avewright/proofwiki-math huggingface dataset:
Theorem: Semiperimeter of Integer Heronian Triangle is Composite

The semiperimeter of an integer Heronian triangle is always a composite number.

Proof:
Let $a, b, c$ be the side lengths of an integer Heronian triangle.
By Heron's Formula, its area is given by:
:$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} } \in \N$
where the semiperimeter $s$ is given by:
:$s = \dfrac {a + b + c} 2$
First we prove that $s$ is indeed an integer.
{{AimForCont}} not.
Since $2 s = a + b + c \in \N$, $2 s$ must be odd.
Hence $2 s - 2 a, 2 s - 2 b, 2 s - 2 c$ are odd as well.
Thus:
{{begin-eqn}}
{{eqn | l = 16 \AA^2
      | r = 16 s \paren {s - a} \paren {s - b} \paren {s - c}
}}
{{eqn | r = 2 s \paren {2 s - 2 a} \paren {2 s - 2 b} \paren {2 s - 2 c}
}}
{{end-eqn}}
Since $16 \AA^2$ is a product of odd numbers, it must be odd.
But then $\AA^2$ is not an integer, a contradiction.
Therefore $s \in \N$.
{{qed|lemma}}
Now we show that $s$ is composite number.
{{AimForCont}} not.
Then $s$ is either $1$ or prime.
Since $a, b, c \ge 1$, $s \ge \dfrac 3 2 > 1$.
Hence $s$ is prime.
Since:
:$\AA^2 = s \paren {s - a} \paren {s - b} \paren {s - c}$
We have $s \divides \AA^2$.
By Prime Divides Power, $s^2 \divides \AA^2$.
Thus $s \divides \paren {s - a} \paren {s - b} \paren {s - c}$.
By Euclid's Lemma, $s$ divides some $s - x$.
However by Absolute Value of Integer is not less than Divisors:
:$s \le s - x$
which is a contradiction.
Therefore $s$ is composite.
{{qed}}
Category:Heronian Triangles
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Theorem: Matroid Bases Iff Satisfies Formulation 1 of Matroid Base Axiom/Necessary Condition

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $\mathscr B$ be the set of bases of the matroid $M$.
Then $\mathscr B$ satisfies formulation $1$ of base axiom:
{{:Axiom:Base Axiom (Matroid)/Formulation 1}}

Proof:
Let $B_1, B_2 \in \mathscr B$.
Let $x \in B_1 \setminus B_2$.
We have:
{{begin-eqn}}
{{eqn | l = \size {B_1 \setminus \set x}
      | r = \size {B_1} - \size {\set x}
      | c = Cardinality of Set Difference with Subset
}}
{{eqn | r = \size {B_2} - \size {\set x}
      | c = All Bases of Matroid have same Cardinality
}}
{{eqn | r = \size {B_2} - 1
      | c = Cardinality of Singleton
}}
{{eqn | o = <
      | r = \size {B_2}
}}
{{end-eqn}}
By matroid axiom $(\text I 3)$:
:$\exists y \in B_2 \setminus \paren{B_1 \setminus \set x} : \paren{ B_1 \setminus \set x} \cup \set y \in \mathscr I$
We have:
{{begin-eqn}}
{{eqn | l = B_2 \setminus \paren{B_1 \setminus \set x}
      | r = \paren{B_2 \setminus B_1} \cup \paren{B_2 \cap \set x}
      | c = Set Difference with Set Difference is Union of Set Difference with Intersection
}}
{{eqn | r = \paren{B_2 \setminus B_1} \cup \O
      | c = Intersection With Singleton is Disjoint if Not Element
}}
{{eqn | r = B_2 \setminus B_1
      | c = Union with Empty Set
}}
{{end-eqn}}
Then:
:$\exists y \in B_2 \setminus B_1 : \paren{ B_1 \setminus \set x} \cup \set y \in \mathscr I$
We have:
{{begin-eqn}}
{{eqn | l = \size{\paren { B_1 \setminus \set x} \cup \set y}
      | r = \size{B_1 \setminus \set x} + \size{\set y}
      | c = {{Corollary|Cardinality of Set Union}}
}}
{{eqn | r = \size{B_1} - \size{\set x} + \size{\set y}
      | c = Cardinality of Set Difference with Subset
}}
{{eqn | r = \size{B_1} - 1 + 1
      | c = Cardinality of Singleton
}}
{{eqn | r = \size{B_1}
}}
{{end-eqn}}
From Independent Subset is Base if Cardinality Equals Cardinality of Base:
:$\paren { B_1 \setminus \set x} \cup \set y \in \mathscr B$
Since $x$, $B_1$ and $B_2$ were arbitrary then the result follows.
Theorem: Independent Subset is Base if Cardinality Equals Rank of Matroid

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $\rho: \powerset S \to \Z$ be the rank function of $M$.
Let $B \in \mathscr I$ such that:
:$\size B = \map \rho S$
Then:
:$B$ is a base of $M$.

