avewright/proofwiki-math huggingface dataset:
Theorem: Difference of Unions is Subset of Union of Differences

Let $I$ be an indexing set.
Let $S_\alpha, T_\alpha$ be sets, for all $\alpha \in I$.
Then:
:$\ds \paren {\bigcup_{\alpha \mathop \in I} S_\alpha} \setminus \paren {\bigcup_{\alpha \mathop \in I} T_\alpha} \subseteq \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \setminus T_\alpha}$
where $S_\alpha \setminus T_\alpha$ denotes set difference.

Proof:
Let $\ds x \in \paren {\bigcup_{\alpha \mathop \in I} S_\alpha} \setminus \paren {\bigcup_{\alpha \mathop \in I} T_\alpha}$.
Then by definition of set difference:
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = \bigcup_{\alpha \mathop \in I} S_\alpha
      | c = 
}}
{{eqn | l = x
      | o = \notin
      | r = \bigcup_{\alpha \mathop \in I} T_\alpha
      | c = 
}}
{{end-eqn}}
By definition of set union, it follows that:
{{begin-eqn}}
{{eqn | q = \exists \beta \in I
      | l = x
      | o = \in
      | r = S_\beta
      | c = 
}}
{{eqn | q = \neg \exists \beta \in I
      | l = x
      | o = \in
      | r = T_\beta
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \forall \beta \in I
      | l = x
      | o = \notin
      | r = T_\beta
      | c = De Morgan's Laws (Predicate Logic)
}}
{{end-eqn}}
and so:
:$\exists \beta \in I: x \in S_\beta \setminus T_\beta$
Hence:
:$\ds x \in \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \setminus T_\alpha}$
by definition of set union.
The result follows by definition of subset.
{{qed}}
Theorem: Symmetric Difference of Unions is Subset of Union of Symmetric Differences

Let $I$ be an indexing set.
Let $S_\alpha, T_\alpha$ be sets, for all $\alpha \in I$.
Then:
:$\ds \bigcup_{\alpha \mathop \in I} S_\alpha \symdif \bigcup_{\alpha \mathop \in I} T_\alpha \subseteq \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \symdif T_\alpha}$
where $S \symdif T$ is the symmetric difference between $S$ and $T$.

Proof:
From Difference of Unions is Subset of Union of Differences, we have:
:$\ds \bigcup_{\alpha \mathop \in I} S_\alpha \setminus \bigcup_{\alpha \mathop \in I} T_\alpha \subseteq \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \setminus T_\alpha}$
:$\ds \bigcup_{\alpha \mathop \in I} T_i \setminus \bigcup_{\alpha \mathop \in I} S_\alpha \subseteq \bigcup_{\alpha \mathop \in I} \paren {T_\alpha \setminus S_\alpha}$
where $\setminus$ denotes set difference.
Thus we have:
{{begin-eqn}}
{{eqn | l = \bigcup_{\alpha \mathop \in I} S_\alpha \symdif \bigcup_{\alpha \mathop \in I} T_\alpha
      | r = \paren {\bigcup_{\alpha \mathop \in I} S_\alpha \setminus \bigcup_{\alpha \mathop \in I} T_\alpha} \cup \paren {\bigcup_{\alpha \mathop \in I} T_\alpha \setminus \bigcup_{\alpha \mathop \in I} S_\alpha}
      | c = {{Defof|Symmetric Difference}}
}}
{{eqn | o = \subseteq
      | r = \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \setminus T_\alpha} \cup \bigcup_{\alpha \mathop \in I} \paren {T_\alpha \setminus S_\alpha}
      | c = Difference of Unions is Subset of Union of Differences
}}
{{eqn | r = \bigcup_{\alpha \mathop \in I} \paren {\paren {S_\alpha \setminus T_\alpha} \cup \paren {T_\alpha \setminus S_\alpha} }
      | c = Union is Associative and Union is Commutative
}}
{{eqn | r = \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \symdif T_\alpha}
      | c = {{Defof|Symmetric Difference}}
}}
{{end-eqn}}
{{qed}}
Category:Set Union
Category:Symmetric Difference
ifszfdajpusadlx5g6277ms9k8qhsme
Theorem: Pseudometric Defines an Equivalence Relation

Let $X$ be a set on which there is a pseudometric $d: X \times X \to \R$.
For any $x, y \in X$, let $x \sim y$ denote that $\map d {x, y} = 0$.
Then $\sim$ is an equivalence relation, and the equivalence classes consist of sets of elements of $X$ at zero distance from each other.

