avewright/proofwiki-math huggingface dataset:
Theorem: Equivalence of Definitions of Order of Entire Function

Let $f: \C \to \C$ be an entire function.
Let $\ln$ denote the natural logarithm.
{{TFAE|def = Order of Entire Function}}

Proof:
Let:
:$\alpha_1 = \ds \limsup_{R \mathop \to \infty} \frac {\ds \ln \ln \max_{\cmod z \mathop \le R} \cmod f} {\ln R}$
:$\alpha_2 = \inf \set {\beta \ge 0: \ds \map \ln {\max_{\cmod z \mathop \le R} \cmod {\map f z} } = \map \OO {R^\beta} }$
:$\alpha_3 = \inf \set {\beta \ge 0: \map f z = \map \OO {\map \exp {\cmod z^\beta} } }$
Theorem: Sum of Big-O Estimates/Sequences

Let $\sequence {a_n},\sequence {b_n},\sequence {c_n},\sequence {d_n}$ be sequences of real or complex numbers.
Let:
:$a_n = \map \OO {b_n}$
:$c_n = \map \OO {d_n}$
where $\OO$ denotes big-$\OO$ notation.
Then:
:$a_n + c_n = \map \OO {\size {b_n} + \size {d_n} }$

Proof:
Since:
:$a_n = \map \OO {b_n}$
there exists a positive real number $C_1$ and natural number $N_1$ such that: 
:$\size {a_n} \le C_1 \size {b_n}$
for all $n \ge N_1$. 
Similarly, since: 
:$c_n = \map \OO {d_n}$
there exists a positive real number $C_2$ and natural number $N_2$ such that: 
:$\size {c_n} \le C_2 \size {d_n}$
for all $n \ge N_2$.
Let:
:$N = \max \set {N_1, N_2}$
Then, for $n \ge N$ we have: 
{{begin-eqn}}
{{eqn | l = \size {a_n + c_n}
      | o = \le
      | r = \size {a_n} + \size {c_n}
      | c = Triangle Inequality
}}
{{eqn | r = C_1 \size {b_n} + \size {c_n}
      | c = since $n \ge N_1$
}}
{{eqn | r = C_1 \size {b_n} + C_2 \size {d_n}
      | c = since $n \ge N_2$
}}
{{end-eqn}}
Let:
:$C = \max \set {C_1, C_2}$
Then:
{{begin-eqn}}
{{eqn | l = C_1 \size {b_n} + C_2 \size {d_n}
      | r = C \size {b_n} + C \size {d_n}
}}
{{eqn | r = C \paren {\size {b_n} + \size {d_n} }
}}
{{eqn | r = C \size {\size {b_n} + \size {d_n} } 
}}
{{end-eqn}}
So:
:$\size {a_n + c_n} \le C \size {\size {b_n} + \size {d_n} }$
for $n \ge N$. 
So:
:$a_n + c_n = \map \OO {\size {b_n} + \size {d_n} }$
{{qed}}
Category:Big-O Notation
doy3hqmejqy99t0ixjj191mijtywew3
Theorem: Little-O Times Big-O is Little-O/Sequences

Let $\sequence {a_n}, \sequence {b_n}, \sequence {c_n}, \sequence {d_n}$ be sequences of real or complex numbers.
Let:
:$a_n = \map \OO {b_n}$
:$c_n = \map {\mathcal o} {d_n}$
where:
:$\OO$ denotes big-$\OO$ notation
:$\mathcal o$ denotes little-$\mathcal o$ notation.
Then:
:$a_n c_n = \map {\mathcal o} {b_n d_n}$

