hcju/proofwikiqa huggingface dataset:
For the first part: {{begin-eqn}} {{eqn | l = \sin x       | o = >       | r = 0       | c = if there exists an integer $n$ such that $2 n \pi < x < \paren {2 n + 1} \pi$       | cc= Sign of Sine }} {{eqn | ll= \leadsto       | l = \frac 1 {\sin x}       | o = >       | r = 0       | c = if there exists an integer $n$ such that $2 n \pi < x < \paren {2 n + 1} \pi$       | cc= Reciprocal of Strictly Positive Real Number is Strictly Positive }} {{eqn | ll= \leadsto       | l = \csc x       | o = >       | r = 0       | c = if there exists an integer $n$ such that $2 n \pi < x < \paren {2 n + 1} \pi$       | cc= Cosecant is Reciprocal of Sine }} {{end-eqn}} For the second part: {{begin-eqn}} {{eqn | l = \sin x       | o = <       | r = 0       | c = if there exists an integer $n$ such that $\paren {2 n + 1} \pi < x < \paren {2 n + 2} \pi$       | cc= Sign of Sine }} {{eqn | ll= \leadsto       | l = \frac 1 {\sin x}       | o = <       | r = 0       | c = if there exists an integer $n$ such that $\paren {2 n + 1} \pi < x < \paren {2 n + 2} \pi$       | cc= Reciprocal of Strictly Negative Real Number is Strictly Negative }} {{eqn | ll= \leadsto       | l = \csc x       | o = <       | r = 0       | c = if there exists an integer $n$ such that $\paren {2 n + 1} \pi < x < \paren {2 n + 2} \pi$       | cc= Cosecant is Reciprocal of Sine }} {{end-eqn}} {{qed}}
For the first part: {{begin-eqn}} {{eqn | l = \cos x       | o = >       | r = 0       | c = if there exists an integer $n$ such that $\paren {2 n - \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 1 2} \pi$       | cc= Sign of Cosine }} {{eqn | ll= \leadsto       | l = \frac 1 {\cos x}       | o = >       | r = 0       | c = if there exists an integer $n$ such that $\paren {2 n - \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 1 2} \pi$       | cc= Reciprocal of Strictly Positive Real Number is Strictly Positive }} {{eqn | ll= \leadsto       | l = \sec x       | o = >       | r = 0       | c = if there exists an integer $n$ such that $\paren {2 n - \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 1 2} \pi$       | cc= Secant is Reciprocal of Cosine }} {{end-eqn}} For the second part: {{begin-eqn}} {{eqn | l = \cos x       | o = <       | r = 0       | c = if there exists an integer $n$ such that $\paren {2 n + \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 3 2} \pi$       | cc= Sign of Cosine }} {{eqn | ll= \leadsto       | l = \frac 1 {\cos x}       | o = <       | r = 0       | c = if there exists an integer $n$ such that $\paren {2 n + \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 3 2} \pi$       | cc= Reciprocal of Strictly Negative Real Number is Strictly Negative }} {{eqn | ll= \leadsto       | l = \sec x       | o = <       | r = 0       | c = if there exists an integer $n$ such that $\paren {2 n + \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 3 2} \pi$       | cc= Secant is Reciprocal of Cosine }} {{end-eqn}} {{qed}}
