hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \frac {355 \cdots 544} {755 \cdots 531}       | r = \paren {3 \times 10^n + \paren {\sum_{i \mathop = 2}^{n - 1} 5 \times 10^i} + 44} \Big / \paren {7 \times 10^n + \paren {\sum_{i \mathop = 2}^{n - 1} 5 \times 10^i} + 31}       | c =  }} {{eqn | r = \paren {3 \times 10^n + \paren {500 \times \sum_{i \mathop = 0}^{n - 3} \times 10^i} + 44} \Big / \paren {7 \times 10^n + \paren {500 \times \sum_{i \mathop = 0}^{n - 3} \times 10^i} + 31}       | c =  }} {{eqn | r = \paren {3 \times 10^n + \paren {500 \times \frac {10^{n - 2} - 1} {10 - 1} } + 44} \Big / \paren {7 \times 10^n + \paren {500 \times \frac {10^{n - 2} - 1} {10 - 1} } + 31}       | c = Sum of Geometric Sequence }} {{eqn | r = \dfrac {3 \times 10^n \times \paren {10 - 1} + 500 \times \paren {10^{n - 2} - 1} + 44 \times \paren {10 - 1} } {7 \times 10^n \times \paren {10 - 1} + 500 \times \paren {10^{n - 2} - 1} + 31 \times \paren {10 - 1} }       | c = simplifying }} {{eqn | r = \dfrac {27 \times 10^n + 500 \times 10^{n - 2} - 104} {63 \times 10^n + 500 \times 10^{n - 2} - 221}       | c = simplifying }} {{eqn | r = \dfrac {2700 \times 10^{n - 2} + 500 \times 10^{n - 2} - 104} {6300 \times 10^{n - 2} + 500 \times 10^{n - 2} - 221}       | c = simplifying }} {{eqn | r = \dfrac {3200 \times 10^{n - 2} - 104} {6800 \times 10^{n - 2} - 221}       | c = simplifying }} {{eqn | r = \dfrac {8 \times \paren {400 \times 10^{n - 2} - 13} } {17 \times \paren {400 \times 10^{n - 2} - 13} }       | c = factoring }} {{eqn | r = \dfrac 8 {17}       | c = factoring }} {{eqn | r = \dfrac {8 \times 43} {17 \times 43}       | c =  }} {{eqn | r = \dfrac {344} {731}       | c =  }} {{end-eqn}} {{qed}}
Let $2^n$ have $m$ digits when expressed in decimal notation. By the Basis Representation Theorem and its implications, a positive integer $x$ has $m$ digits {{iff}}: :$10^{m - 1} \le x < 10^m$ Thus: {{begin-eqn}} {{eqn | l = 10^{m - 1}       | o = \le       | m = 2^n       | mo= <       | r = 10^m       | c =  }} {{eqn | ll= \leadsto       | l = m - 1       | o = \le       | m = \log_{10} \paren {2^n}       | mo= <       | r = m       | c =  }} {{eqn | ll= \leadsto       | l = m - 1       | o = \le       | m = n \log_{10} 2       | mo= <       | r = m       | c =  }} {{end-eqn}} Because a power of $2$ cannot equal a power of $10$, it will always be the case that: :$m - 1 < n \log_{10} 2 < m$ and so: :$m - 1 < n \log_{10} 2 \le m$ Hence from Integer equals Ceiling iff Number between Integer and One Less: :$m = \ceiling {n \log_{10} 2}$ {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{k \mathop = n + 1}^{2 n} \paren {2 k - 1}       | r = \sum_{k \mathop = 1}^{2 n} \paren {2 k - 1} - \sum_{k \mathop = 1}^n \paren {2 k - 1}       | c =  }} {{eqn | r = \paren {2 n}^2 - n^2       | c = Odd Number Theorem }} {{eqn | r = 3 n^2       | c =  }} {{eqn | r = 3 \sum_{k \mathop = 1}^n \paren {2 k - 1}       | c = Odd Number Theorem }} {{end-eqn}} The result follows. {{qed}}
