hcju/proofwikiqa huggingface dataset:
From the Gelfond-Schneider Theorem: If: :$\alpha$ and $\beta$ are algebraic numbers such that $\alpha \notin \set {0, 1}$  :$\beta$ is either irrational or not wholly real then $\alpha^\beta$ is transcendental. We have that: {{begin-eqn}} {{eqn | l = i^{-2 i}       | r = \paren {e^{i \pi / 2} }^{- 2 i}       | c =  }} {{eqn | r = e^{-\pi i^2}       | c =  }} {{eqn | r = e^\pi       | c =  }} {{end-eqn}} As: :$i$ is algebraic :$-2 i$ is algebraic and not wholly real the conditions of the Gelfond-Schneider Theorem are fulfilled. Hence the result. {{qed}}
By definition, a Riemann surface is a complex manifold. Hence it is connected and locally path-connected. {{explain|It needs to be established that a complex manifold indeed has both of these properties.}} {{ProofWanted|A link needed to a page which derives the fact of path-connectedness from connectedness and local path-connectedness.}} Category:Riemann Surfaces euw14tot4aq803m48uc9qnzb4b31o5i
Follows a priori from Riemann Surface admits Metric of Constant Curvature. {{qed}} Category:Riemann Surfaces 66rch1oa45nh22ktrt9tgypjhors75a
{{ProofWanted|A consequence of Hamilton's Principle in the calculus of variations.}}
=== $(1) \implies (2)$ === Assume that :$\forall x \in S, f:S \to S: \left({\forall s \in S: f\left({s}\right) = x \wedge s}\right) \implies f$ is lower adjoint Let $x, t \in S$. Define $f: S \to S$: :$\forall s \in S: f\left({s}\right) = x \wedge s$ By assumption: :$f$ is lower adjoint By definition of lower adjoint: :$\exists g: S \to S: \left({g, f}\right)$ is Galois connection By Galois Connection is Expressed by Maximum: :$\forall s \in S: g\left({s}\right) = \max\left({f^{-1}\left[{s^\preceq}\right]}\right)$ Then :$\max\left({f^{-1}\left[{t^\preceq}\right]}\right)$ exists. By definition of image of set: :$\max\left\{ {s \in S: f\left({s}\right) \in t^\preceq}\right\}$ exists. By definition of lower closure of element: :$\max \left\{ {s \in S: x \wedge s \preceq t}\right\}$ exists. {{qed|lemma}} === $(2) \implies (1)$ === Assume that :$\forall x, t \in S: \max \left\{ {s \in S: x \wedge s \preceq t}\right\}$ exists. Let $x \in S, f: S \to S$ such that :$\forall s \in S: f\left({s}\right) = x \wedge s$ As maxima exist define a mapping $g: S \to S$: :$\forall s \in S: g\left({s}\right) = \max\left({f^{-1}\left[{s^\preceq}\right]}\right)$ We will prove that :$f$ is an increasing mapping Let $y, z \in S$ such that :$y \preceq z$ By definition of $f$: :$f\left({y}\right) = x \wedge y$ and $f\left({z}\right) = x \wedge z$ Thus by Meet Semilattice is Ordered Structure: :$f\left({y}\right) \preceq f\left({z}\right)$ By Galois Connection is Expressed by Maximum: :$\left({g, f}\right)$ is Galois connection By definition :$f$ is a lower adjoint {{qed}}
:$\left({S, \preceq}\right)$ is a Brouwerian lattice {{iff}} :$\forall x, y \in S: x$ has relative pseudocomplement with respect to $y$ by definition of Brouwerian lattice {{iff}} :$\forall x, y \in S: \max \left\{ {s \in S: x \wedge s \preceq y}\right\}$ exists by definition of relative pseudocomplement {{iff}} :$\forall x \in S, f: S \to S: \left({\forall s \in S: f \left({s}\right) = x \wedge s}\right) \implies f$ is a lower adjoint by Shift Mapping is Lower Adjoint iff Appropriate Maxima Exist {{qed}}