For the first part: {{begin-eqn}} {{eqn | l = \tan x       | o = >       | r = 0       | c = if there exists an integer $n$ such that $n \pi < x < \paren {n + \dfrac 1 2} \pi$       | cc= Sign of Tangent }} {{eqn | ll= \leadsto       | l = \frac 1 \tan x       | o = >       | r = 0       | c = if there exists an integer $n$ such that $n \pi < x < \paren {n + \dfrac 1 2} \pi$       | cc= Reciprocal of Strictly Positive Real Number is Strictly Positive }} {{eqn | ll= \leadsto       | l = \cot x       | o = >       | r = 0       | c = if there exists an integer $n$ such that $n \pi < x < \paren {n + \dfrac 1 2} \pi$       | cc= Cotangent is Reciprocal of Tangent }} {{end-eqn}} For the second part: {{begin-eqn}} {{eqn | l = \tan x       | o = <       | r = 0       | c = if there exists an integer $n$ such that $\paren {n + \dfrac 1 2} \pi < x < \paren {n + 1} \pi$       | cc= Sign of Tangent }} {{eqn | ll= \leadsto       | l = \frac 1 {\tan x}       | o = <       | r = 0       | c = if there exists an integer $n$ such that $\paren {n + \dfrac 1 2} \pi < x < \paren {n + 1} \pi$       | cc= Reciprocal of Strictly Negative Real Number is Strictly Negative }} {{eqn | ll= \leadsto       | l = \cot x       | o = <       | r = 0       | c = if there exists an integer $n$ such that $\paren {n + \dfrac 1 2} \pi < x < \paren {n + 1} \pi$       | cc= Cotangent is Reciprocal of Tangent }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sec^2 x - \tan^2 x       | r = 1       | c = Difference of Squares of Secant and Tangent }} {{eqn | ll= \leadsto       | l = \sec^2 x       | r = 1 + \tan ^2 x }} {{eqn | ll= \leadsto       | l = \frac 1 {\cos^2 x}       | r = 1 + \tan^2 x       | c = Secant is Reciprocal of Cosine }} {{eqn | ll= \leadsto       | l = \cos^2 x       | r = \frac 1 {1 + \tan^2 x} }} {{eqn | ll= \leadsto       | l = \cos x       | r = \pm \frac 1 {\sqrt {1 + \tan^2 x} } }} {{end-eqn}} Then from Sign of Cosine: {{begin-eqn}} {{eqn | l = \cos x       | o = >       | r = 0       | c = if there exists an integer $n$ such that $\paren {2 n - \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 1 2} \pi$ }} {{eqn | l = \cos x       | o = <       | r = 0       | c = if there exists an integer $n$ such that $\paren {2 n + \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 3 2} \pi$ }} {{end-eqn}} When $\cos x = 0$, $\tan x$ is undefined. {{qed}}
{{begin-eqn}} {{eqn | l = \sec^2 x - \tan^2 x       | r = 1       | c = Difference of Squares of Secant and Tangent }} {{eqn | ll= \leadsto       | l = \tan^2 x       | r = \sec^2 x - 1 }} {{eqn | ll= \leadsto       | l = \tan x       | r = \pm \sqrt {\sec^2 x - 1} }} {{end-eqn}} Also, from Sign of Tangent: :If there exists integer $n$ such that $n \pi < x < \paren {n + \dfrac 1 2} \pi$, $\tan x > 0$. :If there exists integer $n$ such that $\paren {n + \dfrac 1 2} \pi < x < \paren {n + 1} \pi$, $\tan x < 0$. When $\cos x = 0$, $\tan x$ and $\sec x$ is undefined. {{qed}}