We have by definition of the reciprocal of $e$ that: :$\dfrac 1 e \approx 0 \cdotp 36787 \, 94411 \, 71442 \, 32159 \, 55237 \, 70161 \, 46086 \, 74458 \, 11131 \, 031 \ldots$ By inspection: :$\dfrac 1 3 < \dfrac 1 e < \dfrac 1 2$ Thus: :$\dfrac 1 e - \dfrac 1 3 \approx 0 \cdotp 03454 \, 61078 \, 38109 \, 08826 \, 21904 \, 36828 \, 12753 \, 41124 \, 77797 \, 70 \ldots$ Then: :$\dfrac 1 {29} = 0 \cdotp 03448 \, 27586 \, 20689 \, 65517 \, 24137 \, 931 \ldots$ repeating and: :$\dfrac 1 {28} = 0 \cdotp 03571 \, 42857 \, 14285 \ldots$ repeating and so: :$\dfrac 1 e - \dfrac 1 3 - \dfrac 1 {29} \approx 0 \cdotp 00006 \, 33482 \, 17499 \, 43308 \, 97766 \, 33724 \, 67925 \, 82504 \, 08832 \, 19 \cdots$   Thus one can generate this sequence of denominators $\sequence {D_n}$ by: :$D_n = \ceiling {\paren {\dfrac 1 e - \displaystyle \sum_{i \mathop = 0}^{n - 1} \frac 1 {D_i}}^{-1}}$
Let $a$ and $b$ be two integers chosen at random. For $a$ and $b$ to be coprime, it is necessary and sufficient that no prime number divides both of them. The probability that any particular integer is divisible by a prime number $p$ is $\dfrac 1 p$. Hence the probability that both $a$ and $b$ are divisible by $p$ is $\dfrac 1 {p^2}$. The probability that either $a$ or $b$ or both is not divisible by $p$ is therefore $1 - \dfrac 1 {p^2}$. Whether or not $a$ is divisible by $p$ or divisible by another prime number $q$ is independent of both $p$ and $q$. Thus by the Product Rule for Probabilities, the probability that $a$ and $b$ are not both divisible by either $p$ or $q$ is $\paren {1 - \dfrac 1 {p^2} } \paren {1 - \dfrac 1 {q^2} }$. This independence extends to all prime numbers. That is, the probability that $a$ and $b$ are not both divisible by any prime number is equal to the product of $1 - \dfrac 1 {p^2}$ over all prime numbers: :$\map \Pr {a \perp b} = \displaystyle \prod_{\text {$p$ prime} } \paren {1 - \dfrac 1 {p^2} }$ From Sum of Reciprocals of Powers as Euler Product: :$\displaystyle \map \zeta s = \prod_p \frac 1 {1 - p^{-s} }$ from which: :$\displaystyle \dfrac 1 {\map \zeta 2} = \prod_{\text {$p$ prime} } \paren {1 - \dfrac 1 {p^2} }$ where $\map \zeta 2$ is the Riemann $\zeta$ (zeta) function evaluated at $2$. The result follows from Riemann Zeta Function of 2. {{qed}}
Let $a$ be an integer chosen at random. For $a$ to be square-free, it is necessary and sufficient that for all prime numbers $p$, it is not the case that $p^2$ is a  divisor of $a$. The probability that any particular integer is divisible by $p^2$ is $\dfrac 1 {p^2}$. The probability that $a$ is not divisible by $p^2$ is therefore $1 - \dfrac 1 {p^2}$. Whether or not $a$ is divisible by $p^2$ or divisible by $q^2$ for another prime number $q$ is independent of both $p$ and $q$. Thus by the Product Rule for Probabilities, the probability that $a$ is not divisible by either $p^2$ or $q^2$ is $\paren {1 - \dfrac 1 {p^2} } \paren {1 - \dfrac 1 {q^2} }$. This independence extends