Proof:
Checking in turn each of the criteria for equivalence:
Theorem: Metric Induced by a Pseudometric

Let $X$ be a set on which there is a pseudometric $d: X \times X \to \R$.
For any $x, y \in X$, let $x \sim y$ denote that $\map d {x, y} = 0$.
Let $\eqclass x \sim$ denote the equivalence class of $x$ under $\sim$.
Let $X^*$ be the quotient of $X$ by $\sim$.
Then the mapping $d^*: X^* \times X^* \to \R$ defined by:
:$\map {d^*} {\eqclass x \sim, \eqclass y \sim} = \map d {x, y}$
is a metric.
Hence $\struct {X^*, d^*}$ is a metric space.

Proof:
From Pseudometric Defines an Equivalence Relation we have that $\sim$ is indeed an equivalence relation.
First we verify that $d^*$ is well-defined.
Let $z \in \eqclass x \sim$ and $w \in \eqclass y \sim$.
From Equivalence Class Equivalent Statements:
:$\eqclass z \sim = \eqclass x \sim$ and $\eqclass w \sim = \eqclass y \sim$
{{begin-eqn}}
{{eqn | l = \map d {z, w}
      | o = \le
      | r = \map d {z, x} + \map d {x, y} + \map d {y, w}
}}
{{eqn | r = \map d {x, y}
      | c = Definition of $\sim$
}}
{{eqn | l = \map d {x, y}
      | o = \le
      | r = \map d {x, z} + \map d {z, w} + \map d {w, y}
}}
{{eqn | r = \map d {z, w}
      | c = Definition of $\sim$
}}
{{eqn | ll= \leadsto
      | l = \map d {x, y}
      | r = \map d {z, w}
}}
{{end-eqn}}
So $d^*$ is independent of the elements chosen from the equivalence classes.
Now we need to show that $d^*$ is a metric.
To do that, all we need to do is show that $\map {d^*} {a, b} > 0$ where $a \ne b$.
So let $\map {d^*} {a, b} = 0$ where $a = \eqclass x \sim, b = \eqclass y \sim$.
Then $\map d {x, y} = 0$ so $y \in \eqclass x \sim$ and so $a = b$.
Hence the result.
{{qed}}
Category:Pseudometric Spaces
Category:Metric Spaces
amycv8d4853lfaew2g5islv14d74iyf
Theorem: First-Order Reaction

Let a substance decompose spontaneously in a '''first-order reaction'''.
Let $x_0$ be a measure of the quantity of that substance at time $t = 0$.
Let the quantity of the substance that remains after time $t$ be $x$.
Then:
:$x = x_0 e^{-k t}$
where $k$ is the rate constant.

Proof:
From Differential Equation governing First-Order Reaction, the differential equation governing this reaction is given by:
:$-\dfrac {\d x} {\d t} = k x$
for $k \in \R_{>0}$.
{{begin-eqn}}
{{eqn | l = \int \dfrac {\d x} x
      | r = \int - k \rd t
      | c = Solution to Separable Differential Equation
}}
{{eqn | ll= \leadsto
      | l = \ln x
      | r = -k t + c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = e^{- k t + c}
      | c = 
}}
{{eqn | r = e^c e^{- k t}
      | c = 
}}
{{end-eqn}}
At time $t = 0$ we have that $x = x_0$.
So:
:$x_0 = e^c e^0 = e^c$
and hence the result
:$x = x_0 e^{-k t}$
{{qed}}
Theorem: First-Order Reaction

Let a substance decompose spontaneously in a '''first-order reaction'''.
Let $x_0$ be a measure of the quantity of that substance at time $t = 0$.
Let the quantity of the substance that remains after time $t$ be $x$.
Then:
:$x = x_0 e^{-k t}$
where $k$ is the rate constant.

