hcju/proofwikiqa huggingface dataset:
From Derivative of $n$th Root and Combination Theorem for Sequences: :$\forall n \in \N: \forall x \in J : D_x \map {f_n} x = \dfrac {\sqrt [n] x} x$ From Closed Bounded Subset of Real Numbers is Compact, $J$ is compact. Thus: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} D_x \map {f_n} x       | r = \lim_{n \mathop \to \infty} \dfrac {\sqrt [n] x} x       | c = from above }} {{eqn | r = \frac 1 x \lim_{n \mathop \to \infty} \sqrt [n] x       | c = Multiple Rule for Real Sequences }} {{eqn | r = \frac 1 x \lim_{n \mathop \to \infty} x^{1/n}       | c = {{Defof|Root (Analysis)|$n$th Root}} }} {{eqn | r = \frac 1 x       | c = Power Function tends to One as Power tends to Zero: Rational Number }} {{end-eqn}} In particular, $\sequence { {f_n}'}_n$ is pointwise convergent to a continuous function on $J$. For $x \in \closedint {\frac 1 k} 1$: {{begin-eqn}} {{eqn | l = x^{1/n}       | o = \le       | r = x^{1 / \paren {n + 1} }       | c = Exponential is Monotonic: Rational Number }} {{eqn | ll= \leadsto       | l = \sqrt [n] x       | o = \le       | r = \sqrt [n + 1] x       | c = {{Defof|Root (Analysis)|$n$th Root}} }} {{eqn | ll= \leadsto       | l = \frac {\sqrt [n] x} x       | o = \le       | r = \frac {\sqrt [n + 1] x} x       | c = Real Number Ordering is Compatible with Multiplication }} {{end-eqn}} So $\sequence {\map { {f_n}'} x}_n$ is increasing when $x \in \closedint {\dfrac 1 k} 1$. For $x \in \closedint 1 k$: {{begin-eqn}} {{eqn | l = x^{1/n}       | o = \ge       | r = x^{1 / \paren {n + 1} }       | c = Exponential is Monotonic: Rational Number }} {{eqn | ll= \implies       | l = \sqrt [n] x       | o = \ge       | r = \sqrt [n + 1] x       | c = {{Defof|Root (Analysis)|$n$th Root}} }} {{eqn | ll= \implies       | l = \frac {\sqrt [n] x} x       | o = \ge       | r = \frac {\sqrt [n + 1] x} x       | c = Real Number Ordering is Compatible with Multiplication }} {{end-eqn}} So $\sequence {\map { {f_n}'} x}_n$ is decreasing when $x \in \closedint {\dfrac 1 k} 1$. Thus $\sequence {\map { {f_n}'} x}_n$ is monotone for all $x \in \closedint {\frac 1 k} 1 \cup \closedint 1 k = J$. From Dini's Theorem, $\sequence { {f_n}'}_n$ converges uniformly to $\dfrac 1 x$ on $J$. Hence the result. {{qed}} Category:Derivative of Natural Logarithm Function 4zsg9gzbbr954dzh89w5jzi2ibj1hy3
This proof assumes the series definition of $\exp$. From Series of Power over Factorial Converges: :$\displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$ converges  Hence the result, from Convergent Real Sequence has Unique Limit. {{qed}}
This proof assumes the sequence definition of $\exp$. Let $\sequence {f_n}$ be the sequence of mappings $f_n : \R \to \R$ defined as: :$\map {f_n} x = \paren {1 + \dfrac x n}^n$ Fix $x \in \R$. Then: {{begin-eqn}} {{eqn | l = \map {f_n} x       | r = \paren {1 + \dfrac x n}^n       | c = Definition of $\map {f_n} x$ }} {{eqn | r = \sum_{k \mathop = 0}^n \binom n k \frac {x^k} {n^k}        | c = Binomial Theorem: Integral Index }} {{eqn | r = \sum_{k \mathop = 0}^n \frac {x^k} {k!} \frac {\paren n \times \paren {n - 1} \times \paren {n - 2} \times \cdots \paren {n - k + 1} }{n \times n \times n \times \cdots n}       | c = {{Defof|Factorial}} }} {{eqn | r = \sum_{k \mathop = 0}^n \frac {x^k} {k!} \paren 1 \paren {1 - \frac 1 n} \paren {1 - \frac 2 n} \cdots \paren {1 - \frac {k - 1} n} }} {{eqn | o = \le       | r = \size {\sum_{k \mathop = 0}^n \frac {x^k} {k!