hcju/proofwikiqa huggingface dataset:
We will use a proof by induction on the length $n$. For all $n \in \Z_{>0}$, let $\map P n$ be the proposition: :$\sqbrk {a_0, a_1, \ldots, a_n} = \dfrac {p_n} {q_n}$ === Basis for the Induction === $\map P 0$ is the case:  :$\sqbrk {a_0} = \dfrac {a_0} 1 = \dfrac {p_0} {q_0}$ This holds for any continued fraction. $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \sqbrk {a_0, a_1}       | r = a_0 + \frac 1 {a_1} }} {{eqn | r = \frac {a_0 a_1 + 1} {a_1} }} {{eqn | r = \frac {p_1} {q_1} }} {{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: :$\sqbrk {a_0, a_1, \ldots, a_k} = \dfrac {p_k} {q_k}$ Then we need to show: :$\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} } = \dfrac {p_{k + 1} } {q_{k + 1} }$ === Induction Step === This is our induction step: Consider the continued fraction: :$\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} }$ The numerators are: :$p_0, p_1, \ldots, p_k, p_{k + 1}$ and the denominators are: :$q_0, q_1, \ldots, q_k, q_{k + 1}$ By definition of value of a finite continued fraction: :$\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} } = \sqbrk {a_0, a_1, \ldots, a_{k - 1}, a_k'}$ where $a_k' = a_k + \dfrac 1 {a_{k + 1} }$ Consider the {{RHS}}. Take the continued fraction: :$\sqbrk {a_0, a_1, \ldots, a_{k - 1}, a_k'}$ Its numerators are: :$p_0, p_1, \ldots, p_{k-1}$ and $p_k'$ where $p_k' = \paren {a_k + \dfrac 1 {a_{k + 1} } } p_{k - 1} + p_{k - 2}$ by definition. Its denominators are: :$q_0, q_1, \ldots, p_{k - 1}$ and $q_k'$ where $q_k' = \paren {a_k + \dfrac 1 {a_{k + 1} } } q_{k - 1} + q_{k - 2}$ by definition. As it has just $k$ partial quotients, the induction hypothesis tells us that its value is: :$\dfrac {p_k'} {q_k'}$ So: {{begin-eqn}} {{eqn | l = \sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} }       | r = \frac {p_k'} {q_k'}       | c =  }} {{eqn | r = \frac {\paren {a_k + \frac 1 {a_{k + 1} } } p_{k - 1} + p_{k - 2} } {\paren {a_k + \frac 1 {a_{k + 1} } } q_{k - 1} + q_{k - 2} }       | c =  }} {{eqn | r = \frac {a_{k + 1} \paren {a_k p_{k - 1} + p_{k - 2} } + p_{k - 1} } {a_{k + 1} \paren {a_k q_{k - 1} + q_{k - 2} } + q_{k - 1} }       | c =  }} {{eqn | r = \frac {a_{k + 1} p_k + p_{k - 1} } {a_{k + 1} q_k + q_{k - 1} }       | c =  }} {{eqn | r = \frac {p_{k + 1} } {q_{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: :$\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} } = \dfrac {p_{k + 1} } {q_{k + 1} }$ {{qed}} Category:Continued Fractions Category:Proofs by Induction koba2avyl8b5h1kyqcgccqhkffnryi2
We need to show that for any SICF its sequence of convergents $\sequence {C_n}$ always tends to a limit. Several techniques can be used here, but a quick and easy one is to show that $\sequence {C_n}$ is a Cauchy sequence. Let $\epsilon > 0$. For $m > n \ge \max \set {5, \dfrac 1 \epsilon}$: {{begin-eqn}} {{eqn | l = \size {C_m - C_n}       | o = \le       | r = \size {C_m - C_{m - 1} } + \cdots + \size {C_{n + 1} - C_n}       | c = Triangle Inequality }} {{eqn | r = \frac 1 {q_m q_{m - 1} } + \cdots + \frac 1 {q_{n + 1} q_n}       | c = Difference between Adjacent Convergents of Simple Continued Fraction }} {{eqn | o = <       | r = \frac 1 {m \paren {m - 1} } + \cdots + \frac 1 {\paren {n + 1} n}       | c = Lower Bounds for Denominators of Simple Continued Fraction }} {{eqn | r = \sum_{k \mathop = n}^{m - 1} \frac 1 {k \paren {k + 1} } }} {{eqn | r = \sum_{k \mathop = n}^{m - 1} \paren {\frac 1 k - \frac 1 {k + 1} } }} {{eqn | r = \frac 1 n - \frac 1 m       | c = Telescoping Series/Example 1 }} {{eqn | o = <       | r = \frac 1 n }} {{eqn | o = \le       | r = \epsilon }} {{end-eqn}} So $\sequence {C_n}$ is indeed a Cauchy sequence. Hence the result. {{qed}} Category:Continued Fractions rv4r4kw0uqzbis22s4urrf4nk1jzvbp