to all prime numbers. That is, the probability that $a$ is not divisible by the square of any prime number is equal to the product of $1 - \dfrac 1 {p^2}$ over all prime numbers: :$\map \Pr {\neg \exists b \in \Z: b^2 \divides a} = \displaystyle \prod_{\text {$p$ prime} } \paren {1 - \dfrac 1 {p^2} }$ From Sum of Reciprocals of Powers as Euler Product: :$\displaystyle \map \zeta s = \prod_p \frac 1 {1 - p^{-s} }$ from which: :$\displaystyle \dfrac 1 {\map \zeta 2} = \prod_{\text {$p$ prime} } \paren {1 - \dfrac 1 {p^2} }$ where $\map \zeta 2$ is the Riemann $\zeta$ (zeta) function evaluated at $2$. The result follows from Riemann Zeta Function of 2. {{qed}}
=== Sufficient Condition === Let $I$ be a prime ideal. Let $x, y \in \map {\mathit {Ids} } L$ such that :$x \wedge y \precsim I$ By definition of $\precsim$: :$x \wedge y \subseteq I$ By Meet is Intersection in Set of Ideals: :$x \cap y \subseteq I$ {{AimForCont}}: :$x \not \precsim I$ and $y \not \precsim I$ By definition of $\precsim$: $x \nsubseteq I$ and $y \nsubseteq I$ By definition of subset: :$\exists a \in x: a \notin I$ and :$\exists b \in y: b \notin I$ By Meet Precedes Operands: :$a \wedge b \preceq a$ and $a \wedge b \preceq b$ By definition lower set: :$a \wedge b \in x$ and $a \wedge b \in y$ By definitions subset and intersection: :$a \wedge b \in I$ By Characterization of Prime Ideal: :$a \in I$ or $b \in I$ This contradicts $a \notin I$ and $b \notin I$ Hence the assumption: :$x \not \precsim I$ and $y \not \precsim I$ was false. {{qed|lemma}} === Necessary Condition === Let $I$ be prime element in $\struct {\map {\mathit {Ids} } L, \precsim}$ Let $x, y \in S$ such that: :$x \wedge y \in I$ By Lower Closure of Element is Ideal: :$X := x^\preceq$ and $Y := y^\preceq$ are ideals in $L$ By Meet is Intersection in Set of Ideals: :$X \cap Y = X \wedge Y$ We will prove that: :$X \wedge Y \subseteq I$ Let $a \in X \wedge Y$. By definition of intersection: :$a \in X$ and $a \in Y$ By definition of lower closure of element: :$a \preceq x$ and $a \preceq y$ By definition of infimum: :$a \preceq x \wedge y$ Thus by definition of lower set: :$a \in I$ {{qed|lemma}} By definition of $\precsim$: :$X \wedge Y \precsim I$ By definition of prime element: :$X \precsim I$ or $Y \precsim I$ By definition of $\precsim$: :$X \subseteq I$ or $Y \subseteq I$ By definition of reflexivity: :$x \preceq x$ and $y \preceq y$ By definition of lower closure of element: :$x \in X$ and $y \in Y$ Thus by definition of subset: :$x \in I$ or $y \in I$ Hence by Characterization of Prime Ideal: :$I$ is prime ideal. {{qed}}
