hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \gcd \set {a, b c}       | r = \gcd \set {a, \lcm \set {b, c} }       | c = LCM of Coprime Integers }} {{eqn | r = \lcm \set {\gcd \set {a, b}, \gcd \set {a, c} }       | c = GCD and LCM Distribute Over Each Other }} {{eqn | r = \frac {\gcd \set {a, b} \gcd \set {a, c} } {\gcd \set {\gcd \set {a, b}, \gcd \set {a, c} } }       | c = Product of GCD and LCM }} {{eqn | r = \frac {\gcd \set {a, b} \gcd \set {a, c} } {\gcd \set {a, \gcd \set {b, \gcd \set {a, c} } } }       | c = Greatest Common Divisor is Associative }} {{eqn | r = \frac {\gcd \set {a, b} \gcd \set {a, c} } {\gcd \set {a, \gcd \set {\gcd \set {b, c}, a} } }       | c = Greatest Common Divisor is Associative }} {{eqn | r = \frac {\gcd \set {a, b} \gcd \set {a, c} } {\gcd \set {a, \gcd \set {1, a} } }       | c = Definition of coprimality }} {{eqn | r = \frac {\gcd \set {a, b} \gcd \set {a, c} } {\gcd \set {a, 1} } }} {{eqn | r = \frac {\gcd \set {a, b} \gcd \set {a, c} } 1 }} {{eqn | r = \gcd \set {a, b} \gcd \set {a, c} }} {{end-eqn}} {{qed}} Category:Greatest Common Divisor Category:Examples of Multiplicative Functions t5bzldcx2jtjo5r2g2bsb49f38qlacr
Proof by induction on $n$: For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$a^{n + 1} < a^n$ $\map P 0$ is the case: {{begin-eqn}} {{eqn | l = a^1       | r = a       | c = {{Defof|Integer Power}} }} {{eqn | o = <       | r = 1        | c =  }} {{eqn | r = a^0       | c = {{Defof|Integer Power}} }} {{end-eqn}} === Basis for the Induction === $\map P 1$ is true, since: {{begin-eqn}} {{eqn | l = a       | o = <       | r = 1 }} {{eqn | ll= \leadsto       | l = a \times a       | o = <       | r = 1 \times a       | c = Real Number Ordering is Compatible with Multiplication }} {{eqn | ll= \leadsto       | l = a^2       | o = <       | r = a^1       | c = {{Defof|Integer Power}} }} {{end-eqn}} 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: :$a^{k + 1} < a^k$ Then we need to show: :$a^{k + 2} < a^{k + 1}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = a^{k + 1}       | o = <        | r = a^k       | c = Induction Hypothesis }} {{eqn | ll= \leadsto       | l = a \times a^{k + 1}        | o = <       | r = a \times a^k       | c = Real Number Ordering is Compatible with Multiplication }} {{eqn | ll= \leadsto       | l = a^{k + 2}       | o = <       | r = a^{k + 1}       | c = {{Defof|Integer Power}} }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 0}: a^{n + 1} < a^n$ Hence the result, by definition of strictly decreasing. {{qed}} Category:Power Function on Base between Zero and One is Strictly Decreasing 0t61o4y9snf2r829wdmop57akew0xjl
Let $\dfrac r s, \dfrac t u \in \Q$, where $r, t \in \Z, s, u \in \Z_{>0}$. Let $\dfrac r s < \dfrac t u$. Then: {{begin-eqn}} {{eqn | l = r u       | o = <       | r = t s       | c = Real Number Ordering is Compatible with Multiplication }} {{eqn | ll= \leadsto       | l = a^{r u}       | o = >       | r = a^{t s}       | c = Power Function on Base between Zero and One is Strictly Decreasing: Integer }} {{eqn | ll= \leadsto       | l = \paren {a^r}^u        | o = >        | r = \paren {a^t}^s       | c = Product of Indices of Real Number: Integers }} {{eqn | ll= \leadsto       | l = \sqrt [u] {\paren {a^r}^u}        | o = >       | r = \sqrt [u] {\paren {a^t}^s }       | c = Root is Strictly Increasing }} {{eqn | ll= \leadsto       | l = a^r        | o = >        | r = \sqrt [u] {\paren {a^t}^s }       | c = {{Defof|Root (Analysis)|$u$th Root}} }} {{eqn | ll= \leadsto       | l = \sqrt [s] {\paren {a^r} }        | o = >        | r = \sqrt [s] {\sqrt [u] {\paren {a^t}^s } }       | c = Root is Strictly Increasing }} {{eqn | ll= \leadsto       | l = \sqrt [s] {\paren {a^r} }        | o = >        | r = \sqrt [u] {\sqrt [s] {\paren {a^t}^s } }       | c = Root is Commutative }} {{eqn | ll= \leadsto       | l = \sqrt [s] {\paren {a^r} }        | o = >        | r = \sqrt [u] {\paren {a^t} }       | c = {{Defof|Root (Analysis)|$s$th Root}} }} {{eqn | ll= \leadsto       | l = a^{r / s}       | o = >        | r = a^{t / u}       | c = {{Defof|Rational Power}} }} {{end-eqn}} {{qed}} Category:Power Function on Base between Zero and One is Strictly Decreasing byhapr90wzqu0vwr4t9p7jhfjdogro1
Fix $n \in \Z_{\ge 0}$. From Ordering of Reciprocals: :$0 < \dfrac 1 a < 1$ From Power Function on Base between Zero and One is Strictly Decreasing: Positive Integer: :$\paren {\dfrac 1 a}^{n + 1} < \paren {\dfrac 1 a}^n$ From Real Number to Negative Power: Positive Integer: :$\dfrac 1 {a^{n + 1} } < \dfrac 1 {a^n}$ From Ordering of Reciprocals: :$a^n < a^{n + 1}$ Hence the result. {{qed}} Category:Power Function on Base Greater than One is Strictly Increasing 9ty6jg9ov8atkj9p3yw8oxqprdu02lk
Let $y = \sqrt [n] {\dfrac 1 x}$. {{begin-eqn}} {{eqn | l = \sqrt [n] {\dfrac 1 x} = y       | o = \iff       | r = \dfrac 1 x = y^n       | c = {{Defof|Root (Analysis)|$n$th root}} }} {{eqn | o = \iff       | r = x = \dfrac 1 {y^n}       | c = Reciprocal of Real Number is Unique }} {{eqn | o = \iff       | r = x = \left({\dfrac 1 y}\right)^n       | c = Powers of Group Elements/Negative Index }} {{eqn | o = \iff       | r = \sqrt [n] x  = \dfrac 1 y       | c = {{Defof|Root (Analysis)|$n$th root}} }} {{eqn | o = \iff       | r = \dfrac 1 {\sqrt [n] x} = y       | c = Reciprocal of Real Number is Unique }} {{eqn | o = \iff       | r = \dfrac 1 {\sqrt [n] x} = \sqrt [n] {\dfrac 1 x}       | c = Definition of $y$ }} {{end-eqn}} Hence the result. {{qed}} Category:Roots of Numbers qfegueuaev3965956uzlxnpaxmnozjt
Let $\dfrac r s, \dfrac t u \in \Q$, where $r, t \in \Z$ are integers and $s, u \in \Z_{>0}$ are strictly positive integers. Let $\dfrac r s < \dfrac t u$. From Ordering of Reciprocals: :$0 < \dfrac 1 a < 1$ So: {{begin-eqn}} {{eqn | l = \paren {\frac 1 a}^{t / u}        | o = <        | r = \paren {\frac 1 a}^{r / s}       | c = Power Function on Base between Zero and One is Strictly Decreasing: Rational Number }} {{eqn | ll= \leadsto       | l = \sqrt [u] {\paren {\frac 1 a}^t}       | o = <        | r = \sqrt [s] {\paren {\frac 1 a}^r}       | c =  {{Defof|Rational Power}} }} {{eqn | ll= \leadsto       | l = \sqrt [u] {\paren {\frac 1 {a^t} } }        | o = <        | r = \sqrt [s] {\paren {\frac 1 {a^r} } }       | c = Real Number to Negative Power: Integer }} {{eqn | ll= \leadsto       | l = \frac 1 {\sqrt [u] {a^t} }       | o = <       | r = \frac 1 {\sqrt [s] {a^r} }       | c = Root of Reciprocal is Reciprocal of Root }} {{eqn | ll= \leadsto       | l = \frac 1 {a^{t / u} }       | o = <       | r = \frac 1 {a^{r / s} }       | c = {{Defof|Rational Power}} }} {{eqn | ll= \leadsto       | l = a^{r / s}       | o = <       | r = a^{t / u}       | c = Ordering of Reciprocals }} {{end-eqn}} Hence the result. {{qed}} Category:Power Function on Base Greater than One is Strictly Increasing lm6ydcb44fyxnfu932l30zdwqxas4s4