Let both $\mathcal R_1$ and $\mathcal R_2$ be left-total. Let $\mathcal R = \mathcal R_1 \cup \mathcal R_2$. Let $s \in S_1 \cup S_2$. By the definition of union: :$s \in S_1 \lor s \in S_2$ Thus $s \in S_i$ for $i \in \set {1, 2}$. By definition of left-total relation, there is a $t \in T_i$ such that $\tuple {s, t} \in \mathcal R_i$. We have that $\mathcal R$ is a superset of $\mathcal R_i$. Hence from Union is Smallest Superset: :$\tuple {s, t} \in \mathcal R_i \subseteq \mathcal R \implies \tuple {s, t} \in \mathcal R$ {{qed}}
Let $\left(t,s\right) \in \mathcal R_1^{-1} \cup \mathcal R_2^{-1}$. By the definition of union: $\left(t,s\right) \in \mathcal R_1^{-1} \vee \left(t,s\right) \in \mathcal R_2^{-1}$. Assume $\left(t,s\right) \in \mathcal R_i^{-1}$. By the definition of inverse: $\left(s,t\right) \in \mathcal R_i$. By Disjunction Introduction: $\left(s,t\right) \in \mathcal R_1 \vee \left(s,t\right) \in \mathcal R_2 \iff \left(s,t\right) \in \mathcal R_1 \cup \mathcal R_2$. And by the definition of inverse: $\left(t,s\right) \in \left(\mathcal R_1 \cup \mathcal R_2\right)^{-1}$. {{qed}} Category:Inverse Relations ggcvvyz2qfw2ocefk8br3poepkyybtb
From Set Difference as Intersection with Relative Complement: :$U \setminus V = U \cap \complement_S \left({V}\right)$ Since $\tau$ is a topology: :$U, \complement_S \left({V}\right) \in \tau \implies U \cap \complement_S \left({V}\right) \in \tau \implies U \setminus V \in \tau$ The other statement follows mutatis mutandis. {{qed}}
Define the predicates $L$ and $R$ by: :$\map L X \iff \text {$X$ is left-total}$ :$\map R X \iff \text {$X$ is right-total}$ {{begin-eqn}} {{eqn | l = \map R {\mathcal R_1} \land \map R {\mathcal R_2}       | o = \leadsto       | r = \map L {\mathcal R_1^{-1} } \land \map L {\mathcal R_2^{-1} }       | c = Inverse of Right-Total Relation is Left-Total }} {{eqn | o = \leadsto       | r = \map L {\mathcal R_1^{-1} \cup \mathcal R_2^{-1} }       | c = Union of Left-Total Relations is Left-Total }} {{eqn | o = \leadsto       | r = \map L {\paren {\mathcal R_1 \cup \mathcal R_2}^{-1} }       | c = Union of Inverse is Inverse of Union }} {{eqn | o = \leadsto       | r = \map R {\mathcal R_1 \cup \mathcal R_2}       | c = Inverse of Right-Total Relation is Left-Total }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {x \rd x} {\paren {a x + b}^2 \paren {p x + q} }       | r = \frac 1 {b p - a q} \paren {\frac q {b p - a q} \ln \size {\frac {a x + b} {p x + q} } - \frac b {a \paren {a x + b} } } + C       | c = Primitive of $\dfrac x {\paren {a x + b}^2 \paren {p x + q} }$ }} {{eqn | r = \frac 1 {b p - a q} \paren {\frac q {b p - a q} \ln \size {\frac {a x + b} {p x + q} } - \frac {a x + b} {a \paren {a x + b} } + \frac {a x} {a \paren {a x + b} } } + C       | c =  }} {{eqn | r = \frac 1 {b p - a q} \paren {\frac q {b p - a q} \ln \size {\frac {a x + b} {p x + q} } - \frac 1 a + \frac x {a x + b} } + C       | c =  }} {{eqn | r = \frac 1 {b p - a q} \paren {\frac q {b p - a q} \ln \size {\frac {a x + b} {p x + q} } + \frac x {a x + b} } + \paren {C - \frac 1 {a \paren {b p - a q} } }       | c =  }} {{eqn | r = \frac 1 {b p - a q} \paren {\frac q {b p - a q} \ln \size {\frac {a x + b} {p x + q} } + \frac x {a x + b} } + C       | c = subsuming constant }} {{end-eqn}} {{qed}} Category:Primitives involving a x + b and p x + q 8slaipf6109lrl1iotg7c3y113hiik3
