hcju/proofwikips huggingface dataset:
Almost All Real Numbers are Transcendental
Almost all real numbers are transcendental.
Champernowne Constant is Transcendental
The Champernowne constant: :$0 \cdotp 12345 \, 67891 \, 01112 \, 13141 \, 51617 \, 18192 \, 02122 \ldots$ is transcendental.
Champernowne Constant is Normal
The Champernowne constant: :$0 \cdotp 12345 \, 67891 \, 01112 \, 13141 \, 51617 \, 18192 \, 02122 \ldots$ is normal with respect to base $10$.
Anomalous Cancellation/Examples/3544 over 7531
:$\dfrac {344} {731} = \dfrac {3544} {7531} = \dfrac {35544} {75531} = \cdots$
Anomalous Cancellation/Examples/143 185 over 17 018 560
:$\dfrac {1435} {170 \, 560} = \dfrac {143 \, 185} {17 \, 018 \, 560} = \dfrac {14 \, 318 \, 185} {1 \, 701 \, 818 \, 560} = \cdots$
Number of Digits in Power of 2
Let $n$ be a positive integer. Expressed in conventional decimal notation, the number of digits in the $n$th power of $2$: :$2^n$ is equal to: :$\ceiling {n \log_{10} 2}$ where $\ceiling x$ denotes the ceiling of $x$.
One Third as Quotient of Sequences of Odd Numbers
:$\dfrac 1 3 = \dfrac {1 + 3} {5 + 7} = \dfrac {1 + 3 + 5} {7 + 9 + 11} = \dfrac {1 + 3 + 5 + 7} {9 + 11 + 13 + 15} = \cdots$
Euler's Number as Sum of Egyptian Fractions
The reciprocal of Euler's number $e$ can be approximated by the following sequence of Egyptian fractions: :$\dfrac 1 e = \dfrac 1 3 + \dfrac 1 {29} + \dfrac 1 {15 \, 786} + \dfrac 1 {513 \, 429 \, 610} + \cdots$ {{OEIS|A006526}}
One Half as Pandigital Fraction
There are $12$ ways $\dfrac 1 2$ can be expressed as a pandigital fraction: :$\dfrac 1 2 = \dfrac {6729} {13 \, 458}$ :$\dfrac 1 2 = \dfrac {6792} {13 \, 584}$ :$\dfrac 1 2 = \dfrac {6927} {13 \, 854}$ :$\dfrac 1 2 = \dfrac {7269} {14 \, 538}$ :$\dfrac 1 2 = \dfrac {7293} {14 \, 586}$ :$\dfrac 1 2 = \dfrac {7329} {14 \, 658}$ :$\dfrac 1 2 = \dfrac {7692} {15 \, 384}$ :$\dfrac 1 2 = \dfrac {7923} {15 \, 846}$ :$\dfrac 1 2 = \dfrac {7932} {15 \, 864}$ :$\dfrac 1 2 = \dfrac {9267} {18 \, 534}$ :$\dfrac 1 2 = \dfrac {9273} {18 \, 546}$ :$\dfrac 1 2 = \dfrac {9327} {18 \, 654}$
One Seventh as Pandigital Fraction
There are $7$ ways $\dfrac 1 7$ can be expressed as a pandigital fraction: :$\dfrac 1 2 = \dfrac {2394} {16 \, 758}$ :$\dfrac 1 2 = \dfrac {2637} {18 \, 459}$ :$\dfrac 1 2 = \dfrac {4527} {31 \, 689}$ :$\dfrac 1 2 = \dfrac {5274} {36 \, 918}$ :$\dfrac 1 2 = \dfrac {5418} {37 \, 926}$ :$\dfrac 1 2 = \dfrac {5976} {41 \, 832}$ :$\dfrac 1 2 = \dfrac {7614} {53 \, 298}$
One Third as Pandigital Fraction
There are $2$ ways $\dfrac 1 3$ can be expressed as a pandigital fraction: :$\dfrac 1 3 = \dfrac {5823} {17469}$ :$\dfrac 1 3 = \dfrac {5832} {17496}$
One Quarter as Pandigital Fraction