} \paren 1 \paren {1 - \frac 1 n} \paren {1 - \frac 2 n} \cdots \paren {1 - \frac {k - 1} n} }       | c = Negative of Absolute Value }} {{eqn | r = \sum_{k \mathop = 0}^n \frac {\size x^k} {k!} \paren 1 \paren {1 - \frac 1 n} \paren {1 - \frac 2 n} \cdots \paren {1 - \frac {k - 1} n}       | c = Absolute Value Function is Completely Multiplicative }} {{eqn | o = \le       | r = \sum_{k \mathop = 0}^n \frac {\size x^k} {k!}       | c = Multiplication of Positive Number by Real Number Greater than One }} {{eqn | o = <       | r = \sum_{k \mathop = 0}^\infty \frac {\size x^k} {k!}       | c = Sum of positive terms is increasing }} {{eqn | o = <       | r = \infty       | c = Series of Power over Factorial Converges }} {{end-eqn}} {{MissingLinks|Result about increasing series seems not to be up yet}} Thus, $\sequence {\map {f_n} x}$ is bounded above. From Exponential Sequence is Eventually Increasing: :$\exists N \in \N: \sequence {\map {f_{N + n} } x}$ is increasing From Monotone Convergence Theorem (Real Analysis), $\sequence {\map {f_{N + n} } x}$ converges to some $z \in \R$. From Tail of Convergent Sequence, $\sequence {\map {f_n} x}$ converges to $z$. Hence the result, from Limit of Function Unique. {{qed}}
This proof assumes the continuous extension definition of $\exp$. Let $e$ denote Euler's number. Let $f: \Q \to \R$ be the real-valued function defined as: :$f \left({r}\right) = e^r$ From Euler's Number to Rational Power permits Unique Continuous Extension, there exists a unique continuous extension of $f$ to $\R$. Hence the result. {{qed}}
This proof assumes the definition of the exponential as the inverse of the logarithm. From Logarithm is Strictly Increasing, $\ln$ is strictly monotone on $\R_{>0}$. From Inverse of Strictly Monotone Function, $f$ permits an inverse mapping. Hence the result, from Inverse Mapping is Unique. {{qed}}
First we consider the case where the fiber of truth of $R$ is finite. {{ProofWanted}} === Infinite case === Now consider the case where the fiber of truth of $R$ is infinite. {{:Translation of Index Variable of Summation/Infinite Series}}
=== Sufficient Condition === Let :$\mathscr S$ satisfies axiom of approximation. Let $x, y \in S$ such that :$x \npreceq y$ By assumption: :$x^\ll$ is directed. By definition of up-complete: :$x^\ll$ admits a supremum. By definition of axiom of approximation: :$x = \sup \left({x^\ll}\right)$ By definition of supremum: :if $y$ is upper bound for $x^\ll$, then $x \preceq y$ By definition of upper bound: :$\exists u \in x^\ll: u \npreceq y$ Thus by definition of way below closure: :$\exists u \in S: u \ll x \land u \npreceq y$ {{qed|lemma}} === Necessary Condition === Let :$\forall x, y \in S: x \npreceq y \implies \exists u \in S: u \ll x \land u \npreceq y$ Let $x \in S$. By assumption: :$x^\ll$ is directed. By definition of up-complete: :$x^\ll$ admits a supremum. By Operand is Upper Bound of Way Below Closure: :$x$ is upper bound for $x^\ll$ We will prove that :$\forall y \in S: y$ is upper bound for $x^\ll \implies x \preceq y$ Let $y \in S$ such that :$y$ is upper bound for $x^\ll$ Aiming at a contradiction suppose that :$x \npreceq y$ By assumption: :$\exists u \in S: u \ll x \land u \npreceq y$ By definition of way below closure: :$u \in x^\ll$ By definition of upper bound :$u \preceq y$ This contradicts $u \npreceq y$ {{qed|lemma}} Thus by definition of supremum: :$x = \sup \left({x^\ll}\right)$ Thus by definition: :$\mathscr S$ satisfies axiom of approximation. {{qed}}
