hcju/proofwikiqa huggingface dataset:
Proof by induction: For all $n \in \Z$ such that $n \ge 10$, let $P \left({n}\right)$ be the proposition: :$2^n > n^3$ We note that: :$2^9 = 512 < 729 = 9^3$ so when $n < 10$ the proposition does not hold. === Basis for the Induction === $P \left({10}\right)$ is the case: :$2^{10} = 1024 > 1000 = 10^3$ so $P \left({10}\right)$ is seen to hold. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 10$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is our induction hypothesis: :$2^k > k^3$ We need to show that: :$2^{k + 1} > \left({k + 1}\right)^3$ === Induction Step === This is our induction step: We note that when $k \ge 10$: {{begin-eqn}} {{eqn | l = \left({1 + \dfrac 1 k}\right)^3       | r = 1 + \frac 3 k + \frac 3 {k^2} + \frac 1 {k^3}       | c = Binomial Theorem }} {{eqn | o = \le       | r = 1 + \frac 3 {10} + \frac 3 {100} + \frac 1 {1000}       | c =  }} {{eqn | r = 1 + \frac {300 + 30 + 1} {1000}       | c =  }} {{eqn | n = 1       | o = <       | r = 2       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = 2^{k + 1}       | r = 2 \times 2^k }} {{eqn | o = >       | r = \left({1 + \dfrac 1 k}\right)^3 2^k       | c = from $(1)$ }} {{eqn | o = >       | r = \left({1 + \dfrac 1 k}\right)^3 k^3       | c = Induction Hypothesis }} {{eqn | r = k^3 + \frac {3 k^3} k + \frac {3 k^3} {k^2} +  \frac {k^3} {k^3} }} {{eqn | r = k^3 + 3 k^2 + 3 k + 1 }} {{eqn | r = \left({k + 1}\right)^3 }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. {{qed}} </onlyinclude>
{{AimForCont}} $\Z$ is a well-ordered set. Then by definition, all subsets of $\Z$ has a smallest element. But take $\Z$ itself. Suppose $x \in \Z$ is a smallest element. Then $x - 1 \in \Z$. But $x - 1 < x$, which contradicts the supposition that $x \in \Z$ is a smallest element. Hence there can be no such smallest element. So by Proof by Contradiction, $\Z$ is not well-ordered by $\le$. {{qed}}
Consider the ordering $\preccurlyeq \subseteq \Z \times \Z$ defined as: :$x \preccurlyeq y \iff \left({\left\vert{x}\right\vert < \left\vert{y}\right\vert}\right) \lor \left({\left\vert{x}\right\vert = \left\vert{y}\right\vert \land x \le y}\right)$ {{finish|It remains to be shown that $\preccurlyeq$ is an ordering, and also that it is a well-ordering.}}
{{AimForCont}} $\R_{\ge 0}$ is a well-ordered set. Then by definition, all subsets of $\R_{\ge 0}$ has a smallest element. Take the subset $\R_{> 0}$: :$\R_{> 0} = \left\{ {x \in \R_{\ge 0}: x > 0}\right\} = \R_{\ge 0} \setminus \left\{ {0}\right\}$ Suppose $x \in \R_{> 0}$ is a smallest element. Then $\dfrac x 2 \in \R_{> 0}$. But $\dfrac x 2 < x$, which contradicts the supposition that $x \in \R_{> 0}$ is a smallest element. Hence there can be no such smallest element. So by Proof by Contradiction, $\R_{\ge 0}$ is not well-ordered by $\le$. {{qed}}
By definition, the digital root base $b$ for $n$ is the single digit resulting from: :adding up the digits in $n$, and expressing the result in base $b$ :adding up the digits in that result, and again expressing the result in base $b$ :repeating until down to one digit. Let $n = d_1 + b d_2 + \dotsb + b^{m-1} d_m$ where, for all $i$, $0 \le d_i < b$. Let $S \left({n}\right)$ be the digit sum of $n$. Then: :$S \left({n}\right) = d_1 + d_2 + \dotsb + d_m$ Thus: :$S \left({n}\right) < n$ unless $d_2, d_3, \dotsb, d_m = 0$ in which case $n$ is a one digit number. Similarly:  :$S \left({S \left({n}\right)}\right) < S \left({n}\right)$ Every time the digit sum is taken, the result is at least one less than the previous digit sum. As $n$ is finite, it will take a finite number of steps to reduce the result to a one digit number. Hence the result. {{qed}} Category:Number Bases gcq9t3x04n76dwoaqu1kohobad3hhmd