There are $4$ ways $\dfrac 1 4$ can be expressed as a pandigital fraction: :$\dfrac 1 4 = \dfrac {3942} {15768}$ :$\dfrac 1 4 = \dfrac {4392} {17568}$ :$\dfrac 1 4 = \dfrac {5796} {23184}$ :$\dfrac 1 4 = \dfrac {7956} {31824}$
One Fifth as Pandigital Fraction
There are $12$ ways $\dfrac 1 5$ can be expressed as a pandigital fraction: :$\dfrac 1 5 = \dfrac {2697} {13485}$ :$\dfrac 1 5 = \dfrac {2769} {13845}$ :$\dfrac 1 5 = \dfrac {2937} {14685}$ :$\dfrac 1 5 = \dfrac {2967} {14835}$ :$\dfrac 1 5 = \dfrac {2973} {14865}$ :$\dfrac 1 5 = \dfrac {3297} {16485}$ :$\dfrac 1 5 = \dfrac {3729} {18645}$ :$\dfrac 1 5 = \dfrac {6297} {31485}$ :$\dfrac 1 5 = \dfrac {7629} {38145}$ :$\dfrac 1 5 = \dfrac {9237} {46185}$ :$\dfrac 1 5 = \dfrac {9627} {48135}$ :$\dfrac 1 5 = \dfrac {9723} {48615}$
One Sixth as Pandigital Fraction
There are $3$ ways $\dfrac 1 6$ can be expressed as a pandigital fraction: :$\dfrac 1 6 = \dfrac {2943} {17658}$ :$\dfrac 1 6 = \dfrac {4653} {27918}$ :$\dfrac 1 6 = \dfrac {5697} {34182}$
One Eighth as Pandigital Fraction
There are $46$ ways $\dfrac 1 8$ can be expressed as a pandigital fraction: :$\dfrac 1 8 = \dfrac {3187} {25496}$ :$\dfrac 1 8 = \dfrac {4589} {36712}$ :$\dfrac 1 8 = \dfrac {4591} {36728}$ :$\dfrac 1 8 = \dfrac {4689} {37512}$ :$\dfrac 1 8 = \dfrac {4691} {37528}$ :$\dfrac 1 8 = \dfrac {4769} {38152}$ :$\dfrac 1 8 = \dfrac {5237} {41896}$ :$\dfrac 1 8 = \dfrac {5371} {42968}$ :$\dfrac 1 8 = \dfrac {5789} {46312}$ :$\dfrac 1 8 = \dfrac {5791} {46328}$ :$\dfrac 1 8 = \dfrac {5839} {46712}$ :$\dfrac 1 8 = \dfrac {5892} {47136}$ :$\dfrac 1 8 = \dfrac {5916} {47328}$ :$\dfrac 1 8 = \dfrac {5921} {47368}$ :$\dfrac 1 8 = \dfrac {6479} {51832}$ :$\dfrac 1 8 = \dfrac {6741} {53928}$ :$\dfrac 1 8 = \dfrac {6789} {54312}$ :$\dfrac 1 8 = \dfrac {6791} {54328}$ :$\dfrac 1 8 = \dfrac {6839} {54712}$ :$\dfrac 1 8 = \dfrac {7123} {56984}$ :$\dfrac 1 8 = \dfrac {7312} {58496}$ :$\dfrac 1 8 = \dfrac {7364} {58912}$ :$\dfrac 1 8 = \dfrac {7416} {59328}$ :$\dfrac 1 8 = \dfrac {7421} {59368}$ :$\dfrac 1 8 = \dfrac {7894} {63152}$ :$\dfrac 1 8 = \dfrac {7941} {63528}$ :$\dfrac 1 8 = \dfrac {8174} {65392}$ :$\dfrac 1 8 = \dfrac {8179} {65432}$ :$\dfrac 1 8 = \dfrac {8394} {67152}$ :$\dfrac 1 8 = \dfrac {8419} {67352}$ :$\dfrac 1 8 = \dfrac {8439} {67512}$ :$\dfrac 1 8 = \dfrac {8932} {71456}$ :$\dfrac 1 8 = \dfrac {8942} {71536}$ :$\dfrac 1 8 = \dfrac {8953} {71624}$ :$\dfrac 1 8 = \dfrac {8954} {71632}$ :$\dfrac 1 8 = \dfrac {9156} {73248}$ :$\dfrac 1 8 = \dfrac {9158} {73264}$ :$\dfrac 1 8 = \dfrac {9182} {73456}$ :$\dfrac 1 8 = \dfrac {9316} {74528}$ :$\dfrac 1 8 = \dfrac {9321} {74568}$ :$\dfrac 1 8 = \dfrac {9352} {74816}$ :$\dfrac 1 8 = \dfrac {9416} {75328}$ :$\dfrac 1 8 = \dfrac {9421} {75368}$ :$\dfrac 1 8 = \dfrac {9523} {76184}$ :$\dfrac 1 8 = \dfrac {9531} {76248}$ :$\dfrac 1 8 = \dfrac {9541} {76328}$
One Ninth as Pandigital Fraction