Let $x, y, z \in S$. By Brouwerian Lattice iff Shift Mapping is Lower Adjoint: :$\forall x \in S, f: S \to S: \left({\forall s \in S: f \left({s}\right) = x \wedge s}\right) \implies f$ is a lower adjoint Define a mapping $f: S \to S$: :$\forall s \in S: f \left({s}\right) = x \wedge s$ Then: :$f$ is a lower adjoint By Lower Adjoint Preserves All Suprema: :$f$ preserves all suprema By definition of preserves all suprema: :$f$ preserves the supremum of $\left\{ {y, z}\right\}$ By definition of lattice: :$\left\{ {y, z}\right\}$ admits a supremum By preserves the supremum of set: :$\sup \left({f^\to\left({\left\{ {y, z}\right\} }\right)}\right) = f \left({\sup \left\{ {y, z}\right\} }\right)$ Thus {{begin-eqn}} {{eqn | l = x \wedge \left({y \vee z}\right)       | r = f \left({y \vee z}\right)       | c = definition of $f$ }} {{eqn | r = f \left({\sup \left\{ {y, z}\right\} }\right)       | c = definition of join operation }} {{eqn | r = \sup \left({f^\to\left({\left\{ {y, z}\right\} }\right)}\right) }} {{eqn | r = \sup \left({\left\{ {f\left({y}\right), f\left({z}\right)}\right\} }\right)       | c = definition of image of set }} {{eqn | r = \sup \left\{ {x \wedge y, x \wedge z}\right\}       | c = definition of $f$ }} {{eqn | r = \left({x \wedge y}\right) \vee \left({x \wedge z}\right)       | c = definition of join operation }} {{end-eqn}} Thus by definition: :$\left({S, \preceq}\right)$ is a distributive lattice {{qed}}
{{begin-eqn}} {{eqn | l = \int_0^x \frac {\d t} {\sqrt{1 - t^2} }       | r = \intlimits {\arcsin \frac t 1} 0 x       | c = Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$, {{Defof|Definite Integral}} }} {{eqn | r = \arcsin x - \arcsin 0         | c =  }} {{eqn | r = \arcsin x  }} {{end-eqn}} {{qed}}
=== Sufficient Condition === From Holomorphic Function is Analytic, if $f$ is holomorphic on some open ball $D$ in $\C$, then $f$ is complex analytic on $D$.  However, as, by the definition of an entire function, $f$ is holomorphic everywhere, the radius of $D$ can be made arbitrarily large, meaning that $f$ is complex analytic everywhere. This implies that an entire function is complex analytic everywhere. {{qed|lemma}} === Necessary Condition === It remains to prove that if $f$ is complex analytic everywhere, $f$ is entire.  By Power Series is Termwise Differentiable within Radius of Convergence, if $f$ is complex analytic everywhere, $f$ is holomorphic everywhere.  That is, a function complex analyticeverywhere is entire.  Note that by the Cauchy-Hadamard theorem:  :$\displaystyle \lim_{n \mathop \to \infty} \sqrt [n] {\size {a_n} } = 0$ {{iff}} this series has an infinite radius of convergence.  That is, {{iff}} the series converges for all $z \in \C$, satisfying our second demand.  {{qed}} Category:Entire Functions dowwf37pjjl19876udtgduezr6puxvk