Proof:
Let $\epsilon \in \R_{> 0}$.
Since $a_n = \map \OO {b_n}$:
:$\exists c \in \R: c \ge 0: \exists n_0 \in \N: \paren {n \ge n_0 \implies \size {a_n} \le c \cdot \size {b_n} }$
Since $c_n = \map {\mathcal o} {d_n}$:
:$\exists n_1 \in \N: \paren {n \ge n_1 \implies \size {c_n} \le \dfrac \epsilon {c + 1} \cdot \size {d_n} }$
Thus for $n \ge \max \set {n_0, n_1}$:
{{begin-eqn}}
{{eqn | o = 
      | r = \size {a_n c_n}
}}
{{eqn | r = \size {a_n} \size {c_n}
}}
{{eqn | o = \le
      | r = \paren {c \cdot \size {b_n} } \paren {\frac \epsilon {c + 1} \cdot \size {d_n} }
}}
{{eqn | o = \le
      | r = \size {b_n} \paren {\epsilon \cdot \size {d_n} }
}}
{{eqn | r = \epsilon \cdot \size {b_n d_n}
}}
{{end-eqn}}
Thus $a_n c_n = \map {\mathcal o} {b_n d_n}$.
{{qed}}
Category:Big-O Notation
Category:Little-O Notation
f8vzsycqhrw9funxgqhspxu8tgrvsjf
Theorem: Field Norm of Complex Number Equals Field Norm

Let $z = a + i b$ be a complex number, where $a, b \in \R$.
Then the field norm of $z$ is the field norm with respect to the field extension $\C / \R$.

Proof:
{{ProofWanted}}
Category:Field Extensions
Category:Field Norm of Complex Number
0i5v8vch6mvpd11ja0d3gdrgpmfcvbj
Theorem: Big-O Notation for Sequences Coincides with General Definition

Let $\sequence {a_n}$ and $\sequence {b_n}$ be sequences of real or complex numbers.
Let $\N$ be given the discrete topology.
{{TFAE}}
:$(1): \quad a_n = \map \OO {b_n}$, where $\OO$ denotes big-$\OO$ notation for sequences
:$(2): \quad a_n = \map \OO {b_n}$, where $\OO$ stands for the general definition of big-$\OO$ notation

Proof:
{{ProofWanted}}
Category:Big-O Notation
k2irag1srvyqutfx9i6bgs5dh0bu1mm
Theorem: Bounded iff Big-O of 1/Sequences

Let $\sequence {a_n}$ be a sequence of real or complex numbers.
{{TFAE}}
:$(1): \quad a_n$ is bounded
:$(2): \quad a_n = \map \OO 1$, where $\OO$ denotes big-$\OO$ notation

Proof:
{{begin-eqn}}
{{eqn | l = a_n
      | o = \text {is}
      | r = \text {bounded}
}}
{{eqn | ll= \leadstoandfrom
      | q  = \exists k \in \R
      | l = \size {a_n}
      | o = \le
      | r = k
      | c = {{Defof|Bounded Sequence}}
}}
{{eqn | ll= \leadstoandfrom
      | q = \exists k \in \R
      | l = \size {a_n}
      | o = \le
      | r = k \cdot \size 1
}}
{{eqn | ll= \leadstoandfrom
      | l = a_n
      | r = \map \OO 1
      | c = {{Defof|Big-O Notation for Sequences|Big-$\OO$ Notation}}
}}
{{end-eqn}}
{{qed}}
Category:Big-O Notation
ddv4e94tdzo9hg0hdpmougd8s76ydzt
Theorem: Sequence of 11 Primes by Trebling and Adding 16

The process of multiplication by $3$ and then adding $16$ produces a sequence of $11$ primes when starting from $587$:
:$587, 1777, 5347, 16 \, 057, 48 \, 187, 144 \, 577, 433 \, 747, 1 \, 301 \, 257, 3 \, 903 \, 787, 11 \, 711 \, 377, 35 \, 134 \, 147$