There are $3$ ways $\dfrac 1 9$ can be expressed as a pandigital fraction: :$\dfrac 1 9 = \dfrac {6381} {57429}$ :$\dfrac 1 9 = \dfrac {6471} {58239}$ :$\dfrac 1 9 = \dfrac {8361} {75249}$
Four Fifths as Pandigital Fraction
The fraction $\dfrac 4 5$ can be expressed as a pandigital fraction in the following interesting way: :$\dfrac 4 5 = \dfrac {9876} {12 \, 345}$
Probability of Two Random Integers having no Common Divisor
Let $a$ and $b$ be integers chosen at random. The probability that $a$ and $b$ are coprime is given by: :$\map \Pr {a \perp b} = \dfrac 1 {\map \zeta 2} = \dfrac 6 {\pi^2}$ where $\zeta$ denotes the zeta function. The decimal expansion of $\dfrac 1 {\map \zeta 2}$ starts: :$\dfrac 1 {\map \zeta 2} = 0 \cdotp 60792 \, 71018 \, 54026 \, 6 \ldots$ {{OEIS|A059956}}
Probability of Random Integer being Square-Free
Let $a$ be an integer chosen at random. The probability that $a$ is square-free is given by: :$\map \Pr {\neg \exists b \in \Z: b^2 \divides a} = \dfrac 1 {\map \zeta 2} = \dfrac 6 {\pi^2}$ where $\zeta$ denotes the zeta function. The decimal expansion of $\dfrac 1 {\map \zeta 2}$ starts: :$\dfrac 1 {\map \zeta 2} = 0 \cdotp 60792 \, 71018 \, 54026 \, 6 \ldots$ {{OEIS|A059956}}
Kepler's Conjecture
The densest packing of identical spheres in space is obtained when the spheres are arranged with their centers at the points of a face-centered cubic lattice. This obtains a density of $\dfrac \pi {3 \sqrt 2} = \dfrac \pi {\sqrt {18} }$: :$\dfrac \pi {\sqrt {18} } = 0 \cdotp 74048 \ldots$
Prime Ideal is Prime Element
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a lattice. Let $I$ be an ideal in $L$. Then: :$I$ is a prime ideal : :$I$ is a prime element in $\struct {\map {\mathit {Ids} } L, \precsim}$ where: :$\map {\mathit {Ids} } L$ denotes the set of all ideals in $L$ :$\mathord \precsim := \mathord \subseteq \restriction_{\map {\mathit {Ids} } L \times \map {\mathit {Ids} } L}$
Probability of Three Random Integers having no Common Divisor
Let $a, b$ and $c$ be integers chosen at random. The probability that $a, b$ and $c$ have no common divisor: :$\map \Pr {\map \perp {a, b, c} } = \dfrac 1 {\map \zeta 3}$ where $\zeta$ denotes the zeta function: :$\map \zeta 3 = \dfrac 1 {1^3} + \dfrac 1 {2^3} + \dfrac 1 {3^3} + \dfrac 1 {4^3} + \cdots$ The decimal expansion of $\dfrac 1 {\map \zeta 3}$ starts: :$\dfrac 1 {\map \zeta 3} = 0 \cdotp 83190 \, 73725 \, 80707 \ldots$ {{OEIS|A088453}}
Densest Packing of Identical Circles
The densest packing of identical circles in the plane obtains a density of $\dfrac \pi {2 \sqrt 3} = \dfrac \pi {\sqrt {12} }$: :$\dfrac \pi {2 \sqrt 3} = 0 \cdotp 90689 \, 96821 \ldots$ {{OEIS|A093766}} This happens when they are packed together in a hexagonal array, with each circle touching $6$ others.