By definition of golden mean: :$\phi = \dfrac {1 + \sqrt 5} 2$ By Square Root of Prime is Irrational: :$\sqrt 5$ is irrational. By Rational Number plus Irrational Number is Irrational: :$1 + \sqrt 5$ is irrational. By Irrational Number divided by Rational Number is Irrational: :$\dfrac {1 + \sqrt 5} 2$ is irrational. {{qed}}
=== Sufficient Condition === Let $x \ll y$ Let $X \subseteq S$ such that :$y \preceq \sup X$ Define $F := \left\{ {\sup A: A \in {\it Fin}\left({X}\right)}\right\}$ By definition of union: :$X = \bigcup {\it Fin}\left({X}\right)$ By Supremum of Suprema: :$\sup X = \sup F$ We will prove that :$F$ is directed Let $a, b \in F$ By definition of $F$: :$\exists A \in {\it Fin}\left({X}\right): a = \sup A$ and :$\exists B \in {\it Fin}\left({X}\right): b = \sup B$ By Union of Subsets is Subset: :$A \cup B \subseteq X$ By Finite Union of Finite Sets is Finite: :$A \cup B$ is finite By definition of {\it Fin}: :$A \cup B \in {\it Fin}\left({X}\right)$ By definition of $F$: :$\sup\left({A \cup B}\right) \in F$ By Set is Subset of Union: :$A \subseteq A \cup B$ and $B \subseteq A \cup B$ By Supremum of Subset: :$a \preceq \sup\left({A \cup B}\right)$ and $b \preceq \sup\left({A \cup B}\right)$ Thus by definition :$F$ is directed By definition of way below relation: :$\exists d \in F: x \preceq d$ Thus by definition of $F$: :$\exists A \in {\it Fin}\left({X}\right): x \preceq \sup A$ {{qed|lemma}} === Necessary Condition === Assume that :$\forall X \subseteq S: y \preceq \sup X \implies \exists A \in {\it Fin}\left({X}\right): x \preceq \sup A$ 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: :$\exists h \in D: \forall a \in A: a \preceq h$ By definition: :$h$ is upper bound for $A$ By definition of supremum: :$\sup A \preceq h$ Thus by definition of transitivity: :$x \preceq h$ Thus by definition of way below relation: :$x \ll y$ {{qed}}
Let $x, y \in \R$. From Rational Sequence Decreasing to Real Number, there exist rational sequences $\sequence {x_n}$ and $\sequence {y_n}$ converging to $x$ and $y$, respectively. Then, since Power Function on Strictly Positive Base is Continuous: Real Power: {{begin-eqn}} {{eqn | l = a^{x + y}       | r = a^{\displaystyle \paren {\lim_{n \mathop \to \infty} x_n + \lim_{n \mathop \to \infty} y_n} } }} {{eqn | r = a^{\displaystyle \paren {\lim_{n \mathop \to \infty} \paren {x_n + y_n} } }       | c = Sum Rule for Real Sequences }} {{eqn | r = \lim_{n \mathop \to \infty} a^{x_n + y_n}       | c = Sequential Continuity is Equivalent to Continuity in the Reals }} {{eqn | r = \lim_{n \mathop \to \infty} \paren {a^{x_n} a^{y_n} }       | c = Sum of Indices of Real Number: Rational Numbers }} {{eqn | r = \paren {\lim_{n \mathop \to \infty} a^{x_n} } \paren {\lim_{n \mathop \to \infty} a^{y_n} }       | c = Product Rule for Real Sequences }} {{eqn | r = a^x a^y       | c = Sequential Continuity is Equivalent to Continuity in the Reals }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = a^{x y}       | r = \map \exp {x y \ln a}       | c = {{Defof|Power to Real Number}} }} {{eqn | r = \map \exp {y \, \map \ln {a^x} }       | c = Logarithms of Powers }} {{eqn | r = \paren {a^x}^y       | c = {{Defof|Power to Real Number}} }} {{end-eqn}} {{qed}}