Proof:
{{begin-eqn}}
{{eqn | o = 
      | r = 587
      | c = is prime
}}
{{eqn | l = 3 \times 587 + 16
      | r = 1777
      | c = which is prime
}}
{{eqn | l = 3 \times 1777 + 16
      | r = 5347
      | c = which is prime
}}
{{eqn | l = 3 \times 5347 + 16
      | r = 16 \, 057
      | c = which is prime
}}
{{eqn | l = 3 \times 16 \, 057 + 16
      | r = 48 \, 187
      | c = which is prime
}}
{{eqn | l = 3 \times 48 \, 187 + 16
      | r = 144 \, 577
      | c = which is prime
}}
{{eqn | l = 3 \times 144 \, 577 + 16
      | r = 433 \, 747
      | c = which is prime
}}
{{eqn | l = 3 \times 433 \, 747 + 16
      | r = 1 \, 301 \, 257
      | c = which is prime
}}
{{eqn | l = 3 \times 1 \, 301 \, 257 + 16
      | r = 3 \, 903 \, 787
      | c = which is prime
}}
{{eqn | l = 3 \times 3 \, 903 \, 787 + 16
      | r = 11 \, 711 \, 377
      | c = which is prime
}}
{{eqn | l = 3 \times 11 \, 711 \, 377 + 16
      | r = 35 \, 134 \, 147
      | c = which is prime
}}
{{eqn | l = 3 \times 35 \, 134 \, 147 + 16
      | r = 105 \, 402 \, 457
      | c = which is not prime : $105 \, 402 \, 457 = 67 \times 137 \times 11 \, 483$
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Little-O Estimates/Sequences

Let $\sequence {a_n}, \sequence {b_n}, \sequence {c_n}, \sequence {d_n}$ be sequences of real or complex numbers.
Let:
:$a_n = \map \oo {b_n}$
:$c_n = \map \oo {d_n}$
where $\oo$ denotes little-$\oo$ notation.
Then:
:$a_n + c_n = \map \oo {\size {b_n} + \size {d_n} }$

Proof:
Let $\epsilon > 0$.
Then by definition of little-$\oo$ notation:
:$\exists n_1 \in \N: \paren {n \ge n_1 \implies \size {a_n} \le \epsilon \cdot \size {b_n}}$
:$\exists n_2 \in \N: \paren {n \ge n_2 \implies \size {c_n} \le \epsilon \cdot \size {d_n}}$
For $n \ge \max \set {n_1, n_2}$:
{{begin-eqn}}
{{eqn | l = \size {a_n + c_n}
      | o = \le
      | r = \size {a_n} + \size {c_n}
      | c = Triangle Inequality
}}
{{eqn | o = \le
      | r = \epsilon \cdot \size {b_n} + \epsilon \cdot \size {d_n}
}}
{{eqn | r = \epsilon \cdot \size {\size {b_n} + \size {d_n} }
      | c = Absolute value is positive
}}
{{end-eqn}}
Hence by definition of little-$\oo$ notation:
:$a_n + c_n = \map \oo {\size {b_n} + \size {d_n} }$
{{qed}}
Category:Little-O Notation
1vfbif2lovoz1sxddhognmbwqbwnp5n
Theorem: Equivalence of Definitions of Asymptotically Equal Sequences

Let $\sequence {a_n}$ and $\sequence {b_n}$ be sequences in $\R$.
{{TFAE|def = Asymptotically Equal Sequences}}

Proof:
=== $(1)$ iff $(2)$ ===
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \dfrac {a_n} {b_n}
      | o = \to
      | r = 1
}}
{{eqn | ll = \leadstoandfrom
      | l = \lim_{n \mathop \to \infty} \paren {\dfrac {a_n} {b_n} - \dfrac {b_n} {b_n} }
      | o = \to
      | r = 0
}}
{{eqn | ll = \leadstoandfrom
      | l = \lim_{n \mathop \to \infty} \dfrac {a_n - b_n} {b_n}
      | o = \to
      | r = 0
}}
{{eqn | ll = \leadstoandfrom
      | l = a_n - b_n
      | r = \map \oo {b_n}
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Largest Number not Expressible as Sum of Multiples of 23 and 28

The largest integer $n$ that cannot be expressed in the form:
:$n = 23 x + 28 y$
for $x, y \in \Z_{>0}$ is $593$.