Meet is Intersection in Set of Ideals
Let $\mathscr S = \struct {S, \wedge, \preceq}$ be a meet semilattice. Let $\map {\mathit {Ids} } {\mathscr S}$ be the set of all ideals in $\mathscr S$. Let $P = \struct {\map {\mathit {Ids} } {\mathscr S}, \precsim}$ be an ordered set where $\mathord \precsim = \mathord \subseteq \restriction_{\map {\mathit {Ids} } {\mathscr S} \times \map {\mathit {Ids} } {\mathscr S} }$ Let $I_1, I_2$ be ideals in $\mathscr S$. Then :$I_1 \wedge_P I_2 = I_1 \cap I_2$
Characterization of Prime Ideal by Finite Infima
Let $L = \left({S, \wedge, \preceq}\right)$ be a meet semilattice. Let $I$ be an ideal in $L$. Then :$I$ is a prime ideal  :for all non-empty finite subset $A$ of $S: \left({ \inf A \in I \implies \exists a \in A: a \in I}\right)$
Doubling the Cube/Archytas Curve
The Archytas curve can be used for Doubling the Cube.
Doubling the Cube/Intersection of Parabolas
The problem of Doubling the Cube can be solved by finding the intersection of two parabolas.
Doubling the Cube/Intersection of Parabola and Hyperbola
The problem of Doubling the Cube can be solved by finding the intersection of a parabola and a hyperbola.
Unordered Pair is Finite
Let $x, y$ be arbitrary. Then $\left\{ {x, y}\right\}$ is finite.
Pair is Union of Singletons
Let $x, y$ be arbitrary. {{questionable|Need to state that $x \ne y$.}} Then $\left\{ {x, y}\right\} = \left\{ {x}\right\} \cup \left\{ {y}\right\}$
Doubling the Cube/Conchoid of Nicomedes
The problem of Doubling the Cube can be solved by using a conchoid of Nicomedes.
Doubling the Cube/Cissoid of Diocles
The problem of Doubling the Cube can be solved by using a cissoid of Diocles.
Length of Diagonal of Unit Square
The length of a diagonal of a square of side length $1$ is $\sqrt 2$) (the square root of $2$).
Sequence of Best Rational Approximations to Square Root of 2
A sequence of best rational approximations to the square root of $2$ starts: :$\dfrac 1 1, \dfrac 3 2, \dfrac 7 5, \dfrac {17} {12}, \dfrac {41} {29}, \dfrac {99} {70}, \dfrac {239} {169}, \dfrac {577} {408}, \ldots$ where: :the numerators are half of the Pell-Lucas numbers, $\dfrac 1 2 Q_n$ :the denominators are the Pell numbers $P_n$ starting from $\dfrac {\tfrac12 Q_1} {P_1}$. {{OEIS-Numerators|A001333}} {{OEIS-Denominators|A000129}}
Difference between Adjacent Convergents of Simple Continued Fraction
Then for $k \ge 1$: :$p_k q_{k - 1} - p_{k - 1} q_k = \paren {-1}^{k + 1}$ That is: :$C_k - C_{k - 1} = \dfrac {p_k} {q_k} - \dfrac {p_{k - 1} } {q_{k - 1} } = \dfrac {\paren {-1}^{k + 1} } {q_k q_{k - 1} }$
Difference between Adjacent Convergents But One of Simple Continued Fraction
For $k \ge 2$: :$p_k q_{k - 2} - p_{k - 2} q_k = \paren {-1}^k a_k$ That is: :$C_k - C_{k-2} = \dfrac {p_k} {q_k} - \dfrac {p_{k - 2} } {q_{k - 2} } = \dfrac {\paren {-1}^k a_k} {q_k q_{k - 2} }$
Convergents of Simple Continued Fraction are Rationals in Canonical Form
For all $k \ge 1$, $\dfrac {p_k} {q_k}$ is in canonical form: :$p_k$ and $q_k$ are coprime :$q_k > 0$.