By assumption: :$S \ne \O$ By definition of non-empty set: :$\exists s: s \in S$ By definition of Brouwerian lattice: :$s$ has relative pseudocomplement with respect to $s$ By definition of relative pseudocomplement: :$\max \set {x \in S: s \wedge x \preceq s}$ exists and equals $s \to s$ Let $x \in S$. By Meet Precedes Operands: :$s \wedge x \preceq s$ Then: :$x \in \set {x \in S: s \wedge x \preceq s}$ Thus by definition of maximum: :$x \preceq s \to s$ Thus by definition: :$s \to s$ is an upper bound for $S$ Thus by definition: :$S$ is upper bounded. {{qed}}
By Brouwerian Lattice iff Shift Mapping is Lower Adjoint: :$d$ is lower adjoint By definition of lower adjoint: :$\exists g': S \to S: \left({g', d}\right)$ is Galois connection By Galois Connection is Expressed by Maximum: :$\forall s \in S: g'\left({s}\right) = \max\left({d^{-1}\left[{s^\preceq}\right]}\right)$ By definition of image of set: :$\forall s \in S: g'\left({s}\right) = \max\left({\left\{ {x \in S: d\left({x}\right) \in s^\preceq}\right\} }\right)$ By definition of lower closure of element: :$\forall s \in S: g'\left({s}\right) = \max\left({\left\{ {x \in S: d\left({x}\right) \preceq s}\right\} }\right)$ By definition of $d$: :$\forall s \in S: g'\left({s}\right) = \max\left({\left\{ {x \in S: a \wedge x \preceq s}\right\} }\right)$ By definition of relative pseudocomplement: :$\forall s \in S: g'\left({s}\right) = a \to s = g\left({s}\right)$ By Equality of Mappings: :$g = g'$ Hence : $\left({g, d}\right)$ is Galois connection {{qed}}
Define a mapping $d: S \to S$: :$\forall s \in S: \map d s = a \wedge s$ Define a mapping $g: S \to S$: :$\forall s \in S: \map g s = a \to s$ By Relative Pseudocomplement and Shift Mapping form Galois Connection in Brouwerian Lattice: :$\tuple {g, d}$ is Galois connection. By definition of Galois connection: :$x \preceq \map g y$ {{iff}} $\map d x \preceq y$ Thus by definitions of $g$ and $d$: :$a \wedge x \preceq y$ {{iff}} $x \preceq a \to y$ {{qed}}
The rope will be stretched in the form of a triangle with sides $3$, $4$ and $5$. We have that: :$3^2 + 4^2 = 9 + 16 = 25 = 5^2$ and so by Pythagoras's Theorem the triangle is right angled. {{qed}}
=== Sufficient Condition === Let: :$\top = a \to b$ By definition of reflexivity: :$\top \preceq a \to b$ By Inequality with Meet Operation is Equivalent to Inequality with Relative Pseudocomplement in Brouwerian Lattice: :$\top \wedge a \preceq b$ Thus :$a \preceq b$ {{qed|lemma}} === Necessary Condition === Let: :$a \preceq b$ Then: :$a \wedge \top \preceq b$ By Inequality with Meet Operation is Equivalent to Inequality with Relative Pseudocomplement in Brouwerian Lattice: :$\top \preceq a \to b$ By definition of greatest element: :$a \to b \preceq \top$ By definition of ordered set: :$\top = a \to b$ {{qed}}