Let $y \in x^\ll$ By definition of way below closure: :$y \ll x$ where $\ll$ denotes the way below relation. Thus by Way Below implies Preceding: :$y \preceq x$ Thus by definition: :$x$ is upper bound for $x^\ll$ {{qed}}
:$\displaystyle \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^i$ can be expressed as: :$\displaystyle \sum_{R \left({i}\right)} \sum_{S \left({i, j}\right)} a_{i j}$ where: :$R \left({i}\right)$ is the propositional function $1 \le i \le n$ :$S \left({i, j}\right)$ is the propositional function $1 \le j \le i$ We wish to find a propositional function $S' \left({j}\right)$ which is to be: :there exists an $i$ such that both $1 \le i \le n$ and $1 \le j \le i$ This is satisfied by the propositional function: :$S' \left({j}\right) := 1 \le j \le n$ Next we wish to find a propositional function $R' \left({i, j}\right)$ which is to be: :both $1 \le i \le n$ and $1 \le j \le i$ This is satisfied by the propositional function: :$R' \left({i, j}\right) := j \le i \le n$ Hence the result, from Exchange of Order of Summation with Dependency on Both Indices. {{qed}}
This is a restatemtent of Manipulation of Absolutely Convergent Series: Permutation in the context of summations. {{qed}}
Let $b_i =: a_{i 1}$ and $c_i =: a_{i 2}$. Then: {{begin-eqn}} {{eqn | l = \sum_{\map R i} \paren {b_i + c_i}       | r = \sum_{\map R i} \paren {a_{i 1} + a_{i 2} }       | c = Definition of $b_i$ and $c_i$ }} {{eqn | r = \sum_{\map R i} \paren {\sum_{1 \mathop \le j \mathop \le 2} a_{i j} }       | c = {{Defof|Summation}} }} {{eqn | r = \sum_{1 \mathop \le j \mathop \le 2} \paren {\sum_{\map R i} a_{i j} }       | c = Exchange of Order of Summation }} {{eqn | r = \sum_{\map R i} a_{i 1} + \sum_{\map R i} a_{i 2}       | c = {{Defof|Summation}} }} {{eqn | r = \sum_{\map R i} b_i + \sum_{\map R i} c_i       | c = Definition of $b_i$ and $c_i$ }} {{end-eqn}} {{qed}}
Let $b_i =: a_{i 1}$ and $c_i =: a_{i 2}$. Then: {{begin-eqn}} {{eqn | l = \sum_{\map R i} \paren {b_i + c_i}       | r = \sum_{\map R i} \paren {a_{i 1} + a_{i 2} }       | c = by definition }} {{eqn | r = \sum_{\map R i} \paren {\sum_{1 \mathop \le j \mathop \le 2} a_{i j} }       | c = {{Defof|Summation}} }} {{eqn | r = \sum_{1 \mathop \le j \mathop \le 2} \paren {\sum_{\map R i} a_{i j} }       | c = Exchange of Order of Summation: Finite and Infinite Series }} {{eqn | r = \sum_{\map R i} a_{i 1} + \sum_{\map R i} a_{i 2}       | c = {{Defof|Summation}} }} {{eqn | r = \sum_{\map R i} b_i + \sum_{\map R i} c_i       | c = by definition }} {{end-eqn}} {{qed}}
By definition, $\displaystyle \sum_{\map R i} b_i$ and $\displaystyle \sum_{\map R i} c_i$ are sequences in $\R$. Hence the result as an instance of Sum Rule for Real Sequences. {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{\map R j} a_j + \sum_{\map S j} a_j       | r = \sum_{j \mathop \in \Z} a_j \sqbrk {\map R j} + \sum_{j \mathop \in \Z} a_j \sqbrk {\map R j}       | c = {{Defof|Summation by Iverson's Convention}} }} {{eqn | r = \sum_{j \mathop \in \Z} a_j \paren {\sqbrk {\map R j} + \sqbrk {\map R j} }       | c =  }} {{end-eqn}} Let: :$A := \set {j \in \Z: \map R j}$ :$B := \set {j \in \Z: \map R j}$ The result then follows from Cardinality of Set Union. {{qed}}