{{questionable|this needs Correspondence between Irrational Numbers and Simple Infinite Continued Fractions}} Immediate. Note that: :$\size {x - C_n} < \size {C_{n + 1} - C_n} = \dfrac 1 {q_n q_{n + 1} }$ {{qed}} Category:Continued Fractions 3uolekaum9vp63qh9xj7w2yks1jnw2a
The proof proceeds by induction. For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition: :$\dfrac {p_{n + 1} } {q_{n + 1} } = \dfrac {p_n + 2 q_n} {p_n + q_n}$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \dfrac {p_2} {q_2}       | r = \dfrac 3 2       | c =  }} {{eqn | r = \dfrac {1 + 2 \times 1} {1 + 1}       | c =  }} {{eqn | r = \dfrac {p_1 + 2 \times q_1} {p_1 + q_1}       | c =  }} {{end-eqn}} Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown 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 the induction hypothesis: :$\dfrac {p_{k + 1} } {q_{k + 1} } = \dfrac {p_k + 2 q_k} {p_k + q_k}$ from which it is to be shown that: :$\dfrac {p_{k + 2} } {q_{k + 2} } = \dfrac {p_{k + 1} + 2 q_{k + 1} } {p_{k + 1} + q_{k + 1} }$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \dfrac {p_{k + 2} } {q_{k + 2} }       | r = \dfrac {2 p_{k + 1} + p_k} {2 q_{k + 1} + q_k}       | c = {{Defof|Numerators and Denominators of Continued Fraction}} }} {{eqn | r = \dfrac {2 \paren {p_k + 2 q_k} + p_k} {2 \paren {p_k + q_k} + q_k}       | c = Induction Hypothesis }} {{eqn | r = \dfrac {3 p_k + 4 q_k} {2 p_k + 3 q_k}       | c =  }} {{eqn | r = \dfrac {\paren {p_k + 2 q_k} + 2 \paren {p_k + q_k} } {\paren {p_k + q_k} + \paren {p_k + 2 q_k} }       | c =  }} {{eqn | r = \dfrac {p_{k + 1} + 2 q_{k + 1} } {p_{k + 1} + q_{k + 1} }       | c = Induction Hypothesis }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{>0}: \dfrac {p_{n + 1} } {q_{n + 1} } = \dfrac {p_n + 2 q_n} {p_n + q_n}$ {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$P_n$ has the same parity as $n$. === Basis for the Induction === $P \left({0}\right)$ is the case: :$P_0 = 0$ which is even. $P \left({1}\right)$ is the case: :$P_1 = 1$ which is odd. Thus $P \left({n}\right)$ holds for $n = 0$ and $n = 1$. These comprise the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({j}\right)$ is true for all $j \le k$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\forall j \le k: P_j$ has the same parity as $j$ from which it is to be shown that: :$P_{k + 1}$ has the same parity as $k + 1$. === Induction Step === This is the induction step: By definition of Pell numbers: :$P_{k + 1} = 2 P_k + P_{k - 1}$ $2 P_k$ is even by definition. By Sum of Even Integers is Even, if $P_{k - 1}$ is even then so is $P_{k + 1}$. Similarly, if $P_{k - 1}$ is odd then so is $P_{k + 1}$. Thus the parity of $P_{k + 1}$ matches the parity of $P_{k - 1}$. But by the induction hypothesis, the parity of $P_{k - 1}$ matches $k - 1$. We have that: :$k + 1 = 2 + k - 1$ and so the parity of $k + 1$ matches the parity of $k - 1$. Thus the parity of $P_{k + 1}$ matches $k + 1$. So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 0}: P_n$ has the same parity as $n$. {{qed}} Category:Pell Numbers t7sq0mgq4zfifnhkla1415qrce7jefg