Let $\sequence {x_n}$ denote the sequence defined as: :$\forall n \in \N: x_n = \dfrac {\floor {n x} } n$ where $\floor {n x}$ denotes the floor of $n x$. From Floor Function is Integer, $\floor {n x}$ is an integer. Hence by definition of rational number, $\sequence {x_n}$ is a rational sequence. From Real Number is between Floor Functions: :$n x - 1 < \floor {n x} \le nx $ Thus: :$\dfrac {n x - 1} n < \dfrac {\floor {n x} } n \le x$ Further: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \frac {n x - 1} n       | r = \lim_{n \mathop \to \infty} \paren {x - \frac 1 n} }} {{eqn | r = \lim_{n \mathop \to \infty} x - \lim_{n \mathop \to \infty} \frac 1 n       | c = Combined Sum Rule for Real Sequences }} {{eqn | r = x - 0       | c = Quotient Rule for Limits of Functions }} {{eqn | r = x }} {{end-eqn}} From the Squeeze Theorem for Real Sequences: :$\displaystyle \lim_{n \mathop \to \infty} \frac {\floor {n x} } n = x$ From Peak Point Lemma, there is a monotone subsequence $\sequence {x_{n_k} }$ of $\sequence {x_n}$. We have that $\sequence {x_n}$ is bounded above by $x$. Hence $\sequence {x_{n_k} }$ is increasing. From Limit of Subsequence equals Limit of Real Sequence, $\sequence {x_{n_k} }$ converges to $x$. Hence the result. {{qed}} Category:Limits of Sequences Category:Rational Sequences j50mi1ifpkb72l8s4brjklid4wkap2x
Let $\sequence {x_n}$ denote the sequence defined as: :$\forall n \in \N : x_n = \dfrac {\ceiling {n x} } n$ where $\ceiling {n x}$ denotes the ceiling of $n x$. From Ceiling Function is Integer, $\ceiling {n x}$ is an integer. Hence by definition of rational number, $\sequence {x_n}$ is a rational sequence. From Real Number is between Ceiling Functions: :$n x < \ceiling {n x} \le n x + 1$ Thus: :$x < \dfrac {\ceiling {n x} } n \le \dfrac {n x + 1} n$ Further: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \frac {nx + 1} n       | r = \lim_{n \mathop \to \infty} \paren {x + \frac 1 n} }} {{eqn | r = \lim_{n \mathop \to \infty} x + \lim_{n \mathop \to \infty} \frac 1 n       | c = Combined Sum Rule for Real Sequences }} {{eqn | r = x + 0       | c = Quotient Rule for Limits of Functions }} {{eqn | r = x }} {{end-eqn}} Thus, from the Squeeze Theorem for Real Sequences: :$\displaystyle \lim_{n \mathop \to \infty} \frac {\ceiling {n x} } n = x$ From Peak Point Lemma, there is a monotone subsequence $\sequence {x_{n_k} }$ of $\sequence {x_n}$. We have that $\sequence {x_n}$ is bounded below by $x$. Hence $\sequence {x_{n_k} }$ is decreasing. From Limit of Subsequence equals Limit of Real Sequence, $\sequence {x_{n_k} }$ converges to $x$. Hence the result. {{qed}} Category:Limits of Sequences Category:Rational Sequences b7mtd4vcalkk2zn65niysnv74saecmo
