hcju/proofwikiqa huggingface dataset:
Let $\triangle {ABC}$ have sides whose lengths are $a$, $b$ and $c$. By definition of Heronian triangle, each of $a$, $b$ and $c$ are rational. By definition of rational number, we can express: :$a = \dfrac {p_a} {q_a}$, $b = \dfrac {p_b} {q_b}$ and $c = \dfrac {p_c} {q_c}$ where each of $p_a, q_a, p_b, q_b, p_c, q_c$ are integers. Now let: {{begin-eqn}} {{eqn | l = a'       | r = a q_a q_b q_c       | c =  }} {{eqn | l = b'       | r = b q_a q_b q_c       | c =  }} {{eqn | l = c'       | r = c q_a q_b q_c       | c =  }} {{end-eqn}} Let $\triangle {A'B'C'}$ be the triangle whose sides have lengths $a'$, $b'$ and $c'$. By definition, $\triangle {ABC}$ and $\triangle {A'B'C'}$ are similar. Each of $a'$, $b'$ and $c'$ are integers. Consider the area of triangle $\triangle {A'B'C'}$ Let the area of $\triangle {ABC}$ be $A$. Then the area $\triangle {A'B'C'}$ is $q_a q_b q_c A$, which is rational. Hence $\triangle {A'B'C'}$ is an integer Heronian triangle. {{Qed}}
Suppose one of $a, b, c$ is zero. Then $a b c = 0 = a + b + c$. The remaining two numbers sum to $0$, giving the solution set: :$\set {\tuple {-z, 0, z}: z \in \N}$ {{qed|lemma}} Suppose $a < 0$ and $0 < b \le c$. Then $a b c \le a < a + b + c$. Hence equality never happens. Similarly, for $a \le b < 0$ and $c > 0$: :$a b c \ge c > a + b + c$ Hence equality never happens. {{qed|lemma}} Now it remains the case $0 < a \le b \le c$. Suppose $a \ge 2$. Then: {{begin-eqn}} {{eqn | l = a b c       | o = \ge       | r = 4 c }} {{eqn | o = \ge       | r = a + b + c + c }} {{eqn | o = >       | r = a + b + c }} {{end-eqn}} hence $a = 1$. Suppose $b \ge 3$. Then: {{begin-eqn}} {{eqn | l = b c       | o = \ge       | r = c + c + c }} {{eqn | o = \ge       | r = b + c + c }} {{eqn | o = >       | r = b + c }} {{end-eqn}} hence $b = 1$ or $b = 2$. For $b = 1$: :$c = 1 + 1 + c$ which is a contradiction. For $b = 2$: :$2 c = 1 + 2 + c$ giving $c = 3$. Therefore $\tuple {1, 2, 3}$ is the only solution for strictly positive integer values of $a, b, c$. Similarly, for $a \le b \le c < 0$, we have $0 < -c \le -b \le -a$. Also: {{begin-eqn}} {{eqn | l = \paren {-c} + \paren {-b} + \paren {-a}       | r = -\paren {a + b + c} }} {{eqn | r = -a b c }} {{eqn | r = \paren {-c} \paren {-b} \paren {-a} }} {{end-eqn}} and we have already show that $-c = 1, -b = 2, -a = 3$. Therefore we have $\tuple {a, b, c} = \tuple {-3, -2, -1}$. {{qed|lemma}} We have considered all possible signs of $a, b, c$. Hence the result. {{qed}} Category:6 1qfmk0u0l45yr8r2onnev4hi91jlyuf
Suppose $a \ge 4$. Then: {{begin-eqn}} {{eqn | l = a b c       | o = \ge       | r = 16 c       | c = as $4 \le a \le b$ }} {{eqn | o = \ge       | r = 4 \paren {a + b + c + c}       | c = as $a \le b \le c$ }} {{eqn | o = >       | r = 4 \paren {a + b + c}       | c = as $c > 0$ }} {{end-eqn}} hence $0 \le a \le 3$. For $a = 0$, we have $4 \paren {b + c} = 0$. This forces $b = c = 0$, giving the trivial solution: :$\tuple {0, 0, 0}$ {{qed|lemma}} For $a \ge 1$, note that: {{begin-eqn}} {{eqn | l = 4 \paren {a + b + c}       | r = a b c }} {{eqn | ll= \leadstoandfrom       | l = 4 \paren {a + b}       | r = \paren {a b - 4} c }} {{eqn | ll= \leadstoandfrom       | l = c       | r = \frac {4 \paren {a + b} } {a b - 4} }} {{end-eqn}} In the second step, {{LHS}} is strictly positive. Hence $a b > 4$ and the final step is justified. For $a = 1$, we have $b c = 4 \paren {1 + b + c}$. Since $a b > 4$, $b > 4$. Suppose $b \ge 