Let $a$, $b$ and $c$ be three integers chosen at random. For $a$, $b$ and $c$ to be coprime, it is necessary and sufficient that no prime number divides all of them. The probability that any particular integer is divisible by a prime number $p$ is $\dfrac 1 p$. Hence the probability that $a$, $b$ and $c$ are divisible by $p$ is $\dfrac 1 {p^3}$. The probability that at least one of $a$, $b$ and $c$ is not divisible by $p$ is therefore $1 - \dfrac 1 {p^3}$. Whether or not $a$ is divisible by $p$ or divisible by another prime number $q$ is independent of both $p$ and $q$. Thus by the Product Rule for Probabilities, the probability that $a$, $b$ and $c$ are not all divisible by either $p$ or $q$ is $\paren {1 - \dfrac 1 {p^3} } \paren {1 - \dfrac 1 {q^3} }$. This independence extends to all prime numbers. That is, the probability that $a$, $b$ and $c$ are not all divisible by any prime number is equal to the product of $1 - \dfrac 1 {p^3}$ over all prime numbers: :$\map \Pr {\map \perp {a, b, c} } = \displaystyle \prod_{\text {$p$ prime} } \paren {1 - \dfrac 1 {p^3} }$ From Sum of Reciprocals of Powers as Euler Product: :$\displaystyle \map \zeta s = \prod_p \frac 1 {1 - p^{-s} }$ from which: :$\displaystyle \dfrac 1 {\map \zeta 3} = \prod_{\text {$p$ prime} } \paren {1 - \dfrac 1 {p^3} }$ where $\map \zeta 3$ is the Riemann $\zeta$ (zeta) function evaluated at $3$. The result follows. {{qed}}
{{ProofWanted|It remains to be proved that the hexagonal packing is in fact the densest that can be achieved.}} Consider the rectangular area $ABCD$ of the densest packing of circles. :300px Let the radius of one circle be $1$. The length $AB$ is $2$. The length $AC$ is $2 \sqrt 3$. Thus, from Area of Rectangle, the area of $\Box ABCD$ is $4 \sqrt 3$. Within $ABCD$ there is one complete circle and one quarter of each of $4$ other circles. That makes a total of $2$ circles. Thus, from Area of Circle, the area of $ABCD$ which is inside a circle is $2 \pi$. So the density is: {{begin-eqn}} {{eqn | l = \dfrac {\text {Area of Circles} } {\text {Area of Rectangle} }       | r = \dfrac {2 \pi} {4 \sqrt 3}       | c =  }} {{eqn | r = \dfrac {\pi} {2 \sqrt 3}       | c =  }} {{eqn | r = \dfrac {\pi} {12}       | c =  }} {{end-eqn}} as required.
Let $b$ be the base in question. Let $F = \dfrac {101 \, 010 \, 101} {110 \, 010 \, 011}$. Then: :$F = \dfrac {1 + b^2 + b^4 + b^6 + b^8} {1 + b + b^4 + b^7 + b^8}$ It is necessary to prove that for all $k \in \Z_{>0}$: :$F = \dfrac {1 + b^2 + b^4 + b^5 + \cdots + b^{2 k + 2} + b^{2 k + 4} + b^{2 k + 6} } {1 + b + b^4 + b^5 + \cdots + b^{2 k + 2} + b^{2 k + 5} + b^{2 k + 6} }$ This is done by: : multiplying the numerator of one by the denominator of the other and then: : multiplying the denominator of one by the numerator of the other and checking that they are equal. Thus we proceed: {{begin-eqn}} {{eqn | o =        | r = 1 + b^2 + b^4 + b^6 + b^8       | c =  }} {{eqn | o =        | ro= \times       | r = 1 + b + b^4 + b^5 + \cdots + b^{2 k + 2} + b^{2 k + 5} + b^{2 k + 6}       | c =  }} {{eqn | o = =       | r = 1 + b + b^4 + b^5 + \cdots + b^{2 k + 2} + b^{2 k + 5} + b^{2 k + 6}       | c =  }} {{eqn | o =        | ro= +       | r = b^2 + b^3 + b^6 + b^7 + \cdots + b^{2 k + 4} + b^{2 k + 7} + b^{2 k + 8}       | c =  }} {{eqn | o =        | ro= +       | r = b^4 + b^5 + b^8 + b^9 + \cdots + b^{2 