hcju/proofwikiqa huggingface dataset:
The surface area for a smooth surface embedded in $3$-dimensional Euclidean space is given by: :$\displaystyle A \sqbrk z = \iint_\Gamma \sqrt {1 + z_x^2 + z_y^2} \rd x \rd y$ It follows that: {{begin-eqn}} {{eqn | l = \dfrac \d {\d x} \frac \partial {\partial z_x} \sqrt {1 + z_x^2 + z_y^2}       | r = \dfrac \d {\d x} \frac {z_x} {\sqrt {1 + z_x^2 + z_y^2} } }} {{eqn | r = \frac {z_{xx} + z_y^2 z_{xx} - z_x z_y z_{xy} } {\paren {1 + z_x^2 + z_y^2}^{\frac 3 2} } }} {{eqn | r = \frac {r \paren {1 + q^2} - p q s } {\paren {1 + p^2 + q^2}^{\frac 3 2} } }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \dfrac \d {\d y} \frac \partial {\partial z_y} \sqrt {1 + z_x^2 + z_y^2}       | r = \dfrac \d {\d y} \frac {z_y} {\sqrt {1 + z_x^2 + z_y^2} } }} {{eqn | r = \frac {z_{yy} + z_x^2 z_{yy} - z_x z_y z_{xy} } {\paren {1 + z_x^2 + z_y^2}^{\frac 3 2} } }} {{eqn | r = \frac {t \paren {1 + p^2} - p q s } {\paren {1 + p^2 + q^2}^{\frac 3 2} } }} {{end-eqn}} By Euler's equation: :$\dfrac {r \paren {1 + q^2} - 2 s p q + t \paren {1 + p^2} } {\paren {1 + p^2 + q^2}^{\frac 3 2} } = 0$ Due to the smoothness of the surface, $1 + p^2 + q^2$ is bounded. Hence, the following equation is sufficient: :$r \paren {1 + q^2} - 2 s p q + t \paren {1 + p^2} = 0$ Introduce the following change of variables: {{begin-eqn}} {{eqn | l = E       | r = 1 + p^2 }} {{eqn | l = F       | r = p q }} {{eqn | l = G       | r = 1 + q^2 }} {{eqn | l = e       | r = \dfrac r {\sqrt {1 + p^2 + q ^2} } }} {{eqn | l = f       | r = \dfrac s {\sqrt {1 + p^2 + q^2} } }} {{eqn | l = g       | r = \dfrac t {\sqrt {1 + p^2 + q^2} } }} {{end-eqn}} Then Euler's equation can be rewritten as: :$\dfrac {E g - 2 F f + G e} {2 \paren {E G - F^2} } = 0$ By definition, mean curvature is: :$M = \dfrac {E g - 2 F f + G e} {2 \paren {E G - F^2} }$ Hence: :$M = 0$ {{qed}}
By definition, $J$ satisfies the following conditions: :$(1): \quad n, m \in J \implies m + n \in J, m - n \in J$ :$(2): \quad n \in J, r \in \Z \implies r n \in J$ First note that the null ideal $\set 0$ is an integral ideal. This is of the form $0 \Z$. Let $J \ne \set 0$. Then $\exists a \in J: a \ne 0$. As $0 \in \Z$ it follows from $(2)$ that: :$0 \times a = 0 \in J$ Thus from $(1)$ we have that: :$0 - a \in J$ so both $a \in J$ and $-a \in J$ Thus there exists $a \in J: a > 0$. Consider the set $S = \set {x \in J: x > 0}$. $S$ is bounded below by $0$, for example. $S$ is not empty, because it has been established that $a \in S$. By Set of Integers Bounded Below has Smallest Element, $S$ has a smallest element, $m$, say. Consider $y \in m \Z$. By definition of $m \Z$, $y = r m$ for some $r \in \Z$. Thus, by $(2)$, $y \in J$. So by definition of subset, $m \Z \subseteq J$. It remains to be demonstrated that $J \subseteq m \Z$. {{AimForCont}} $k \in J: k \notin m \Z$. As $k \in J$, it follows from $(1)$ that $0 - k = -k \in J$ also. Either $k > 0$ or $-k > 0$. {{WLOG}}, let $k > 0$. If $k < m$, then that contradicts the statement that $m \in J$ is the smallest element of $J$ that is (strictly) positive. So $k > m$. Consider the set $T = \set {x \in m \Z: x > k}$. $T$ is not empty, as $m \paren {k + 1}$, for example, is bound to be in $T$ (even if $m = 1$, the smallest it can be). By Set of Integers Bounded Below has Smallest Element, $T$ has a smallest