This proof depends on the series definition of $\exp$. That is, let: :$\displaystyle \exp x = \sum_{k \mathop = 0}^ \infty \frac {x^k} {k!}$ From Series of Power over Factorial Converges, the radius of convergence of $\exp$ is $\infty$. Thus, from Power Series Converges to Continuous Function, $\exp$ is continuous on $\R$. {{qed}}
This proof assumes the definition of $\exp$ as the solution to an initial value problem. That is, suppose $\exp$ solves: :$ (1): \quad \dfrac \d {\d x} y = \map f {x, y}$ :$ (2): \quad \map \exp 0 = 1$ on $\R$, where $\map f {x, y} = y$. From Derivative of Exponential Function: Proof 4, the function $\phi : \R \to \R$ defined as: :$\displaystyle \map \phi x = \sum_{k \mathop = 0}^\infty \frac {x^k} {k!}$ satisfies $\map {\phi'} x = \map \phi x$. So $\phi$ satisfies $(1)$. From Exponential of Zero: Proof 3: :$\map \phi 0 = 1$ So $\phi$ satisfies $(2)$. Thus, $\phi$ is a solution to the initial value problem given. From Exponential Function is Continuous: Proof 5 and $(1)$: :$\phi$ is continuously differentiable on $\R$. Because $\map f {x, \phi} = \phi$: :$f$ is continuously differentiable on $\R^2$. Thus, from Uniqueness of Continuously Differentiable Solution to Initial Value Problem, this solution is unique. {{qed}}
To prove that $\sequence {S_k}_k$ is exhausting $\R_{>0}$, it is sufficient to show: :$(1): \quad \forall k \in \N: S_k \subseteq S_{k + 1}$ :$(2): \quad \ds \bigcup_{k \mathop \in \N} S_k = \R_{>0}$ === $\sequence {S_k}_k$ is increasing === Let $k \in \N$. Let $k = 1$. Then: :$S_k = \openint {\dfrac 1 k} k = \O$ Thus, by Empty Set is Subset of All Sets: :$\openint {\dfrac 1 k} k \subseteq \openint {\dfrac 1 {k + 1} } {k + 1}$ If $k \ge 2$: {{begin-eqn}} {{eqn | o =        | m = x       | mo= \in       | r = S_k }} {{eqn | ll= \leadsto       | l = \frac 1 k       | o = <       | m = x       | mo= <       | r = k       | c = {{Defof|Open Real Interval}} }} {{eqn | ll= \leadsto       | l = \frac 1 {k + 1}       | o = <       | m = x       | mo= <       | r = k       | c = Ordering of Reciprocals }} {{eqn | ll= \leadsto       | l = \frac 1 {k + 1}       | o = <       | m = x       | mo= <       | r = k + 1 }} {{eqn | ll= \leadsto       | o =        | m = x       | mo= \in       | r = S_{k + 1}       | c = {{Defof|Open Real Interval}} }} {{end-eqn}} It follows that $\sequence {S_k}_k$ is increasing. {{qed|lemma}} === $\ds \bigcup_{k \mathop \in \N} S_k = \R_{>0}$ === Let $x \in \R_{>0}$. ==== Case 1: $1 < x$ ==== Let $1 < x$. Let $k = \ceiling x$. From Real Number is between Ceiling Functions: :$x < k$ From Ordering of Reciprocals: :$1 < k \implies \dfrac 1 k < 1$ So: :$\dfrac 1 k < x < k$ and so $x \in S_k$. ==== Case 2: $x = 1$ ==== Let $x = 1$. Let $k = 2$. Then: :$\dfrac 1 2 < 1 < 2$ and hence $x \in S_2$. ==== Case 3: $0 < x < 1$ ==== Let $0 < x < 1$. Let $k = \ceiling {\dfrac 1 x}$. From Ordering of Reciprocals: :$0 < x < 1 \implies 1 < \dfrac 1 x$ From Real Number is between Ceiling Functions: :$1 < \dfrac 1 x < k$ From Ordering of Reciprocals: :$\dfrac 1 k < x < 1$ So: :$\dfrac 1 k < x < k$ and so $x \in S_k$. Hence: :$\forall x \in \R_{>0} : \exists k \in \N : x \in S_k$ {{qed|lemma}} The result follows from the definition of exhausting sequence of sets. {{qed}} Category:Real Numbers i0x820ehi7pux4ldqc3afafq0cwwykx