Let $S = \set {x \in \Q: x^2 < r}$. As $0 \in S$, it follows that $S$ is non-empty. To show that $S$ is bounded above, we note that $n + 1$ is an upper bound: :$y > n + 1 \implies y^2 > n^2 + 2 n + 1 > n$ and so $y \notin S$. Thus $x \in S \implies x < n + 1$. By the Completeness Axiom, $S$ has a supremum, say: :$u = \sup S$ We already have that $u \ge 1$, as $0 \in S$ as seen. It remains to demonstrate that $u^2 = r$. {{AimForCont}} $u^2 \ne r$. Then either $u^2 > r$ or $u^2 < r$. Suppose that $u^2 > r$. Then: :$\dfrac {u^2 - r} {2 u} > 0$ So there exists $n \in \N$ such that: :$0 < \dfrac 1 n < \dfrac {u^2 - r} {2 u}$ Then: {{begin-eqn}} {{eqn | l = \paren {u - \dfrac 1 n}^2       | r = u^2 - \dfrac {2 u} n + \dfrac 1 {n^2}       | c =  }} {{eqn | o = >       | r = u^2 - \dfrac {2 u} n       | c =  }} {{eqn | o = >       | r = u^2 - \paren {u^2 - r}       | c =  }} {{eqn | r = r       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = S       | c =  }} {{eqn | ll= \leadsto       | l = x^2       | o = <       | r = r       | c =  }} {{eqn | o = <       | r = \paren {u - \dfrac 1 n}^2       | c =  }} {{eqn | ll= \leadsto       | l = x       | o = <       | r = u - \dfrac 1 n       | c =  }} {{end-eqn}} which contradicts the leastness of $u$. Suppose instead that $u^2 < r$. Then $\exists n \in \N$ such that: :$0 < \dfrac 1 n \le \dfrac {r - u^2} {4 u}$ and: :$\dfrac 1 n < 2 u$ Then: {{begin-eqn}} {{eqn | l = \paren {u + \dfrac 1 n}^2       | r = u^2 + \dfrac {2 u} n + \dfrac 1 {n^2}       | c =  }} {{eqn | o = <       | r = u^2 + \dfrac {2 u} n + \dfrac {2 u} n       | c = as $\dfrac 1 n < 2 u$ }} {{eqn | o = \le       | r = u^2 + r - u^2       | c = as $r - u^2 \ge \dfrac {4 u} n$ }} {{eqn | r = r       | c =  }} {{end-eqn}} Hence: :$u + \dfrac 1 n \in S$ which contradicts the fact that $u$ is an upper bound of $S$. Hence by Proof by Contradiction it follows that $u^2 = r$. {{qed}}
{{begin-eqn}} {{eqn | l = x < y       | o = \leadstoandfrom       | r = y > x       | c = {{Defof|Dual Ordering}} }} {{eqn | o = \leadstoandfrom       | r = y + \paren {-x} > x + \paren {-x}       | c = Real Number Ordering is Compatible with Addition }} {{eqn | o = \leadstoandfrom       | r = y + \paren {-x} > 0       | c = Real Number Axioms: $\R \text A 4$: Inverse Elements }} {{eqn | o = \leadstoandfrom       | r = y - x > 0       | c = {{Defof|Field Subtraction}} }} {{end-eqn}} Hence the result. {{qed}}
Let $0 < a < 1$. By Power Function on Base between Zero and One is Strictly Decreasing: Positive Integer, the theorem is already proven for positive integers. It remains to be proven over the negative integers. Let $i, j$ be integers such that $i < j < 0$. From Order of Real Numbers is Dual of Order of their Negatives: :$0 < -j < -i$ So: {{begin-eqn}} {{eqn | l = a^{-j}       | o = >        | r = a^{-i}       | c = Power Function on Base between Zero and One is Strictly Decreasing: Positive Integer }} {{eqn | ll= \leadsto       | l = \frac 1 {a^j}       | o = >       | r = \frac 1 {a^i}       | c = Real Number to Negative Power: Positive Integer }} {{eqn | ll= \leadsto       | l = a^i       | o = >       | r = a^j       | c = Ordering of Reciprocals }} {{end-eqn}} Hence the result. {{qed}} Category:Power Function on Base between Zero and One is Strictly Decreasing rtbdssg7n11m3w8hg6poyxc849q5hfi