Let $x, y \in \R$ be such that $x < y$. Let $\delta = \dfrac {y - x} 2$. From Rational Sequence Decreasing to Real Number, there is some rational sequence $\sequence {x_n}$ that decreases to $x$. From Rational Sequence Increasing to Real Number, there is some rational sequence $\sequence {y_n}$ that increases to $y$. From Convergent Real Sequence is Bounded: :$\exists N_1 \in \N: n \ge N_1 \implies x - \delta < x_n < x + \delta$ Since $\sequence {x_n}$ is decreasing: :$n \ge N_1 \implies x \le x_n < x + \delta$ From Convergent Real Sequence is Bounded: :$\exists N_2 \in \N : n \ge N_2 \implies y - \delta < y_n < y + \delta$ Since $\sequence {y_n}$ is increasing: :$n \ge N_2 \implies y - \delta < y_n \le y$ Let $N = \max \set {N_1, N_2}$. Then, for $n \ge N$: {{begin-eqn}} {{eqn | l = x       | o = \le       | r = x_n }} {{eqn | o = <       | r = x + \delta }} {{eqn | r = y - \delta }} {{eqn | o = <       | r = y_n }} {{eqn | o = \le       | r = y }} {{end-eqn}}       From Power Function on Strictly Positive Base is Continuous and Sequential Continuity is Equivalent to Continuity in the Reals: :$x_n \to x \implies a^{x_n} \to a^x$ :$y_n \to y \implies a^{y_n} \to a^y$ Also, from Power Function on Base Greater than One is Strictly Increasing: Rational Number: :$\sequence {a^{x_n} }$ decreases to $a^x$ and: :$\sequence {a^{y_n} }$ increases to $a^y$. So, for $n \ge N$: {{begin-eqn}} {{eqn | l = a^x       | o = \le       | r = a^{x_n}       | c = as $\sequence {a^{x_n} }$ decreases to $a^x$ }} {{eqn | o = <       | r = a^{y_n}       | c = Power Function on Base Greater than One is Strictly Increasing: Rational Number }} {{eqn | o = \le       | r = a^y       | c = as $\sequence {a^{y_n} }$ increases to $a^y$ }} {{end-eqn}} Hence the result. {{qed}} Category:Power Function on Base Greater than One is Strictly Increasing dtnnkqxpiiafd566txd5qdaq52eo9sp
By definition, $a^x$ is the unique continuous extension of $a^r$, for rational $r$. By definition, continuous extensions are continuous. Hence the result. {{qed}}
Fix $x \in \R_{>0}$. From Defining Sequence of Natural Logarithm is Strictly Decreasing, $\left\langle{ f_n \left({ x }\right) }\right\rangle$ is  strictly decreasing. From Lower Bound of Natural Logarithm/Proof 3, $\left\langle{ f_n \left({ x }\right) }\right\rangle$ is bounded below. From Monotone Convergence Theorem, $\left\langle{ f_n \left({ x }\right) }\right\rangle$ is convergent Hence the result, by definition of pointwise convergence. {{qed}} Category:Natural Logarithms g613kd5i1wa3ssljn1w1bfk4gqetbha
Let $x = \dfrac p q, y = \dfrac u v$. Then: {{begin-eqn}} {{eqn | l = r^\paren {x + y}       | r = r^\paren {\paren {p / q} + \paren {u / v} }       | c =  }} {{eqn | r = r^\paren {\paren {p v + u q} / q v}       | c =  }} {{eqn | r = \paren {r^\paren {1 / q v} }^\paren {p v + u q}       | c = {{Defof|Rational Power}} }} {{eqn | r = \paren {r^\paren {1 / q v} }^\paren {p v} \times \paren {r^\paren {1 / q v} }^\paren {u q}       | c = Sum of Indices of Real Number: Integers }} {{eqn | r = r^\paren {p v / q v} \times r^\paren {u q / q v}       | c = {{Defof|Rational Power}} }} {{eqn | r = r^\paren {p / q} \times r^\paren {u / v}       | c =  }} {{eqn | r = r^x \times r^y       | c =  }} {{end-eqn}} {{qed}}
Let $x = \dfrac p q, y = \dfrac u v$. Consider $\paren {\paren {r^x}^y}^{q v}$. Then: {{begin-eqn}} {{eqn | l = \paren {\paren {r^x}^y}^{q v}       | r = \paren {\paren {r^\paren {p / q} }^\paren {u / v} }^{q v}       | c =  }} {{eqn | r = \paren {\paren {\paren {r^\paren {p / q} }^\paren {u / v} }^v}^q       | c = Product of Indices of Real Number: Integers }} {{eqn | r = \paren {\paren {r^\paren {p / q} }^u}^q       | c = {{Defof|Rational Power}} }} {{eqn | r = \paren {r^\paren {p / q} }^{q u}       | c = Product of Indices of Real Number: Integers }} {{eqn | r = \paren {\paren {r^\paren {p / q} }^q}^u       | c = Product of Indices of Real Number: Integers }} {{eqn | r = \paren {r^p}^u       | c = {{Defof|Rational Power}} }} {{eqn | r = r^{p u}       | c = Product of Indices of Real Number: Integers }} {{eqn | ll= \leadsto       | l = \paren {\paren {r^\paren {p / q} }^\paren {u / v} }       | r = r^{\paren {p u} / \paren {q v} }       | c = taking $q v$th root of both sides }} {{eqn | ll= \leadsto       | l = \paren {r^x}^y       | r = r^{x y}       | c =  }} {{end-eqn}} {{qed}}