Proof:
By Largest Number not Expressible as Sum of Multiples of Coprime Integers, the largest such number is:
:$\paren {23 - 1} \times \paren {28 - 1} - 1 = 593$
{{ProofWanted|This is a specific "historical" example of a general result which I read in Polya and Szego some time back, which still needs to be added into {{ProofWiki}}. All in due course.  I may already have it on here, I misremember.}}
Theorem: Open implies There Exists Way Below Element

Let $L = \struct {S, \preceq, \tau}$ be a continuous topological lattice with Scott topology.
Let $p \in S, A \subseteq S$ such that:
:$A$ is open and $p \in A$.
Then:
:$\exists q \in A: q \ll p$
where $q \ll p$ denotes $q$ is way below $p$.

Proof:
By definition of continuous ordered set:
:$p^\ll$ is directed
and
:$L$ satisfies the axiom of approximation.
By the axiom of approximation:
:$p = \map \sup {p^\ll}$
By definition of Scott topology:
:$A$ is inaccessible by directed suprema.
By definition of inaccessible by directed suprema:
:$A \cap p^\ll \ne \O$
By definition of non-empty set:
:$\exists q: q \in A \cap p^\ll$
Thus by definition of intersection:
:$q \in A$
By definition of intersection:
:$q \in p^\ll$
By definition of way below closure:
:$q \ll p$
{{qed}}
Theorem: Smallest Fourth Power which is Sum of 5 Fourth Powers

$625$ is the smallest fourth power which is the sum of $5$ fourth powers:
:$625 = 5^4 = 2^4 + 2^4 + 3^4 + 4^4 + 4^4$

Proof:
We check that for $n = 2, 3, 4$, $n^4$ is not a sum of $5$ smaller fourth powers.
We have:
:$5 \times 1^4 = 5 < 16 = 2^4$
:$5 \times 2^4 = 80 < 81 = 3^4$
so $2^4, 3^4$ are not sums of $5$ fourth powers.
For $n = 4$:
:$\dfrac {4^4} {3^4} < 4$
so such a sum can include at most $3$ $3^4$'s. 
However:
:$3 \times 3^4 + 2^4 + 1^4 = 260 > 256 = 4^4$
:$3 \times 3^4 + 2 \ \ \times 1^4 = 245 < 256 = 4^4$
:$2 \times 3^4 + 3 \ \ \times 2^4 = 220 < 256 = 4^4$
therefore $4^4$ is not a sum of $5$ smaller fourth powers.
This shows that $5^4$ is the smallest fourth power which is the sum of $5$ fourth powers.
{{qed}}
Theorem: Interior is Union of Way Above Closures

Let $\left({S, \preceq, \tau}\right)$ be a complete continuous topological lattice with Scott topology.
Let $X \subseteq S$.
Then $X^\circ = \bigcup \left\{ {x^\gg: x \in S \land x^\gg \subseteq X}\right\}$
where
:$X^\circ$ denotes the interior of $X$,
:$x^\gg$ denotes the way above closure of $x$.

Proof:
We have:
:$\left\{ {G \in \left\{ {g^\gg: g \in S}\right\}: G \subseteq X}\right\} = \left\{ {x^\gg: x \in S \land x^\gg \subseteq X}\right\}$
By Way Above Closures Form Basis:
:$\left\{ {x^\gg: x \in S}\right\}$ is basis of $\left({S, \tau}\right)$.
By Interior is Union of Elements of Basis:
:$X^\circ = \bigcup \left\{ {x^\gg: x \in S \land x^\gg \subseteq X}\right\}$
{{qed}}
Theorem: Way Above Closures Form Basis

Let $L = \struct {S, \preceq, \tau}$ be a complete continuous topological lattice with Scott topology.
Then $\set {x^\gg: x \in S}$ is an (analytic) basis of $L$.