Proof:
From Differential Equation governing First-Order Reaction, the differential equation governing this reaction is given by:
:$-\dfrac {\d x} {\d t} = k x$
for $k \in \R_{>0}$.
This is an instance of the Decay Equation, and has the solution:
:$x = x_0 e^{-k t}$
{{qed}}
Theorem: Solution to First Order ODE

Let:
:$\Phi = \dfrac {\d y} {\d x} = \map f {x, y}$
be a first order ordinary differential equation.
Then $\Phi$ has a general solution which can be expressed in terms of an indefinite integral of $\map f x$:
:$\ds y = \int \map f {x, y} \rd x + C$
where $C$ is an arbitrary constant.

Proof:
Integrating both sides with respect to $x$:
{{begin-eqn}}
{{eqn | l = \int \paren {\frac {\d y} {\d x} } \rd x
      | r = \int \map f {x, y} \rd x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y + C_1
      | r = \int \map f {x, y} \rd x
      | c = {{Defof|Indefinite Integral}}: $C_1$ is arbitrary
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \int \map f {x, y} \rd x + C
      | c = replacing $-C_1$ with $C$
}}
{{end-eqn}}
The validity of this follows from Picard's Existence Theorem.
{{qed}}
Category:First Order ODEs
341o13cwb5gvkaszzxrzrbel1lzbchy
Theorem: Solution to First Order Initial Value Problem

Let $\map y x$ be a solution to the first order ordinary differential equation:
:$\dfrac {\d y} {\d x} = \map f {x, y}$
which is subject to an initial condition: $\tuple {a, b}$.
Then this problem is equivalent to the integral equation:
:$\ds y = b + \int_a^x \map f {t, \map y t} \rd t$

Proof:
From Solution to First Order ODE, the general solution of:
:$\dfrac {\d y} {\d x} = \map f {x, y}$
is:
:$\ds y = \int \map f {x, \map y x} \rd x + C$
When $x = a$, we have $y = b$.
Thus:
:$\ds b = \valueat {\int \map f {x, \map y x} \rd x + C} a$
{{MissingLinks|to the notation $[..]_a$}}
which gives:
:$\ds C = b - \valueat {\int \map f {x, \map y x} \rd x} a$
and so:
$\ds y = b + \int \map f {x, \map y x} \rd x - \valueat {\int \map f {x, \map y x} \rd x} a$
whence the result, by the Fundamental Theorem of Calculus.
{{qed}}
Category:First Order ODEs
i0p26rgmkp607ipzi7ff4o3gnhc6tiv
Theorem: Condition for Lipschitz Condition to be Satisfied

Let $\phi$ be a real function.
Then $\phi$ satisfies the Lipschitz condition on a closed real interval $\closedint a b$ if:
:$\forall y \in \closedint a b: \exists A \in \R: \size {\map {\phi'} y} \le A$

Proof:
Integrating both sides of $\size {\map {\phi'} y} \le A$ gives us:
{{begin-eqn}}     
{{eqn | l = \size {\map {\phi'} y}
      | o = \le
      | r = A
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -A
      | o = \le
      | r = \map {\phi'} y \le A
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int {-A} \rd y
      | o = \le
      | r = \map \phi y \le \int A \rd y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -A y
      | o = \le
      | r = \map \phi y \le y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \size {\map \phi y}
      | o = \le
      | r = A y
      | c = 
}}
{{end-eqn}}
On the interval $\closedint a b$ it follows that $\size {\map \phi y}$ is bounded by the greater of $A a$ and $A b$.
Hence the result.
{{qed}}
Category:Calculus
76a11q7p0pufuwmztwjht5uak92mi0g
Theorem: Quotient of Homogeneous Functions

Let $\map M {x, y}$ and $\map N {x, y}$ be homogeneous functions of the same degree.
Then:
:$\dfrac {\map M {x, y} } {\map N {x, y} }$
is homogeneous of zero degree.