=== Sufficient Condition === Let :$x \ll y$ Let $F$ be a set of open subsets of $S$ such that :$y \subseteq \bigcup F$ By definition of subset: :$F \subseteq \tau$ By proof of Topology forms Complete Lattice: :$\sup F = \bigcup F$ By assumption: :$y \preceq \sup F$ By Topology forms Complete Lattice: :$\left({\tau, \preceq}\right)$ is complete lattice. By Way Below in Complete Lattice: :$\exists G \in {\it Fin}\left({F}\right): x \preceq \sup G$ where ${\it Fin}\left({F}\right)$ denotes the set of all finite subsets of $F$. Thus by proof of Topology forms Complete Lattice: :$x \subseteq \bigcup G$ {{qed|lemma}} === Necessary Condition === Assume that :for every set $F$ of open subsets of $S$: if $y \subseteq \bigcup F$, then ::there exists a finite subset $G$ of $F$: $x \subseteq \bigcup G$ Let $I$ be an ideal in $\left({\tau, \preceq}\right)$ such that :$y \preceq \sup I$ By proof of Topology forms Complete Lattice: :$y \subseteq \bigcup I$ By assumption: :there exists a finite subset $G$ of $I$: $x \subseteq \bigcup G$ By proof of Topology forms Complete Lattice: :$x \preceq \sup I$ By definition of ideal: :$I$ is directed and lower. By Directed iff Finite Subsets have Upper Bounds: :$\exists u \in I: u$ is upper bound for $G$ By definition of supremum: :$\sup G \preceq u$ By definition of transitivity: :$x \preceq u$ Thus by definition of lower set: :$x \in I$ Thus by Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal: :$x \ll y$ {{qed}}
From Strictly Monotone Real Function is Bijective, $f$ is a bijection. From Inverse of Strictly Monotone Function, $f^{-1} : J \to I$ exists and is strictly monotone. From Surjective Monotone Function is Continuous, $f^{-1}$ is continuous. Hence the result. {{qed}}
From Logarithm is Strictly Increasing, $\ln x$ is strictly increasing on $x > 0$. From Second Derivative of Natural Logarithm Function: : $D^2 \ln x = -\dfrac 1 {x^2}$ Thus $D^2 \ln x$ is strictly negative on $x > 0$ (in fact is strictly negative for all $x \ne 0$). Thus from Derivative of Monotone Function, $D \dfrac 1 x$ is strictly decreasing on $x > 0$. So from Real Function is Strictly Concave iff Derivative is Strictly Decreasing, $\ln x$ is strictly concave on $x > 0$. {{qed}}
By definition, the exponential function is the inverse of the natural logarithm function. From Logarithm is Strictly Increasing, $\ln x$ is strictly increasing. From Logarithm is Strictly Concave, $\ln x$ is strictly concave. The result follows from Inverse of Strictly Increasing Strictly Concave Real Function is Strictly Convex. {{qed}} Category:Exponential Function pa119jqwoaw6rc4k41qso9nu8eayuv8
=== Sufficient Condition === Let :$x \ll y$ Let $X \subseteq S$ such that :$y \preceq \sup X$ By Set of Finite Suprema is Directed: :$\left\{ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}$ is directed. By definition of union: :$\bigcup {\it Fin}\left({X}\right) \setminus \left\{ {\varnothing}\right\} = X$ By Supremum of Suprema: :$\sup X = \sup \left\{ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}$ By definition of way below relation: :$\exists z \in \left\{ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}: x \preceq z$ Thus :$\exists A \in {\it Fin}\left({X}\right): x \preceq \sup A$ {{qed|lemma}} === Necessary Condition === Let :$\forall X \subseteq S: y \preceq \sup X \implies \exists A \in {\it Fin}\left({X}\right): x \preceq \sup A$ Let $D$ be a directed subset of $S$ such that :$y \preceq \sup D$ By assumption: :$\exists A \in {\it Fin}\left({D}\right): x \preceq \sup A$ By Directed iff Finite Subsets have Upper Bounds: :$\exists u \in D: u$ is upper bound for $A$ By definition of supremum: :$\sup A \preceq u$ Thus by definition of transitivity: :$x \preceq u$ Thus by definition of way below relation: :$x \ll y$ {{qed}}