Proof:
Let $Z \in \mathscr I$ such that:
:$B \subseteq Z$
From Cardinality of Subset of Finite Set:
:$\size B \le \size Z$
By definition of the rank function:
:$\size Z \le \map \rho S$
Then:
:$\size Z = \size B$
From the contrapositive statement of Cardinality of Proper Subset of Finite Set:
:$B = Z$
It follows that $B$ is a maximal independent subset by definition.
That is, $B$ is a base by definition.
{{qed}}
Category:Matroid Independent Subsets
Category:Matroid Bases
Category:Matroid Rank Functions
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Theorem: Matroid Bases Iff Satisfies Formulation 1 of Matroid Base Axiom/Sufficient Condition

Let $S$ be a finite set.
Let $\mathscr B$ be a non-empty set of subsets of $S$ satisfying formulation $1$ of base axiom:
{{:Axiom:Base Axiom (Matroid)/Formulation 1}}
Then $\mathscr B$ is the set of bases of a matroid on $S$.

Proof:
Let $\mathscr I = \set{X \subseteq S : \exists B \in \mathscr B : X \subseteq B}$
It is to be shown that:
:* $\quad \mathscr I$ satisfies the matroid axioms
and
:* $\quad \mathscr B$ is the set of bases of the matroid $M = \struct{S, \mathscr I}$
From Independence System Induced from Set of Subsets:
:$\mathscr I$ is an independence system
==== Matroid Axiom $(\text I 3)$ ====
Let $U, V \in \mathscr I$ such that:
:$\card V < \card U$
By definition of $\mathscr I$:
:$\exists B_1, B_2 \in \mathscr B : U \subseteq B_1, V \subseteq B_2$
From Max Operation Equals an Operand:
:$\exists B_1, B_2 \in \mathscr B : U \subseteq B_1, V \subseteq B_2 : \card{B_1 \cap B_2} = \max \set{\card{C_1 \cap C_2} : U \subseteq C_1, V \subseteq C_2 \text{ and } C_1, C_2 \in \mathscr B}$
{{AimForCont}}:
:$B_2 \cap \paren{U \setminus V} = \O$
===== Lemma 1 =====
{{:Matroid Bases Iff Satisfies Formulation 1 of Matroid Base Axiom/Lemma 1}}{{qed|lemma}}
This contradicts the choice of $B_1, B_2$ such that:
:$\card{B_1 \cap B_2} = \max \set{\card{C_1 \cap C_2} : U \subseteq C_1, V \subseteq C_2 \text{ and } C_1, C_2 \in \mathscr B}$
Hence:
:$B_2 \cap \paren{U \setminus V} \ne \O$
Let $x \in B_2 \cap \paren{U \setminus V}$.
From Union of Subsets is Subset:
:$V \cup \set x \subseteq B_2$
By definition of $\mathscr I$:
:$V \cup \set x \in \mathscr I$
It follows that $\mathscr I$ satisfies the matroid axiom $(\text I 3)$ by definition.
{{qed|lemma}}
This completes the proof that $M = \struct{S, \mathscr I}$ forms a matroid. 
{{qed|lemma}}
==== $\mathscr B$ is Set of Bases ====
Let $B \in \mathscr B$.
From Set is Subset of Itself:
:$B \in \mathscr I$
Let $U \in \mathscr I$ such that:
:$B \subseteq U$
By definition of $\mathscr I$:
:$\exists B' \in \mathscr B : I \subseteq B'$
From Subset Relation is Transitive:
:$B \subseteq B'$
===== Lemma 2 =====
{{:Matroid Bases Iff Satisfies Formulation 1 of Matroid Base Axiom/Lemma 2}}{{qed|lemma}}
From Lemma 2:
:$\card B = \card {B'}$
From Cardinality of Proper Subset of Finite Set:
:$B = B'$
By definition of set equality:
:$U = B$
It has been shown that $B$ is a maximal subset of the ordered set $\struct{\mathscr I, \subseteq}$.
It follows that $\mathscr B$ is the set of bases of the matroid $M = \struct{S, I}$ by definition.
Theorem: Mean Number of Elements Fixed by Self-Map