element, $m s$, say. Then: :$m \paren {s - 1} < k < m s$ As $k > m$ it follows also that $m \paren {s - 1} > 0$. Equality cannot happen because $k \notin m \Z$. But we have that $k \in J$ and so: :$m s - k \in J$ and: :$k - m \paren {s - 1} \in J$ But $m s - k < m$, otherwise $m \paren {s - 1} > k$ which contradicts the statement that $m s \in J$ is the smallest element of $T$ Thus $m s - k \in J$ and $0 < m s - k < m$. This contradicts the statement that $m$ is the smallest element of $J$ that is (strictly) positive. Hence by Proof by Contradiction there is no such $k$. Hence $J \subseteq m \Z$ and the result follows. {{qed}}
Let $a = 2 m + 1$, $b = 2 n + 1$. Then: {{begin-eqn}} {{eqn | l = a^2 - b^2        | r = \paren {2 m + 1}^2 - \paren {2 n + 1}^2       | c =  }} {{eqn | r = \paren {4 m^2 + 4 m + 1} - \paren {4 n^2 + 4 n - 1}       | c =  }} {{eqn | r = 4 \paren {m^2 - n^2} + 4 \paren {m - n}       | c =  }} {{eqn | r = 4 \paren {m + n} \paren {m - n} + 4 \paren {m - n}       | c = Difference of Two Squares }} {{eqn | r = 4 \paren {m + n + 1} \paren {m - n}       | c =  }} {{end-eqn}} Suppose $m - n$ is even such that $m - n = 2 k$. Then: :$a^2 - b^2 = 4 \paren {2 k} \paren {m + n + 1} = 8 k \paren {m + n + 1}$ and so is divisible by $8$. Suppose $m - n$ is odd such that $m - n = 2 k + 1$. Then: :$m + n + 1 = m + \paren {2 k + 1 + m} + 1 = 2 m + 2 k = 2 \paren {m + k}$ and so: :$a^2 - b^2 = 4 \paren {2 k + 1} 2 \paren {m + k} = 8 \paren {2 k + 1} \paren {m + k}$ and so is again divisible by $8$. The result follows. {{qed}}
=== $(1)$ implies $(2)$ === Let $\gcd \set{a, b}$ be the greatest common divisor of integers $a$ and $b$ by definition 1. Then by definition: :$\gcd \set{a, b}$ is the largest $d \in \Z_{>0}$ such that $d \divides a$ and $d \divides b$. From Common Divisor Divides GCD: :$c \divides a \land c \divides b \implies c \divides d$ Thus $\gcd \set{a, b}$ is the greatest common divisor of integers $a$ and $b$ by definition 2. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $\gcd \set{a, b}$ be the greatest common divisor of integers $a$ and $b$ by definition 2. Then by definition $\gcd \set{a, b}$ is  the (strictly) positive integer $d \in \Z_{>0}$ such that: : $(1): \quad d \divides a \land d \divides b$ : $(2): \quad c \divides a \land c \divides b \implies c \divides d$ From $d \divides a \land d \divides b$, we see that $d$ is a common divisor of $a$ and $b$. From $c \divides a \land c \divides b$, we see that $c$ is also a common divisor of $a$ and $b$. Also, we have that $c \divides d$. From Absolute Value of Integer is not less than Divisors, we see that (in the domain of $\Z_{>0}$):  :$c \divides d \implies c \le d$ Thus, whatever $c$ may be, it is no larger than $d$. Therefore, $d$ must be the greatest of all the common divisors of $a$ and $b$. Thus $\gcd \set{a, b}$ is the greatest common divisor of integers $a$ and $b$ by definition 1. {{qed}}
The contrapositive is proved. Let $\alpha \gt 1$. The norm axiom (N3) (Triangle Inequality) is not satisfied: {{begin-eqn}} {{eqn|l= \norm {1 + 1}      |r= \size {1 + 1}^\alpha }} {{eqn|r= 2^\alpha }} {{eqn|o= \gt      |r= 2      |c= Real Power Function on Base Greater than One is Strictly Increasing }} {{eqn|r= \size{1}^\alpha + \size{1}^\alpha }} {{eqn|r= \norm{1} + \norm{1} }} {{end-eqn}} By Rule of Transposition the result follows.