Let $y = \map {\tan^{-1} } {i x}$. Let $x = \tanh \theta$, then $\theta = \tanh^{-1} x$. {{begin-eqn}} {{eqn | l = \tan y       | r = i x       | c =  }} {{eqn | l = \tan y       | r = i \tanh \theta       | c =  }} {{eqn | l = \tan y       | r = \map \tan {i \theta}       | c = Hyperbolic Tangent in terms of Tangent }} {{eqn | ll= \leadsto       | l = y       | r = i \theta       | c =  }} {{eqn | l = y       | r = i \tanh^{-1} x       | c =  }} {{eqn | l = y       | r = \frac i 2 \map \ln {\frac {1 + x} {1 - x} }       | c = {{Defof|Real Hyperbolic Arctangent}} }} {{end-eqn}} {{qed}} Category:Complex Numbers Category:Tangent Function ltvuizouywxeukypjt6rxq3tg1fc6c2
=== Definition 1 implies Definition 2 === Let $x = \tanh y$. Then: {{begin-eqn}} {{eqn | l = x       | r = \frac {e^{2 y} - 1} {e^{2 y} + 1}       | c = Definition of Hyperbolic Tangent }} {{eqn | ll = \implies       | l = x e^{2 y} + x       | r = e^{2 y} - 1       | c =  }} {{eqn | ll = \implies       | l = e^{2 y} - x e^{2 y}       | r = 1 + x       | c =  }} {{eqn | ll = \implies       | l = e^{2 y}       | r = \frac {1 + x} {1 - x}       | c =  }} {{eqn | ll = \implies       | l = 2 y       | r = \ln \left({ \frac {1 + x} {1 - x} }\right)       | c =  }} {{eqn | ll = \implies       | l = y       | r = \frac 1 2 \ln \left({ \frac {1 + x} {1 - x} }\right)       | c =  }} {{end-eqn}} {{qed|lemma}} === Definition 2 implies Definition 1 === Let $y = \dfrac {1 + x} {1 - x}$. {{begin-eqn}} {{eqn | l = \tanh \left({\frac 1 2 \ln \left({ \frac {1 + x} {1 - x} }\right)}\right)       | r = \tanh \left({\frac 1 2 \ln y}\right)       | c =  }} {{eqn | r = \frac {e^{2 \left({\frac 1 2 \ln y}\right) - 1} } {e^{2 \left({\frac 1 2 \ln y}\right) + 1} }       | c = Definition of Hyperbolic Tangent }} {{eqn | r = \frac {e^{\ln y} - 1} {e^{\ln y} + 1}       | c =  }} {{eqn | r = \frac {y - 1} {y + 1}       | c = Exponential of Natural Logarithm }} {{eqn | r = \frac {\frac {1 + x} {1 - x} - 1} {\frac {1 + x} {1 - x} + 1}       | c =  }} {{eqn | r = \frac {\left({1 + x}\right) - \left({1 - x}\right)} {\left({1 + x}\right) + \left({1 - x}\right)}       | c =  }} {{eqn | r = \frac {2x} 2       | c =  }} {{eqn | r = x       | c =  }} {{end-eqn}} {{qed|lemma}} Therefore: {{begin-eqn}} {{eqn | n = 1       | l = x = \tanh \left({y}\right)       | o = \implies       | r = y = \frac 1 2 \ln \left({ \frac {1 + x} {1 - x} }\right)       | c = Definition 1 implies Definition 2 }} {{eqn | n = 2       | l = y = \frac 1 2 \ln \left({ \frac {1 + x} {1 - x} }\right)       | o = \implies       | r = x = \tanh \left({y}\right)       | c = Definition 2 implies Definition 1 }} {{eqn | ll = \implies       | l = x = \tanh \left({y}\right)       | o = \iff       | r = y = \frac 1 2 \ln \left({ \frac {1 + x} {1 - x} }\right)       | c =  }} {{end-eqn}} {{qed}}