Dual of Preordered Set is Preordered Set
Let $P = \left({S, \preceq}\right)$ be a preordered set. Then dual of $P$, $P^{-1} = \left({S, \succeq}\right)$ is also a preordered set. {{finish|Dual Ordered Set $\ne$ Dual Preordered Set}}
Dual Ordered Set is Ordered Set
Let $P = \left({S, \preceq}\right)$ be an ordered set. Then its dual, $P^{-1} = \left({S, \succeq}\right)$, is also an ordered set.
Value of Finite Continued Fraction equals Numerator Divided by Denominator
Let $F$ be a field. Let $\tuple {a_0, a_1, \ldots, a_n}$ be a finite continued fraction of length $n \ge 0$. Let $p_n$ and $q_n$ be its $n$th numerator and denominator. Then the value $\sqbrk {a_0, a_1, \ldots, a_n}$ equals $\dfrac {p_n} {q_n}$.
Simple Infinite Continued Fraction Converges
Let $C = (a_0, a_1, \ldots)$ be a simple infinite continued fraction in $\R$. Then $C$ converges.
Continued Fraction Identities/First/Infinite
Let $\left[{a_0, a_1, a_2, \ldots}\right]$ be a simple infinite continued fraction. Then: :$\left[{a_1, a_2, a_3, \ldots}\right] = a_1 + \dfrac 1 {\left[{a_2, a_3, \ldots}\right]}$
Relation between Adjacent Best Rational Approximations to Root 2
Consider the Sequence of Best Rational Approximations to Square Root of 2: :$\sequence S := \dfrac 1 1, \dfrac 3 2, \dfrac 7 5, \dfrac {17} {12}, \dfrac {41} {29}, \dfrac {99} {70}, \dfrac {239} {169}, \dfrac {577} {408}, \ldots$ Let $\dfrac {p_n} {q_n}$ and $\dfrac {p_{n + 1} } {q_{n + 1} }$ be adjacent terms of $\sequence S$. Then: :$\dfrac {p_{n + 1} } {q_{n + 1} } = \dfrac {p_n + 2 q_n} {p_n + q_n}$
Parity of Pell Numbers
Consider the Pell numbers $P_0, P_1, P_2, \ldots$ :$0, 1, 2, 5, 12, 29, \ldots$ $P_n$ has the same parity as $n$. That is: :if $n$ is odd then $P_n$ is odd :if $n$ is even then $P_n$ is even.
Parity of Best Rational Approximations to Root 2
Consider the Sequence of Best Rational Approximations to Square Root of 2: :$\sequence S := \dfrac 1 1, \dfrac 3 2, \dfrac 7 5, \dfrac {17} {12}, \dfrac {41} {29}, \dfrac {99} {70}, \dfrac {239} {169}, \dfrac {577} {408}, \ldots$ where $S_1 := \dfrac 1 1$. 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 $n$.
Prime Ideal is Prime Filter in Dual Lattice
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a lattice. Let $X$ be a subset of $S$. Then :$X$ is a prime ideal in $L$ : :$X$ is a prime filter in $L^{-1}$ where $L^{-1} = \struct {S, \succeq}$ denotes the dual of $L$.
Ideal is Filter in Dual Ordered Set
Let $P = \left({S, \preceq}\right)$ be an ordered set. Let $X$ be a subset of $S$. Then :$X$ is ideal in $P$  :$X$ is filter in $P^{-1}$ where $P^{-1} = \left({S, \succeq}\right)$ denotes the dual of $P$.