Suppose $\alpha \le 1$. It is shown that $\norm{\,\cdot\,}$ satisfies the norm axioms (N1)-(N3). ==== (N1) Positive Definiteness ==== Let $x \in \Q$. {{begin-eqn}} {{eqn|l= \norm {x} = 0      |o= \iff      |r= \size {x}^\alpha = 0      |c = Definition of $\norm{\,\cdot\,}$ }} {{eqn|o= \iff      |r= \size {x} = 0      |c= Definition of $a$ to the power of $r$ for $a \in \R_{\ge 0}$ and $r \in \R_{\gt 0}$ }} {{eqn|o= \iff      |r= x = 0      |c= Absolute Value is Norm and norm axiom (N1) (Positive Definiteness) }} {{end-eqn}} {{qed|lemma}} ==== (N2) Multiplicativity ==== Let $x, y \in \Q$. Then: {{begin-eqn}} {{eqn|l= \norm {x y}      |r= \size {x y}^\alpha      |c = Definition of $\norm{\,\cdot\,}$ }} {{eqn|r= \paren {\size {x} \size {y} }^\alpha      |c= Absolute Value is Norm and norm axiom (N2) (Multiplicativity) }} {{eqn|r= \size {x}^\alpha \size {y}^\alpha      |c= Power of product }} {{eqn|r= \norm {x} \norm {y}      |c = Definition of $\norm{\,\cdot\,}$ }} {{end-eqn}} {{qed|lemma}} ==== (N3) Triangle Inequality ==== Let $x, y \in \Q$. {{WLOG}} let $\norm y \lt \norm x$. If $\norm x = 0$ then $\norm y = 0$. By (N1) above, $x = y = 0$. Hence: {{begin-eqn}} {{eqn|l= \norm{x + y}      |r= \norm 0 }} {{eqn|r= 0 }} {{eqn|r= 0 + 0 }} {{eqn|r= \norm x + \norm y }} {{end-eqn}} If $\norm x \gt 0$ then: :$\norm x \gt 0 \iff \size {x}^\alpha \gt 0 \iff \size x \gt 0$ Hence: {{begin-eqn}} {{eqn|l= \norm{x + y}      |r= \size{x + y}^\alpha }} {{eqn|o= \le      |r= \paren{\size x + \size y}^\alpha      |c= Norm axiom (N3) (Triangle Inequality) }} {{eqn|r= \size {x}^\alpha \paren{1 + \dfrac {\size y} {\size x} }^\alpha }} {{eqn|o= \le      |r= \size {x}^\alpha \paren{1 + \dfrac {\size y} {\size x} }      |c= Real Power Function on base Greater than One is Strictly Increasing }} {{eqn|o= \le      |r= \size {x}^\alpha \paren{1 + \dfrac {\size {y}^\alpha} {\size {x}^\alpha} }      |c= Real Power Function on base between Zero and One is Strictly Decreasing }} {{eqn|o= \le      |r= \size {x}^\alpha + \size {y}^\alpha }} {{eqn|r= \norm {x} + \norm {y} }} {{end-eqn}}
We have that: {{begin-eqn}} {{eqn | l = b       | o = \divides       | r = \lcm \set {b, d}       | c = {{Defof|Lowest Common Multiple of Integers}} }} {{eqn | lo= \land       | l = d       | o = \divides       | r = \lcm \set {b, d}       | c =  }} {{eqn | ll= \leadsto       | l = r b       | r = \lcm \set {b, d}       | c = for some $r \in \Z$ }} {{eqn | lo= \land       | l = s d       | r = \lcm \set {b, d}       | c = for some $s \in \Z$ }} {{eqn | ll= \leadsto       | l = r \paren {a b}       | r = a \lcm \set {b, d}       | c =  }} {{eqn | lo= \land       | l = s \paren {a d}       | r = a \lcm \set {b, d}       | c =  }} {{eqn | ll= \leadsto       | l = a b       | o = \divides       | r = a \lcm \set {b, d}       | c = {{Defof|Divisor of Integer}} }} {{eqn | lo= \land       | l = a d       | o = \divides       | r = a \lcm \set {b, d}       | c =  }} {{end-eqn}} Suppose $n \in \Z$ such that $a b \divides n$ and $a d \divides n$. It will be shown that $a \lcm \set {b, d} \divides n$. So: {{begin-eqn}} {{eqn | l = a b       | o = \divides       | r = n       | c = by hypothesis }} {{eqn | lo= \land       | l = a d       | o = \divides       | r = n       | c =  }} {{eqn | ll= \leadsto       | l = a r b       | r = n       | c = for some $r \in \Z$ }} {{eqn | lo= \land       | l = a s d       | r = n       | c = for some $s \in \Z$ }} {{eqn | ll= \leadsto       | l = r b       | r = \dfrac n a       | c =  }} {{eqn | lo= \land       | l = s d       | r = \dfrac n a       | c =  }} {{eqn | ll= \leadsto       | l = b       | o = \divides       | r = \dfrac n a       | c = {{Defof|Divisor of Integer}} }} {{eqn | lo= \land       | l = d       | o = \divides       | r = \dfrac n a       | c =  }} {{eqn | ll= \leadsto       | l = \lcm \set {b, d}       | o = \divides       | r = \dfrac n a       | c = LCM Divides Common Multiple }} {{eqn | lo= \land       | l = a \lcm \set {b, d}       | o = \divides       | r = n       | c =  }} {{end-eqn}} Thus we have: :$a b \divides a \lcm \set {b, d} \land a d \divides a \lcm \set {b, d}$ and: :$a b \divides n \land a d \divides n \implies a \lcm \set {b, d} \divides n$ It follows from LCM iff Divides All Common Multiples that: :$\lcm \set {a b, a d} = a \lcm \set {b, d}$ {{qed}}
{{begin-eqn}} {{eqn | l = \dfrac a b + \dfrac c d       | r = \dfrac {a d} {b d} + \dfrac {b c} {b d}       | c =  }} {{eqn | r = \dfrac {a d + b c} {b d}       | c =  }} {{eqn | r = \dfrac {a d + b c} {\gcd \set {b, d} \lcm \set {b, d} }       | c = Product of GCD and LCM }} {{eqn | r = \dfrac {a D \gcd \set {b, d} + B \gcd \set {b, d} c} {\gcd \set {b, d} \lcm \set {b, d} }       | c = substituting for $b$ and $d$ as defined above }} {{eqn | r = \dfrac {a D + B c} {\lcm \set {b, d} }       | c = dividing top and bottom by $\gcd \set {b, d}$ }} {{end-eqn}} {{qed}}
Let $d = \gcd \set {a, b}$. Then by definition of greatest common divisor: :$d \divides a \land d \divides b$ From Common Divisor Divides Integer Combination: :$d \divides \paren {a + b} \land d \divides \paren {a - b}$ By definition of common divisor: :$d \divides \gcd \set {a + b, a - b}$ Hence from Absolute Value of Integer is not less than Divisors: :$d \le \gcd \set{a + b, a - b}$ {{qed}}
We have that: :$\gcd \set {a, b} \divides a$ and: :$a \divides \lcm \set {a, b}$ The result follows from Divisor Relation is Transitive. {{qed}}