=== Sufficient Condition === Assume that: :$\mathscr S$ is meet-continuous. We will prove that :for every an element $x$ of $S$, a directed subset $D$ of $S$ if $x \preceq \sup D$, then $x = \sup \left\{ {x \wedge d: d \in D}\right\}$ Let $x \in S$, $D$ be a directed subset of $S$ such that :$x \preceq \sup D$ Thus {{begin-eqn}} {{eqn | l = x       | r = x \wedge \sup D       | c = Preceding iff Meet equals Less Operand }} {{eqn | r = \sup \left\{ {x \wedge d: d \in D}\right\}       | c = {{Defof|Meet-Continuous Lattice}} }} {{end-eqn}} {{qed|lemma}} By definition of reflexivity: :for every an element $x$ of $S$, a directed subset $D$ of $S$ if $x \preceq \sup D$, then $x \preceq \sup \left\{ {x \wedge d: d \in D}\right\}$ By Meet is Directed Suprema Preserving: :$f$ preserves directed suprema. {{qed|lemma}} === Necessary Condition === Assume that: :$f$ preserves directed suprema. By Meet is Directed Suprema Preserving implies Meet of Suprema equals Supremum of Meet of Directed Subsets:  :$\left({\sup D_1}\right) \wedge \left({\sup D_2}\right) = \sup \left\{ {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}\right\}$ Thus by Meet-Continuous iff Meet of Suprema equals Supremum of Meet of Directed Subsets: :$\mathscr S$ is meet-continuous. {{qed}}
Let $a > 1$. By Power Function on Base Greater than One is Strictly Increasing: Positive Integer, the theorem is already proven for positive integers. It remains to be proven over the strictly negative integers. Let $i, j$ be integers such that $i < j < 0$. From Order of Real Numbers is Dual of Order of their Negatives: :$0 < -j < -i$ So: {{begin-eqn}} {{eqn | l = a^{-j}       | o = <        | r = a^{-i}       | c = Power Function on Base Greater than One is Strictly Increasing: Positive Integer }} {{eqn | ll= \leadsto       | l = \frac 1 {a^j}       | o = <        | r = \frac 1 {a^i}       | c = Real Number to Negative Power: Positive Integer }} {{eqn | ll= \leadsto       | l = a^i       | o = <        | r = a^j       | c = Ordering of Reciprocals }} {{end-eqn}} {{qed}} Category:Power Function on Base Greater than One is Strictly Increasing sjlmfdufptgv70wy09w64dxget1hs7b
{{begin-eqn}} {{eqn | l = a \oplus b       | r = \left( a \oplus b \right) ^{*}       | c = Definition of real *-algebra }} {{eqn | r = b^{*} \oplus a^{*}       | c = Definition of conjugation }} {{eqn | r = b \oplus a       | c = Definition of real *-algebra }} {{end-eqn}} {{qed}} Category:Algebra i3wknk52uzvwo406a5i4lui4ojcz541
Let $D$ be a directed subset of $S$ such that :$D$ admits a supremum. By Singleton is Directed and Filtered Subset: :$\left\{ {x}\right\}$ is directed. By Up-Complete Product/Lemma 1: :$\left\{ {x}\right\} \times D$ is directed in Cartesian product $\left({S \times S, \precsim}\right)$ of $\mathscr S$ and $\mathscr S$. Define a mapping $g: S \times S \to S$: :$\forall s, t \in S: g\left({s, t}\right) = s \wedge t$ By Meet-Continuous iff Meet Preserves Directed Suprema: :$g$ preserves directed suprema. By definition of meet-continuous: :$\mathscr S$ is up-complete. By Up-Complete Product: :$\left({S \times S, \precsim}\right)$ is up-complete. By definition of up-complete: :$\left\{ {x}\right\} \times D$ admits a supremum. Thus {{begin-eqn}} {{eqn | l = \sup \left({f^\to \left({D}\right)}\right)       | r = \sup \left\{ {f \left({d}\right): d \in D}\right\}       | c = {{Defof|Image of Subset under Mapping}} }} {{eqn | r = \sup \left\{ {x \wedge d: d \in D}\right\}       | c = definition of $f$ }} {{eqn | r = \sup \left\{ {g \left({y, d}\right): y \in \left\{ {x}\right\}, d \in D}\right\}       | c = definition of $g$ }} {{eqn | r = \sup \left({g^\to \left({\left\{ {x}\right\} \times D}\right)}\right)       | c = {{Defof|Image of Subset under Mapping}} }} {{eqn | r = g \left({\sup \left({\left\{ {x}\right\} \times D}\right)}\right)       | c = definition of mapping preserves the supremum of set }} {{eqn | r = g \left({\sup \left\{ {x}\right\}, \sup D}\right)       | c = Supremum of Cartesian Product }} {{eqn | r = g \left({x, \sup D}\right)       | c = Supremum of Singleton }} {{eqn | r = x \wedge \sup D       | c = Definition of $g$ }} {{eqn | r = f \left({\sup D}\right)       | c = Definition of $f$ }} {{end-eqn}} {{qed}}
Proof by Mathematical Induction: For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\forall x \in \R_{>0}: x^n > 0$ $\map P 0$ is true, as this just says: {{begin-eqn}} {{eqn | l = x^0       | r = 1       | c = {{Defof|Integer Power}} }} {{eqn | o = >       | r = 0 }} {{end-eqn}} === Basis for the Induction === $\map P 1$ true, as this just says: {{begin-eqn}} {{eqn | l = x^1       | r = x       | c = {{Defof|Integer Power}} }} {{eqn | o = >       | r = 0 }} {{end-eqn}} This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 0 $, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\forall x \in \R_{>0}: x^k > 0$ Then we need to show: :$\forall x \in \R_{>0}: x^{k + 1} > 0$ === Inductive Step === This is our induction step: {{begin-eqn}} {{eqn | l = x^k       | o = >       | r = 0       | c = Induction Hypothesis }} {{eqn | ll= \leadsto       | l = x^{k + 1}       | o = >       | r = 0       | c = Multiply both sides by $x > 0$ }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 0}: \forall x \in \R_{>0}: x^n > 0$ {{qed}} Category:Power of Positive Real Number is Positive tmg67c2ywdvxvu01dumrsl4rv3rxrj3
By Power of Positive Real Number is Positive: Natural Number, the theorem is already proven for non-negative integers. Suppose $n \in \Z_{< 0}$. When $n < 0$, by Real Number Ordering is Compatible with Multiplication: Negative Factor: :$-n > 0$ Then, by Power of Positive Real Number is Positive: Natural Number: :$x^{-n} > 0$ Therefore, by Reciprocal of Strictly Positive Real Number is Strictly Positive: :$x^n = \dfrac 1 {x^{-n} } > 0$ {{qed}} Category:Power of Positive Real Number is Positive q6y71g607yse61jxsqnx1yreip7soz1
Let $q = \dfrac r s$, where $r \in \Z$, $s \in \Z \setminus \set 0$. Then: {{begin-eqn}} {{eqn | l = x > 0       | o = \leadsto       | r = x^r > 0       | c = Power of Positive Real Number is Positive: Integer }} {{eqn | o = \leadsto       | r = \sqrt [s] {\paren {x^r} } > 0       | c = Existence of Positive Root of Positive Real Number }} {{eqn | o = \leadsto       | r = x^{r / s}       | c = {{Defof|Rational Power|Rational Power}} }} {{end-eqn}} Hence the result. {{qed}} Category:Power of Positive Real Number is Positive 82zj1qzhj73vryu2ve2f4ricm3ohqr4