k + 6} + b^{2 k + 9} + b^{2 k + 10}       | c =  }} {{eqn | o =        | ro= +       | r = b^6 + b^7 + b^{10} + b^{11} + \cdots + b^{2 k + 8} + b^{2 k + 11} + b^{2 k + 12}       | c =  }} {{eqn | o =        | ro= +       | r = b^8 + b^9 + b^{12} + b^{13} + \cdots + b^{2 k + 10} + b^{2 k + 13} + b^{2 k + 14}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | o =        | r = 1 + b + b^4 + b^7 + b^8       | c =  }} {{eqn | o =        | ro= \times       | r = 1 + b^2 + b^4 + b^5 + \cdots + b^{2 k + 2} + b^{2 k + 4} + b^{2 k + 6}       | c =  }} {{eqn | o = =       | r = 1 + b^2 + b^4 + b^5 + \cdots + b^{2 k + 2} + b^{2 k + 4} + b^{2 k + 6}       | c =  }} {{eqn | o =        | ro= +       | r = b + b^3 + b^5 + b^6 + \cdots + b^{2 k + 3} + b^{2 k + 5} + b^{2 k + 7}       | c =  }} {{eqn | o =        | ro= +       | r = b^4 + b^6 + b^8 + b^9 + \cdots + b^{2 k + 6} + b^{2 k + 8} + b^{2 k + 10}       | c =  }} {{eqn | o =        | ro= +       | r = b^7 + b^9 + b^{11} + b^{12} + \cdots + b^{2 k + 9} + b^{2 k + 11} + b^{2 k + 13}       | c =  }} {{eqn | o =        | ro= +       | r = b^8 + b^{10} + b^{12} + b^{13} + \cdots + b^{2 k + 10} + b^{2 k + 12} + b^{2 k + 14}       | c =  }} {{end-eqn}} Equality can be demonstrated. {{qed}} {{Namedfor|Norman Herbert Anning|cat = Anning}}
By Intersection of Semilattice Ideals is Ideal: :$I_1 \cap I_2 \in \map {\mathit {Ids} } {\mathscr S}$ We will prove that :$I_1 \cap I_2$ is lower bound for $\set {I_1, I_2}$ Let $x \in \set {I_1, I_2}$. Then by definition of unordered tuple: :$x = I_1$ or $x = I_2$ By Intersection is Subset :$I_1 \cap I_2 \subseteq x$ Thus by definition of $\precsim$: :$I_1 \cap I_2 \precsim x$ {{qed|lemma}} We will prove that :$\forall I \in \map {\mathit {Ids} } {\mathscr S}: I$ is lower bound for $\set {I_1, I_2} \implies I \precsim I_1 \cap I_2$ Let $I \in \map {\mathit {Ids} } {\mathscr S}$ such that :$I$ is lower bound for $\set {I_1, I_2}$ By definition of lower bound: :$I \precsim I_1$ and $I \precsim I_2$ By definition of $\precsim$: :$I \subseteq I_1$ and $I \subseteq I_2$ By Intersection is Largest Subset: :$I \subseteq I_1 \cap I_2$ Thus by definition of $\precsim$ :$I \precsim I_1 \cap I_2$ {{qed|lemma}} By definition of infimum: :$\inf \set {I_1, I_2} = I_1 \cap I_2$ Hence by definition of meet: :$I_1 \wedge I_2 = I_1 \cap I_2$ {{qed}}
=== Sufficient Condition === Let $I$ be a prime ideal. Define $\mathcal P\left({X}\right) :\equiv X \ne \varnothing \land \inf X \in I \implies \exists x \in X: x \in I$ where $X$ is subset of $S$. Let $A$ be a non-empty finite subset of $S$. By definition of empty set: :$\mathcal P\left({\varnothing}\right)$ We will prove that :$\forall x \in A, B \subseteq A: \mathcal P\left({B}\right) \implies \mathcal P\left({B \cup \left\{ {x}\right\} }\right)$ Let $x \in A, B \subseteq A$ such that :$\mathcal P\left({B}\right)$ (Induction Hypothesis) Assume that :$B \cup \left\{ {x}\right\} \ne \varnothing$ and $\inf \left({B \cup \left\{ {x}\right\} }\right) \in I$ Case $B = \varnothing$: By Union with Empty Set: :$B \cup \left\{ {x}\right\} = \left\{ {x}\right\}$ By Infimum of Singleton: :$\inf \left\{ {x}\right\} = x$ By definition of singleton: :$x \in \left\{ {x}\right\}$ Thus :$\exists a \in B \cup \left\{ {x}\right\}: a \in I$ {{qed|lemma}} Case $B \ne \varnothing$: By Subset of Finite Set is Finite: :$B$ is finite. By Existence of Non-Empty Finite Infima in Meet Semilattice: :$B$ admits an infimum. By Infimum of Singleton: :$\left\{ {x}\right\}$ admits an infimum. {{begin-eqn}} {{eqn | l = \inf\left({B \cup \left\{ {x}\right\} }\right) | r = \inf\left({\bigcup\left\{ {B, \left\{ {x}\right\} }\right\} }\right) | c = definition of union }} {{eqn | r = \inf \left\{ {\inf B, x}\right\} | c = Infimum of Infima: }} {{eqn | r = \left({\inf B}\right) \wedge x | c = definition of meet }} {{end-eqn}} By Characterization of Prime Ideal: :$\inf B \in I$ or $x \in I$ Case $\inf B \in I$: By Induction Hypothesis: :$\exists a \in B: a \in I$ By definition of union: :$a \in B \cup \left\{ {x}\right\}$ Thus :$\exists a \in B \cup \left\{ {x}\right\}: a \in I$ {{qed|lemma}} Case $x \in I$: By definition of union: :$x \in B \cup \left\{ {x}\right\}$ Thus :$\exists a \in B \cup \left\{ {x}\right\}: a \in I$ {{qed|lemma}} By Induction of Finite Set: :$\mathcal P\left({A}\right)$ Thus the result. {{qed|lemma}} === Necessary Condition === Suppose that :for all non-empty finite subset $A$ of $S: \left({ \inf A \in I \implies \exists a \in A: a \in I}\right)$ Let $x, y \in S$ such that :$x \wedge y \in I$ By Unordered Pair is Finite: :$\left\{ {x, y}\right\}$ is a finite set. By definition of unordered tuple: :$x \in \left\{ {x, y}\right\}$ By definition of non-empty set: :$\left\{ {x, y}\right\}$ is a non-empty set. By definition of meet: :$\inf \left\{ {x, y}\right\} = x \wedge y$ By assumption: :$\exists a \in \left\{ {x, y}\right\}: a \in I$ Thus by definition of unordered tuple: :$x \in I$ or $y \in I$ Hence by Characterization of Prime Ideal: :$I$ is prime ideal. {{qed}}
By Pair is Union of Singletons: :$\left\{ {x, y}\right\} = \left\{ {x}\right\} \cup \left\{ {y}\right\}$ By Singleton is Finite: :$\left\{ {x}\right\}$ and $\left\{ {y}\right\}$ are finite. Thus by Union of Finite Sets is Finite: :$\left\{ {x, y}\right\}$ is finite. {{qed}}
:300px Two adjacent sides $AB$, $BC$ and the diagonal $AC$ of square $ABCD$ form a right triangle. The hypotenuse of triangle $\triangle ABC$ can be found by using Pythagoras's Theorem: :$AC^2 = AB^2 + BC^2$ from which: :$AC^2 = 2$ and so: :$AC = \sqrt 2$ {{qed}}