=== Reflexivity === Let $\left({s, t}\right) \in S_1 \times S_2$. By definition of reflexivity: :$s \preceq_1 s$ and $t \preceq_2 t$ Thus by definition of Cartesian product: :$\left({s, t}\right) \preceq \left({s, t}\right)$ === Transitivity === Let $\left({s_1, t_1}\right), \left({s_2, t_2}\right), \left({s_3, t_3}\right) \in S_1 \times S_2$ such that :$\left({s_1, t_1}\right) \preceq \left({s_2, t_2}\right) \preceq \left({s_3, t_3}\right)$ By definition of Cartesian product: :$s_1 \preceq_1 s_2 \preceq_1 s_3$ and $t_1 \preceq_2 t_2 \preceq_2 t_3$ By definition of transitivity: :$s_1 \preceq_1 s_3$ and $t_1 \preceq_2 t_3$ Thus by definition of Cartesian product: :$\left({s_1, t_1}\right) \preceq \left({s_3, t_3}\right)$ === Antisymmetry === Let $\left({s_1, t_1}\right), \left({s_2, t_2}\right) \in S_1 \times S_2$ such that :$\left({s_1, t_1}\right) \preceq \left({s_2, t_2}\right)$ and $\left({s_2, t_2}\right) \preceq \left({s_1, t_1}\right)$ By definition of Cartesian product: :$s_1 \preceq_1 s_2$ and $s_2 \preceq_1 s_1$ and $t_1 \preceq_2 t_2$ and $t_2 \preceq_2 t_1$ By definition of antisymmetry: :$s_1 = s_2$ and $t_1 = t_2$ Thus: :$\left({s_1, t_1}\right) = \left({s_2, t_2}\right)$ {{qed}}
Consider a strictly positive integer $n$. By Definition 1, $n$ is a perfect number {{iff}} $n$ equals the sum of its aliquot parts. By definition of sigma function, $\sigma \left({n}\right)$ equals the sum of all the divisors of $n$. Thus $\sigma \left({n}\right) - n$ equals the aliquot sum of $n$. So by Definition 2, $n$ is a perfect number {{iff}} $\sigma \left({n}\right) - n = n$. Hence the definitions are equivalent. The equivalence of definition 4 to definition 2 follows directly. The equivalence of definition 3 to definition 2 follows from the definition of abundance. {{qed}}
From Motion of Body with Variable Mass: :$(1): \quad \mathbf w \dfrac {\mathrm d m} {\mathrm d t} + \mathbf F = m \dfrac {\mathrm d \mathbf v} {\mathrm d t}$ where: :$\mathbf F$ is the external force being applied :$\mathbf w$ is the velocity of the added mass relative to $B$. In this scenario: :there is no external force and so $\mathbf F = \mathbf 0$ :the velocity of the added mass relative to $B$ is $-\mathbf b$. Thus $(1)$ becomes: :$-\mathbf b \dfrac {\mathrm d m} {\mathrm d t} = m \dfrac {\mathrm d \mathbf v} {\mathrm d t}$ {{qed}}
From Newton's Second Law of Motion: :$\mathbf F = \dfrac \d {\d t} \paren {m \mathbf v}$ At time $t + \Delta t$, let: : the mass of $B$ be $m + \Delta m$ : the velocity of $B$ be $\mathbf v + \Delta \mathbf v$. The fuel is being consumed, so the increase in mass of the fuel during time $\Delta t$ is $-\Delta m$. Thus the exhaust products, therefore of mass $-\Delta m$, are expelled at a velocity $-\mathbf b$ relative to $B$. Thus this material is actually moving at a velocity $\mathbf v - \mathbf b$. By Conservation of Momentum, the total linear momentum is constant. Thus: {{begin-eqn}} {{eqn | l = m \mathbf v       | r = \paren {m + \Delta m} \paren {\mathbf v + \Delta \mathbf v} + \paren {-\Delta m} \paren {\mathbf v - \mathbf b}       | c =  }} {{eqn | r = m \mathbf v + m \Delta \mathbf v + \paren {\Delta m} \mathbf v + \Delta m \Delta \mathbf v - \paren {-\Delta m} \mathbf v + \Delta m \mathbf b       | c =  }} {{eqn | ll= \leadsto       | l = m \Delta \mathbf v       | r = - \Delta m \paren {\mathbf b + \Delta \mathbf v}       | c =  }} {{eqn | ll= \leadsto       | l = m \frac {\Delta \mathbf v} {\Delta t}       | r = - \frac {\Delta m} {\Delta t} \paren {\mathbf b + \Delta \mathbf v}       | c =  }} {{eqn | ll= \leadsto       | l = m \frac {\d \mathbf v} {\d t}       | r = - \mathbf b \frac {\d m} {\d t}       | c = letting $\Delta t \to 0$ }} {{end-eqn}} {{qed}}