Let $x \in m \Z \cap n \Z$. Then by definition of set intersection: :$m \divides x$ and $n \divides x$ So from LCM Divides Common Multiple: :$\lcm \set {m, n} \divides x$ and so $x \in \lcm \set {m, n} \Z$ That is: :$m \Z \cap n \Z \subseteq \lcm \set {m, n} \Z$ {{qed|lemma}} Now suppose $x \in \lcm \set {m, n} \Z$. Then $\lcm \set {m, n} \divides x$. Thus by definition of lowest common multiple: :$m \divides x$  and: :$n \divides x$ and so: :$x \in m \Z \land x \in n \Z$ That is: :$x \in \Z \cap n \Z$ and so: :$\lcm \set {m, n} \Z \subseteq m \Z \cap n \Z$ {{qed|lemma}} The result follows by definition of set equality. {{qed}}
Let $x \in m \Z \cup n \Z$. Then either: :$m \divides x$ or: :$n \divides x$ In both cases: :$\gcd \set {m, n} \divides x$ and so: :$x \in \gcd \set {m, n} \Z$ Hence by definition of subset: :$m \Z \cup n \Z \subseteq \gcd \set {m, n} \Z$ {{qed}}
From Set of Integer Multiples is Integral Ideal, each of $m \Z$, $n \Z$, $r \Z$ and $\gcd \set {m, n} \Z$ are integral ideals. Let $c \in \gcd \set {m, n} \Z$. By definition of integral ideal: :$\gcd \set {m, n} \divides c$ By Set of Integer Combinations equals Set of Multiples of GCD: :$\exists x, y \in \Z: c = x m + y n$ But as $m \Z \subseteq r \Z$ and $n \Z \subseteq r \Z$: :$m \in r \Z$ and $n \in \r Z$ Thus by definition of integral ideal: :$x m + y n \in r \Z$ So: :$c \in \gcd \set {m, n} \Z \implies c \in r \Z$ and the result follows by definition of subset. {{qed}}
From the construction of a general Fibonacci sequence, $a_n$ is an integer combination of $r$ and $s$. From Set of Integer Combinations equals Set of Multiples of GCD, $a_n$ is divisible by $\gcd \set {r, s}$. Hence the result. {{qed}}
By Corollary 2 then: :$\norm {x + y} = \norm {x - y} = \norm {y - x} = \max \set {\norm x, \norm y} = \norm y$ {{qed}}
The contrapositive statements are proved. Let $\norm x \ne \norm y$ By Corollary 2 then: :$\norm {x + y} = \norm {x - y} = \norm {y - x} = \max \set {\norm x, \norm y} \ge \norm y$ The result follows. {{qed}}
By Characterisation of Non-Archimedean Division Ring Norms then: :$\norm {\,\cdot\,}_1$ is non-Archimedean $\iff \forall n \in \N_{>0}: \norm{n \cdot 1_R}_1 \le 1$. By the definition of norm equivalence then: :$\forall n \in \N: \norm {n \cdot 1_R}_1 \le 1 \iff \norm {n \cdot 1_R}_2 \le 1$ Similarly, by Characterisation of Non-Archimedean Division Ring Norms then: :$\forall n \in \N_{>0}: \norm {n \cdot 1_R}_2 \le 1 \iff \norm {\,\cdot\,}_2$ is non-Archimedean. The result follows. {{qed}}
{{WLOG}} let one of the vertices of $T$ be at $\tuple {0, 0}$. Let the other $2$ vertices be at $\tuple {a, b}$ and $\tuple {x, y}$. By Area of Triangle in Determinant Form with Vertex at Origin: :$\map \Area T = \dfrac {\size {b y - a x} } 2$ As the vertices of $T$ are non-collinear, $\map \Area T \ge 0$. Thus $\size {b y - a x} > 0$. As $\tuple {a, b}$ and $\tuple {x, y}$ are lattice points, all of $a, b, x, y \in \Z$. Thus $\size {b y - a x} \ge 1$. Hence the result. {{Qed}}
