hcju/proofwikiqa huggingface dataset:
The straight line boundaries of $R$ are given as $\theta_1$ and $\theta_2$. The corners of $R$ are located where: :$\theta = \theta_1 + 2 n \pi$ :$\theta = \theta_2 + 2 n \pi$ :$\theta = \theta_1 + 2 \left({n + 1}\right) \pi$ :$\theta = \theta_2 + 2 \left({n + 1}\right) \pi$ 500px {{begin-eqn}} {{eqn | l = \mathcal A       | r = \int_{\theta_1 + 2 \left({n + 1}\right) \pi}^{\theta_2 + 2 \left({n + 1}\right) \pi} \frac {\left({a \theta}\right)^2 \, \mathrm d \theta} 2 - \int_{\theta_1 + 2 n \pi}^{\theta_2 + 2 n \pi} \frac {\left({a \theta}\right)^2 \, \mathrm d \theta} 2       | c = Area between Radii and Curve in Polar Coordinates }} {{eqn | r = \left[{\frac {a^2 \theta^3} 6}\right]_{\theta_1 + 2 \left({n + 1}\right) \pi}^{\theta_2 + 2 \left({n + 1}\right) \pi} - \left[{\frac {a^2 \theta^3} 6}\right]_{\theta_1 + 2 n \pi}^{\theta_2 + 2 n \pi}       | c = Primitive of Power }} {{eqn | r = \frac {a^2} 6 \left({\left({\theta_2 + 2 \left({n + 1}\right) \pi}\right)^3 - \left({\theta_1 + 2 \left({n + 1}\right) \pi}\right)^3 - \left({\theta_2 + 2 n \pi}\right)^3 + \left({\theta_1 + 2 n \pi}\right)^3}\right)       | c =  }} {{eqn | r = \frac {a^2} 6 \left({\theta_2^3 + 3 \left({2 \left({n + 1}\right) \pi}\right)\theta_2^2 + 3 \left({2 \left({n + 1}\right) \pi}\right)^2 \theta_2 + \left({2 \left({n + 1}\right) \pi}\right)^3}\right)       | c =  }} {{eqn | o =        | ro= -       | r = \frac {a^2} 6 \left({\theta_1^3 + 3 \left({2 \left({n + 1}\right) \pi}\right)\theta_1^2 + 3 \left({2 \left({n + 1}\right) \pi}\right)^2 \theta_1 + \left({2 \left({n + 1}\right) \pi}\right)^3}\right)       | c =  }} {{eqn | o =        | ro= -       | r = \frac {a^2} 6 \left({\theta_2^3 + 3 \left({2 n \pi}\right)\theta_2^2 + 3 \left({2 n \pi}\right)^2 \theta_2 + \left({2 n \pi}\right)^3}\right)       | c =  }} {{eqn | o =        | ro= +       | r = \frac {a^2} 6 \left({\theta_1^3 + 3 \left({2 n \pi}\right)\theta_1^2 + 3 \left({2 n \pi}\right)^2 \theta_1 + \left({2 n \pi}\right)^3}\right)       | c =  }} {{eqn | r = a^2 \pi \left({\left({n + 1}\right) \theta_2^2 + 2 \pi \left({n + 1}\right)^2 \theta_2 - \left({n \theta_2^2 + 2 \pi n^2 \theta_2}\right)}\right)       | c =  }} {{eqn | o =        | ro= -       | r = a^2 \pi \left({\left({n + 1}\right)\theta_1^2 + 2 \pi \left({n + 1}\right)^2 \theta_1 - \left({n \theta_1^2 + 2 \pi n^2 \theta_1}\right)}\right)       | c =  }} {{eqn | r = a^2 \pi \left({\theta_2^2 + 2 \pi \left({2 n + 1}\right) \theta_2}\right)       | c =  }} {{eqn | o =        | ro= -       | r = a^2 \pi \left({\theta_1^2 + 2 \pi \left({2 n + 1}\right) \theta_1}\right)       | c =  }} {{eqn | r = a^2 \pi \left({\theta_2 - \theta_1}\right) \left({\theta_2 + \theta_1 + 2 \pi \left({2 n + 1}\right)}\right)       | c =  }} {{end-eqn}} {{proofread}}