Let $n \in \Z_{>0}$ be a strictly positive integer.
Let $S$ be a finite set of cardinality $n$.
Let $S^S$ be the set of all mappings from $S$ to itself.
Let $\map \mu n$ denote the arithmetic mean of the number of fixed points of all the mappings in $S^S$.
Then:
:$\map \mu n = 1$

Proof:
Let $f \in S^S$ be an arbitrary mapping from $S$ to itself.
Let $s \in S$ be an arbitrary element of $S$.
$s$ has an equal probability of being mapped to any element of $S$.
Hence the probability that $\map f s = s$ is equal to $\dfrac 1 n$.
There are $n$ elements of $S$.
By the above argument, each one has a probability of $\dfrac 1 n$ that it is a fixed point.
Thus the expectation of the number of fixed points is $n \times \dfrac 1 n = 1$.
{{qed}}
Theorem: Divisor of Product

Let $a, b, c \in \Z$ be integers.
Let the symbol $\divides$ denote the divisibility relation.
Let $a \divides b c$.
Then there exist integers $r, s$ such that:
:$a = r s$, where $r \divides b$ and $s \divides c$.

Proof:
Let $r = \gcd \set {a, b}$.
By Integers Divided by GCD are Coprime:
:$\exists s, t \in \Z: a = r s \land b = r t \land \gcd \set {s, t} = 1$
So we have written $a = r s$ where $r$ divides $b$.
We now show that $s$ divides $c$.
Since $a$ divides $b c$ there exists $k$ such that $b c = k a$.
Substituting for $a$ and $b$:
:$r t c = k r s$
which gives:
:$t c = k s$
So $s$ divides $t c$.
But we have that:
:$s \perp t$
Hence by Euclid's Lemma $s \divides c$ as required.
{{qed}}
Category:Divisors
3tjwbapcmnp3ys5ykp601oz57g71k9i
Theorem: Equivalence of Definitions of Matroid

Let $M = \struct {S, \mathscr I}$ be an independence system.
{{TFAE|def = Matroid}}

Proof:
=== Definition 1 implies Definition 2 ===
{{:Equivalence of Definitions of Matroid/Definition 1 implies Definition 2}}{{qed|lemma}}
Theorem: P-adic Norm is Well Defined

P-adic norm $\norm {\, \cdot \,}_p$ is well defined.

Proof:
{{AimForCont}} $\norm {\, \cdot \,}_p$ is not well defined.
Then, given $r \in \Q$, for two equivalent representations of $r$, $\norm r_p$ will yield two different results. 
Let $k_1, k_2, m_1, m_2 \in \Z, n_1, n_2 \in \Z_{\ne 0} : p \nmid m_1, m_2, n_1, n_2$.
Let $\ds r = p^{k_1} \frac {m_1} {n_1} = p^{k_2} \frac {m_2} {n_2}$, with $k_1 \ne k_2$.
Suppose $k_2 < k_1$.
Then:
:$p^{k_1 - k_2} m_1 n_2 = m_2 n_1$
Therefore:
:$p \divides m_2 n_1$
Since $p$ is prime, it cannot be expressed as a product of selected divisors of both $m_2$ and $n_1$.
Hence $p \divides m_2$ or $p \divides n_1$.
This is a contradiction.
Thus, $k_1 \le k_2$.
Similarly, assuming $k_1 < k_2$ leads to a contradiction.
Hence, $k_1 \ge k_2$.
Since $k_1 \ne k_2$, both $k_1$ and $k_2$ have to be such that:
:$k_1 < k_2$
:$k_1 > k_2$
are satisfied.
No integers satisfy this.
{{help|because Integers form Totally Ordered Ring. How to use this argument rigorously?}}
Hence, we reached a contradiction.
{{qed}}
Theorem: Equivalence of Definitions of Matroid/Definition 1 implies Definition 2