Let $\tuple {a_0, a_1, \ldots}$ be the continued fraction expansion of $\sqrt 2$. By Continued Fraction Expansion of Root 2: :$\sqrt 2 = \sqbrk {1, \sequence 2} = \sqbrk {1, 2, 2, 2, \ldots}$ From Convergents are Best Approximations, the convergents of $\sqbrk {1, \sequence 2}$ are the best rational approximations of $\sqrt 2$. Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be the numerators and denominators of the continued fraction expansion of $\sqrt 2$. Then $\dfrac {p_n} {q_n}$ is the $n$th convergent of $\sqbrk {1, \sequence 2}$. By Convergents of Simple Continued Fraction are Rationals in Canonical Form, $p_n$ and $q_n$ are coprime for all $n \ge 0$. It remains to show that for all $n \ge 1$: :$Q_n = 2 p_{n - 1}$ :$P_n = q_{n - 1}$ It is sufficient to prove that they satisfy the same recurrence relation. By definition: {{begin-eqn}} {{eqn | l = p_0       | r = a_0 }} {{eqn | r = 1 }} {{eqn | l = q_0       | r = 1 }} {{end-eqn}} so that: :$\tuple {Q_1, P_1} = \tuple {2, 1} = \tuple {2 p_0, q_0}$ {{begin-eqn}} {{eqn | l = p_1       | r = a_0 a_1 + 1 }} {{eqn | r = 1 \times 2 + 1 }} {{eqn | r = 3 }} {{eqn | l = q_1       | r = a_1 }} {{eqn | r = 2 }} {{end-eqn}} so that: :$\tuple {Q_2, P_2} = \tuple {6, 2} = \tuple {2 p_1, q_1}$ {{begin-eqn}} {{eqn | l = p_k       | r = a_k p_{k - 1} + p_{k - 2} }} {{eqn | r = 2 p_{k - 1} + p_{k - 2} }} {{eqn | l = q_k       | r = a_k q_{k - 1} + q_{k - 2} }} {{eqn | r = 2 q_{k - 1} + q_{k - 2} }} {{end-eqn}} The result follows by definition of Pell numbers and Pell-Lucas numbers. {{qed}}
Proof by induction: For all $n \in \Z: n \ge 2$, let $\map P n$ be the proposition: :$p_n q_{n - 1} - p_{n - 1} q_n = \paren {-1}^{n + 1}$ === Basis for the Induction === $\map P 1$ is the case: :$p_0 = a_0, p_1 = a_0 a_1 + 1, q_0 = 1, q_1 = a_1$ So: :$p_1 q_0 - p_0 q_1 = \paren {a_0 a_1 + 1} \times 1 - a_0 a_1 = 1 = \paren {-1}^2$ So $P(2)$ is seen to hold. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$p_k q_{k - 1} - p_{k - 1} q_k = \paren {-1}^{k + 1}$ Then we need to show: :$p_{k + 1} q_k - p_k q_{k + 1} = \paren {-1}^{k + 2}$ === Induction Step === This is our induction step: Consider $\sqbrk {a_0, a_1, a_2, \ldots, a_k, a_{k + 1} }$. Its final numerator and denominator are by definition: :$p_{k + 1} = a_{k + 1} p_k + p_{k - 1}, q_{k + 1} = a_{k + 1} q_k + q_{k - 1}$ Therefore: {{begin-eqn}} {{eqn | l = p_{k + 1} q_k - p_k q_{k + 1}       | r = \paren {a_{k + 1} p_k + p_{k - 1} } q_k - p_k \paren {a_{k + 1} q_k + q_{k - 1} }       | c =  }} {{eqn | r = p_{k - 1} q_k - p_k q_{k - 1}       | c =  }} {{eqn | r = \paren {-1} \paren {p_k q_{k - 1} - p_{k - 1} q_k}       | c =  }} {{eqn | r = \paren {-1} \paren {-1}^{k + 1}       | c = Induction Hypothesis }} {{eqn | r = \paren {-1}^{k + 2}       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \ge 1: p_n q_{n-1} - p_{n-1} q_n = \paren {-1}^{n + 1}$ {{qed}} Category:Continued Fractions adgeudgpif0gfqi87f5hkm6uavpfct8
Let $k \ge 2$. {{begin-eqn}} {{eqn | l = p_k q_{k - 2} - p_{k - 2} q_k       | r = \paren {a_k p_{k - 1} + p_{k - 2} } q_{k - 2} - p_{k - 2} \paren {a_k q_{k - 1} + q_{k - 2} }       | c =  }} {{eqn | r = a_k \paren {p_{k - 1} q_{k - 2} - p_{k - 2} q_{k - 1} }       | c =  }} {{eqn | r = a_k \paren {-1}^k       | c = Difference between Adjacent Convergents of Simple Continued Fraction }} {{end-eqn}} {{qed}} Category:Continued Fractions cr2z76m7r823waql0ap13l9jameiac2