=== Case 1: $a > 1$ === Let $a > 1$. Then by Power Function on Base Greater than One is Strictly Increasing: :$f$ is strictly increasing. By Strictly Increasing Mapping is Increasing: :$f$ is increasing. === Case 2 : $a = 1$ === Let $a = 1$. Then: :$\forall r \in \Q: a^r = 1$ and so $f$ is constant. From Mapping is Constant iff Increasing and Decreasing: :$f$ is both increasing and decreasing. === Case 3: $0 < a < 1$ === Let $0 < a < 1$. Then by Power Function on Base between Zero and One is Strictly Decreasing: :$f$ is strictly decreasing. By Strictly Decreasing Mapping is Decreasing: :$f$ is decreasing. === Conclusion === It has been shown that in all cases $f$ is either increasing or decreasing or both. Thus by definition of monotone function, $f$ is monotone. In either of the cases where $a \ne 1$, It has been shown that in all cases $f$ is either strictly increasing or strictly decreasing. Thus by definition of strictly monotone function, $f$ is strictly monotone. Hence the result. {{qed}} Category:Powers gr1gfcbgcpbpct12u4hd79apbasu0nh
=== Case 1: $a > 1$ === If $a > 1$, then: :$\displaystyle \lim_{x \mathop \to 0} \map f x = 1$ from Power Function on Base greater than One tends to One as Power tends to Zero/Rational Number. === Case 2: $a = 1$ === If $a = 1$, then: {{begin-eqn}} {{eqn | l = \lim_{x \mathop \to 0} \map f x       | r = \lim_{x \mathop \to 0} 1^x       | c = Definition of $f$ }} {{eqn | r = \lim_{x \mathop \to 0} 1       | c = Exponential with Base One is Constant/Rational Number }} {{eqn | r = 1       | c = Real Polynomial Function is Continuous }} {{end-eqn}} === Case 3: $0 < a < 1$ === If $0 < a < 1$, then: :$\displaystyle \lim_{x \mathop \to 0} \map f x = 1$ from Power Function on Base between Zero and One Tends to One as Power Tends to Zero/Rational Number. Hence the result. {{qed}} Category:Powers 4cfp0fmmaqxg4pnvnse02adeir3ajq5
Define a real function $g_r: \R_{> 0} \to \R$ as: :$\map {g_r} a = 1 + a r - a^r$ Then differentiating {{WRT|Differentiation}} $a$ gives: :$D_a \map {g_r} a = r \paren {1 - a^{r - 1} }$ We show now that the derivative of $g_r$ is positive for all $a > 1$: {{begin-eqn}} {{eqn | l = r       | o = <       | r = 1 }} {{eqn | ll= \leadsto       | l = r - 1       | o = <       | r = 0       | c = Subtract $1$ from both sides }} {{eqn | ll= \leadsto       | l = a^{r - 1}       | o = <       | r = a^0       | c = Power Function on Base Greater than One is Strictly Increasing: Rational Number }} {{eqn | ll= \leadsto       | l = a^{r - 1}       | o = <       | r = 1       | c = {{Defof|Integer Power}} }} {{eqn | ll= \leadsto       | l = -1        | o = <       | r = -a^{r - 1}       | c = Order of Real Numbers is Dual of Order of their Negatives }} {{eqn | ll= \leadsto       | l = 0        | o = <       | r = 1 - a^{r - 1}       | c = adding $1$ to both sides }} {{eqn | ll= \leadsto       | l = 0        | o = <       | r = r \left({1 - a^{r - 1} }\right)       | c = multiplying both sides by $r > 0$ }} {{eqn | ll= \leadsto       | l = 0        | o = <       | r = D_a \map {g_r} a       | c = from the formula found for $D_a \map {g_r} a$ above }} {{end-eqn}} So $D_a \map {g_r} a$ is positive for all $a > 1$. Whence, by Derivative of Monotone Function, $g_r$ is increasing for all $a > 1$. Now: :$\map {g_r} 1 = r > 0$ So $\map {g_r} a$ is positive for all $a > 1$. That is: :$0 < 1 + a r - a^r$ Adding $a^r$ to both sides of the above yields: :$a^r < 1 + a r$ Finally: {{begin-eqn}} {{eqn | l = 0       | o = <       | r = r }} {{eqn | ll= \leadsto       | l = a^0       | o = <       | r = a^r       | c = Power Function on Base Greater than One is Strictly Increasing: Rational Number }} {{eqn | ll= \leadsto       | l = 1       | o = <       | r = a^r       | c = {{Defof|Integer Power}} }} {{end-eqn}} So, for $0 < r < 1$: :$1 < a^r < 1 + a r$ {{qed}} Category:Powers ee733l081xaewexuydgpzitbjpw99in