Let $M = \struct {S, \mathscr I}$ be an independence system.
Let $M$ also satisfy:
{{begin-axiom}}
{{axiom | n = \text I 3
        | q = \forall U, V \in \mathscr I
        | mr= \size V < \size U \implies \exists x \in U \setminus V : V \cup \set x \in \mathscr I
}}
{{end-axiom}}
Then $M$ satisfies:
{{begin-axiom}}
{{axiom | n = \text I 4
        | q = \forall U, V \in \mathscr I
        | mr= \size U = \size V + 1 \implies \exists x \in U \setminus V : V \cup \set x \in \mathscr I
}}
{{end-axiom}}

Proof:
Since:
:$\forall U, V \in \mathscr I : \size U = \size V + 1 \implies \size V < \size U$
If follows that if $M$ satisfies condition $(\text I 3)$ then $M$ satisfies condition $(\text I 4)$.
Theorem: Equivalence of Definitions of Matroid/Definition 2 implies Definition 3

Let $M = \struct {S, \mathscr I}$ be an independence system.
Let $M$ also satisfy:
{{begin-axiom}}
{{axiom | n = \text I 4
        | q = \forall U, V \in \mathscr I
        | mr= \size U = \size V + 1 \implies \exists x \in U \setminus V : V \cup \set x \in \mathscr I
}}
{{end-axiom}}
Then $M$ satisfies:
{{begin-axiom}}
{{axiom | n = \text I 5
        | q = \forall U, V \in \mathscr I
        | mr= \size V < \size U \implies \exists Z \subseteq U \setminus V : \paren{V \cup Z \in \mathscr I} \land \paren{ \size {V \cup Z} = \size U}
}}
{{end-axiom}}

Proof:
From Independent Set can be Augmented by Larger Independent Set it follows that if $M$ satisfies condition $(\text I 4)$ then $M$ satisfies condition $(\text I 5)$.
Theorem: Equivalence of Definitions of Matroid/Definition 3 implies Definition 1

Let $M = \struct {S, \mathscr I}$ be an independence system.
Let $M$ also satisfy:
{{begin-axiom}}
{{axiom | n = \text I 5
        | q = \forall U, V \in \mathscr I
        | mr= \size V < \size U \implies \exists Z \subseteq U \setminus V : \paren {V \cup Z \in \mathscr I} \land \paren {\size {V \cup Z} = \size U}
}}
{{end-axiom}}
Then $M$ satisfies:
{{begin-axiom}}
{{axiom | n = \text I 3
        | q = \forall U, V \in \mathscr I
        | mr= \size V < \size U \implies \exists x \in U \setminus V : V \cup \set x \in \mathscr I
}}
{{end-axiom}}

Proof:
Let $M$ satisfy condition $(\text I 5)$.
Let $U, V \in \mathscr I$ such that $\size V < \size U$.
By condition $(\text I 5)$:
:$\exists Z : \exists Z \subseteq U \setminus V : \paren {V \cup Z \in \mathscr I} \land \paren {\size {V \cup Z} = \size U}$
Then:
:$V \cup Z \ne V$
From Union with Empty Set:
:$Z \ne \O$
Then:
:$\exists x : x \in Z$
From Singleton of Element is Subset:
:$\set x \subseteq Z$
From Set Union Preserves Subsets:
:$V \cup \set x \subseteq V \cup Z$
From independence system axiom $(\text I 2)$:
:$V \cup \set x \in \mathscr I$
By definition of a subset:
:$x \in U \setminus V$
It follows that $M$ satisfies condition $(\text I 3)$.
Theorem: Equivalence of Definitions of Matroid/Definition 1 implies Definition 4

Let $M = \struct {S, \mathscr I}$ be an independence system.
Let $M$ also satisfy:
{{begin-axiom}}
{{axiom | n = \text I 3
        | q = \forall U, V \in \mathscr I
        | mr= \size V < \size U \implies \exists x \in U \setminus V : V \cup \set x \in \mathscr I
}}
{{end-axiom}}
Then $M$ satisfies:
{{begin-axiom}}
{{axiom | n = \text I 6
        | q = \forall A \subseteq S
        | mr= \text{ all maximal subsets } Y \subseteq A \text{ with } Y \in \mathscr I \text{ have the same cardinality}
}}
{{end-axiom}}