From Sum of Indices of Real Number: Positive Integers, we have that: :$m \in \Z_{\ge 0}: \forall n \in \Z: r^{n + m} = r^n \times r^m$ It remains to be shown that: :$\forall m \in \Z_{<0}: \forall n \in \Z: r^{n + m} = r^n \times r^m$ The proof will proceed by induction on $m$. As $m < 0$ we have that $m = -p$ for some $p \in \Z_{> 0}$. For all $p \in \Z_{>0}$, let $\map P p$ be the proposition: :$\forall n \in \Z: r^{n + \paren {-p} } = r^n \times r^{-p}$ that is: :$\forall n \in \Z: r^{n - p} = r^n \times r^{-p}$ === Basis for the Induction === $\map P 1$ is true, as follows: When $n > 0$: {{begin-eqn}} {{eqn | l = r^{n - 1} \times r       | r = r^n       | c = {{Defof|Integer Power}} }} {{eqn | ll= \leadsto       | l = r^{n - 1}       | r = r^n \dfrac 1 r       | c = multiplying both sides by $\dfrac 1 r$ }} {{eqn | r = r^n \times r^{-1}       | c = Real Number to Negative Power: Integer }} {{end-eqn}} When $n \le 0$: {{begin-eqn}} {{eqn | l = r^{n - 1}       | r = \dfrac {r^n} r       | c = {{Defof|Integer Power}} }} {{eqn | r = r^n \times r^{-1}       | c = Real Number to Negative Power: Integer }} {{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: :$\forall n \in \Z: r^{n - k} = r^n \times r^{- k}$ Then we need to show: :$\forall n \in \Z: r^{n - \paren {k + 1} } = r^n \times r^{-\paren {k + 1} }$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = r^n \times r^{-\paren {k + 1} }       | r = r^n \times \dfrac {r^{-k} } r       | c = {{Defof|Integer Power}} }} {{eqn | r = \paren {r^n \times r^{-k} } \times \dfrac 1 r       | c = Real Multiplication is Associative }} {{eqn | r = r^{n - k} \times \dfrac 1 r       | c = Induction Hypothesis }} {{eqn | r = r^{n - k} \times r^{-1}       | c = Real Number to Negative Power: Integer }} {{eqn | r = r^{n - k - 1}       | c = Basis for the Induction }} {{eqn | r = r^{n - \paren {k + 1} }       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n, m \in \Z: r^{n + m} = r^n \times r^m$ {{qed}}
From Product of Indices of Real Number: Positive Integers, we have that: :$m \in \Z_{\ge 0}: \forall n \in \Z: \paren {r^n}^m = r^{n m}$ It remains to be shown that: :$\forall m \in \Z_{<0}: \forall n \in \Z: \paren {r^n}^m = r^{n m}$ As $m < 0$ we have that $m = -p$ for some $p \in \Z_{> 0}$. Thus: {{begin-eqn}} {{eqn | l = \paren {r^n}^{- p}       | r = \dfrac 1 {\paren {r^n}^p}       | c = Real Number to Negative Power: Integer }} {{eqn | r = \dfrac 1 {r^{n p} }       | c = from $(1)$ }} {{eqn | r = r^{-n p}       | c = Real Number to Negative Power: Integer }} {{end-eqn}} Hence the result, by replacing $-p$ with $m$. {{qed}}