9$. Then: {{begin-eqn}} {{eqn | l = b c       | o = \ge       | r = 9 c }} {{eqn | o = \ge       | r = 4 \paren {b + c} + c       | c = as $b \le c$ }} {{eqn | o = >       | r = 4 \paren {1 + b + c}       | c = as $c \ge b > 4$ }} {{end-eqn}} hence $b \le 8$. We find the value of $c$ by substituting $a$ and $b$: :$b = 5: c = \dfrac {4 \paren {1 + 5} } {5 - 4} = 24$ :$b = 6: c = \dfrac {4 \paren {1 + 6} } {6 - 4} = 14$ :$b = 7: c = \dfrac {4 \paren {1 + 7} } {7 - 4} = \dfrac {32} 3$ :$b = 8: c = \dfrac {4 \paren {1 + 8} } {8 - 4} = 9$ and we see that $\tuple {1, 5, 24}, \tuple {1, 6, 14}, \tuple {1, 8, 9}$ are valid solutions. {{qed|lemma}} For $a = 2$, we have $2 b c = 4 \paren {2 + b + c}$. Since $a b > 4$, $b > 2$. Suppose $b \ge 5$. Then: {{begin-eqn}} {{eqn | l = 2 b c       | o = \ge       | r = 10 c }} {{eqn | o = \ge       | r = 4 \paren {b + c} + 2 c       | c = as $b \le c$ }} {{eqn | o = >       | r = 4 \paren {2 + b + c}       | c = as $c \ge b > 4$ }} {{end-eqn}} hence $b \le 4$. We find the value of $c$ by substituting $a$ and $b$: :$b = 3: c = \dfrac {4 \paren {2 + 3} } {2 \times 3 - 4} = 10$ :$b = 4: c = \dfrac {4 \paren {2 + 4} } {2 \times 4 - 4} = 6$ and we see that $\tuple {2, 3, 10}, \tuple {2, 4, 6}$ are valid solutions. {{qed|lemma}} For $a = 3$, we have $3 b c = 4 \paren {3 + b + c}$. Suppose $b \ge 4$. Then: {{begin-eqn}} {{eqn | l = 3 b c       | o = \ge       | r = 12 c }} {{eqn | o = \ge       | r = 4 \paren {b + c} + 4 c       | c = as $b \le c$ }} {{eqn | o = >       | r = 4 \paren {3 + b + c}       | c = as $c \ge b > 1$ }} {{end-eqn}} hence $b \le 3$. Since $3 = a \le b$, this forces $b = 3$ We find the value of $c$ by substituting $a$ and $b$: :$c = \dfrac {4 \paren {3 + 3} } {3 \times 3 - 4} = \dfrac {24} 5$ and we see that it is not a valid solution. {{qed|lemma}} We have considered all possible values of $a$. Hence the result. {{qed}} mgja9o3yt8d0kuze46853enzwrydbf2
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 9rjbrxvgau2ac4m16h5h5g1n9j9h3ym
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 to 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.
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 Theory 4qx1rhugi5cnbb38luslx8mvxpmnhlu
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 $M = \struct{S, \mathscr I}$ === Matroid Axioms === ==== Matroid Axiom $(\text I 1)$ ==== We have $\mathscr B$ is non-empty. Let $B \in \mathscr B$. From Empty Set is Subset of All Sets: :$\O \subseteq B$ By definition of $\mathscr I$: :$\O \in \mathscr I$ It follows that $\mathscr I$ satisfies the matroid axiom $(\text I 1)$ by definition. {{qed|lemma}} ==== Matroid Axiom $(\text I 2)$ ==== Let $X \in \mathscr I$. Let $Y \subseteq X$. By definition of $\mathscr I$: :$\exists B \in \mathscr B : X \subseteq B$ From Subset Relation is Transitive: :$Y \subseteq B$ By definition of $\mathscr I$: :$Y \in \mathscr I$ It follows that $\mathscr I$ satisfies the matroid axiom $(\text I 2)$ by definition. {{qed|lemma}} ==== 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 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 ===== {{:Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Sufficient Condition/Lemma}}{{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 $M = \struct{S, I}$ 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: :$B \subseteq B'$ From Leigh.Samphier/Sandbox/Matroid Base Axiom Implies Sets Have Same Cardinality: :$\card B = \card {B'}$ 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.