Proof:
Let $M$ satisfy condition $(\text I 3)$.
Let $A \subseteq S$.
Let $Y_1, Y_2$ be maximal independent subsets of $A$.
{{WLOG}}, let:
:$\size {Y_2} \le \size {Y_1}$
{{AimForCont}}:
:$\size {Y_2} < \size {Y_1}$
By condition $(\text I 3)$:
:$\exists y \in Y_1 \setminus Y_2 : Y_2 \cup \set y \in \mathscr I$
From Union of Subsets is Subset:
:$Y_2 \cup \set y \subseteq A$
This contradicts the maximality of $Y_2$.
Then:
:$\size {Y_2} = \size {Y_1}$
It follows that $M$ satisifies $(\text I 6)$.
Theorem: Equivalence of Definitions of Matroid/Definition 4 implies Definition 1

Let $M = \struct {S, \mathscr I}$ be an independence system.
Let $M$ also satisfy:
{{begin-axiom}}
{{axiom | n = \text I 6
        | q = \forall A \subseteq S
        | mr= \text{ all maximal subsets } Y \subseteq A \text{ with } Y \in \mathscr I \text{ have the same cardinality}
}}
{{end-axiom}}
Then $M$ satisfies:
{{begin-axiom}}
{{axiom | n = \text I 3
        | q = \forall U, V \in \mathscr I
        | mr= \size V < \size U \implies \exists x \in U \setminus V : V \cup \set x \in \mathscr I
}}
{{end-axiom}}

Proof:
Let $M$ satisfy condition $(\text I 6)$.
Let $U, V \in \mathscr I$ such that $\size V < \size U$.
Let $W$ be a maximal independent subset of $U \cup V$ containing $U$.
Then:
:$\size U \le \size W$
By condition $(\text I 6)$:
:$V$ is not a maximal independent subset of $U \cup V$
Then:
:$\exists x \in \paren {U \cup V} \setminus V$ such that $V \cup \set x \in \mathscr I$
From Set Difference with Union is Set Difference:
:$\exists x \in U \setminus V$ such that $V \cup \set x \in \mathscr I$
It follows that $M$ satisifies $(\text I 3)$
Theorem: Cardinality of Set Difference

Let $S$ and $T$ be sets such that $T$ is finite.
Then:
:$\card {S \setminus T} = \card S - \card {S \cap T}$
where $\card S$ denotes the cardinality of $S$.

Proof:
From Intersection is Subset:
:$S \cap T \subseteq S$
:$S \cap T \subseteq T$
From Subset of Finite Set is Finite:
:$S \cap T$ is finite.
We have:
{{begin-eqn}}
{{eqn | l = \card {S \setminus T}
      | r = \card {S \setminus \paren {S \cap T} }
      | c = Set Difference with Intersection is Difference
}}
{{eqn | r = \card S -  \card {S \cap T}
      | c = Cardinality of Set Difference with Subset
}}
{{end-eqn}}
{{qed}}
Category:Set Difference
Category:Cardinality
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Theorem: Set Difference and Intersection are Disjoint

Let $S$ and $T$ be sets.
Then: 
:$S \setminus T$ and $S \cap T$ are disjoint 
where $S \setminus T$ denotes set difference and $S \cap T$ denotes set intersection.

Proof:
From Set Difference Intersection with Second Set is Empty Set:
:$\paren {S \setminus T} \cap T = \O$
and hence immediately from Intersection with Empty Set:
:$\paren {S \setminus T} \cap \paren {S \cap T} = \O$
So $S \setminus T$ and $S \cap T$ are disjoint.
{{qed}}
Category:Set Difference
Category:Set Intersection
5bipag1bp4tpinkf1sdxe4njjr7mdn9
Theorem: Straight Line has Zero Curvature

A straight line has zero curvature.

Proof:
From Equation of Straight Line in Plane: Slope-Intercept Form, a straight line has the equation:
:$y = m x + c$
Differentiating twice {{WRT|Differentiation}} $x$:
{{begin-eqn}}
{{eqn | l = \dfrac {\d y} {\d x}
      | r = m
      | c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\d^2 y} {\d x^2}
      | r = 0
      | c = 
}}
{{end-eqn}}
By definition, the curvature of a curve is defined as:
:$\kappa = \dfrac {y' '} {\paren {1 + y'^2}^{3/2} }$
But we have that:
:$y' ' := \dfrac {\d^2 y} {\d x^2} = 0$
and so, as in general $y' := \dfrac {\d y} {\d x} = m \ne 0$:
:$\kappa = \dfrac 0 {\paren {1 + m^2}^{3/2} }$
the curvature is zero.
{{qed}}
Theorem: Partial Differential Equation of Spheres in 3-Space