From Perpendicular Distance from Straight Line in Plane to Point, the perpendicular distance $d$ between $L$ and the point $\tuple {x_0, y_0}$ is gven by: :$d = \dfrac {\size {a x_0 + b y_0 + c} } {\sqrt {a^2 + b^2} }$ The result follows by setting $x_0 = 0$ and $y_0 = 0$. {{qed}}
From Condition for Differentiable Functional to have Extremum we have :$\bigvalueat {\delta J \sqbrk {z; h} } {z \mathop = \hat z} = 0$ The variation exists if $J$ is a differentiable functional. Since $\map z \Gamma = 0$, $\map h {x, y}$ vanishes on the boundary $\Gamma$: :$\bigvalueat h \Gamma = 0$ From the definition of increment of a functional: :{{begin-eqn}} {{eqn | l = \Delta J \sqbrk {z; h}       | r = J \sqbrk {z + h} - J \sqbrk z       | c = definition }} {{eqn | r = \iint_D \map F {x, y, z + h, z_x + h_x, z_y + h_y} \rd x \rd y - \iint_D \map F {x, y, z, z_x, z_y} \rd x \rd y       | c = form of considered functional }} {{eqn | r = \iint_D \paren {\map F {x, y, z + h, z_x + h_x, z_y + h_y} - \map F {x, y, z, z_x, z_y} } \rd x \rd y       | c = bringing under the same integral }} {{end-eqn}} Using multivariate Taylor's theorem, expand $\map F {x, y, z + h, z_x + h_x, z_y + h_y}$ {{WRT}} $h$, $h_x$ and $h_y$. Denote the ordered tuples $\tuple {h, h_x, h_y}$ as $\mathbf h$ and $\tuple {z, z_x, z_y}$ as $\mathbf z$ respectively. Then: :$\ds \map F {x, y, \mathbf z + \mathbf h} = \valueat {\map F {x, y, \mathbf z + \mathbf h} } {\mathbf h \mathop = 0} + \valueat {\frac {\partial {\map F {x, y, \mathbf z + \mathbf h} } } {\partial z} } {\mathbf h \mathop = 0} h + \valueat {\frac {\partial {\map F {x, y, \mathbf z + \mathbf h} } } {\partial z_x} } {\mathbf h \mathop = 0} z_x + \valueat {\frac {\partial {\map F {x, y, \mathbf z + \mathbf h} } } {\partial z_y} } {\mathbf h \mathop = 0} z_y + \map \OO {\mathbf h^2}$ where the last term includes all terms of the order not lesser than 2 {{WRT}} the elements of $\mathbf h$. Substitute this back into the integral:  :$\ds \Delta J \sqbrk {y; h} = \iint_D \paren {\map F {x, y, \mathbf h}_z h + \map F {x, y, \mathbf z}_{z_x} h_x + \map F {x, y, \mathbf z}_{z_y} h_y + \map \OO {\mathbf h^2} } \rd x \rd y$ Terms in $\map \OO {\mathbf h^2}$ represent terms of order higher than 1 with respect to elements of $\mathbf h$. {{Explain|Precise way to drop higher order terms}} By definition, the integral not counting in $\map \OO {\mathbf h^2}$ is a variation of functional: {{begin-eqn}} {{eqn | l = \delta J \sqbrk {z; h}       | r = \iint_D \paren {F_z h + \dfrac \partial {\partial x} F_{z_x} h + \dfrac \partial {\partial y} F_{z_y} h } \rd x \rd y }} {{eqn | r = \iint_D \paren {F_z h + \dfrac \partial {\partial x} \paren {F_{z_x} h} + \dfrac \partial {\partial y} \paren {F_{z_y} h} } \rd x \rd y - \iint_D \paren {\dfrac \partial {\partial x} F_{z_x} + \dfrac \partial {\partial y} F_{z_y} } h \rd x \rd y }} {{eqn | r = \int_\Gamma \paren {F_{z_x} h \rd y - F_{z_y} h \rd x} - \iint_D \paren {\dfrac \partial {\partial x} F_{z_x} + \dfrac \partial {\partial y} F_{z_y} -F_z} h \rd x \rd y       | c = Green's Theorem }} {{eqn | r = \iint_D \paren {F_z - \dfrac \partial {\partial x} F_{z_x} - \dfrac \partial {\partial y} F_{z_y} } h \rd x \rd y       | c = $\map h \Gamma = 0$ }} {{end-eqn}} Then, by the use of the lemma: :$F_z - \dfrac \partial {\partial x} F_{z_x} - \dfrac \partial {\partial y} F_{z_y} = 0$ {{qed}}
Let $L$ be embedded in the cartesian plane. The slope of $\mathcal L$ at a point $p = \tuple {x, y}$ is defined as being its derivative at $p$ {{WRT|Differentiation}} $x$. $\grad p = \dfrac {\d y} {\d x}$ :500px Let $\mathcal L$ be a straight line. Let $\triangle ABC$ and $\triangle DEF$ be right triangles constructed so that: :$A, B, D, E$ are on $\mathcal L$ :$AC$ and $DF$ are parallel to the $x$-axis :$BC$ and $EF$ are parallel to the $y$-axis. From Parallelism implies Equal Corresponding Angles: :$\angle ABC = \angle DEF$ and: :$\angle BAC = \angle EDF$ Also we have that $\angle ACB = \angle DFE$ and are right angles. Thus $\triangle ABC$ and $\triangle DEF$ are similar. Thus: :$\dfrac {BC} {AC} = \dfrac {EF} {DF}$ That is, the slope of $\mathcal L$ between $A$ and $B$ is the same as the slope of $\mathcal L$ between $D$ and $E$. The argument reverses. {{Qed}}
Let $y = \map f x$ be the equation of a straight line $\mathcal L$. From Line in Plane is Straight iff Gradient is Constant, $\mathcal L$ has constant slope. Thus the derivative of $y$ {{WRT|Differentiation}} $x$ will be of the form: :$y' = c$ Thus: {{begin-eqn}} {{eqn | l = y       | r = \int c \rd x       | c = Fundamental Theorem of Calculus }} {{eqn | r = c x + K       | c = Primitive of Constant }} {{end-eqn}} where $K$ is arbitrary. Taking the equation: :$\alpha_1 x + \alpha_2 y = \beta$ it can be seen that this can be expressed as: :$y = - \dfrac {\alpha_1} {\alpha_2} x + \dfrac {\beta} {\alpha_2}$ thus demonstrating that $\alpha_1 x + \alpha_2 y = \beta$ is of the form $y = c x + K$ for some $c, K \in \R$. {{qed}}
:500px The slope of a straight line is defined as the change in $y$ divided by the change in $x$. The change in $y$ from $p_1$ to $p_2$ is $y_2 - y_1$. The change in $x$ from $p_1$ to $p_2$ is $x_2 - x_1$. By definition of tangent of $\theta$: :$\tan \theta = \dfrac {y_2 - y_1} {x_2 - x_1}$ Hence the result. {{qed}}