Proof:
Define $B = \set {x^\gg: x \in S}$.
Thus by Way Above Closure is Open:
:$B \subseteq \tau$
We will prove that:
:for all $x \in S$: there exists a local basis $Q$ of $x$: $Q \subseteq B$
Let $x \in S$.
By Way Above Closures that Way Below Form Local Basis:
:$Q := \set {g^\gg: g \in S \land g \ll x}$ is a local basis at $x$.
Thus by definition of subset:
:$Q \subseteq B$
{{qed|lemma}}
Thus by Characterization of Analytic Basis by Local Bases:
:$B$ is an (analytic) basis of $L$.
{{qed}}
Theorem: Numbers whose Product with Reverse are Equal

:$651 \times 156 = 372 \times 273$

Proof:
{{begin-eqn}}
{{eqn | l = 651 \times 156
      | r = \paren {3 \times 7 \times 31} \times \paren {2^2 \times 3 \times 13}
      | c = 
}}
{{eqn | r = 2^2 \times 3^2 \times 7 \times 13 \times 31
      | c = 
}}
{{eqn | r = \paren {2^2 \times 3 \times 31} \times \paren {3 \times 7 \times 13}
      | c = 
}}
{{eqn | r = 372 \times 273
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Record Gaps between Twin Primes

The gaps between the following pairs of twin primes are larger than those for all smaller pairs:
{{begin-eqn}}
{{eqn | l = \tuple {3, 5}
      | o = \to
      | r = \tuple {5, 7}
      | c = a gap of $0$
}}
{{eqn | l = \tuple {5, 7}
      | o = \to
      | r = \tuple {11, 13}
      | c = a gap of $4$
}}
{{eqn | l = \tuple {17, 19}
      | o = \to
      | r = \tuple {29, 31}
      | c = a gap of $10$
}}
{{eqn | l = \tuple {41, 43}
      | o = \to
      | r = \tuple {59, 61}
      | c = a gap of $16$
}}
{{eqn | l = \tuple {69, 73}
      | o = \to
      | r = \tuple {101, 103}
      | c = a gap of $28$
}}
{{eqn | l = \tuple {311, 313}
      | o = \to
      | r = \tuple {347, 349}
      | c = a gap of $34$
}}
{{eqn | l = \tuple {347, 349}
      | o = \to
      | r = \tuple {419, 421}
      | c = a gap of $70$
}}
{{eqn | l = \tuple {659, 661}
      | o = \to
      | r = \tuple {809, 811}
      | c = a gap of $148$
}}
{{end-eqn}}
{{OEIS|A036061}}
{{finish|Add this list to the Number Sequence}}

Proof:
By cases and inspection.
Theorem: Consecutive Sophie Germain Primes cannot be Pair of Twin Primes

Let $p$ and $p + 2$ be twin primes.
Then unless $p = 3$ it is not possible for both $p$ and $p + 2$ to be Sophie Germain primes.

Proof:
First it is noted that $3$ and $5$ twin primes which are both Sophie Germain.
Prime numbers greater than $3$ are of the form $6 n - 1$ and $6 n + 1$.
Thus a pair of twin primes is of the form $\left({6 n - 1, 6 n + 1}\right)$.
The result follows from Sophie Germain Prime cannot be 6n+1.
{{qed}}
Category:Twin Primes
Category:Sophie Germain Primes
snxstobc2zqlmxfufczkwl31v2642ye
Theorem: Sophie Germain Prime cannot be 6n+1

Let $p$ be a Sophie Germain prime.
Then $p$ cannot be of the form $6 n + 1$, where $n$ is a positive integer.