=== Sufficient Condition === Let :$x \ll x$ Let $F \subseteq  \tau_x$ be a open cover of $x$. By definition of cover: :$x \subseteq \bigcup F$ By definition of topological space: :$\forall y \in \tau: x \cap y \in \tau$ By definition of subset: :$\tau_x \subseteq \tau$ By Subset Relation is Transitive: :$F \subseteq \tau$ By Way Below in Ordered Set of Topology: :there exists a finite subset $G$ of $F$: $x \subseteq \bigcup G$ By definition: :$G$ is cover of $x$ Thus by definition: :$x$ has finite subcover. Thus by definition: :$T_x$ is compact. {{qed|lemma}} === Necessary Condition === Let :$T_x$ is compact. Let $F$ be a set of open subsets of $S$ such that :$x \subseteq \bigcup F$ Define $Y := \left\{ {x \cap y: y \in F}\right\}$ By definition of subset: :$Y \subseteq \tau_x$ By definition of union: :$x \subseteq \bigcup Y$ By definition :$Y$ is open cover of $x$. By definition of compact: :$x$ has finite subcover $G$. By Axiom of Choice: :$\exists g:G \to F: \forall y \in G: g\left({y}\right) \cap x = y$ By Cardinality of Image of Set not greater than Cardinality of Set: :$\left\vert{g^\to\left({G}\right)}\right\vert \le \left\vert{G}\right\vert$ where $\left\vert{G}\right\vert$ denotes the cardinality of $G$. Thus :$g^\to\left({G}\right)$ is finite. Thus by definitions of subset and image of set: :$g^\to\left({G}\right) \subseteq F$ By Intersection is Subset: :$\forall y \in G: y \subseteq g\left({y}\right)$ By Set Union Preserves Subsets/Families of Sets: :$\displaystyle \bigcup G = \bigcup_{y \mathop \in G}y \subseteq \bigcup_{y \mathop \in G} g \left({y}\right) = \bigcup g^\to \left({G}\right)$ By definition of cover: :$x \subseteq \bigcup G$ Thus by Subset Relation is Transitive: :$x \subseteq \bigcup g^\to\left({G}\right)$ Thus by Way Below in Ordered Set of Topology; :$x \ll x$ {{qed}}
Let $x, y \in \R$. Note that, from Power of Positive Real Number is Positive: Real Number: :$\forall t \in \R: a^t > 0$. So: {{begin-eqn}} {{eqn | l = a^{\paren {x + y} / 2}       | r = \sqrt {a^{x + y} }       | c = Exponent Combination Laws: Power of Power: Proof 2 }} {{eqn | r = \sqrt {a^x a^y}       | c = Exponent Combination Laws: Product of Powers: Proof 2 }} {{eqn | o = \le       | r = \frac {a^x + a^y} 2       | c = Cauchy's Mean Theorem }} {{end-eqn}} Hence $a^x$ is midpoint-convex. Further, from Power Function on Strictly Positive Base is Continuous: Real Power, $a^x$ is continuous. Thus, from Continuous Midpoint-Convex Function is Convex, $a^x$ is convex. {{qed}} Category:Real Analysis Category:Powers ozrx4xqacs1l03nudbcq362bc9qbixd
For all $n \in \N$, let $\map P n$ be the proposition: :$0 < x < 1 \implies 0 < x^n \le x$ === Basis for the Induction === $\map P 1$ is true, since $0 < x < 1 \implies 0 < x^1 \le x$ by definition of exponent of $1$. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$0 < x < 1 \implies 0 < x^k \le x$ Then we need to show: :$0 < x < 1 \implies 0 < x^{k + 1} \le x$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = 0 < x < 1       | o = \leadsto       | r = 0 < x^k \le x       | c = Induction Hypothesis }} {{eqn | o = \leadsto       | r = 0 < x^{k + 1} \le x \cdot x       | c = Real Number Ordering is Compatible with Multiplication }} {{eqn | o = \leadsto       | r = 0 < x^{k + 1} \le x       | c = Multiple of Positive Real Number with Number Less Than One is Less Than Real Number }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N: 0 < x < 1 \implies 0 < x^n \le x$ Hence the result. {{qed}}