First the parity of the numerators of the terms of $\sequence S$ is established. Let $\dfrac {p_n} {q_n}$ be a general term of $\sequence S$. By Relation between Adjacent Best Rational Approximations to Root 2: :$p_{n + 1} = p_n + 2 q_n$ Thus if $p_n$ is odd then so is $p_{n + 1}$. But $p_1 = 1$ is odd. So $p_n$ is odd for all $n$, by Principle of Mathematical Induction. The denominators are the Pell numbers. The result follows from Parity of Pell Numbers. {{qed}} Category:Square Root of 2 i9dhdw5j9ztimw3f2vzovtljpv3jeui
From Parity of Best Rational Approximations to Root 2: :The numerators of the terms of $\sequence S$ are all odd. :For all $n$, the parity of the denominator of term $S_n$ is the same as the parity of $n$. Thus it follows that every other term of $\sequence S$ has a numerator and a denominator which are both odd. {{ProofWanted|To be completed once several other results about Pell numbers established. Proof by {{AuthorRef|Artemas Martin}} establishes the triple $2 P_n P_{n + 1}, P_{n + 1}^2 - P_n^2, P_{n + 1}^2 + P_n^2 + P_{2 n + 1}$.}}
=== Sufficient Condition === Let $X$ be a prime ideal in $L$. Then :$X$ is an ideal in $L$. By Ideal is Filter in Dual Ordered Set: :$X$ is a filter in $L^{-1}$. Let $x, y \in S$ such that :$x \vee' y \in X$ where $\vee'$ denotes the join in $L^{-1}$. By Join is Dual to Meet: :$x \wedge y \in X$ Thus by Characterization of Prime Ideal: :$x \in X$ or $y \in X$ Hence $X$ is a prime filter {{qed|lemma}} === Necessary Condition === This follows by mutatis mutandis. {{qed}}
=== Sufficient Condition === Let $X$ be ideal in $P$. By definition of ideal in ordered set: :$X$ is non-empty directed lower. By definition of directed: :$\forall x, y \in X: \exists z \in X: x \preceq z \land y \preceq z$ Then :$\forall x, y \in X: \exists z \in X: z \succeq x \land z \succeq y$ By definition :$X$ is filtered in $P^{-1}$ By definition of lower set: :$\forall x \in X, y \in S: y \preceq x \implies y \in X$ Then :$\forall x \in X, y \in S: x \succeq y \implies y \in X$ By definition :$X$ is an upper set in $P^{-1}$. Thus by definition of filter in ordered set :$X$ is a filter in $P^{-1}$. {{qed|lemma}} === Necessary Condition === Let $X$ be a filter in $P^{-1}$. By definition of filter in ordered set :$X$ is non-empty, filtered in $P^{-1}$, and upper in $P^{-1}$. By definition of filtered in $P^{-1}$:  :$\forall x, y \in X: \exists z \in X: z \succeq x \land z \succeq y$ Then :$\forall x, y \in X: \exists z \in X: x \preceq z \land y \preceq z$ By definition :$X$ is directed. By definition of upper set in $P^{-1}$: :$\forall x \in X, y \in S: x \succeq y \implies y \in X$ Then :$\forall x \in X, y \in S: y \preceq x \implies y \in X$ By definition :$X$ is a lower set. Thus by definition of ideal in ordered set: :$X$ is an ideal in $P$. {{qed}}