Proof:
Let:
:$\map Q {x, y} = \dfrac {\map M {x, y} } {\map N {x, y} }$
where $M$ and $N$ are homogeneous functions of degree $n$.
Let $t \in \R$.
Then:
{{begin-eqn}}
{{eqn | l = \map Q {t x, t y}
      | r = \frac {\map M {t x, t y} } {\map N {t x, t y} }
      | c = 
}}
{{eqn | r = \frac {t^n \map M {x, y} } {t^n \map N {x, y} }
      | c = as $M$ and $N$ are homogeneous of degree $n$
}}
{{eqn | r = t^0 \frac {\map M {x, y} } {\map N {x, y} }
      | c = 
}}
{{eqn | r = t^0 \map Q {x, y}
      | c = 
}}
{{eqn | r = \map Q {x, y}
      | c = 
}}
{{end-eqn}}
The result follows from the definition.
{{qed}}
Theorem: Solution to Homogeneous Differential Equation

Let:
:$\map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
be a homogeneous differential equation.
It can be solved by making the substitution $z = \dfrac y x$.
Its solution is:
:$\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
where:
:$\map f {x, y} = -\dfrac {\map M {x, y} } {\map N {x, y} }$

Proof:
From the original equation, we see:
:$\dfrac {\d y} {\d x} = \map f {x, y} = -\dfrac {\map M {x, y} } {\map N {x, y} }$
From Quotient of Homogeneous Functions it follows that $\map f {x, y}$ is homogeneous of degree zero.
Thus:
:$\map f {t x, t y} = t^0 \map f {x, y} = \map f {x, y}$
Set $t = \dfrac 1 x$ in this equation:
{{begin-eqn}}
{{eqn | l = \map f {x, y}
      | r = \map f {\paren {\frac 1 x} x, \paren {\frac 1 x} y}
      | c =
}}
{{eqn | r = \map f {1, \frac y x}
      | c =
}}
{{eqn | r = \map f {1, z}
      | c =
}}
{{end-eqn}}
where $z = \dfrac y x$.
Then:
{{begin-eqn}}
{{eqn | l = z
      | r = \frac y x
      | c =
}}
{{eqn | ll= \leadsto
      | l = y
      | r = z x
      | c =
}}
{{eqn | ll= \leadsto
      | l = \frac {\d y} {\d x}
      | r = z + x \frac {\d z} {\d x}
      | c = Product Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = z + x \frac {\d z} {\d x}
      | r = \map f {1, z}
      | c =
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d z} {\map f {1, z} - z}
      | r = \int \frac {\d x} x
      | c =
}}
{{end-eqn}}
This is seen to be a separable differential equation.
On performing the required integrations and simplifying as necessary, the final step is to substitute $\dfrac y x$ back for $z$.
{{qed}}
Theorem: Solution to Exact Differential Equation

The first order ordinary differential equation:
:$F = \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
is an exact differential equation {{iff}}:
:$\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$
The general solution of such an equation is:
:$\map f {x, y} = C$
where:
:$\dfrac {\partial f} {\partial x} = M$
:$\dfrac {\partial f} {\partial y} = N$

Proof:
=== Necessary Condition ===
Let $F$ be exact.
Then by definition there exists a function whose second partial derivatives of $f$ exist and are continuous:
:$\map f {x, y}$
where:
:$\dfrac {\partial f} {\partial x} = M$
:$\dfrac {\partial f} {\partial y} = N$
Differentiating $M$ and $N$ partially {{WRT|Differentiation}} $y$ and $x$ respectively:
:$\dfrac {\partial M} {\partial y} = \dfrac {\partial^2 f} {\partial x \partial y}$
:$\dfrac {\partial N} {\partial x} = \dfrac {\partial^2 f} {\partial y \partial x}$
From Mixed Second Partial Derivatives are Equal:
:$\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$
{{qed|lemma}}
Theorem: Existence of Integrating Factor

Let the first order ordinary differential equation:
:$(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
be such that $M$ and $N$ are real functions of two variables which are ''not'' homogeneous functions of the same degree.
Suppose also that:
:$\dfrac {\partial M} {\partial y} \ne \dfrac {\partial N} {\partial x}$
that is, $(1)$ is not exact.
Finally, suppose that $(1)$ has a general solution.
Then it is always possible to find an integrating factor $\map \mu {x, y}$ such that:
:$\map \mu {x, y} \paren {\map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} } = 0$
is an exact differential equation.
Hence it is possible to find that solution by Solution to Exact Differential Equation.