Let $k \ge 1$. Let $d = \gcd \set {p_k, q_k}$. From Common Divisor Divides Integer Combination: :$p_k q_{k - 1} - p_{k - 1} q_k$ is a multiple of $d$. From Difference between Adjacent Convergents of Simple Continued Fraction: :$d \divides \paren {-1}^{k + 1}$ where $\divides$ denotes divisibility. It follows that: :$d = 1$ By: :$q_0 = 1$ :Denominators of Simple Continued Fraction are Strictly Increasing It follows that $q_k > 0$ for all $k \ge 0$. {{qed}} Category:Continued Fractions kpjquz66bpv06cdtbedgi40mne9f7t2
=== Sufficient Condition === Let $F$ be a prime ideal. Define $\map \PP X: \equiv X \ne \O \land \sup X \in F \implies \exists x \in X: x \in F$ where $X$ is subset of $S$. Let $A$ be a non-empty finite subset of $S$. By definition of empty set: :$\map \PP \O$ We will prove that: :$\forall x \in A, B \subseteq A: \map \PP B \implies \map \PP {B \cup \set x}$ Let $x \in A, B \subseteq A$ such that: :$\map \PP B$ (Induction Hypothesis) Assume that: :$B \cup \set x \ne \O$ and $\map \sup {B \cup \set x} \in F$ Case $B = \O$: By Union with Empty Set: :$B \cup \set x = \set x$ By Supremum of Singleton: :$\sup \set x = x$ By definition of singleton: :$x \in \set x$ Thus :$\exists a \in B \cup \set x: a \in F$ {{qed|lemma}} Case $B \ne \O$: By Subset of Finite Set is Finite: :$B$ is finite. By Existence of Non-Empty Finite Suprema in Join Semilattice: :$B$ admits a supremum. By Supremum of Singleton: :$\set x$ admits a supremum. {{begin-eqn}} {{eqn | l = \map \sup {B \cup \set x}       | r = \map \sup {\bigcup \set {B, \set x} }       | c = {{Defof|Set Union}} }} {{eqn | r = \sup \set {\sup B, x}       | c = Supremum of Suprema: }} {{eqn | r = \paren {\sup B} \vee x       | c = {{Defof|Join (Order Theory)}} }} {{end-eqn}} By definition of prime filter: :$\sup B \in F$ or $x \in F$ Case $\sup B \in F$: By Induction Hypothesis: :$\exists a \in B: a \in F$ By definition of union: :$a \in B \cup \set x$ Thus: :$\exists a \in B \cup \set x: a \in F$ {{qed|lemma}} Case $x \in F$: By definition of union: :$x \in B \cup \set x$ Thus: :$\exists a \in B \cup \set x: a \in F$ {{qed|lemma}} By Induction of Finite Set: :$\map \PP A$ Thus the result. {{qed|lemma}} === Necessary Condition === Suppose that :for all non-empty finite subset $A$ of $S: \paren {\sup A \in F \implies \exists a \in A: a \in F}$ Let $x, y \in S$ such that :$x \vee y \in F$ By Unordered Pair is Finite: :$\set {x, y}$ is a finite set. By definition of unordered tuple: :$x \in \set {x, y}$ By definition of non-empty set: :$\set {x, y}$ is a non-empty set. By definition of join: :$\sup \set {x, y} = x \vee y$ By assumption: :$\exists a \in \set {x, y}: a \in F$ Thus by definition of unordered tuple: :$x \in F$ or $y \in F$ Hence $I$ is prime filter. {{qed}}
By Inverse of Reflexive Relation is Reflexive: :$\succeq$ is reflexive. By Inverse of Transitive Relation is Transitive: :$\succeq$ is transitive. Hence $\succeq$ is a preordering. {{qed}}