Denote the numerators of the terms of $S$ as $\sequence {N_n}$. Denote the denominators of the terms of $S$ as $\sequence {D_n}$. From the definition: :$\sequence {N_n}$ is a Beatty sequence, where $\sequence {N_n} = \BB_{\sqrt 2} = \sequence{\floor{n \sqrt 2} }_{n \mathop \in \Z_{> 0} }$ :$\sequence {N_n}$ and $\sequence {D_n}$ are complementary Beatty sequences. Then by Beatty's Theorem, $\sequence {D_n}$ is a Beatty sequence. Define $\sequence {D_n} = \BB_y = \sequence{\floor{n y} }_{n \mathop \in \Z_{> 0} }$. Then we have:  {{begin-eqn}} {{eqn | l = \dfrac 1 {\sqrt 2} + \dfrac 1 y       | r = 1       | c = Beatty's Theorem }} {{eqn | ll = \leadsto       | l = \dfrac 1 y       | r = 1 - \dfrac 1 {\sqrt 2} }} {{eqn | r = \dfrac {\sqrt 2 - 1} {\sqrt 2} }} {{eqn | ll = \leadsto       | l = y       | r = \dfrac {\sqrt 2} {\sqrt 2 - 1}  }} {{eqn | r = \dfrac {\sqrt 2 \paren {\sqrt 2 + 1} } {\paren {\sqrt 2 - 1} \paren {\sqrt 2 + 1} } }} {{eqn | r = 2 + \sqrt 2 }} {{end-eqn}} So we have $s_n = \dfrac {N_n} {D_n} = \dfrac {\floor {n \sqrt 2} } {\floor {n \paren {2 + \sqrt 2} } }$. The difference between the numerator and denominator of $s_n$ is $\floor {n \paren {2 + \sqrt 2} } - \floor {n \sqrt 2}$. From Integer equals Floor iff Number between Integer and One More: {{begin-eqn}} {{eqn | l = n \paren {2 + \sqrt 2}        | o = <       | m = \floor {n \paren {2 + \sqrt 2} }       | mo = <       | r = n \paren {2 + \sqrt 2} + 1 }} {{eqn | l = n \sqrt 2       | o = <       | m = \floor {n \sqrt 2}       | mo = <       | r = n \sqrt 2 + 1 }} {{eqn | ll = \leadsto       | l = n \paren {2 + \sqrt 2} - \paren {n \sqrt 2 + 1}       | o = <       | m = \floor {n \paren {2 + \sqrt 2} } - \floor {n \sqrt 2}       | mo = <       | r = n \paren {2 + \sqrt 2} + 1 - n \sqrt 2 }} {{eqn | ll = \leadsto       | l = 2 n - 1       | o = <       | m = \floor {n \paren {2 + \sqrt 2} } - \floor {n \sqrt 2}       | mo = <       | r = 2 n + 1 }} {{end-eqn}} The only integer strictly between $2 n - 1$ and $2 n + 1$ is $2 n$, and $\floor {n \paren {2 + \sqrt 2} } - \floor {n \sqrt 2}$ is an integer. Therefore $\floor {n \paren {2 + \sqrt 2} } - \floor {n \sqrt 2} = 2 n$. {{qed}} {{Namedfor|Roland Percival Sprague|cat = Sprague, Roland}}
{{begin-eqn}} {{eqn | l = \map {f'} x       | r = \frac \d {\d x} x^{1/x} }} {{eqn | r = \frac \d {\d x} e^{\ln x / x} }} {{eqn | r = e^{\ln x / x} \paren {\frac 1 {x^2} - \frac {\ln x} {x^2} } }} {{eqn | r = \frac {x^{1/x} } {x^2} \paren {1 - \ln x} }} {{end-eqn}} $\dfrac {x^{1/x} } {x^2}$ is always greater than $0$. Therefore: :$\map {f'} x > 0$ for $\ln x < 1$ :$\map {f'} x = 0$ for $\ln x = 1$ :$\map {f'} x < 0$ for $\ln x > 1$ By Derivative at Maximum or Minimum, maximum is obtained when $\ln x = 1$, that is, when $x = e$. {{qed}} {{Namedfor|Jakob Steiner|cat = Steiner}}
Follows directly from Straight Lines Subtending Two Consecutive Angles in Regular Pentagon cut in Extreme and Mean Ratio. {{qed}} {{ProofWanted|Flesh this out}}
By Dual of Dual Ordering: :dual of $P^{-1}$ is $P$. Hence by Ideal is Filter in Dual Ordered Set: :the result fallows. {{qed}}
By Dual of Dual Ordering: :dual of $L^{-1}$ is $L$. Hence by Prime Ideal is Prime Filter in Dual Lattice: :the result follows. {{qed}}