Square Root of 2 as Sum of Egyptian Fractions
The square root of $2$ can be approximated by the following sequence of Egyptian fractions: :$\sqrt 2 = 1 + \dfrac 1 3 + \dfrac 1 {253} + \dfrac 1 {218 \, 201} + \dfrac 1 {61 \, 323 \, 543 \, 802} + \cdots$ {{OEIS|A006487}}
Sprague's Property of Root 2
Let $S = \sequence {s_n}$ be the sequence of fractions defined as follows: Let the numerator of $s_n$ be: :$\floor {n \sqrt 2}$ where $\floor x$ denotes the floor of $x$. Let the denominators of the terms of $S$ be the (strictly) positive integers missing from the numerators of $S$: :$S := \dfrac 1 3, \dfrac 2 6, \dfrac 4 {10}, \dfrac 5 {13}, \dfrac 7 {17}, \dfrac 8 {20}, \ldots$ Then the difference between the numerator and denominator of $s_n$ is equal to $2 n$.
Steiner's Calculus Problem
Let $f: \R_{>0} \to \R$ be the real function defined as: :$\forall x \in \R_{>0}: \map f x = x^{1/x}$ Then $\map f x$ reaches its maximum at $x = e$ where $e$ is Euler's number .
Ratio of Lengths of Arms of Pentagram
Consider a pentagram. :400px Let $AC$ be the length of one of the lines which span the pentagram and define it. Let $B$ be one of the points where $AC$ intersects one of the other such lines such that $AB > AC$. Then: :$\dfrac {AC} {AB} = \phi$ where $\phi$ denotes the golden mean.
Filter is Ideal in Dual Ordered Set
Let $P = \left({S, \preceq}\right)$ be an ordered set. Let $X$ be a subset of $S$. Then :$X$ is filter in $P$  :$X$ is ideal in $P^{-1}$ where $P^{-1} = \left({S, \succeq}\right)$ denotes the dual of $P$.
Prime Filter is Prime Ideal in Dual Lattice
Let $L = \struct {S, \preceq}$ be a lattice. Let $X$ be a subset of $S$. Then :$X$ is a prime filter in $L$ : :$X$ is a prime ideal in $L^{-1}$ where $L^{-1} = \struct {S, \succeq}$ denotes the dual of $L$.
Trefoil Knot in Paper forms Pentagon
Take a strip of paper with parallel edges. Tie a simple trefoil knot in it. Flatten the knot and pull the paper tight. The knot will be in the shape of a regular pentagon. The edges of the paper strip trace out a pentagram. :500px
Sequence of Golden Rectangles
A golden rectangle can be divided into a square and another golden rectangle. That golden rectangle can in turn be divided into a square and another golden rectangle. The sequence can be continued indefinitely. :700px
Sequence of Golden Rectangles/Equiangular Spiral
The points where the vertices of successive squares of this sequence meet can be joined together by an equiangular spiral. This equiangular spiral can be approximated by quarter circles constructed as shown. The equiangular spiral tends towards the point of intersection of the diagonals of the golden rectangles.
Continued Fraction Expansion of Golden Mean/Rate of Convergence
This continued fraction expansion has the slowest rate of convergence of all simple infinite continued fractions.
Continued Fraction Expansion of Golden Mean/Successive Convergents
The $n$th convergent is given by: :$C_n = \dfrac {F_{n + 1} } {F_n}$ where $F_n$ denotes the $n$th Fibonacci number.