Let $x$ and $y$ be positive real numbers such that $x < y$. {{AimForCont}} $\sqrt x \ge \sqrt y$. {{begin-eqn}} {{eqn | n = 1       | l = \sqrt x       | o = \ge       | r = \sqrt y       | c =  }} {{eqn | n = 2       | l = \sqrt x       | o = \ge       | r = \sqrt y       | c =  }} {{eqn | l = x       | o = \ge       | r = y       | c = Real Number Axioms: $\R O2$: compatibility with multiplication, $(1) \times (2)$ }} {{end-eqn}} Thus a contradiction is created. Therefore: :$\forall x, y \in \R_{>0}: x < y \implies \sqrt x < \sqrt y$ {{qed}} Category:Real Numbers Category:Square Roots fdvd6p7g0m570jrvlsjkqp533pis13w
=== Definition 1 implies Definition 2 === Let $x = \sinh y$. Let $z = e^y$. Then: {{begin-eqn}} {{eqn | l = x       | r = \frac {e^y - e^{- y} } 2       | c = {{Defof|Hyperbolic Sine}} }} {{eqn | ll= \leadsto       | l = 2 x       | r = e^y - e^{- y}       | c =  }} {{eqn | ll= \leadsto       | l = 2 x e^y       | r = e^{2 y} - 1       | c =  }} {{eqn | ll= \leadsto       | l = z^2 - 2 x z - 1       | r = 0       | c = Power of Power }} {{eqn | ll= \leadsto       | l = z       | r = \frac {2 x \pm \sqrt {\paren {-2 x}^2 + 4} } 2       | c = Quadratic Formula }} {{eqn | r = x \pm \sqrt {x^2 + 1}       | c =  }} {{eqn | ll= \leadsto       | l = e^y       | r = x \pm \sqrt {x^2 + 1}       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \map \ln {x \pm \sqrt {x^2 + 1} }       | c =  }} {{end-eqn}} If $x \ge 0$, then: {{begin-eqn}} {{eqn | l = \sqrt {x^2 + 1}       | o = >       | r = \sqrt {x^2}       | c = Square Root is Strictly Increasing }} {{eqn | r = x       | c =  }} {{eqn | ll= \leadsto       | l = \sqrt {x^2 + 1}       | o = >       | r = x       | c =  }} {{eqn | ll= \leadsto       | l = x - \sqrt {x^2 + 1}       | o = <       | r = 0       | c =  }} {{end-eqn}} If $x < 0$, then: {{begin-eqn}} {{eqn | n = 1       | l = -\sqrt {x^2 + 1}       | o = <       | r = 0       | c = Positive Square Root is Positive }} {{eqn | n = 2       | l = x       | o = <       | r = 0       | c = Assumption }} {{eqn | l = x - \sqrt {x^2 + 1}       | o = <       | r = 0       | c = $(1) + (2)$ }} {{end-eqn}} Since the natural logarithm of a negative number is not defined: :$y = \map \ln {x + \sqrt {x^2 + 1} }$ {{qed|lemma}} === Definition 2 implies Definition 1 === Let $y = x + \sqrt {x^2 + 1}$. {{begin-eqn}} {{eqn | l = \map \sinh {\map \ln {x + \sqrt {x^2 + 1} } }       | r = \map \sinh {\ln y}       | c =  }} {{eqn | r = \frac {e^{\ln y} - e^{-\ln y} } 2       | c = {{Defof|Hyperbolic Sine}} }} {{eqn | r = \frac {y - \frac 1 y} 2       | c =  }} {{eqn | r = \frac {y^2 - 1} {2 y}       | c =  }} {{eqn | r = \frac {\paren {x + \sqrt {x^2 + 1} }^2 - 1} {2 x + 2 \sqrt {x^2 + 1} }       | c =  }} {{eqn | r = \frac {x^2 + 2 x \sqrt {x^2 + 1} + \paren {x^2 + 1} - 1} {2 x + 2 \sqrt {x^2 + 1} }       | c =  }} {{eqn | r = \frac {2 x^2 + 2 x \sqrt {x^2 + 1} } {2 x + 2 \sqrt {x^2 + 1} }       | c =  }} {{eqn | r = \frac {\paren {2 x + 2 \sqrt {x^2 + 1} } x} {2 x + 2 \sqrt {x^2 + 1} }       | c =  }} {{eqn | r = x       | c =  }} {{end-eqn}} {{qed|lemma}} Therefore: {{begin-eqn}} {{eqn | n = 1       | l = x = \sinh y       | o = \implies       | r = y = \map \ln {x + \sqrt {x^2 + 1} }       | c = Definition 1 implies Definition 2 }} {{eqn | n = 2       | l = y = \map \ln {x + \sqrt {x^2 + 1} }       | o = \implies       | r = x = \sinh y       | c = Definition 2 implies Definition 1 }} {{eqn | ll= \leadsto       | l = x = \sinh y       | o = \iff       | r = y = \map \ln {x + \sqrt {x^2 + 1} }       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cosh^2 x - \sinh^2 x       | r = 1       | c = Difference of Squares of Hyperbolic Cosine and Sine }} {{eqn | ll= \leadsto       | l = \cosh^2 x       | r = 1 + \sinh^2 x       | c =  }} {{eqn | o = \ge       | r = 1       | c = Square of Real Number is Non-Negative }} {{end-eqn}} Furthermore, $\cosh x = 1$ when $x = 0$, satisfying the equality case. {{qed}} Category:Hyperbolic Cosine Function 1c4szrr57lywuqxqhmd77xy3o014eh0
This proof assumes the series definition of $\exp$. That is, let: :$\ds \exp x = \sum_{n \mathop = 0}^\infty \dfrac {x^n} {n!}$ First, suppose $0 < x$. Then: {{begin-eqn}} {{eqn | l = 0       | o = <       | r = x^n       | c = Power Function is Strictly Increasing over Positive Reals: Natural Exponent }} {{eqn | ll= \leadsto       | l = 0       | o = <       | r = \frac {x^n} {n!}       | c = Real Number Ordering is Compatible with Multiplication }} {{eqn | ll= \leadsto       | l = 0       | o = <       | r = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}       | c = Ordering of Series of Ordered Sequences }} {{eqn | ll= \leadsto       | l = 0       | o = <       | r = \exp x       | c = Definition of $\exp$ }} {{end-eqn}} So $\exp$ is strictly positive on $\R_{>0}$. From Exponential of Zero, $\exp 0 = 1$. Finally, suppose that $x < 0$. Then: {{begin-eqn}} {{eqn | l = 0       | o = <       | r = -x       | c = Order of Real Numbers is Dual of Order of their Negatives }} {{eqn | ll= \leadsto       | l = 0       | o = <       | r = \map \exp {-x}       | c = from above }} {{eqn | ll= \leadsto       | l = 0       | o = <       | r = \frac 1 {\exp x}       | c = Reciprocal of Real Exponential }} {{eqn | ll= \leadsto       | l = 0       | o = <       | r = \exp x       | c = Ordering of Reciprocals }} {{end-eqn}} So $\exp$ is strictly positive on $\R_{<0}$. Hence the result. {{qed}}
This proof assumes the limit definition of $\exp$. That is, let: :$\displaystyle \exp x = \lim_{n \mathop \to \infty} \map {f_n} x$ where $\map {f_n} x = \paren {1 + \dfrac x n}^n$ First, fix $x \in \R$. Let $N = \ceiling {\size x}$, where $\ceiling {\, \cdot \,}$ denotes the ceiling function. Then: {{begin-eqn}} {{eqn | l = \exp x       | r = \lim_{n \mathop \to \infty} \map {f_n} x }} {{eqn | r = \lim_{n \mathop \to \infty} \map {f_{n + N} } x       | c = Tail of Convergent Sequence }} {{eqn | o = \ge       | r = \map {f_{n + N} } x       | c = Exponential Sequence is Eventually Increasing and Limit of Bounded Convergent Sequence is Bounded }} {{eqn | o = >       | r = 0       | c = Corollary to Exponential Sequence is Eventually Increasing }} {{end-eqn}} {{MissingLinks|Corollary to Exponential Sequence is Eventually Increasing does not actually exist. The page it gets sent to does not give that result.}} {{qed}}