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}}
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:Coprime Integers k9j3wdqnu5y7uqj2ft8uko6w8dii5n7
=== Definition 1 implies Definition 2 === {{:Equivalence of Definitions of Matroid/Definition 1 implies Definition 2}}{{qed|lemma}} === Definition 2 implies Definition 3 === {{:Equivalence of Definitions of Matroid/Definition 2 implies Definition 3}}{{qed|lemma}} === Definition 3 implies Definition 1 === {{:Equivalence of Definitions of Matroid/Definition 3 implies Definition 1}}{{qed|lemma}} === Definition 1 implies Definition 4 === {{:Equivalence of Definitions of Matroid/Definition 1 implies Definition 4}}{{qed|lemma}} === Definition 4 implies Definition 1 === {{:Equivalence of Definitions of Matroid/Definition 4 implies Definition 1}}{{qed}}
{{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 $\displaystyle r = p^{k_1} \frac {m_1} {n_1} = p^{k_2} \frac {m_2} {n_2}$, 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}}
From Independent Set can be Augmented by Larger Independent Set it follows that if $M$ satisfies condition $(\text I 3')$ then $M$ satisfies condition $(\text I 3'')$.
Let $M$ satisfy condition $(\text I 3'')$. Let $U, V \in \mathscr I$ such that $\size V < \size U$. By condition $(\text I 3'')$: :$\exists Z : \exists Z \subseteq U \setminus V : \paren {V \cup Z \in \mathscr I} \land \paren {\size {V \cup Z} = \size U}$ 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 $(\text I 3)$.
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 $(\text I 3''')$
Let $M$ satisfy condition $(\text I 3''')$. 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 $(\text I 3''')$: :$V$ 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 $(\text I 3)$
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 ev0k49h3v3w0sf3xazascpr8dqudj59
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
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}}
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 $z - c$: :$\dfrac {1 + z_x^2} {z_{xx} } = \dfrac {z_x z_x} {z_{xy} } = \dfrac {1 + z_y^2} {z_{yy} }$ 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}}
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
We define $F$ and $G$ as follows: {{begin-eqn}} {{eqn | l = \map F t       | r = \map f t + \map {f'} t \paren {x - t} + \dfrac {\map {f''} t} {2!} \paren {x - t}^2 + \dotsb + \dfrac {\map {f^{\paren n} } t} {n!} \paren {x - t}^n }} {{eqn | l = \map G t       | r = \map g t + \map {g'} t \paren {x - t} + \dfrac {\map {g''} t} {2!} \paren {x - t}^2 + \dotsb + \dfrac {\map {g^{\paren k} } t} {k!} \paren {x - t}^k }} {{end-eqn}} Then $F$ and $G$ are continuous on $\closedint a x$ and differentiable on $\openint a x$. Differentiating {{WRT|Differentiation}} $t$: {{begin-eqn}} {{eqn | l = \map {F'} t       | r = \map {f'} t + \paren {\map {f''} t \paren {x - t} - \map {f'} t} + \paren {\frac {\map {f^{\paren 3} } t} {2!} \paren {x - t}^2 - \frac {\map {f^{\paren 2} } t} {1!} \paren {x - t} } + \dotsb + \paren {\frac {\map {f^{\paren {n + 1} } } t} {n!} \paren {x - t}^n - \frac {\map {f^{\paren n} } t} {\paren {n - 1}!} \paren {x - t}^{n - 1} } }} {{eqn | r = \frac {\map {f^{\paren {n + 1} } } t} {n!} \paren {x - t}^n }} {{eqn | l = \map {G'} t       | r = \frac {\map {f^{\paren {k + 1} } } t} {k!} \paren {x - t}^k }} {{end-eqn}} By condition $(5)$ of the statement of the theorem, it follows that $G$ does not vanish on $\openint a x$. By Cauchy Mean Value Theorem, there exists a real number $\xi \in \openint a x$ such that: :$\dfrac {\map {F'} \xi} {\map {G'} \xi} = \dfrac {\map F x - \map F a} {\map G x - \map G a}$ That is: :$\dfrac {\map {f^{\paren {n + 1} } } \xi / n!} {\map {g^{\paren {k + 1} } } \xi /k!} \paren {x - \xi}^{n - k} = \dfrac {\map f x - \map f a - \map {f'} a \paren {x - a} - \dfrac {\map {f''} a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {f^{\paren n} } a} {n!} \paren {x - a}^n} {\map g x - \map g a - \map {g'} a \paren {x - a} - \dfrac {\map {g''} a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {g^{\paren k}} a} {k!} \paren {x - a}^k}$ {{qed}}
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}}
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: :$\displaystyle \int_a^b \paren {\map f t \map G t + \map F t \map g t} \rd t = \bigintlimits {\map F t \map G t} a b$ The result follows. {{qed}}