Fix $x \in \R$. Then: {{begin-eqn}} {{eqn | l = n       | o = \ge       | r = \ceiling {\size x} }} {{eqn | ll= \leadsto       | l = n       | o = >       | r = -x       | c = Real Number is between Ceiling Functions and Negative of Absolute Value }} {{eqn | ll= \leadsto       | l = 1       | o = >       | r = \frac {-x} n       | c = Divide both sides by $n$ }} {{eqn | ll= \leadsto       | l = 1 + \frac x n       | o = >       | r = 0 }} {{end-eqn}} So we may apply the AM-GM inequality, with $x_1 := 1$ and $x_2 := \ldots := x_{n + 1} = 1 + \dfrac x n$, to obtain that: {{explain|The above sentence is unclear.}} :$\dfrac {1 + n \paren {1 + \dfrac x n} } {n + 1} > \paren {\paren {1 + \dfrac x n}^n}^{1 / \paren {n + 1} }$ After simplification: :$1 + \dfrac x {n + 1} > \paren {1 + \dfrac x n}^{n / \paren {n + 1} }$ From Power Function is Strictly Increasing over Positive Reals: Natural Exponent: :$\paren {1 + \dfrac x {n + 1} }^{n + 1} > \paren {1 + \dfrac x n}^n$ where we have raised both sides to the power of $n + 1$. Hence the result. {{qed}} Category:Exponential Function c9b26890rtj1exfkejzxhbfk25q4mt5
First we prove that $e$ is a solution to $\ln \left({x}\right) = 1$: {{begin-eqn}} {{eqn | l = \ln \left({e}\right)       | r = \ln \left({\lim_{n \mathop \to \infty} \left({1 + \frac 1 n}\right)^n}\right)       | c = Definition:Euler's Number/Limit of Sequence }} {{eqn | r = \lim_{n \mathop \to \infty} \left({\ln \left({1 + \frac 1 n}\right)^n}\right)       | c = Sequential Continuity is Equivalent to Continuity in Metric Space }} {{eqn | r = \lim_{n \mathop \to \infty} \left({n \ln \left({1 + \frac 1 n}\right)}\right)       | c = Logarithm of Power }} {{eqn | r = \lim_{n \mathop \to \infty} \frac {\ln \left({1 + \frac 1 n}\right)} {1 / n}       | c = Inverse of Group Inverse }} {{eqn | r = \lim_{x \mathop \to \infty} \frac {\ln \left({1 + \frac 1 n}\right)} {1 / x}       | c = Limit of Sequence is Limit of Real Function }} {{eqn | r = \lim_{x \mathop \to \infty} \frac{\left({-\frac 1 {x^2} }\right) \left({1 + \frac 1 x}\right)} {-\frac 1 {x^2} }       | c = L'Hôpital's Rule }} {{eqn | r = \lim_{x \mathop \to \infty} \frac x {x + 1} }} {{eqn | r = 1 }} {{end-eqn}} {{improve|Find a better link than Inverse of Group Inverse -- Reciprocal of Reciprocal is worth creating. Then it will remove the need to invoke group-theoretical results in what is a purely real-analytical theorem.}} From Logarithm is Strictly Increasing, $\ln$ is strictly monotone. By Strictly Monotone Mapping with Totally Ordered Domain is Injective it follows that $\ln$ is an injection. So the solution to $\ln \left({x}\right) = 1$ is unique. Hence the result. {{qed}} Category:Logarithms 4pwnm4m0tgj1lbyhjni276cjdx9ospj
Let $G$ be an Ore graph. {{AimForCont}} $G$ is not connected. Then it has at least two components. Call these components $C_1$ and $C_2$. Thus, there exist non-adjacent vertices $u$ and $v$ such that $u$ is in $C_1$ and $v$ is in $C_2$. Let $m_1$ and $m_2$ be the number of vertices in $C_1$ and $C_2$ respectively. It is clear that: :$m_1 + m_2 \le n$ By definition of Ore graph, $G$ is simple. Thus it follows that: :$\deg_G \left({u}\right) \le m_1 - 1$ and $\deg_G \left({v}\right) \le m_2 - 1$ Thus: {{begin-eqn}} {{eqn | l = \deg_G \left({u}\right) + \deg_G \left({v}\right)       | o = \le       | r = m_1 - 1 + m_2 - 1       | c =  }} {{eqn | r = m_1 + m_2 - 2       | c =  }} {{eqn | o = <       | r = m_1 + m_2       | c =  }} {{eqn | o = \le       | r = n       | c =  }} {{end-eqn}} That is: :$\deg_G \left({u}\right) + \deg_G \left({v}\right) < n$ But by definition of Ore graph: :$\deg_G \left({u}\right) + \deg_G \left({v}\right) \ge n$ By Proof by Contradiction, it follows that our assumption that $G$ is not connected was false. Hence the result. {{Qed}}
