x = n \pi$.
The fact of its discontinuity at these points follows from the definition of discontinuity.
$\map \cot {n + \dfrac 1 2} \pi = 0$:
Follows directly from Sine and Cosine are Periodic on Reals:
:$\forall n \in \Z: \map \cos {n + \dfrac 1 2} \pi = 0$
{{qed}}
Theorem: Change of Base of Logarithm

:$\log_b x = \dfrac {\log_a x} {\log_a b}$

Proof:
Let:
:$y = \log_b x \iff b^y = x$
:$z = \log_a x \iff a^z = x$
Then:
{{begin-eqn}}
{{eqn | l = z
      | r = \map {\log_a} {b^y}
}}
{{eqn | r = y \log_a b
      | c = Logarithms of Powers
}}
{{eqn | r = \log_b x \log_a b
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Change of Base of Logarithm

:$\log_b x = \dfrac {\log_a x} {\log_a b}$

Proof:
We begin with an equation that relates two exponentials with two different bases, $a$ and $b$, both positive and neither equal to one:
:$b^y = a^z$
{{WLOG}}, select and solve for $b$:
:$b = a^{z/y}$
Take logarithm's base $a$ of both sides:
{{begin-eqn}}
{{eqn | l = \log_a b
      | r = \map {\log_a} {a^{z/y} }
}}
{{eqn | r = \dfrac z y \, \log_a a
      | c = Logarithm of Power
}}
{{eqn | l = \log_a b
      | r = \dfrac z y
      | c = using $\log_a a = 1$
}}
{{end-eqn}}
Now define $x = b^y = a^z$, and note that:
{{begin-eqn}}
{{eqn | l = \quad b^y = x
      | o = \implies
      | r = y = \log_b x
      | c = {{Defof|General Logarithm}}
}}
{{eqn | l = \quad a^z = x
      | o = \implies
      | r = z = \log_a x
      | c = {{Defof|General Logarithm}}
}}
{{end-eqn}}
Substituting these values of $y$ and $z$ into our expression, $\log_a b = z / y$, yields the desired version of the change-of-base formula:
:$\log_a b = \dfrac {\log_a x} {\log_b x} \implies \log_b x = \dfrac {\log_a x} {\log_a b}$
{{qed}}
Theorem: Derivative of Function to Power of Function

Let $\map u x, \map v x$ be real functions which are differentiable on $\R$.
Then:
:$\map {\dfrac \d {\d x} } {u^v} = v u^{v - 1} \map {\dfrac \d {\d x} } u + u^v \paren {\ln u} \map {\dfrac \d {\d x} } v$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d x} } {u^v}
      | r = \map {\dfrac \d {\d x} } {\map \exp {v \ln u} }
      | c = {{Defof|Power to Real Number}}
}}
{{eqn | r = \map \exp {v \ln u} \map {\dfrac \d {\d x} } {v \ln u}
      | c = Chain Rule for Derivatives and Derivative of Exponential Function
}}
{{eqn | r = \map \exp {v \ln u} \paren {\paren {\ln u} \map {\dfrac \d {\d x} } v + v \map {\dfrac \d {\d x} } {\ln u} }
      | c = Product Rule for Derivatives
}}
{{eqn | r = u^v \paren {\paren {\ln u} \map {\dfrac \d {\d x} } v + \frac v u \map {\dfrac \d {\d x} } u}
      | c = Chain Rule for Derivatives
}}
{{eqn | r = v u^{v - 1} \map {\dfrac \d {\d x} } u + u^v \paren {\ln u} \map {\dfrac \d {\d x} } v
      | c = gathering terms
}}
{{end-eqn}}
{{qed}}
Theorem: Liouville's Theorem (Complex Analysis)

Let $f: \C \to \C$ be a bounded entire function.
Then $f$ is constant.