This proof assumes the definition of $\exp x$ as the unique continuous extension of $e^x$. Since $e > 0$, the result follows immediately from Power of Positive Real Number is Positive over Rationals. {{qed}}
This proof assumes the definition of $\exp$ as the inverse mapping of extension of $\ln$, where $\ln$ denotes the natural logarithm. Recall that the domain of $\ln$ is $\R_{>0}$. From the definition of inverse mapping, the image of $\exp$ is the domain of $\ln$. That is, the image of $\exp$ is $\R_{>0}$. Hence the result. {{qed}}
This proof assumes the definition of $\exp$ as the solution to an initial value problem. That is, suppose $\exp$ satisfies: :$ (1): \quad D_x \exp x = \exp x$ :$ (2): \quad \map \exp 0 = 1$ on $\R$. === Lemma === {{:Exponential of Real Number is Strictly Positive/Proof 5/Lemma}}{{qed|lemma}} {{AimForCont}} that $\exists \alpha \in \R: \exp \alpha < 0$. Then $0 \in \openint {\exp \alpha} 1$. From Intermediate Value Theorem: :$\exists \zeta \in \openint \alpha 0: \map f \zeta = 0$ This contradicts the lemma. {{qed}}
From the definition of power: :$x^y = \exp \left({y \ln x}\right)$ From Exponential of Real Number is Strictly Positive: :$x^y = \exp \left({y \ln x}\right) > 0$ {{qed}} Category:Real Analysis oa8bjlxwrg5kas1mcyixm085z7x58vm
{{begin-eqn}} {{eqn | l = \map {\dfrac \d {\d x} } {\map u x^n}       | r = \lim_{h \mathop \to 0} \frac {\paren {\map u {x + h} }^n - \paren {\map u x}^n} h       | c =  }} {{eqn | r = \paren {\map u x}^n \lim_{h \mathop \to 0} \frac {\paren {\frac {\map u {x + h} } {\map u x} }^n - 1} h       | c = Exponent Combination Laws/Power of Product }} {{eqn | r = \paren {\map u x}^n \lim_{h \mathop \to 0} \frac {\exp \paren {n \ln \frac {\map u {x + h} } {\map u x} } - 1} h       | c = {{Defof|Power to Real Number}} }} {{eqn | r = \paren {\map u x}^n \lim_{h \mathop \to 0} \paren {\frac {\map \exp {n \ln \frac {\map u {x + h} } {\map u x} } - 1} {n \ln \frac {\map u {x + h} } {\map u x} } } \paren {\frac {n \ln \frac {\map u {x + h} } {\map u x} } h}       | c =  }} {{eqn | r = \paren {\map u x}^n \lim_{h \mathop \to 0} \frac {n \ln \frac {\map u {x + h} } {\map u x} } h       | c = Derivative of Exponential at Zero }} {{eqn | r = n \paren {\map u x}^n \lim_{h \mathop \to 0} \frac {\ln \frac {\map u {x + h} } {\map u x} } h       | c =  }} {{eqn | r = n \paren {\map u x}^n \lim_{h \mathop \to 0} \frac {\map \ln {1 + \frac {\map u {x + h} - \map u x} {\map u x} } } h       | c =  }} {{eqn | r = n \paren {\map u x}^n \lim_{h \mathop \to 0} \paren {\frac {\map \ln {1 + \frac {\map u {x + h} - \map u x} {\map u x} } } {\frac {\map u {x + h} - \map u x} {\map u x} } } \paren {\frac {\frac {\map u {x + h} - \map u x} {\map u x} } h }       | c =  }} {{eqn | r = n \paren {\map u x}^n \lim_{h \mathop \to 0} \frac {\paren {\frac {\map u {x + h} - \map u x} {\map u x} } } h       | c = Derivative