From Motion of Rocket in Outer Space, the equation of motion of $B$ is given by: :$m \dfrac {\d \mathbf v} {\d t} = -\mathbf b \dfrac {\d m} {\d t}$ Hence: {{begin-eqn}} {{eqn | l = \int_0^t \dfrac {\d \mathbf v} {\d t} \rd t       | r = -\int_0^t \mathbf b \frac 1 m \dfrac {\d m} {\d t} \rd t       | c =  }} {{eqn | ll= \leadsto       | l = \int_{t \mathop = 0}^t \rd \mathbf v       | r = -\int_{t \mathop = 0}^t \mathbf b \dfrac {\d m} m       | c =  }} {{eqn | ll= \leadsto       | l = \bigintlimits {\mathbf v} 0 t       | r = -\mathbf b \bigintlimits {\ln m} 0 t       | c =  }} {{eqn | ll= \leadsto       | l = \map {\mathbf v} t - \map {\mathbf v} 0       | r = -\mathbf b \paren {\map {\ln m} t - \map {\ln m} 0}       | c =  }} {{eqn | ll= \leadsto       | l = \map {\mathbf v} t       | r = \map {\mathbf v} 0 + \mathbf b \ln \dfrac {\map m 0} {\map m t}       | c =  }} {{end-eqn}} {{qed}}
From Motion of Rocket in Outer Space, the equation of motion of $B$ is given by: :$(1): \quad m \dfrac {\mathrm d \mathbf v} {\mathrm d t} = - \mathbf b \dfrac {\mathrm d m} {\mathrm d t}$ By definition, the acceleration of $B$ is its rate of change of velocity: :$\mathbf a = \dfrac {\mathrm d \mathbf v} {\mathrm d t}$ The result follows by substituting $\mathbf a$ for $\dfrac {\mathrm d \mathbf v} {\mathrm d t}$ and dividing by $m$. {{qed}}
Let $x, y \in \map {\pr_1^\to} X$. By definitions of image of set and projections: :$\exists x' \in T: \tuple {x, x'} \in X$ and :$\exists y' \in T: \tuple {y, y'} \in X$ By definition of directed: :$\exists \tuple {a, b} \in X: \tuple {x, x'} \preceq \tuple {a, b} \land \tuple {y, y'} \preceq \tuple {a, b}$ By definition of Cartesian product of ordered sets: :$\exists a \in \map {\pr_1^\to} X: x \preceq_1 a \land y \preceq_1 a$ Thus by definition :$\map {\pr_1^\to} X$ is directed. By mutatis mutandis: :$\map {\pr_2^\to} X$ is directed. {{qed}}
By the Simpson's Formula for Cosine by Sine: :$2 \cos \alpha \sin \beta = \sin \left({\alpha + \beta}\right) - \sin \left({\alpha - \beta}\right)$ Thus we establish the following sequence of identities: {{begin-eqn}} {{eqn | l = 2 \cdot \frac 1 2 \sin \frac x 2       | r = \sin \frac x 2       | c =  }} {{eqn | l = 2 \cos x \sin \frac x 2       | r = \sin \frac {3 x} 2 - \sin \frac x 2       | c =  }} {{eqn | l = 2 \cos 2 x \sin \frac x 2       | r = \sin \frac {5 x} 2 - \sin \frac {3 x} 2       | c =  }} {{eqn | o = \cdots       | c =  }} {{eqn | l = 2 \cos n x \sin \frac x 2       | r = \sin \frac {\left({2 n + 1}\right) x} 2 - \sin \frac {\left({2 n - 1}\right) x} 2       | c =  }} {{end-eqn}} Summing the above: :$\displaystyle 2 \sin \frac x 2 \left({\frac 1 2 + \sum_{k \mathop = 1}^n \cos \left({k x}\right)}\right) = \sin \frac {\left({2 n + 1}\right) x} 2$ as the sums on the {{RHS}} form a telescoping series. The result follows by dividing both sides by $2 \sin \dfrac x 2$. It is noted that when $x$ is a multiple of $2 \pi$ then: :$\sin \dfrac x 2 = 0$ leaving the {{RHS}} undefined. {{qed}}