The set of spheres in real Cartesian $3$-dimensional space can be described by the system of partial differential equations:
:$\dfrac {1 + z_x^2} {z_{xx} } = \dfrac {z_x z_x} {z_{xy} } = \dfrac {1 + z_y^2} {z_{yy} }$
and if the radii of these spheres are expected to be real:
:$z_{xx} z_{yy} > z_{xy}$

Proof:
From Equation of Sphere, we have that the equation defining a general sphere $S$ is:
:$\paren {x - a}^2 + \paren {y - b}^2 + \paren {z - c}^2 = R^2$
where $a$, $b$ and $c$ are arbitrary constants.
We use the technique of Elimination of Constants by Partial Differentiation.
Taking the partial first derivatives {{WRT|Differentiation}} $x$ and $y$ and simplifying, we get:
{{begin-eqn}}
{{eqn | l = \paren {x - a} + \paren {z - c} \dfrac {\partial z} {\partial x}
      | r = 0
}}
{{eqn | l = \paren {y - b} + \paren {z - c} \dfrac {\partial z} {\partial y}
      | r = b
}}
{{end-eqn}}
$2$ equations are insufficient to dispose of $3$ constants, so the process continues by taking the partial second derivatives {{WRT|Differentiation}} $x$ and $y$:
{{begin-eqn}}
{{eqn | l = 1 + \paren {\dfrac {\partial z} {\partial x} }^2 + \paren {z - c} \dfrac {\partial^2 z} {\partial x^2}
      | r = 0
}}
{{eqn | l = \dfrac {\partial z} {\partial x} \dfrac {\partial z} {\partial y} + \paren {z - c} \dfrac {\partial^2 z} {\partial x \partial y}
      | r = 0
}}
{{eqn | l = 1 + \paren {\dfrac {\partial z} {\partial y} }^2 + \paren {z - c} \dfrac {\partial^2 z} {\partial y^2}
      | r = 0
}}
{{end-eqn}}
Eliminating $z - c$:
:$\dfrac {1 + z_x^2} {z_{xx} } = \dfrac {z_x z_x} {z_{xy} } = \dfrac {1 + z_y^2} {z_{yy} }$
This is our system of partial differential equations describing the set of spheres in real Cartesian $3$-dimensional space.
{{qed|lemma}}
We now investigate the conditions under which the radii of these spheres are real.
Let $\lambda = \dfrac {1 + z_x^2} {z_{xx} } = \dfrac {z_x z_y} {z_{xy} } = \dfrac {1 + z_y^2} {z_{yy} }$.
Then:
{{begin-eqn}}
{{eqn | l = \lambda^2
      | r = \dfrac {1 + z_x^2} {z_{xx} } \dfrac {1 + z_y^2} {z_{yy} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \lambda^2 \paren {z_{xx} z_{yy} }
      | r = 1 + z_x^2 + z_y^2 + z_x^2 z_y^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \lambda^2 \paren {z_{xx} z_{yy} } - z_{xy} \dfrac {z_x z_y} {z_{xy} }
      | r = 1 + z_x^2 + z_y^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \lambda^2 \paren {z_{xx} z_{yy} - z_{xy} }
      | r = 1 + z_x^2 + z_y^2
      | c = 
}}
{{eqn | o = >
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = z_{xx} z_{yy} - z_{xy}
      | o = >
      | r = 0
      | c = 
}}
{{end-eqn}}
and so:
:$z_{xx} z_{yy} > z_{xy}$
{{qed}}
Theorem: Fermat's Right Triangle Theorem

$x^4 + y^4 = z^2$ has no solutions in the (strictly) positive integers.