Proof:
By assumption, there is $M \ge 0$ such that $\cmod {\map f z} \le  M$ for all $z \in \C$. 
For any $R \in \R: R > 0$, consider the function:
:$\map {f_R} z := \map f {R z}$
Using the Cauchy Integral Formula, we see that:
:$\ds \cmod {\map {f_R'} z} = \frac 1 {2 \pi} \cmod {\int_{\map {C_1} z} \frac {\map {f_R} w} {\paren {w - z}^2} \rd w} \le \frac 1 {2 \pi} \int_{\map {C_1} z} M \cmod {\d w} = M$
where $\map {C_1} z$ denotes the circle of radius $1$ around $z$.
{{Improve|$\int_{\map {C_1} z} M \cmod {\d w}$ must be defined.}}
Hence:
:$\ds \cmod {\map {f'} z} = \cmod {\map {f_R'} z} / R \le M / R$
Since $R$ was arbitrary, it follows that $\cmod {\map {f'} z} = 0$ for all $z \in \C$.
Thus $f$ is constant.
{{qed}}
Theorem: Liouville's Theorem (Complex Analysis)

Let $f: \C \to \C$ be a bounded entire function.
Then $f$ is constant.

Proof:
By assumption, there is $M \ge 0$ such that $\cmod {\map f z} \le  M$ for all $z \in \C$. 
Let $r > 0$.
consider:
:$D_r = \set {z \in \C: \cmod z \le r}$
Then, consider the parameterization $\gamma_r : \closedint 0 {2 \pi} \to \partial D_r$ given by:
:$\map  {\gamma_r} t := r e^{2 \pi i t}$
For all $z \in \C$, we have:
{{begin-eqn}}
{{eqn | l = \map {f'} z
      | r = \map {g_z'} 0
}}
{{eqn | r = \frac 1 {2 \pi} \oint_{\partial D_r} \frac {\map {g_z} w} {w^2} \rd w
      | c = Cauchy Integral Formula
}}
{{eqn | r = \frac 1 {2 \pi} \int_0^{2 \pi} \frac {\map {g_z} {\map {\gamma_r} t} } { {\map {\gamma_r} t}^2} {\map {\gamma_r '} t} \rd t
      | c = {{Defof|Complex Contour Integral}}
}}
{{eqn | r = \frac 1 {2 \pi} \int_0^{2 \pi} \frac {\map {g_z} {\map {\gamma_r} t} } { {\paren {r e^{2 \pi i t} } }^2} {2 \pi i r e^{2 \pi i t} } \rd t
}}
{{eqn | r = \frac i r \int_0^{2 \pi} \frac {\map {g_z} {\map {\gamma_r} t} } { e^{2 \pi i t} } \rd t
}}
{{eqn | r = \frac i r \int_0^{2 \pi} \frac {\map f { {\map {\gamma_r} t} + z} } { e^{2 \pi i t} } \rd t
}}
{{end-eqn}}
where:
:$\map {g_z} w := \map f {w + z}$
Hence, for all $z \in \C$:
{{begin-eqn}}
{{eqn | l = \cmod {\map {f'} z}
      | r = \cmod {\frac i r \int_0^{2 \pi} \frac {\map f { {\map {\gamma_r} t} + z} } { e^{2 \pi i t} } \rd t}
}}
{{eqn | o = \le
      | r = \frac 1 r \int_0^{2 \pi} \cmod {\frac {\map f {\map {\gamma_r} t + z} } { e^{2 \pi i t} } } \rd t
      | c = Modulus of Complex Integral
}}
{{eqn | o = \le
      | r = \frac 1 r \int_0^{2 \pi} M \rd t
}}
{{eqn | r = \dfrac {2 \pi M} r
}}
{{eqn | o = \to
      | r = 0
      | c = as $r \to \infty$
}}
{{end-eqn}}
Thus it follows that $\map {f'} z = 0$ for all $z \in \C$.
By Zero Derivative implies Constant Complex Function, $f$ is constant.
{{qed}}