Consider $I_k := \left({-k \,.\,.\, k }\right)$ for arbitrary $k \in \N$. Let $I_k' = I_k \cap \Q$. Note that, for all $x, y \in I_k'$: {{begin-eqn}} {{eqn | l = \size {a^x - a^y}       | r = \size {a^{x - y + y} - a^y} }} {{eqn | r = \size {a^{x - y} a^y - a^y}       | c = Powers of Group Elements }} {{eqn | r = \size {a^y} \size {a^{x - y} - 1}       | c = Absolute Value Function is Completely Multiplicative }} {{eqn | r = a^y \size {a^{x - y} - 1}       | c = Power of Positive Real Number is Positive over Rationals }} {{eqn | o = \le       | r = M \size {a^{x - y} - 1}       | c = Power Function is Monotone over Rationals }} {{end-eqn}} where $M = \map \max {a^{-k}, a^k}$. Fix $\epsilon \in \R_{> 0}$. From Power Function tends to One as Power tends to Zero: Rational Number: {{begin-eqn}} {{eqn | ll= \exists \delta \in \R_{> 0}:       | l = 0       | o = <       | m = \left\vert{x - y}\right\vert       | mo= <       | r = \delta }} {{eqn | lll= \leadsto       | o =        | m = \size {a^{x - y} - a^0}       | mo= <       | r = \dfrac \epsilon M }} {{eqn | lll= \leadsto       | o =        | m = \size {a^{x - y} - 1}       | mo= <       | r = \dfrac \epsilon M       | c = {{Defof|Integer Power|Power of Zero}} }} {{end-eqn}} where $M$ is defined as above. Therefore, $\exists \delta \in \R_{> 0}$ such that: {{begin-eqn}} {{eqn | l = \size {x - y}       | o = <       | r = \delta }} {{eqn | ll= \leadsto       | l = \size {a^x - a^y}       | o = \le       | r = M \size {a^{x - y} - 1} }} {{eqn | o = <       | r = M \dfrac \epsilon M }} {{eqn | r = \epsilon }} {{end-eqn}} That is, $f$ is uniformly continuous on $I_k'$. From Rationals are Everywhere Dense in Topological Space of Reals, $I_k'$ is everywhere dense in $I_k$. From Continuous Extension from Dense Subset, there exists a unique continuous extension of $f$ to $I_k$. Call this function $f_k$. Define: : $\displaystyle F = \bigcup \set {f_k : k \in \N}$ Note that $\sequence {I_k}$ is an exhausting sequence of sets in $\R$. By the Union of Functions Theorem, $F$ defines a function $\R \to \R$ From the Pasting Lemma, $F$ is continuous on $\R$. Hence the result. {{qed}} Category:Powers 4z686v4j71hiajz8ium9ffl6m5v7wuj
This proof depends on the continuous extension definition of the exponential function. Let $\exp$ be the unique continuous extension of $e^x$ from $\Q$ to $\R$. By definition, $\exp$ is continuous. Hence the result. {{qed}}
Let $n \in \N_{>0}$. Thus, let $\map f x = x^{1 / n}$. From the definition of the power to a rational number, or alternatively from the definition of the root of a number, $\map f x$ is defined when $x \ge 0$. (However, see the special case where $x = 0$.) From Continuity of Root Function, $f \left({x}\right)$ is continuous over the open interval $\openint 0 \infty$, but not at $x = 0$ where it is continuous only on the right. Let $y > x$. From Inequalities Concerning Roots: :$\forall n \in \N_{>0}: X Y^{1 / n} \, \size {x - y} \le n X Y \, \size {x^{1 / n} - y^{1 / n} } \le Y X^{1 / n} \, \size {x - y}$ where $x, y \in \closedint X Y$. Setting $X = x$ and $Y = y$, this reduces (after algebra) to: :$\dfrac 1 {n y} y^{1 / n} \le \dfrac {y^{1 / n} - x^{1 / n} } {y - x} \le \dfrac 1 {n x} x^{1 / n}$ From the Squeeze Theorem, it follows that: :$\displaystyle \lim_{y \mathop \to x^+} \dfrac {y^{1 / n} - x^{1 / n} } {y - x} = \dfrac 1 {n x} x^{1 / n} = \dfrac 1 n x^{\dfrac 1 n - 1}$ A similar argument shows that the left hand limit is the same. Thus the result holds for $\map f x = x^{1 / n}$. {{qed}}