Let $n \in \Z_{\ge 0}$. Then from Real Number to Negative Power: Positive Integer: :$r^{-n} = \dfrac 1 {r^n}$ It remains to show that this holds when $n < 0$. Let $n \in \Z_{<0}$. Then $n = - m$ for some $m \in \Z_{> 0}$. Thus: {{begin-eqn}} {{eqn | l = r^{-m}       | r = \dfrac 1 {r^m}       | c = Real Number to Negative Power: Positive Integer }} {{eqn | ll= \leadsto       | l = \dfrac 1 {r^{-m} }       | r = \dfrac 1 {1 / r^m}       | c = taking reciprocal of both sides }} {{eqn | r = r^m       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 {r^n}       | r = r^{-n}       | c =  }} {{end-eqn}} {{qed}} Category:Powers rcdclehr2jtyzdkxmwo7chsg18957rf
=== Sufficient Condition === Let $x \ll y$ Let $I \in {\it Ids}\left({\mathscr S}\right)$ such that :$y \preceq \sup I$ By definition of ideal: :$I$ is directed and lower. By definition of up-complete: :$I$ admits a supremum. By definition of way below relation: :$\exists i \in I: x \preceq i$ Thus by definition of lower set: :$x \in I$ {{qed|lemma}} === Necessary Condition === Assume that :$\forall I \in {\it Ids}\left({\mathscr S}\right): y \preceq \sup I \implies x \in I$ Let $D$ be a directed subset of $S$ such that :$D$ admits a supremum and $y \preceq \sup D$ By Supremum of Lower Closure of Set: :$D^\preceq$ admits a supremum and $\sup D^\preceq = \sup D$ where $D^\preceq$ denotes the lower closure of $D$. By Lower Closure of Directed Subset is Ideal: :$D^\preceq$ is an ideal. By assumption: :$x \in D^\preceq$ Thus by definition of lower closure of set: :$\exists d \in D: x \preceq d$ Thus by definition of way below relation: :$x \ll y$ {{qed}}
By Directed iff Lower Closure Directed: :$D^\preceq$ is directed. By Lower Closure is Lower Set: :$D^\preceq$ is a lower set. Thus by definition :$D^\preceq$ is an ideal in $\mathscr S$ {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{j \mathop = 0}^{n - 1} j x^{j - 1}       | r = \sum_{j \mathop = 0}^{n - 1} D_x x^j       | c = Power Rule for Derivatives }} {{eqn | r = D_x \sum_{j \mathop = 0}^{n - 1} x^j       | c = Sum Rule for Derivatives }} {{eqn | r = D_x \frac {x^n - 1} {x - 1}       | c = Sum of Geometric Sequence }} {{eqn | r = \frac {\paren {n x^{n - 1} } \paren {x - 1} - \paren 1 \paren {x^n - 1} } {\paren {x - 1}^2}       | c = Quotient Rule for Derivatives }} {{eqn | r = \frac {\paren {n - 1} x^n - n x^{n - 1} + 1} {\paren {x - 1}^2} }} {{eqn | ll= \leadsto       | l = \sum_{j \mathop = 0}^{n - 1} j x^j       | r = x \sum_{j \mathop = 0}^{n - 1} j x^{j - 1}       | c = Multiplication of Numbers Distributes over Addition }} {{eqn | r = x \frac {\paren {n - 1} x^n - n x^{n - 1} + 1} {\paren {x - 1}^2}       | c = by the above result }}     {{eqn | r = \frac {\paren {n - 1} x^{n + 1} - n x^n + x} {\paren {x - 1}^2} }} {{end-eqn}} {{qed}} Category:Geometric Sequences Category:Sum of Geometric Sequence eprp4tj6qovtipfbbjp6fic4ipk970e
Fix $t \in R_{>0}$. Then: {{begin-eqn}} {{eqn | l = n \paren {t^{n + 1} - 1} - \paren {n + 1} \paren {t^n - 1}       | r = \paren {t - 1}^2 \paren {1 + 2 t + 3 t^2 + \ldots + n t^{n - 1} }       | c = Sum of Geometric Sequence: Corollary 2 }} {{eqn | o = >       | r = 0       | c = Product of Real Numbers is Positive iff Numbers have Same Sign }} {{end-eqn}} Thus: :$n \paren {t^{n + 1} - 1} > \paren {n + 1} \paren {t^n - 1}$ Fix $x \in \R_{>0}$. From Power of Positive Real Number is Positive: Rational Number: :$\forall n \in \N : x^{1 / \paren {n \paren {n + 1} } } \in \R_{>0}$ Thus: {{begin-eqn}} {{eqn | l = \forall t \in \R_{>0} : n \paren {t^{n + 1} - 1}       | o = >       | r = \paren {n + 1} \paren {t^n - 1}       | c = by the result obtained above }} {{eqn | ll= \leadsto       | l = n \paren {\paren {x^{1 / \paren {n \paren {n + 1} } } }^{n + 1} - 1}       | o = >        | r = \paren {n + 1} \paren {\paren {x^{1 / \paren {n \paren {n + 1} } } }^n - 1 }       | c = as $x^{1 / \paren {n \paren {n + 1} } } \in \R_{>0}$ }} {{eqn | ll= \leadsto       | l = n \paren {x^{1 / n} - 1}       | o = >       | r = \paren {n + 1} \paren {x^{1 / \paren {n + 1} } - 1}       | c = Product of Indices of Real Number: Rational Numbers }} {{eqn | ll= \leadsto       | l = n \paren {\sqrt [n] x - 1}       | o = >       | r = \paren {n + 1} \paren {\map {f_n} x - 1 }       | c = {{Defof|Definition:Root (Analysis)|$n$th root}} }} {{end-eqn}} Hence the result, by definition of strictly decreasing sequence. {{qed}} Category:Natural Logarithms iyqyhorbip7hkcij2oytplaw2931uc1
{{AimForCont}} $\log_{10} 2$ is rational. Then: {{begin-eqn}} {{eqn | l = \log_{10} 2       | r = \frac p q       | c = for some $p, q \in \Z_{\ne 0}$ }} {{eqn | ll= \leadsto       | l = 2       | r = 10^{p / q}       | c = {{Defof|General Logarithm}} }} {{eqn | ll= \leadsto       | l = 2^q       | r = 10^p       | c = raising both sides to the power of $q$ }} {{end-eqn}} Both $10^p$ and $2^q$ are integers, by construction. But $10^p$ is divisible by $5$, while $2^p$, which has only $2$ as a prime factor, is not. So $10^p \ne 2^q$. So, by Proof by Contradiction, it follows that $\log_{10} 2$ is irrational. {{qed}}
Because $5$ is a divisor of $10$ but not $2$, it cannot be the case that $2^a = 10^b$ for $a, b \in \Z_{>0}$. Hence this is a special case of Irrationality of Logarithm. {{qed}}
From Bernoulli's Inequality: :$\left({1 + y}\right)^n \ge 1 + n y$ which holds for: :$y \in \R$ where $y > -1$ :$n \in \Z_{\ge 0}$ Thus it holds for $y \in \R_{> 0}$ and  $n \in \Z_{> 0}$. So: {{begin-eqn}} {{eqn | l = 1 + n y       | o = \le       | r = \left({1 + y}\right)^n       | c =  }} {{eqn | ll= \leadsto       | l = 1 + n \frac x n       | o = \le       | r = \left({1 + \frac x n}\right)^n       | c = substituting $y = \dfrac x n$ }} {{eqn | ll= \leadsto       | l = 1 + x       | o = \le       | r = \left({1 + \frac x n}\right)^n       | c =  }} {{eqn | ll= \leadsto       | l = \sqrt [n] {1 + x}       | o = \le       | r = 1 + \dfrac x n       | c = Root is Strictly Increasing }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \log_x y \log_y x       | r = \log_y y       | c = Change of Base of Logarithm }} {{eqn | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \log_y x       | r = \dfrac 1 {\log_x y}       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = s \ln x       | r = \ln {x^s}       | c = Logarithm of Power }} {{eqn | o = \le       | r = x^s - 1       | c = Upper Bound of Natural Logarithm }} {{eqn | o = \le       | r = x^s       | c = }} {{end-eqn}} The result follows by dividing both sides by $s$. {{qed}}
Put $y = \dfrac 1 x$ in the Powers Drown Logarithms. {{qed}}
{{begin-eqn}} {{eqn | l = \map {\log_b} {\log_b x}       | r = \map {\log_b} {\frac {\ln x} {\ln b} }       | c = Change of Base of Logarithm }} {{eqn | r = \map {\log_b} {\ln x} - \map {\log_b} {\ln b}       | c = Difference of Logarithms }} {{eqn | r = \frac {\map \ln {\ln x} } {\ln b} - \frac {\map \ln {\ln b} } {\ln b}       | c = Change of Base of Logarithm }} {{eqn | r = \frac {\map \ln {\ln x} - \map \ln {\ln b} } {\ln b}       | c = simplifying }} {{end-eqn}} {{qed}}