The proof proceeds by induction. Listing the first few convergents, which can be calculated: :$C_1 = \dfrac 1 1$ :$C_2 = \dfrac 2 1$ :$C_3 = \dfrac 3 2$ :$C_4 = \dfrac 5 3$ and so on. For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition: :$C_n = \dfrac {F_{n + 1} } {F_n}$ $\map P 1$ is the case: :$C_1 = \dfrac {F_{n + 1} } {F_n}$ {{begin-eqn}} {{eqn | l = C_1       | r = \dfrac 1 1       | c =  }} {{eqn | r = \dfrac {F_2} {F_1}       | c = {{Defof|Fibonacci Numbers}} }} {{end-eqn}} Thus $\map P 1$ is seen to hold. === Basis for the Induction === $\map P 2$ is the case: {{begin-eqn}} {{eqn | l = C_2       | r = 1 + \cfrac 1 1       | c =  }} {{eqn | r = \dfrac 2 1       | c =  }} {{eqn | r = \dfrac {F_3} {F_2}       | c = {{Defof|Fibonacci Numbers}} }} {{end-eqn}} Thus $\map P 2$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown 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 the induction hypothesis: :$C_k = \dfrac {F_{k + 1} } {F_k}$ from which it is to be shown that: :$C_{k + 1} = \dfrac {F_{k + 2} } {F_{k + 1} }$ === Induction Step === This is the induction step: Let $C_n$ be expressed as $\dfrac {p_n} {q_n}$ for any given $n$. {{begin-eqn}} {{eqn | l = C_{k + 1}       | r = \dfrac {p_{k + 1} } {q_{k + 1} }       | c =  }} {{eqn | r = \dfrac {p_k + p_{k - 1} } {q_k + q_{k - 1} }       | c = {{Defof|Numerators and Denominators of Continued Fraction}} }} {{eqn | r = \dfrac {F_{k + 1} + F_k} {F_k + F_{k - 1} }       | c = Induction Hypothesis }} {{eqn | r = \dfrac {F_{k + 2} } {F_{k + 1} }       | c = {{Defof|Fibonacci Numbers}} }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{>0}: C_n = \dfrac {F_{n + 1} } {F_n}$ {{qed}}
=== Sufficient Condition === Let $F$ be prime. Let $x \in S$. By definition of Boolean lattice: :$x \vee \neg x = \top$ where $\top$ denotes the top of $B$. By definition of non-empty set: :$\exists y: y \in F$ By definition of greatest element: :$y \preceq \top$ By definition of upper set: :$\top \in F$ Thus by definition of prime filter: :$x \in F$ or $\neg x \in F$ {{qed|lemma}} === Necessary Condition === Assume that :$\forall x \in S: x \in F \lor \left({\neg x}\right) \in F$ Let $a, b \in S$ such that :$a \vee b \in F$ Aiming for a contradiction suppose that :$a \notin F$ and $b \notin F$ By assumption: :$\neg a \in F$ and $\neg b \in F$ By Filtered in Meet Semilattice: :$\left({\neg a}\right) \wedge \left({\neg b}\right) \in F$ By De Morgan's Laws (Boolean Algebras): :$\neg\left({a \vee b}\right) \in F$ By Filtered in Meet Semilattice: :$\left({a \vee b}\right) \wedge \neg\left({a \vee b}\right) \in F$ By definition of Boolean lattice: :$\bot \in F$ where $\bot$ denotes the bottom of $B$. By definition of smallest element: :$\bot \preceq a$ By definition of upper set: :$a \in F$ This contradicts $a \notin F$ Thus by Proof by Contradiction :the result holds. {{qed}}
We have been given that $r > 1$. Let $\dfrac 1 r + \dfrac 1 s = 1$. Then: :$s = \dfrac r {r - 1}$ It is to be shown that every positive integer lies in exactly one of the two Beatty sequences $\BB_r$ and $\BB_s$. Consider the ordinal positions occupied by all the fractions $\dfrac j r$ and $\dfrac k s$ when they are jointly listed in nondecreasing order for positive integers $j$ and $k$. {{AimForCont}} that $\dfrac j r = \dfrac k s$ for some $j, k \in \Z_{>0}$. Then: :$\dfrac r s = \dfrac j k$ which is rational. But also: :$\dfrac r s = r \paren {1 - \dfrac 1 r} = r - 1$ which is not rational. Therefore, no two of the numbers occupy the same position. Consider some $\dfrac j r$. There are $j$ numbers $\dfrac i r \le \dfrac j r$. There are also $\floor {\dfrac {j s} r}$ numbers $\dfrac k s \le \dfrac j r$. So the position of $\dfrac j r$ in the list is $j + \floor {\dfrac {j s} r}$. The equation $\dfrac 1 r + \dfrac 1 s = 1$ implies: :$j + \floor {\dfrac {j s} r} = j + \floor {j \paren {s - 1} } = \floor {j s}$ Likewise, the position of $\dfrac k s$ in the list is $\floor {k r}$. It is concluded that every positive integer corresponding to every position in the list is of the form $\floor {n r}$ or of the form $\floor {n r}$, but not both. The converse statement is also true: if $p$ and $q$ are two real numbers such that every positive integer occurs precisely once in the above list, then $p$ and $q$ are irrational and the sum of their reciprocals is $1$. {{qed}}
=== Collisions === {{AimForCont}} there exist integers $j > 0, k, m$ such that: :$j = \floor {k \cdot r} = \floor {m \cdot s}$ This is equivalent to the inequalities: :$j \le k \cdot r < j + 1$ and: :$j \le m \cdot s < j + 1$ As $r$ and $s$ are irrational, equality cannot happen. So: :$j < k \cdot r < j + 1$ and: :$j < m \cdot s < j + 1$ which leads to: :$\dfrac j r < k < \dfrac {j + 1} r$ and: :$\dfrac j s < m < \dfrac {j + 1} s$ Adding these together and using the by hypothesis: :$j < k + m < j + 1$ Thus there is an integer strictly between two adjacent integers. This is impossible. Thus the supposition must be false. {{qed|lemma}} === Anti-collisions === {{AimForCont}} that there exist integers $j > 0, k, m$ such that: :$k \cdot r < j$ and: :$j + 1 \le \paren {k + 1} \cdot r$ and: :$m \cdot s < j$ and: :$j + 1 \le \paren {m + 1} \cdot s$ Since $j + 1 \ne 0$, and $r$ and $s$ are irrational, equality cannot happen. So: :$k \cdot r < j$ and: :$j + 1 < \paren {k + 1} \cdot r$ and: :$m \cdot s < j$ and: :$j + 1 < \paren {m + 1} \cdot s$ Then: :$k < \dfrac j r$: and: :$\dfrac {j + 1} r < k + 1$ and: :$m < \dfrac j s$ and: :$\dfrac {j + 1} s < m + 1$ Adding corresponding inequalities: :$k + m < j$ and: :$j + 1 < k + m + 2$ :$k + m < j < k + m + 1$ which is also impossible. Thus the supposition is false. {{qed}}
=== Sufficient Condition === Let us assume :$F$ is a proper subset of $S$ and $F$ is a prime filter in $B$. Thus :$F$ is a proper subset of $S$. Let $G$ be a filter in $B$ such that :$F \subseteq G$ and $F \ne G$. By definitions of subset and set equality: :$\exists x: x \in G \land x \notin F$ By definition of Boolean algebra: :$x \vee \neg x = \top$ where $\top$ denotes the top of $B$. By Top in Filter: :$\top \in F$ By definition of prime filter: :$\neg x \in F$ By definition of subset: :$\neg x \in G$ By Filtered in Meet Semilattice: :$x \wedge \neg x \in G$ By definition of Boolean algebra: :$\bot \in G$ Thus by definition of subset: :$G \subseteq S$ By definition of set equality it remains to prove that :$S \subseteq G$ Let $y \in S$. By definition of smallest element: :$\bot \preceq y$ Thus by definition of upper set: :$y \in G$ {{qed|lemma}} === Necessary Condition === Let $F$ be ultrafilter on $B$. Thus by definition of ultrafilter: :$F$ is proper subset of $S$. Let $x \in S$. Aiming for a contradiction suppose that :$x \notin F$ and $\neg x \notin F$ By Finite Infima Set and Upper Closure is Smallest Filter: :$F \cup \left\{ {x}\right\} \subseteq {\operatorname{fininfs}\left({F \cup \left\{ {x}\right\} }\right)}^\succeq$ By Set is Subset of Union: :$\left\{ {x}\right\} \subseteq F \cup \left\{ {x}\right\}$ and $F \subseteq F \cup \left\{ {x}\right\}$ By definition of singleton: :$x \in \left\{ {x}\right\}$ By definition of subset: :$x \in {\operatorname{fininfs}\left({F \cup \left\{ {x}\right\} }\right)}^\succeq$ By Finite Infima Set and Upper Closure is Filter: :${\operatorname{fininfs}\left({F \cup \left\{ {x}\right\} }\right)}^\succeq$ is filter in $L$. By Subset Relation is Transitive: :$F \subseteq {\operatorname{fininfs}\left({F \cup \left\{ {x}\right\} }\right)}^\succeq$ By definition of ultrafilter: :${\operatorname{fininfs}\left({F \cup \left\{ {x}\right\} }\right)}^\succeq = S$ By Finite Subset Bounds Element of Finite Infima Set and Upper Closure: :$\exists a \in S: a \in F \land \neg x \succeq a \wedge \inf \left\{ {x}\right\}$ By definition of greatest element: :$a \preceq \top$ {{begin-eqn}} {{eqn | l = a | r = a \wedge \top | c = Preceding iff Meet equals Less Operand }} {{eqn | r = a \wedge \left({x \vee \neg x}\right) | c = definition of Boolean algebra }} {{eqn | r = a \wedge x \vee a \wedge \neg x | c = definition of distributive lattice }} {{end-eqn}} By Infimum of Singleton: :$\inf \left\{ {x}\right\} = x$ By Meet Precedes Operands: :$a \wedge \neg x \preceq \neg x$ By definition of infimum: :$a \preceq \neg x$ By definition of upper set: :$\neg x \in F$ This contradicts $\neg x \notin F$ Thus by Proof by Contradiction: :$F$ is a prime filter by Filter is Prime iff For Every Element Element either Negation Belongs to Filter in Boolean Lattice. {{qed}}
From Beatty's Theorem, the Beatty sequences $\mathcal B_r$ and $\mathcal B_s$ are complementary {{iff}}: :$\dfrac 1 r + \dfrac 1 s = 1$ It remains to be demonstrated that this holds for $r = \phi$ and $s = \phi^2$. Thus: {{begin-eqn}} {{eqn | l = \dfrac 1 \phi + \dfrac 1 {\phi^2}       | r = \dfrac {\phi + 1} {\phi^2}       | c =  }} {{eqn | r = \dfrac {\phi^2} {\phi^2}       | c = Square of Golden Mean equals One plus Golden Mean }} {{eqn | r = 1       | c =  }} {{end-eqn}} Hence the result. {{qed}} Category:Beatty Sequences jxlkrqozmwpp9tlf1o4ypx3ti503jl8
{{begin-eqn}} {{eqn | l = \floor {k \phi^2}       | r = \floor {k \paren {1 + \phi} }       | c = Square of Golden Mean equals One plus Golden Mean }} {{eqn | r = \floor {k + k \phi}       | c =  }} {{eqn | r = k + \floor {k \phi}       | c =  }} {{end-eqn}} {{qed}}
By Set is Subset of Finite Infima Set: :$X \subseteq \operatorname{fininfs}\left({X}\right)$ By Upper Closure of Subset is Subset of Upper Closure: :$X^\succeq \subseteq {\operatorname{fininfs}\left({X}\right)}^\succeq$ By Set is Subset of Upper Closure: :$X \subseteq X^\succeq$ Thus by Subset Relation is Transitive: :$X \subseteq {\operatorname{fininfs}\left({X}\right)}^\succeq$ Let $F$ be a filter in $L$ such that :$X \subseteq F$ Let $x \in {\operatorname{fininfs}\left({X}\right)}^\succeq$ By definition of upper closure of subset: :$\exists y \in \operatorname{fininfs}\left({X}\right): y \preceq x$ By definition of finite infima set: :$\exists A \in \mathit{Fin}\left({X}\right): y = \inf A \land A$ admits an infimum where $\mathit{Fin}\left({X}\right)$ denotes the set of all finite subsets of $S$. By Subset Relation is Transitive: :$A \subseteq F$ By Filtered in Meet Semilattice with Finite Infima: :$A \ne \varnothing \implies \inf A \in F$ By Infimum of Empty Set is Greatest Element and Top in Filter: :$A = \varnothing \implies \inf A = \top \in F$ where $\top$ denotes the greatest element of $S$. So :$y \in F$ Thus by definition of upper set: :$x \in F$ {{qed}}