Proof:
This proof using Method of Infinite Descent was created by {{AuthorRef|Pierre de Fermat}}.
Suppose there is such a solution.
Then there is one with $\gcd \set {x, y, z} = 1$.
By Parity of Smaller Elements of Primitive Pythagorean Triple we can assume that $x^2$ is even and $y^2$ is odd.
By Primitive Solutions of Pythagorean Equation, we can write:
:$x^2 = 2 m n$
:$y^2 = m^2 - n^2$
:$z = m^2 + n^2$
where $m, n$ are coprime positive integers.
Similarly we can write:
:$n = 2 r s$
:$y = r^2 - s^2$
:$m = r^2 + s^2$
where $r, s$ are coprime positive integers, since $y$ is odd, forcing $n$ to be even.
We have:
:$\paren {\dfrac x 2}^2 = m \paren {\dfrac n 2}$
Since $m$ and $\dfrac n 2$ are coprime, they are both squares.
Similarly we have:
:$\dfrac n 2 = r s$
Since $r$ and $s$ are coprime, they are both squares.
Therefore $m = r^2 + s^2$ becomes an equation of the form $u^4 + v^4 = w^2$.
Moreover:
:$z^2 > m^4 > m$
and so we have found a smaller set of solutions.
By Method of Infinite Descent, no solutions can exist.
{{qed}}
{{Namedfor|Pierre de Fermat|cat = Fermat}}
Category:Number Theory
ewnxr39df8505pmfqljt4eegmfj4f5m
Theorem: Set of Even Integers is Countably Infinite

Let $\Bbb E$ be the set of even integers.
Then $\Bbb E$ is countably infinite.

Proof:
Let $f: \Bbb E \to \Z$ be the mapping defined as:
:$\forall x \in \Bbb E: \map f x = \dfrac x 2$
$f$ is well-defined as $x$ is even and so $\dfrac x 2 \in \Z$.
Let $x, y \in \Bbb E$ such that $\map f x = \map f y$.
Then:
{{begin-eqn}}
{{eqn | l = \map f x
      | r = \map f y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac x 2
      | r = \dfrac y 2
      | c = Definition of $f$
}}
{{eqn | ll= \leadsto
      | l = x
      | r = y
      | c = 
}}
{{end-eqn}}
Thus $f$ is injective by definition.
Consider the inverse $f^{-1}$.
By inspection:
:$\forall x \in \Z: \map {f^{-1} } x = 2 x$
$f^{-1}$ is well-defined, and $2 x$ is even.
Thus $f^{-1}$ is a mapping from $\Z$ to $\Bbb E$.
Then:
{{begin-eqn}}
{{eqn | l = \map {f^{-1} } x
      | r = \map {f^{-1} } y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2 x
      | r = 2 y
      | c = Definition of $f^{-1}$
}}
{{eqn | ll= \leadsto
      | l = x
      | r = y
      | c = 
}}
{{end-eqn}}
Thus $f^{-1}$ is injective by definition.
It follows by the Cantor-Bernstein-Schröder Theorem that there exists a bijection between $\Z$ and $\Bbb E$.
{{qed}}
Theorem: Basis Expansion of Rational Number

Let $b$ be a number base.
Let $x$ be a real number.
Then $x$ is a rational number {{iff}} the basis expansion of $x$ in base $b$ terminates or recurs.

Proof:
{{ProofWanted}}
Category:Basis Expansions
Category:Rational Numbers
cjine5yordcl9st3b5sphsip4414gyj
Theorem: Integration by Parts/Definite Integral

:$\ds \int_a^b \map f t \map G t \rd t = \bigintlimits {\map F t \map G t} a b - \int_a^b \map F t \map g t \rd t$

Proof:
By Product Rule for Derivatives:
:$\map D {F G} = f G + F g$
Thus $F G$ is a primitive of $f G + F g$ on $\closedint a b$.
Hence, by the Fundamental Theorem of Calculus:
:$\ds \int_a^b \paren {\map f t \map G t + \map F t \map g t} \rd t = \bigintlimits {\map F t \map G t} a b$
The result follows.
{{qed}}
Theorem: Integration by Parts/Primitive

:$\ds \int \map f t \map G t \rd t = \map F t \map G t - \int \map F t \map g t \rd t$
on $\closedint a b$.

Proof:
{{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d t} } {\map F t \map G t}
      | r = \map f t \map G t + \map F t \map g t
      | c = Product Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \int \paren {\map f t \map G t + \map F t \map g t} \rd t
      | r = \map F t \map G t
      | c = Fundamental Theorem of Calculus: integrating both sides {{WRT|Integration}} $t$
}}
{{eqn | ll= \leadsto
      | l = \int \map f t \map G t \rd t + \int \map F t \map g t \rd t
      | r = \map F t \map G t
      | c = Linear Combination of Primitives
}}
{{eqn | ll= \leadsto
      | l = \int \map f t \map G t \rd t
      | r = \map F t \map G t - \int \map F t \map g t \rd t
      | c = rearranging
}}
{{end-eqn}}
{{qed}}