Continued Fraction Expansion of Pi
The constant $\pi$ (pi) has the continued fraction expansion: :$\pi = \sqbrk {3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, \ldots}$
Continued Fraction Expansion of Pi/Convergents
The convergents of the continued fraction expansion to $\pi$ (pi) are: :$3, \dfrac {22} 7, \dfrac {333} {106}, \dfrac {355} {113}, \dfrac {103993} {33102}, \dfrac {104348} {33215}$ {{OEIS-Numerators|A002485}} {{OEIS-Denominators|A002486}} These best rational approximations are accurate to $0, 2, 4, 6, 9, 9, 9, 10, 11, 11, 12, 13, \ldots$ decimals. {{OEIS|A114526}} === Zu Chongzhi Fraction === {{:Definition:Zu Chongzhi Fraction}}
Filter is Prime iff For Every Element Element either Negation Belongs to Filter in Boolean Lattice
Let $B = \struct {S, \vee, \wedge, \neg, \preceq}$ be a Boolean lattice. Let $F$ be a filter in $B$. Then :$F$ is prime  :$\forall x \in S: x \in F \lor \left({\neg x}\right) \in F$
Beatty's Theorem
Let $r, s \in \R \setminus \Q$ be an irrational number such that $r > 1$ and $s > 1$. Let $\BB_r$ and $\BB_s$ be the Beatty sequences on $r$ and $s$ respectively. Then $\BB_r$ and $\BB_s$ are complementary Beatty sequences : :$\dfrac 1 r + \dfrac 1 s = 1$
Proper and Prime iff Ultrafilter in Boolean Lattice
Let $B = \left({S, \vee, \wedge, \neg, \preceq}\right)$ be a Boolean lattice. Let $F$ be a filter in $B$. Then :$F$ is a proper subset of $S$ and $F$ is a prime filter in $B$ : :$F$ is ultrafilter on $B$
Difference between Terms of Wythoff Pair
Let $\tuple {\floor {k \phi}, \floor {k \phi^2} }$ be a Wythoff pair. The difference between the coordinates of this Wythoff pair is $k$. That is: :$\floor {k \phi^2} - \floor {k \phi} = k$
Finite Infima Set and Upper Closure is Smallest Filter
Let $L = \left({S, \wedge, \preceq}\right)$ be a meet semilattice. Let $X$ be a non-empty subset of $S$. Then :$X \subseteq {\operatorname{fininfs}\left({X}\right)}^\succeq$ and :for every a filter $F$ in $L$: $\left({X \subseteq F \implies {\operatorname{fininfs}\left({X}\right)}^\succeq \subseteq F}\right)$ where : $\operatorname{fininfs}\left({X}\right)$ denotes the finite infima set of $X$, : $X^\succeq$ denotes the upper closure of $X$.
Smaller Number of Wythoff Pair is Smallest Number not yet in Sequence
Consider the sequence of Wythoff pairs arranged in sequential order: :$\tuple {0, 0}, \tuple {1, 2}, \tuple {3, 5}, \tuple {4, 7}, \tuple {6, 10}, \tuple {8, 13}, \ldots$ The first coordinate of each Wythoff pair is the smallest positive integer which has not yet appeared in the sequence.
Sequence of Wythoff Pairs contains all Positive Integers exactly Once Each
Consider the sequence of Wythoff pairs arranged in sequential order: :$\tuple {0, 0}, \tuple {1, 2}, \tuple {3, 5}, \tuple {4, 7}, \tuple {6, 10}, \tuple {8, 13}, \ldots$ Apart from the first Wythoff pair $\tuple {0, 0}$, every positive integer appears in this sequence exactly once.
Wythoff Pairs are Winning Positions in Wythoff's Game
The Wythoff pairs are the winning positions in Wythoff's game.
Zeta of 2 as Product of Fractions with Prime Numerators
{{begin-eqn}} {{eqn | l = \map \zeta 2       | r = \prod_p \paren {\frac p {p - 1} } \paren {\frac p {p + 1} }       | c =  }} {{eqn | r = \dfrac 2 1 \times \dfrac 2 3 \times \dfrac 3 2 \times \dfrac 3 4 \times \dfrac 5 4 \times \dfrac 5 6 \times \dfrac 7 6 \times \dfrac 7 8 \times \dfrac {11} {10} \times \dfrac {11} {12} \times \dfrac {13} {12} \times \dfrac {13} {14} \times \cdots       | c =  }} {{end-eqn}} where: : $\zeta$ denotes the Riemann zeta function : $\displaystyle \prod_p$ denotes the product over all prime numbers.