of Logarithm at One }} {{eqn | r = n \paren {\map u x}^n \lim_{h \mathop \to 0} \frac 1 {\map u x} \frac {\map u {x + h} - \map u x} h       | c =  }} {{eqn | r = n \paren {\map u x}^{n - 1} \lim_{h \mathop \to 0} \frac {\map u {x + h} - \map u x} h       | c = Exponent Combination Laws/Product of Powers }} {{eqn | r = n \paren {\map u x}^{n - 1} \map {\dfrac \d {\d x} } {\map u x}       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {\frac \d {\d x} } {\map u x^n}       | r = \map {\frac \d {\d u} } {\map u x^n} \map {\frac \d {\d x} } {\map u x}       | c = Chain Rule for Derivatives }} {{eqn | r = n \map u x^{n - 1} \map {\frac {\d u} {\d x} } {\map u x}       | c = Derivative of Hyperbolic Sine }} {{end-eqn}} {{qed}}
The definition of the Euler's number as the Base of Logarithm will be used. Then the result follows directly. {{qed}}
{{begin-eqn}} {{eqn | n = 1       | l = x       | o = <       | r = y       | c = Assumption }} {{eqn | ll= \leadsto       | l = x^2       | o = <       | r = y^2       | c = Axiom $\R \text O 2$: Usual ordering is compatible with multiplication }} {{eqn | ll= \leadsto       | l = x^2 - 1       | o = <       | r = y^2 - 1       | c =  }} {{eqn | n = 2       | ll= \leadsto       | l = \sqrt {x^2 - 1}       | o = <       | r = \sqrt {y^2 - 1}       | c = Square Root is Strictly Increasing }} {{eqn | ll= \leadsto       | l = x + \sqrt {x^2 - 1}       | o = <       | r = y + \sqrt {y^2 - 1}       | c = $(1) + (2)$ }} {{eqn | ll= \leadsto       | l = \cosh^{-1} x       | o = <       | r = \cosh^{-1} y       | c = {{Defof|Real Inverse Hyperbolic Cosine}} }} {{end-eqn}} {{qed}} Category:Inverse Hyperbolic Cosine 8erlqsh2xzkc6eesg1sz2xfizza49i3
{{begin-eqn}} {{eqn | l = a \laptrans {\map f {a t} }       | r = a \int_0^{\to + \infty} e^{-s t} \map f {a t} \rd t       | c = {{Defof|Laplace Transform}} }} {{eqn | r = a \paren {\frac 1 a} \int_0^{\to + \infty} e^{-s t} \map f {a t} \rd \paren {a t}       | c = Primitive of Function of Constant Multiple }} {{eqn | r = \int_0^{\to + \infty} e^{-u a t} \map f {a t} \rd \paren {a t}       | c = where $u = \dfrac s a$ }} {{eqn | r = \int_0^{\to + \infty} e^{-u a t} \map f {a t} \rd \paren {a t} }} {{eqn | r = \map F u       | c = {{Defof|Laplace Transform}} }} {{eqn | r = \map F {\dfrac s a} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \laptrans {\map g t}       | r = \int_0^\infty e^{-s t} \map g t \rd t       | c = {{Defof|Laplace Transform}} }} {{eqn | r = \int_0^a e^{-s t} \map g t \rd t + \int_a^\infty e^{-s t} \map g t \rd t       | c =  }} {{eqn | r = \int_0^a 0 \times e^{-s t} \rd t + \int_a^\infty e^{-s t} \map f {t - a} \rd t       | c = Definition of $\map g t$ }} {{eqn | r = \int_a^\infty e^{-s t} \map f {t - a} \rd t       | c =  }} {{eqn | r = \int_0^\infty e^{-s \paren {u + a} } \map f u \rd u       | c = Integration by Substitution: $t = u + a$ }} {{eqn | r = e^{-a s} \int_0^\infty e^{-s u} \map f u \rd u       | c =  }} {{eqn | r = e^{-a s} \map F s       | c = {{Defof|Laplace Transform}} }} {{end-eqn}} {{qed}}