Proof:
Let us for ease of manipulation express $(1)$ in the form of differentials:
:$(2): \quad \map M {x, y} \rd x + \map N {x, y} \rd y = 0$
Suppose that $(2)$ has a general solution:
:$(3): \quad \map f {x, y} = C$
where $C$ is some constant.
We can eliminate $C$ by differentiating:
:$\dfrac {\partial f} {\partial x} \rd x + \dfrac {\partial f} {\partial y} \rd y = 0$
It follows from $(2)$ and $(3)$ that:
:$\dfrac {\d y} {\d x} = - \dfrac M N = -\dfrac {\partial f / \partial x} {\partial f / \partial y}$
and so:
:$(4): \quad \dfrac {\partial f / \partial x} M = \dfrac {\partial f / \partial y} N$
Let this common ratio in $(4)$ be denoted $\map \mu {x, y}$.
Then:
:$\dfrac {\partial f} {\partial x} = \mu M$
:$\dfrac {\partial f} {\partial y} = \mu N$
So, if we multiply $(2)$ by $\mu$, we get:
:$\mu M \rd x + \mu N \rd y = 0$
or:
:$\dfrac {\partial f} {\partial x} \rd x + \dfrac {\partial f} {\partial y} \rd y = 0$
which is exact.
So, if $(2)$ has a general solution, it has at least one integrating factor $\map \mu {x, y}$.
{{qed}}
Theorem: Integrating Factor for First Order ODE

Let the first order ordinary differential equation:
:$(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
be non-homogeneous and not exact.
By Existence of Integrating Factor, if $(1)$ has a general solution, there exists an integrating factor $\map \mu {x, y}$ such that:
:$\ds \map \mu {x, y} \paren {\map M {x, y} + \map N {x, y} \frac {\d y} {\d x} } = 0$
is an exact differential equation.
Unfortunately, there is no systematic method of finding such a $\map \mu {x, y}$ for all such equations $(1)$.<ref>"In general this is quite difficult."
::: -- {{BookReference|Differential Equations|1972|George F. Simmons}}: $\S 2.9$: Integrating Factors</ref>
However, there are certain types of first order ODE for which an integrating factor ''can'' be found procedurally.

Proof:
We have one of these:
: Integrating Factor for First Order ODE: Function of One Variable: $x$ or $y$ only
: Integrating Factor for First Order ODE: Function of $x + y$
: Integrating Factor for First Order ODE: Function of $x y$
Theorem: Solution to Linear First Order Ordinary Differential Equation

A linear first order ordinary differential equation in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
has the general solution:
:$\ds y = e^{-\int P \rd x} \paren {\int Q e^{\int P \rd x} \rd x + C}$

Proof:
Consider the first order ordinary differential equation:
:$\map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
We can put our equation:
:$(1): \quad \dfrac {\d y} {\d x} + \map P x y = \map Q x$
into this format by identifying:
:$\map M {x, y} \equiv \map P x y - \map Q x, \map N {x, y} \equiv 1$
We see that:
:$\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x} = \map P x$
and hence:
:$\map P x = \dfrac {\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x} } N$
is a function of $x$ only.
It immediately follows from Integrating Factor for First Order ODE that:
:$e^{\int \map P x \rd x}$
is an integrating factor for $(1)$.
So, multiplying $(1)$ by this factor:
:$e^{\int \map P x \rd x} \dfrac {\d y} {\d x} + e^{\int \map P x \rd x} \map P x y = e^{\int \map P x \rd x} \map Q x$
The result follows by an application of Solution to Exact Differential Equation.
{{qed}}
Theorem: Solution to Linear First Order Ordinary Differential Equation