Let $n \in \N_{>0}$. Thus, let $f \left({x}\right) = y = x^{1/n}$. Thus $f^{-1} \left({y}\right) = x = y^n$ from the definition of root. So: {{begin-eqn}} {{eqn | l = D x^{1/n}       | r = \frac 1 {D y^n}       | c = Derivative of Inverse Function }} {{eqn | r = \frac 1 {n y^{n-1} }       | c = Power Rule for Derivatives: Integer Index }} {{eqn | r = \frac 1 {n \left({x^{1/n} }\right)^{n-1} }       | c =  }} {{eqn | r = \frac 1 n x^{\frac 1 n - 1}       | c =  }} {{end-eqn}} {{qed}}
Let $f \left({x}\right) = x^n$ for $x \in \R, n \in \N$. By the definition of the derivative: : $\displaystyle \dfrac \d {\d x} f \left({x}\right) = \lim_{h \mathop \to 0} \dfrac {f \left({x + h}\right) - f \left({x}\right)} h = \lim_{h \mathop \to 0} \dfrac{(x + h)^n - x^n} h$ Using the binomial theorem this simplifies to: {{begin-eqn}} {{eqn | o =        | r = \lim_{h \mathop \to 0} \left({\frac { {n \choose 0} x^n + {n \choose 1} x^{n - 1} h + {n \choose 2} x^{n - 2} h^2 + \cdots + {n \choose n-1} x h^{n-1} + {n \choose n} h^n - x^n} h}\right)       | c =  }} {{eqn | r = \lim_{h \mathop \to 0} \left({\frac { {n \choose 1} x^{n - 1} h + {n \choose 2} x^{n - 2} h^2 + \cdots + {n \choose n - 1} x h^{n - 1} + {n \choose n} h^n} h}\right)       | c =  }} {{eqn | r = \lim_{h \mathop \to 0} \left({ {n \choose 1} x^{n - 1} + {n \choose 2} x^{n - 2} h^1 + \cdots + {n \choose n - 1} x h^{n - 2} + {n \choose n} h^{n - 1} }\right)       | c =  }} {{eqn | r = {n \choose 1} x^{n - 1}       | c = evaluating the limit }} {{eqn | r = n x^{n - 1}       | c = Binomial Coefficient with One: $\dbinom r 1 = r$ }} {{end-eqn}} {{qed}}
We will use the notation $D f \left({x}\right) = f^{\prime} \left({x}\right)$ as it is convenient. Let $n = 0$. Then $\forall x \in \R: x^n = 1$. Thus $f \left({x}\right)$ is the constant function $f_1 \left({x}\right)$ on $\R$. Thus from Derivative of Constant, $D f \left({x}\right) = D \left({x^0}\right) = 0 x^{-1}$, except where $x = 0$. So the result holds for $n = 0$. Let $n = 1$. Then: : $\forall x \in \R: f \left({x}\right) = x^n = x$ Then from Derivative of Identity Function: : $D \left({x}\right) = 1 = 1 \cdot x^{1-1}$ So the result holds for $n = 1$. Now assume $D \left({x^k}\right) = k x^{k-1}$. Then by the Product Rule for Derivatives: : $D \left({x^{k+1}}\right) = D \left({x^k x}\right) = x^k D \left({x}\right) + D \left({x^k}\right) x = x^k \cdot 1 + k x^{k-1} x = \left({k+1}\right) x^k$ The result follows by induction. {{qed}}
=== Sufficient Condition === Let $\mathscr S$ be meet-continuous. By Meet-Continuous iff Meet of Suprema equals Supremum of Meet of Ideals: :for every ideals $I, J$ in $\mathscr S$: $\paren {\sup I} \wedge \paren {\sup J} = \sup \set {i \wedge j: i \in I, j \in J}$ Let $D_1, D_2$ directed subsets of $S$. By definition of up-complete: :$D_1$ and $D_2$ admit suprema By Supremum of Lower Closure of Set: :$D_1^\preceq$ and $D_2^\preceq$ admit suprema where :$D_1^\preceq$ denotes the lower closure of $D_1$. Thus {{begin-eqn}} {{eqn | l = \paren {\sup D_1} \wedge \paren {\sup D_2}       | r = \paren {\sup D_1^\preceq} \wedge \paren {\sup D_2^\preceq}       | c = Supremum of Lower Closure of Set }} {{eqn | r = \sup \set {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}       | c = $D_1^\preceq$ is ideal in $\mathscr S$ }} {{eqn | r = \sup \set {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}^\preceq       | c = Supremum of Lower Closure of Set }} {{eqn | r = \sup \set {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}^\preceq       | c = Lower Closure of Meet of Lower Closures }} {{eqn | r = \sup \set {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}       | c = Supremum of Lower Closure of Set }} {{end-eqn}} {{qed|lemma}} === Necessary Condition === Assume that :for every directed subsets $D_1, D_2$ of $S$: $\paren {\sup D_1} \wedge \paren {\sup D_2} = \sup \set {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}$ By exemplification: :for every ideals $I_1, I_2$ of $S$: $\paren {\sup I_1} \wedge \paren {\sup I_2} = \sup \set {d_1 \wedge d_2: d_1 \in I_1, d_2 \in I_2}$ Thus by Meet-Continuous iff Meet of Suprema equals Supremum of Meet of Ideals: :$\mathscr S$ is meet-continuous. {{qed}}
=== Sufficient Condition === Let $\mathscr S$ be a Brouwerian lattice. By Relative Pseudocomplement and Shift Mapping form Galois Connection in Brouwerian Lattice: :for every $x \in S$, a mapping $f: S \to S$ if $\forall y \in S: f\left({y}\right) = x \wedge y$, then $f$ is lower adjoint of Galois connection By Shift Mapping is Lower Adjoint iff Meet is Distributive with Supremum: :$\forall x \in S, X \subseteq S: x \wedge \sup X = \sup \left\{ {x \wedge y: y \in X}\right\}$ Thus by definition of complete lattice: :$\mathscr S$ is up-complete. Thus :for every an element $x \in S$ and a directed subset $D$ of $S$: $x \wedge \sup D = \sup \left\{ {x \wedge d: d \in D}\right\}$ Thus by definition :$\mathscr S$ is meet-continuous. Thus by Brouwerian Lattice is Distributive: :$\mathscr S$ is distributive. {{qed|lemma}} === Necessary Condition === Assume that: :$\mathscr S$ is meet-continuous and distributive. Let $x \in S$. Define a mapping $f: S \to S$: :$\forall y \in S: f\left({y}\right) = x \wedge y$ By Meet-Continuous and Distributive implies Shift Mapping Preserves Finite Suprema: :$f$ preserves finite suprema. By Meet-Continuous implies Shift Mapping Preserves Directed Suprema: :$f$ preserves directed suprema. By Mapping Preserves Finite and Directed Suprema: :$f$ preserves all suprema. By All Suprema Preserving Mapping is Lower Adjoint of Galois Connection: :$f$ is lower adjoint of Galois connection. Thus by Brouwerian Lattice iff Shift Mapping is Lower Adjoint: :$\mathscr S$ is Brouwerian lattice. {{qed}}
Fix $x \in \R \setminus \left\{{ 0 }\right\}$. === Positive Integers === We first prove the theorem for positive integers by induction: For all $n \in \N$, let $P \left({n}\right)$ be the proposition: : $x^n$ is well-defined ==== Basis for the Induction ==== $P(1)$ is true, since $x^1 = 1$ by definition. This is our basis for the induction. ==== Induction Hypothesis ==== Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$x^k$ is well-defined Then we need to show: :$x^{k + 1}$ is well-defined ==== Induction Step ==== This is our induction step: By definition of integer power: :$x^{k + 1} = x \cdot x^k$ From Real