Consider the number: :$N := \paren {n + 1}!$ where $!$ denotes the factorial. Let $i \in I$ where: :$I = \set {i \in \N: 2 \le i \le n + 1}$ We have that $\size I = n$. Then for all $i \in I$: {{begin-eqn}} {{eqn | l = i       | o = \divides       | r = N       | c = where $\divides$ denotes divisibility }} {{eqn | ll= \leadsto       | l = i       | o = \divides       | r = N + i       | c =  }} {{end-eqn}} and so $N + i$ is composite. That is: :$N + 2, N + 3, \ldots, N + n, N + n + 1$ are all composite. {{qed}}
This will be demonstrated by showing that: :$\displaystyle \lim_{n \mathop \to \infty} \dfrac {p_n} {n \ln n} = 1$ where $p_n$ denotes the $n$th prime number. By definition of prime-counting function: :$\map \pi {p_n} = n$ The Prime Number Theorem gives: :$\displaystyle \lim_{x \mathop \to \infty} \dfrac {\map \pi x} {x / \ln x} = 1$ Thus: :$\displaystyle \lim_{x \mathop \to \infty} \dfrac n {p_n / \ln p_n} = 1$ {{begin-eqn}} {{eqn | l = \lim_{x \mathop \to \infty} \dfrac {\map \pi x} {x / \ln x}       | r = 1       | c = Prime Number Theorem }} {{eqn | ll= \leadsto       | l = \lim_{n \mathop \to \infty} \dfrac n {p_n / \ln p_n}       | r = 1       | c = Definition of Prime-Counting Function }} {{eqn | n = 1       | ll= \leadsto       | l = \lim_{n \mathop \to \infty} \dfrac {p_n} {n \ln p_n}       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \lim_{n \mathop \to \infty} \paren {\ln p_n - \ln n - \ln \ln p_n}       | r = 0       | c = Logarithm of both sides }} {{eqn | ll= \leadsto       | l = \lim_{n \mathop \to \infty} \ln p_n \paren {1 - \frac {\ln n} {\ln p_n} - \frac {\ln \ln p_n} {\ln p_n} }       | r = 0       | c = multiplying argument through by $\ln p_n$ }} {{eqn | ll= \leadsto       | l = \lim_{n \mathop \to \infty} \ln p_n \paren {1 - \frac {\ln n} {\ln p_n} }       | r = 0       | c = as $\dfrac {\ln n} n \to 0$ }} {{eqn | ll= \leadsto       | l = \lim_{n \mathop \to \infty} \paren {1 - \frac {\ln n} {\ln p_n} }       | r = 0       | c = as $\ln p_n \ne 0$ }} {{eqn | ll= \leadsto       | l = \lim_{n \mathop \to \infty} \frac {\ln n} {\ln p_n}       | r = 1       | c = as $\ln p_n \ne 0$ }} {{eqn | ll= \leadsto       | l = \lim_{n \mathop \to \infty} \dfrac {p_n} {n \ln n}        | r = \lim_{n \mathop \to \infty} \dfrac {p_n} {n \ln p_n} \frac {\ln p_n} {\ln n}       | c = from $(1)$ }} {{eqn | r = 1       | c =  }} {{end-eqn}} {{qed}}