From the slope-intercept form of the equation of the straight line: :$(1): \quad y = m x + c$ which is to be satisfied by both $\tuple {x_1, y_1}$ and $\tuple {x_2, y_2}$. We express $m$ and $c$ in terms of $\paren {x_1, y_1}$ and $\paren {x_2, y_2}$: {{begin-eqn}} {{eqn | l = y_1       | r = m x_1 + c       | c =  }} {{eqn | l = y_2       | r = m x_2 + c       | c =  }} {{eqn | ll= \leadsto       | l = c       | r = y_1 - m x_1       | c =  }} {{eqn | ll= \leadsto       | l = y_2       | r = m x_2 + y_1 - m x_1       | c =  }} {{eqn | n = 2       | ll= \leadsto       | l = m       | r = \dfrac {y_2 - y_1} {x_2 - x_1}       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = y_1       | r = m x_1 + c       | c =  }} {{eqn | l = y_2       | r = m x_2 + c       | c =  }} {{eqn | ll= \leadsto       | l = m       | r = \dfrac {y_2 - c} {x_2}       | c =  }} {{eqn | ll= \leadsto       | l = y_1       | r = \dfrac {y_2 - c} {x_2} x_1 + c       | c =  }} {{eqn | ll= \leadsto       | l = y_1 x_2       | r = x_1 y_2  + c \paren {x_2 - x_1}       | c =  }} {{eqn | n = 3       | ll= \leadsto       | l = c       | r = \dfrac {y_1 x_2 - x_1 y_2} {x_2 - x_1}       | c =  }} {{end-eqn}} Substituting for $m$ and $c$ in $(1)$: {{begin-eqn}} {{eqn | l = y       | r = m x + c       | c = which is $(1)$ }} {{eqn | ll= \leadsto       | l = y       | r = \dfrac {y_2 - y_1} {x_2 - x_1} x + \dfrac {y_1 x_2 - x_1 y_2} {x_2 - x_1}       | c = from $(2)$ and $(3)$ }} {{eqn | ll= \leadsto       | l = y \paren {x_2 - x_1} + x_1 y_2       | r = x \paren {y_2 - y_1} + y_1 x_2       | c =  }} {{eqn | ll= \leadsto       | l = y \paren {x_2 - x_1} + x_1 y_2 - y_1 x_1       | r = x \paren {y_2 - y_1} + y_1 x_2 - x_1 y_1       | c = adding $y_1 x_1 = x_1 y_1$ to both sides }} {{eqn | ll= \leadsto       | l = y \paren {x_2 - x_1} - y_1 \paren {x_2 - x_1}       | r = x \paren {y_2 - y_1} - x_1 \paren {y_2 - y_1}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {y - y_1} \paren {x_2 - x_1}        | r = \paren {x - x_1} \paren {y_2 - y_1}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {y - y_1} {x - x_1}       | r = \dfrac {y_2 - y_1} {x_2 - x_1}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {x - x_1} {x_2 - x_1}       | r = \dfrac {y - y_1} {y_2 - y_1}       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \alpha_1 x + \alpha_2 y       | r = \beta       | c =  }} {{eqn | ll= \leadsto       | l = \alpha_2 y       | r = y_1 - \alpha_1 x + \beta       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = y       | r = -\dfrac {\alpha_1} {\alpha_2} x + \dfrac {\beta} {\alpha_2}       | c =  }} {{end-eqn}} Setting $x = 0$ we obtain: :$y = \dfrac {\beta} {\alpha_2}$ which is the $y$-intercept. Differentiating $(1)$ {{WRT|Differentiation}} $x$ gives: :$y' = -\dfrac {\alpha_1} {\alpha_2}$ By definition, this is the slope of $\mathcal L$ and is seen to be constant. The result follows by setting: {{begin-eqn}} {{eqn | l = m       | r = -\dfrac {\alpha_1} {\alpha_2}       | c =  }} {{eqn | l = c       | r = \dfrac {\beta} {\alpha_2}       | c =  }} {{end-eqn}} {{qed}}