Multiplication is Well-Defined and the Induction Hypothesis: :$x \cdot x^k$ is well-defined So $x^{k + 1}$ is well-defined. So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: : $\displaystyle \forall n \in \N: x^n$ is well-defined {{qed|lemma}} === Non-Positive Integers === Fix $k \in \Z_{< 0}$ First: :$k \in \Z_{< 0} \implies -k \in \N$ From the development above, $x^{-k}$ is well-defined. From Powers of Group Elements/Negative Index, $\dfrac 1 {x^k}$ is well-defined. From Reciprocal of Real Number is Unique, $x^k$ is well-defined. Finally: :$\forall x \in \R \setminus \left\{{0}\right\} : x^0 = 1$ So $x^k$ is well-defined for $k = 0$. Hence the result. {{qed}} Category:Powers 2aqjx1fi6383i67ea8ix5h7kbhd52jk
Let $x \in \R_{>0}$ be fixed. Let $q \in \Q \setminus \set 0$. Let $\dfrac r s$ and $\dfrac t u$ be two representations of $q$. That is, $r, s, t, u$ are non-zero integers. We now show that: : $\dfrac r s = \dfrac t u \implies x^{r / s} = x^{t / u}$ So: {{begin-eqn}} {{eqn | l = \dfrac r s = \dfrac t u       | o = \implies       | r = r u = t s }} {{eqn | o = \implies       | r = x^{r u} = x^{s t}       | c = Power is Well-Defined: Integer }} {{eqn | o = \implies       | r = \paren {x^r}^u = \paren {x^t}^s       | c = Product of Indices of Real Number: Integers }} {{eqn | o = \implies       | r = \sqrt [u] {\paren {x^r}^u} = \sqrt [u] {\paren {x^t}^s}       | c = Existence and Uniqueness of Positive Root of Positive Real Number }} {{eqn | o = \implies       | r = x^r = \sqrt [u] {\paren {x^t}^s}       | c = {{Defof|Root (Analysis)|Root}} }} {{eqn | o = \implies       | r = \sqrt [s] {\paren {x^r} } = \sqrt [s] {\sqrt [u] {\paren {x^t}^s} }       | c = Existence and Uniqueness of Positive Root of Positive Real Number }} {{eqn | o = \implies       | r = \sqrt [s] {\paren {x^r} } = \sqrt [u] {\sqrt [s] {\paren {x^t}^s} }       | c = Root is Commutative }} {{eqn | o = \implies       | r = \sqrt [s] {\paren {x^r} } = \sqrt [u] {\paren {x^t} }       | c = {{Defof|Root (Analysis)|Root}} }} {{eqn | o = \implies       | r = x^{r / s} = x^{t / u}       | c = {{Defof|Rational Power}} }} {{end-eqn}} Hence the result. {{qed}} Category:Powers 7xrtveddkhi280uir0dbcnwyfxcq8l1
Proof by induction on $n$: Let $x, y \in \R_{>0}$ be strictly positive real numbers. For all $n \in \Z_{>0}$, let $P \left({n}\right)$ be the proposition: :$x < y \implies x^n < y^n$ === Basis for the Induction === $P \left({1}\right)$ is true, as this just says: {{begin-eqn}} {{eqn | l = x^1       | r = x       | c = Definition of integer power }} {{eqn | o = <       | r = y       | c = By assumption }} {{eqn | r = y^1       | c = Definition of integer power }} {{end-eqn}} This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$x < y \implies x^k < y^k$ Then we need to show: :$x < y \implies x^{k+1} < y^{k+1}$ === Induction Step === This is our induction step: First: {{begin-eqn}} {{eqn | l = x       | o = <       | r = y }} {{eqn | ll = \implies       | l = x^k       | o = <       | r = y^k       | c = Induction Hypothesis }} {{eqn | ll = \implies       | l = x^{k+1}       | o = <       | r = x \times y^k       | c = Multiply both sides by $x > 0$ }} {{eqn | o = <       | r = y \times y^k       | c = Multiply both sides of $x < y$ by $y^k > 0$ }} {{eqn | r = y^{k+1}       | c = Definition of integer power }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{>0}: x < y \implies x^n < y^n$ {{qed}} Category:Powers dggzpwgfijhmelesjckhfrz3l7zc2u3