First: {{begin-eqn}} {{eqn | l = n       | o = >       | r = \left\lceil{\left\vert{x}\right\vert}\right\rceil }} {{eqn | ll= \implies       | l = n        | o = >       | r = -x       | c = Negative of Absolute Value and Real Number is between Ceiling Functions }} {{eqn | ll= \implies       | l = \frac n {n + x}       | o = >       | r = 0 }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = n       | o = >       | m = \left\lceil{\left\vert{x}\right\vert}\right\rceil }} {{eqn | ll= \implies       | l = n       | o = >       | m = \left\vert{x}\right\vert       | c = Real Number is between Ceiling Functions }} {{eqn | ll= \implies       | l = \left\vert{\frac x n}\right\vert       | o = <       | m = 1       | c = dividing both sides by $n$ }} {{eqn | ll= \implies       | l = -1       | o = <       | m = \frac x n       | mo= <       | r = 1       | c = Negative of Absolute Value: Corollary 1 }} {{eqn | ll= \implies       | l = 0       | o = <       | m = 1 + \frac x n }} {{eqn | ll= \implies       | l = 0       | o = <       | m = \left({1 + \frac x n}\right)^n       | c = Power of Positive Real Number is Positive: Natural Number }} {{end-eqn}} So, for $n \ge \left\lceil{\left\vert{x}\right\vert}\right\rceil$: :$\left\langle{\dfrac n {n + x} }\right\rangle$ and: :$\left\langle{\left({1 + \dfrac x n}\right)^n}\right\rangle$ are  positive. Now let $n \ge \left\lceil{\left\vert{x}\right\vert}\right\rceil$. Suppose first that $x \in \R_{\ge 0}$. Then: {{begin-eqn}} {{eqn | l = \frac n {n + x}       | o = \le       | r = \frac {n + 1} {n + x + 1} }} {{eqn | ll= \iff       | l = n \left({n + x + 1}\right)       | o = \le       | r = \left({n + 1}\right) \left({n + x}\right)       | c = Real Number Ordering is Compatible with Multiplication }} {{eqn | ll= \iff       | l = n^2 + n x + n       | o = \le       | r = n^2 + n x + n + x }} {{eqn | ll= \iff       | l = 0       | o = \le       | r = x }} {{end-eqn}} So $\left\langle{\dfrac n {n + x} }\right\rangle$ is increasing. Further, from Exponential Sequence is Eventually Increasing: :$\left\langle{\left({1 + \dfrac x n}\right)^n}\right\rangle$ is increasing. Hence, from Product of Positive Increasing Functions is Increasing: :$n \ge \left\lceil{\left\vert{x}\right\vert }\right\rceil \implies \left\langle{\dfrac n {n + x} \left({1 + \dfrac x n}\right)^n}\right\rangle$ is increasing. Suppose instead that $x \in \R_{<0}$. {{AimForCont}} that: :$\left\langle{\dfrac n {n + x} \left({1 + \dfrac x n}\right)^n}\right\rangle$ is decreasing. From above: $\left\langle{1 + \dfrac x n}\right\rangle = \left\langle{\dfrac {n + x} n}\right\rangle$ is decreasing. Thus, from Product of Positive Increasing and Decreasing Functions is Decreasing: :$\left\langle{\dfrac {n + x} n \dfrac n {n + x} \left({1 + \dfrac x n}\right)^n}\right\rangle = \left\langle{\left({1 + \dfrac x n}\right)^n}\right\rangle$ is decreasing. This contradicts Exponential Sequence is Eventually Increasing. Hence the result, by Proof by Contradiction. {{qed}} Category:Derivative of Exponential Function ckqbhfr7zj6nuyrsadq9uq2v596uvtk