Snub Cube Inscribed in Octahedron
A snub cube inscribed in an octahedron involves the Tribonacci constant. {{explain|a bit more detail could help here}}
Snub Cube Inscribed in Cube
A snub cube inscribed in a cube involves the Tribonacci constant. {{explain|a bit more detail could help here}}
Icosahedron Inscribed in Octahedron
An icosahedron inscribed in an octahedron involves the Golden Ratio. {{stub|a bit more detail could help here}}
Top in Filter
Let $\left({S, \preceq}\right)$ be a bounded above ordered set. Let $F$ be a filter on $S$. Then $\top \in F$ where $\top$ denotes the greatest element of $S$.
Finite Infima Set and Upper Closure is Filter
Let $P = \left({S, \wedge, \preceq}\right)$ be a meet semilattice. Let $X$ be a non-empty subset of $S$. Then :${\operatorname{fininfs}\left({X}\right)}^\succeq$ is filter in $P$. where :$\operatorname{fininfs}\left({X}\right)$ denotes the finite infima set of $X$, :$X^\succeq$ denotes the upper closure of $X$.
Brun's Theorem
The sum of the reciprocals of all the twin primes converges.
Divisibility by 2
An integer $N$ expressed in decimal notation is divisible by $2$  the least significant digit of $N$ is divisible by $2$. That is: :$N = [a_n \ldots a_2 a_1 a_0]_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $2$  : :$a_0$ is divisible by $2$.
Divisibility by Power of 2
Let $r \in \Z_{\ge 1}$ be a strictly positive integer. An integer $N$ expressed in decimal notation is divisible by $2^r$  the last $r$ digits of $N$ form an integer divisible by $2^r$. That is: :$N = [a_n \ldots a_2 a_1 a_0]_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $2^r$  : :$a_0 + a_1 10 + a_2 10^2 + \cdots + a_r 10^r$ is divisible by $2^r$.
Equivalence of Definitions of Deficient Number
The following definitions of a deficient number are equivalent:
Power of Prime is Deficient
Let $n \in \Z_{>0}$ be a power of a prime number $p$: :$n = p^k$ for some $k \in \Z_{>0}$. Then $n$ is deficient.
Set is Subset of Finite Infima Set
Let $\left({S, \preceq}\right)$ be an ordered set. Let $X$ be a subset of $S$. Then $X \subseteq \operatorname{fininfs}\left({X}\right)$ where $\operatorname{fininfs}\left({X}\right)$ denotes the finite infima set of $X$.
Set is Subset of Upper Closure
Let $\left({S, \preceq}\right)$ be an ordered set. Let $X$ be a subset of $S$. Then $X \subseteq X^\succeq$ where $X^\succeq$ denotes the upper closure of $X$.
Tamref's Last Theorem
The Diophantine equation: :$n^x + n^y = n^z$ has exactly one form of solutions in integers: :$2^x + 2^x = 2^{x + 1}$ for all $x \in \Z$.
3^x + 4^y equals 5^z has Unique Solution
The Diophantine equation: :$3^x + 4^y = 5^z$ has exactly one solution in integers: :$3^2 + 4^2 = 5^2$
5^x + 12^y equals 13^z has Unique Solution
The Diophantine equation: :$5^x + 12^y = 13^z$ has exactly one solution in integers: :$5^2 + 12^2 = 13^2$
Upper Closure of Subset is Subset of Upper Closure
Let $\left({S, \preceq}\right)$ be an ordered set. Let $X, Y$ be subsets of $S$. Then :$X \subseteq Y \implies X^\succeq \subseteq Y^\succeq$ where $X^\succeq$ denotes the upper closure of $X$.
Euler's Number as Limit of n over nth Root of n Factorial
:$\displaystyle e = \lim_{n \mathop \to \infty} \dfrac n {\sqrt [n] {n!} }$ where: :$e$ denotes Euler's number :$n!$ denotes $n$ factorial.
Continued Fraction Expansion of Euler's Number
The constant Euler's number $e$ has the continued fraction expansion: {{begin-eqn}} {{eqn | l = e       | r = \sqbrk {2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, \ldots }       | c = }} {{eqn | l =        | r = \sqbrk {1, 0, 1, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, \ldots }       | c = }} {{end-eqn}}