Using Integration by Parts: {{begin-eqn}} {{eqn | l = \map \Li x       | r = \int_2^x \dfrac {\d t} {\ln t}       | c =  }} {{eqn | n = 1       | r = \dfrac x {\ln x} - \dfrac 2 {\ln 2} + \int_2^x \dfrac {\d t} {\paren {\ln t}^2}       | c =  }} {{end-eqn}} We have that $\dfrac 1 {\paren {\ln t}^2}$ is positive and decreasing for $t > 1$. Let $x \ge 4$. Then: {{begin-eqn}} {{eqn | l = 0       | o = <       | r = \int_2^x \dfrac {\d t} {\paren {\ln t}^2}       | c =  }} {{eqn | r = \int_2^{\sqrt x} \dfrac {\d t} {\paren {\ln t}^2} + \int_{\sqrt x}^x \dfrac {\d t} {\paren {\ln t}^2}       | c =  }} {{eqn | o = <       | r = \dfrac {\sqrt x - 2} {\paren {\ln 2}^2} + \dfrac {x - \sqrt x} {\paren {\ln \sqrt x}^2}       | c =  }} {{eqn | o = <       | r = \dfrac {\sqrt x} {\paren {\ln 2}^2} + \dfrac {4 x} {\paren {\ln \sqrt x}^2}       | c =  }} {{eqn | ll= \leadsto       | l = 0       | o = <       | r = \dfrac {\displaystyle \int_2^x \frac {\d t} {\paren {\ln t}^2} } {x / \ln x}       | c =  }} {{eqn | o = <       | r = \dfrac {\ln x} {\sqrt x \paren {\ln 2}^2} + \frac 4 {\ln x}       | c =  }} {{eqn | n = 2       | ll= \leadsto       | l = \lim_{x \mathop \to \infty} \dfrac {\displaystyle \int_2^x \frac {\d t} {\paren {\ln t}^2} } {x / \ln x}       | r = 0       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \frac {\map \Li x} {x / \ln x}       | r = \dfrac {\dfrac x {\ln x} - \dfrac 2 {\ln 2} + \displaystyle \int_2^x \dfrac {\d t} {\paren {\ln t}^2} } {x / \ln x}       | c = from $(1)$ }} {{eqn | r = 1 - \dfrac {2 \ln x} {x \ln 2} + \dfrac {\displaystyle \int_2^x \frac {\d t} {\paren {\ln t}^2} } {x / \ln x}       | c =  }} {{eqn | ll= \leadsto       | l = \lim_{x \mathop \to \infty} \frac {\map \Li x} {x / \ln x}       | r = 1 - \lim_{x \mathop \to \infty} \dfrac 2 {\ln 2} \dfrac x {\ln x} - 0       | c = from $(2)$ }} {{eqn | n = 3       | ll= \leadsto       | l = \lim_{x \mathop \to \infty} \frac {\map \Li x} {x / \ln x}       | r = 1       | c = as $\displaystyle \lim_{x \mathop \to \infty} \dfrac x {\ln x} = 0$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \lim_{x \mathop \to \infty} \frac {\map \pi x} {x / \ln x}       | r = \lim_{x \mathop \to \infty} \frac {\map \pi x} {\map \Li x} \frac {\map \Li x} {x / \ln x}       | c =  }} {{eqn | r = \lim_{x \mathop \to \infty} \frac {\map \pi x} {\map \Li x}       | c = from $(3)$ }} {{end-eqn}} {{qed}}
Let $r = \dfrac p q$ be rational such that $r \ne 0$. {{AimForCont}} $e^r$ is rational. Then $\paren {e^r}^q = e^p$ is also rational. Then if $e^{-p}$ is rational, it follows that $e^p$ is rational. It is therefore sufficient to derive a contradiction from the supposition that $e^p$ is rational for every $p \in \Z_{>0}$. Let $e^p = \dfrac a b$ for $a, b \in \Z_{>0}$. Let $n \in \Z_{>0}$ be a strictly positive integer. Let $\map f x$ be the function defined as: {{begin-eqn}} {{eqn | n = 1       | l = \map f x       | o = :=       | r = \frac {x^n \paren {1 - x}^n} {n!}       | c =  }} {{eqn | r = \frac 1 {n!} \sum_{k \mathop = n}^{2 n} c_k x^k       | c = where the $c_k$'s are some integers }} {{end-eqn}} Note that: :$(2): \quad 0 < x < 1 \implies 0 < \map f x < \dfrac 1 {n!}$ We also have: :$\map f 0 = 0$ :$\map {f^{\paren m} } 0 = 0$ if $m < n$ or $m > 2 n$ and: :$n \le m \le 2 n \implies \map {f^{\paren m} } 0 = \dfrac {m!