Let $x, y \in \R$ such that $0 < x < y$. {{AimForCont}} that: :$\sqrt [n] x \ge \sqrt [n] y$ We have: {{begin-eqn}} {{eqn | l = \sqrt [n] x       | o = \ge       | r = \sqrt [n] y       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\sqrt [n] x}^n       | o = \ge       | r = \paren {\sqrt [n] y}^n       | c = Power Function is Strictly Increasing over Positive Reals: Natural Exponent }} {{eqn | ll= \leadsto       | l = x       | o = \ge       | r = y       | c ={{Defof|Root (Analysis)}} }} {{end-eqn}} This contradicts the hypothesis that $x < y$. Therefore, by Proof by Contradiction: :$0 < x < y \implies \sqrt [n] x \le \sqrt [n] y$ Hence the result by definition of strictly increasing real function. {{qed}} Category:Roots of Numbers dayc12o1tyy9uobmrjhoftknykfuqxl
=== Sufficient Condition === Assume that :$\forall x \in S, f: S \to S:\left({\forall y \in S: f\left({y}\right) = x \wedge y}\right) \implies f$ is lower adjoint of Galois connection Let $x \in S, X \subseteq S$. Define a mapping $f: S \to S$: :$\forall y \in S: f\left({y}\right) := x \wedge y$ By assumption: :$f$ is lower adjoint of a Galois connection. By Lower Adjoint Preserves All Suprema: :$f$ preserves all suprema. By definition of mapping preserves all suprema :$f$ preserves the supremum of $X$. By definition of complete lattice: :$X$ admits a supremum. Thus {{begin-eqn}} {{eqn | l = x \wedge \sup X       | r = f\left({\sup X}\right)       | c = definition of $f$ }} {{eqn | r = \sup \left({f^\to\left({X}\right)}\right)       | c = definition of mapping preserves the supremum of set }} {{eqn | r = \sup \left\{ {f\left({y}\right): y \in X}\right\}       | c = definition of image of set }} {{eqn | r = \sup \left\{ {x \wedge y: y \in X}\right\}       | c = definition of $f$ }} {{end-eqn}} {{qed|lemma}} === Necessary Condition ===  Assume that :$\forall x \in S, X \subseteq S: x \wedge \sup X = \sup \left\{ {x \wedge y: y \in X}\right\}$ Let $x \in S$, $f: S \to S$ such that :$\forall y \in S: f\left({y}\right) = x \wedge y$ We will prove that :$f$ preserves all suprema. Let $X \subseteq S$ such that :$X$ admits a supremum. Thus by definition of complete lattice: :$f^\to\left({X}\right)$ admits a supremum. Thus {{begin-eqn}} {{eqn | l = \sup \left({f^\to\left({X}\right)}\right)       | r = \sup \left\{ {f\left({y}\right): y \in X}\right\}       | c = definition of image of set }} {{eqn | r = \sup \left\{ {x \wedge y: y \in X}\right\}       | c = definition of $f$ }} {{eqn | r = x \wedge \sup X       | c = assumption }} {{eqn | r = f\left({\sup X}\right)       | c = definition of $f$ }} {{end-eqn}} Thus by definition :$f$ preserves the supremum of $X$. {{qed|lemma}} Thus by All Suprema Preserving Mapping is Lower Adjoint of Galois Connection: :$f$ is lower adjoint of a Galois connection.  {{qed}}
{{begin-eqn}} {{eqn | l = \lcm \set {a, b}       | r = \frac {a b} {\gcd \set {a, b} }       | c = Product of GCD and LCM }} {{eqn | r = \frac {a b} 1       | c = {{Defof|Coprime Integers}} }} {{eqn | r = a b }} {{end-eqn}} {{qed}} Category:Lowest Common Multiple Category:Coprime Integers fcl75w8sx5a69le8gxkj1593ls9hup4