We start by treating the right-sided limit. Let $0 < r < 1$. === Lemma === {{:Power Function on Base greater than One tends to One as Power tends to Zero/Rational Number/Lemma}}{{qed|lemma}} From the lemma: :$1 < a^r < 1 + a r$ Also: {{begin-eqn}} {{eqn | l = \lim_{r \mathop \to 0} \paren {1 + a r}       | r = 1 + a \times 0       | c = Real Polynomial Function is Continuous }} {{eqn | r = 1       | c = Real Zero is Zero Element }} {{eqn | r = \lim_{r \mathop \to 0} 1       | c = Real Polynomial Function is Continuous }} {{end-eqn}} So from the Squeeze Theorem: :$\displaystyle \lim_{r \mathop \to 0^+} a^r = 1$ We now treat the left-sided limit. Let $-1 < r < 0$. {{begin-eqn}} {{eqn | l = -1        | o = <        | m = r       | mo= <        | r = 0 }} {{eqn | ll= \leadsto       | l = 0       | o = <        | m = -r       | mo= <        | r = 1       | c = Order of Real Numbers is Dual of Order of their Negatives }} {{eqn | ll= \leadsto       | l = 1       | o = <        | m = a^{-r}       | mo= <        | r = 1 - a r       | c = Lemma }} {{eqn | ll= \leadsto       | l = 1       | o = <        | m = \frac 1 {a^r}       | mo= <        | r = 1 - a r       | c = Real Number to Negative Power: Rational Number }} {{eqn | ll= \leadsto       | l = \frac 1 {1 - a r}       | o = <       | m = a^r       | mo= <       | r = 1       | c = Ordering of Reciprocals }} {{end-eqn}} Also: {{begin-eqn}} {{eqn | l = \lim_{r \mathop \to 0} \dfrac 1 {1 - a r}       | r = \frac 1 {\displaystyle \lim_{r \mathop \to 0} \paren {1 - a r} }       | c = Quotient Rule for Limits of Functions }} {{eqn | r = \frac 1 {1 - a \times 0}       | c = Real Polynomial Function is Continuous }} {{eqn | r = 1       | c = Real Zero is Zero Element }} {{eqn | r = \lim_{r \mathop \to 0} 1       | c = Real Polynomial Function is Continuous }} {{end-eqn}} So from the Squeeze Theorem: :$\displaystyle \lim_{r \mathop \to 0^-} a^r = 1$ From Limit iff Limits from Left and Right: :$\displaystyle \lim_{r \mathop \to 0} a^r = 1$ Hence the result. {{qed}} Category:Powers 7akolrmpim9lbuhon3cb87n7o01jav3
From Ordering of Reciprocals: :$0 < a < 1 \implies 1 < \dfrac 1 a$ So: {{begin-eqn}} {{eqn | l = \lim_{r \mathop \to 0} \paren {\frac 1 a}^r       | r = 1       | c = Power Function on Base greater than One tends to One as Power tends to Zero: Rational Number }} {{eqn | ll= \leadsto       | l = \lim_{r \mathop \to 0} \frac 1 {a^r}       | r = 1       | c = Power of Quotient: Rational Numbers }} {{eqn | ll= \leadsto       | l = \frac 1 {\displaystyle \lim_{r \mathop \to 0} a^r}       | r = 1       | c = Quotient Rule for Limits of Functions }} {{eqn | ll= \leadsto       | l = \lim_{r \mathop \to 0} a^r        | r = 1       | c = taking reciprocal of each side }} {{end-eqn}} {{qed}} Category:Powers cy9oml498lz0mynci3npt3k51qjck6y
Since $e > 0$, we may apply Power Function to Rational Power permits Unique Continuous Extension. Hence the result. {{qed}} Category:Exponential Function j3aj4ucitdfj9juy3c0f2ti85dhfkmm