} {n!} c_m$ and this number is an integer. Thus at $x = 0$, $\map f x$ and all its derivatives are integers. Since $\map f {1 - x} = \map f x$, the same is true for $x = 1$. Let $\map F x$ be the function defined as: {{begin-eqn}} {{eqn | n = 3       | l = \map F x       | o = :=       | r = \sum_{k \mathop = 0}^{2 n} \paren {-1}^k p^{2 n - k} \map {f^{\paren k} } x       | c =  }} {{eqn | r = p^{2 n} \map f x - p^{2 n - 1} \map {f'} x + p^{2 n - 2} \map {f''} x - \cdots - p \map {f^{\paren {2 n - 1} } } x + \map {f^{\paren {2 n} } } x       | c =  }} {{end-eqn}} Because of the properties of $\map f x$ and its derivatives above, $\map F 0$ and $\map F 1$ are integers. Next we have: {{begin-eqn}} {{eqn | n = 4       | l = \map {\frac {\d} {\d x} } {e^{p x} \map F x}       | r = e^{p x} \paren {\map {F'} x + p \, \map F x}       | c =  }} {{eqn | r = p^{2 n + 1} e^{p x} \map f x       | c = from $\map {F'} x + p \, \map F x$ based on $(3)$ }} {{end-eqn}} $(4)$ leads to: {{begin-eqn}} {{eqn | l = b \int_0^1 p^{2 n + 1} e^{p x} \map f x \d x       | r = b \bigintlimits {e^{p x} \map F x} 0 1       | c =  }} {{eqn | r = a \, \map F 1 - b \, \map F 0       | c =  }} {{end-eqn}} which is an integer. But from $(2)$: {{begin-eqn}} {{eqn | l = 0       | o = <       | r = b \int_0^1 p^{2 n + 1} e^{p x} \map f x \rd x       | c =  }} {{eqn | o = <       | r = \frac {b p^{2 n + 1} e^p} {n!}       | c =  }} {{eqn | r = b p e^p \frac {\paren {p^2}^n e^p} {n!}       | c =  }} {{end-eqn}} The expression on the {{RHS}} tends to $0$ as $n$ tends to $\infty$. Hence: :$0 < a \, \map F 1 - b \, \map F 0 < 1$ for sufficiently large $n$. But there exists no integer strictly between $0$ and $1$. From this contradiction it follows that our original assumption, that is, that $e^r$ is rational, must have been false. {{qed}}
From Exponential of Rational Number is Irrational: :$r \in \Q_{\ne 0} \implies e^r \in \R - \Q$ Thus, apart from the point $\tuple {0, 1}$, when $x$ is rational, $e^x$ is not. Hence the result. {{qed}}
Consider the graph of the exponential function in the real Cartesian plane $\R^2$: :$g := \set {\tuple {x, y} \in \R^2: y = e^x}$ From Rational Points on Graph of Exponential Function, the only rational point of $g$ is $\tuple {0, 1}$. By definition of the exponential function, $f$ and $g$ are inverses. Thus: :$\tuple {x, y} \in g \iff \tuple {y, x} \in f$ Thus for $\tuple {x, y} \in g$ to be rational, $\tuple {y, x} = \tuple {0, 1}$. Hence the result. {{qed}}
Let $r$ be expressed in the form: :$r = \dfrac p q$ Consider the linear polynomial in $x$: :$q x - p = 0$ which has the solution: :$x = \dfrac p q$ Hence the result, by definition of algebraic number. {{qed}}
Let $r$ be expressed in the form: :$r = \dfrac p q$ By Rational Number is Algebraic, $r$ can be expressed as the root of the linear polynomial: :$q x - p = 0$ A polynomial of degree zero is a constant polynomial, and has no roots. Hence the result by definition of degree of algebraic number. {{qed}}