A linear first order ordinary differential equation in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
has the general solution:
:$\ds y = e^{-\int P \rd x} \paren {\int Q e^{\int P \rd x} \rd x + C}$

Proof:
From the Product Rule for Derivatives:
{{begin-eqn}}
{{eqn | l = \frac \d {\d x} \paren {e^{\int \map P x \rd x} y}
      | r = e^{\int \map P x \rd x} \dfrac {\d y} {\d x} + y e^{\int \map P x \rd x} \map P x
      | c = 
}}
{{eqn | r = e^{\int \map P x \rd x} \paren {\dfrac {\d y} {\d x} + \map P x y}
      | c = 
}}
{{end-eqn}}
Hence, multiplying $(1)$ all through by $e^{\int \map P x \rd x}$:
:$\dfrac \d {\d x} \paren {e^{\int \map P x \rd x} y} = \map Q x e^{\int \map P x \rd x}$
Integrating {{WRT|Integration}} $x$ now gives:
:$\ds e^{\int \map P x \rd x} y = \int \map Q x e^{\int \map P x \rd x} \rd x + C$
whence we get the result by dividing by $e^{\int \map P x \rd x}$.
{{qed}}
Theorem: Extremal Length of Composition

Let $\Gamma_1$ and $\Gamma_2$ be families of (unions of) rectifiable curves on a Riemann surface $X$.
Let $\Gamma_1$ and $\Gamma_2$ be disjoint in the sense that there exist disjoint Borel sets $E_1 \subseteq X$ and $E_2 \subseteq X$ with $\bigcup \Gamma_1 \subset E_1$ and $\bigcup \Gamma_2 \subset E_2$.
The extremal length of the family:
:$\Gamma := \set {\gamma_1 \cup \gamma_2:\ \gamma_1 \in \Gamma_1 \text{ and }\gamma_2 \in \Gamma_2}$
satisfies:
:$\map \lambda \Gamma = \map \lambda {\Gamma_1} + \map \lambda {\Gamma_2}$

Proof:
By the Series Law for Extremal Length:
:$\map \lambda \Gamma \ge \map \lambda {\Gamma_1} + \map \lambda {\Gamma_2}$
Hence it remains only to prove the opposite inequality.
Let $\rho$ be a metric as in the definition of extremal length, normalized such that $\map A \rho = 1$.
The claim to be proved is that:
:$\paren {\map L {\Gamma, \rho} }^2 \le \map \lambda {\Gamma_1} + \map \lambda {\Gamma_2}$
Define $\alpha_j := \sqrt {\map A {\rho {\restriction_{E_j} } } }$ for $j = 1, 2$.
Suppose either $\map \lambda {\Gamma_1}$ or $\map \lambda {\Gamma_2}$ is infinite.
It trivially follows that both $\alpha_1$ and $\alpha_2$ are positive.
Assume both extremal lengths are finite.
Let $\alpha_j = 0$.
Then:
:$\map L {\Gamma_j, \rho} = 0$
Indeed, we can choose a metric $\tilde \rho$ that agrees with $\rho$ on $A_j$ and has total area $\epsilon > 0$.
Then:
:$\paren {\map L {\Gamma_j, \rho} }^2 = \paren {\map L {\Gamma_j, \tilde \rho} }^2 \le \epsilon \cdot \map \lambda {\Gamma_j}$
Because $\epsilon$ was arbitrary, the claim follows.
{{WLOG}}, suppose $\alpha_2 = 0$.
Then:
:$\paren {\map L {\Gamma_j, \rho} }^2 = \paren {\map L {\Gamma_1, \rho} + \map L {\Gamma_2, \rho} }^2 = \paren {\map L {\Gamma_j, \rho} }^2 \le \map \lambda {\Gamma_1} \le \map \lambda {\Gamma_1} + \map \lambda {\Gamma_2}$
Thus both $\alpha_1$ and $\alpha_2$ are positive.
Let:
:$\rho_j := \dfrac {\rho {\restriction_{E_j} } } {\alpha_j}$
Then $\map A \rho = 1$.
So:
{{begin-eqn}}
{{eqn | l = \paren {\map L {\Gamma, \rho} }^2
      | r = \paren {\map L {\Gamma_1, \rho} + \map L {\Gamma_2, \rho} }^2
      | c = Definition of $\Gamma$
}}
{{eqn | r = \paren {\map L {\Gamma_1, \alpha_1 \rho_1} + \map L {\Gamma_2, \alpha_2 \rho_2} }^2
      | c = Definition of $\rho_j$ and disjointness of $E_1$ and $E_2$
}}
{{eqn | r = \paren {\alpha_1 \map L {\Gamma_1, \rho_1} + \alpha_2 \map L {\Gamma_2, \rho_2} }^2
      | c = Definition of $\map L {\Gamma_j, \rho_j}$)
}}
{{eqn | o = \le
      | r = \paren {\alpha_1^2 + \alpha_2^2} \paren {\map L {\Gamma_1, \rho_1}^2 + \map L {\Gamma_2, \rho_2}^2}
      | c = Cauchy-Bunyakovsky-Schwarz Inequality
}}
{{eqn | o = \le
      | r = \map L {\Gamma_1, \rho_1}^2 + \map L {\Gamma_2, \rho_2}^2
      | c = as $\alpha_1^2 + \alpha_2^2 \le \map A \rho = 1$
}}
{{eqn | o = \le
      | r = \map \lambda {\Gamma_1} + \map \lambda {\Gamma_2}
      | c = {{Defof|Extremal Length}}
}}
{{end-eqn}}
In conclusion:
:$\map \lambda \Gamma = \sup_\rho \map L {\Gamma, \rho} \le \map \lambda {\Gamma_1} + \map \lambda {\Gamma_2}$
as claimed.
{{qed}}
Category:Functional Analysis
qhz0ox9fd1o74dx6q3nttg9f8srqtg7
Theorem: Squares with All Odd Digits

The only squares whose digits (when written in base $10$ notation) are all odd are $1$ and $9$.

Proof:
If $n$ is even, then at least the last digit of $n^2$ is even.
So for $n^2$ to have its digits all odd, $n$ itself must be odd.
We can see immediately that $1 = 1^2$ and $9 = 3^2$ fit the criterion.
Of the other $1$-digit odd integers, we have $5^2 = 25, 7^2 = 49, 9^2 = 81$, all of which have an even digit.
Now, let $n > 10$ be an odd integer.
There are $5$ cases to consider:
* $n = 10 p + 1$: we have $\paren {10 p + 1}^2 = 100 p^2 + 20 p + 1 = 10 \paren {10 p^2 + 2 p} + 1$.
* $n = 10 p + 3$: we have $\paren {10 p + 3}^2 = 100 p^2 + 60 p + 9 = 10 \paren {10 p^2 + 6 p} + 9$.
* $n = 10 p + 5$: we have $\paren {10 p + 5}^2 = 100 p^2 + 100 p + 25 = 10 \paren {10 p^2 + 10 p + 2} + 5$.
* $n = 10 p + 7$: we have $\paren {10 p + 7}^2 = 100 p^2 + 140 p + 49 = 10 \paren {10 p^2 + 14 p + 4} + 9$.
* $n = 10 p + 9$: we have $\paren {10 p + 9}^2 = 100 p^2 + 180 p + 81 = 10 \paren {10 p^2 + 18 p + 8} + 1$.
It is clear that in all cases the $10$s digit is even.
So the square of every odd integer greater than $3$ always has at least one even digit.
Hence the result.
{{qed}}
Category:Square Numbers
9kr13bemkuh1maygdgbikblbyrwd39u
Theorem: Parity of Integer equals Parity of its Square

Let $p \in \Z$ be an integer.
Then $p$ is even {{iff}} $p^2$ is even.

Proof:
Let $p$ be an integer.
By the Division Theorem, there are unique integers $k$ and $r$ such that $p = 2k + r$ and $0 \le r < 2$.
That is, $r = 0$ or $r = 1$, where $r = 0$ corresponds to the case of $p$ being even and $r = 1$ corresponds to the case of $p$ being odd.