Proof:
Let $p$ be a Sophie Germain prime.
Then, by definition, $2 p + 1$ is prime.
{{AimForCont}} $p = 6 n + 1$ for some $n \in \Z_{>0}$.
Then:
{{begin-eqn}}
{{eqn | l = 2 p + 1
      | r = 2 \paren {6 n + 1} + 1
      | c = 
}}
{{eqn | r = 12 n + 3
      | c = 
}}
{{eqn | r = 3 \paren {4 n + 1}
      | c = 
}}
{{end-eqn}}
and so $2 p + 1$ is not prime.
The result follows by Proof by Contradiction.
{{qed}}
Category:Sophie Germain Primes
runb03zh6lpwhgzstdneqn105w7lp87
Theorem: Way Above Closure is Open

Let $L = \struct {S, \preceq, \tau}$ be a complete continuous topological lattice with Scott topology.
Let $x \in S$.
Then $x^\gg$ is open
where $x^\gg$ denotes the way above closure of $x$.

Proof:
By Way Above Closure is Upper:
:$x^\gg$ is upper.
We will prove that
:$x^\gg$ is inaccessible by directed suprema.
Let $D$ be a directed subset of $S$ such that
:$\sup D \in x^\gg$
By definition of way above closure:
:$x \ll \sup D$
By Way Below iff Second Operand Preceding Supremum of Directed Set There Exists Element of Directed Set First Operand Way Below Element:
:$\exists d \in D: x \ll d$
By definition of way above closure:
:$d \in x^\gg$
By definitions of intersection and non-empty set:
:$x^\gg \cap D \ne \O$
{{qed|lemma}}
Thus by definition of Scott topology:
:$x^\gg$ is open.
{{qed}}
Theorem: Way Above Closure is Upper

Let $\struct {S, \preceq}$ be an ordered set.
Let $x \in S$.
Then $x^\gg$ is upper
where $x^\gg$ denotes the way above closure of $x$.

Proof:
Let $y \in x^\gg$, $z \in S$ such that
:$y \preceq z$
By definition of way above closure:
:$x \ll y$
By Preceding and Way Below implies Way Below:
:$x \ll z$
Thus by definition of way above closure:
:$z \in x^\gg$
{{qed}}
Theorem: Way Above Closures that Way Below Form Local Basis

{{rename|more descriptive of what the statement of the theorem says}}
Let $L = \struct {S, \preceq, \tau}$ be a complete continuous topological lattice with Scott topology.
Let $p \in S$.
Then $\set {q^\gg: q \in S \land q \ll p}$ is a local basis at $p$.

Proof:
Define $B := \set {q^\gg: q \in S \land q \ll p}$
By Way Above Closure is Open:
:$B \subseteq \tau$
By definition of way above closure:
:$\forall X \in B: p \in X$
Thus by definition:
:$B$ is set of open neighborhoods.
{{explain|open neighborhoods of what?}}
Let $U$ be an open subset of $S$ such that
:$p \in U$
By Open implies There Exists Way Below Element:
:$\exists u \in U: u \ll p$
Thus by definition of $B$:
:$u^\gg \in B$
By definition of Scott topology:
:$U$ is upper.
We will prove that
:$u^\gg \subseteq U$
Let $z \in u^\gg$
By definition of way above closure:
:$u \ll z$
By Way Below implies Preceding:
:$u \preceq z$
Thus by definition of upper section:
:$z \in U$
{{qed}}
Theorem: Characterization of Analytic Basis by Local Bases

Let $T = \struct {S, \tau}$ be a topological space.
Let $P$ be a set of subsets of $S$ such that
:$P \subseteq \tau$
and
:for all $p \in S$: there exists local basis $B$ at $p: B \subseteq P$
Then $P$ is basis of $T$.

Proof:
By assumption:
:$P \subseteq \tau$
Let $U$ be an open subset of $S$.
Define:
:$X := \set {V \in P: V \subseteq U}$
By definition of subset:
:$X \subseteq P$
We will prove that:
:$\forall u \in S: u \in U \iff \exists Z \in X: u \in Z$
Let $u \in S$.
We will prove that:
:$u \in U \implies \exists Z \in X: u \in Z$
Assume that:
:$u \in U$
By assumption:
:there exists local basis $B$ at $u: B \subseteq P$.
By definition of local basis:
:$\exists V \in B: V \subseteq U$
Thus by definitions of subset and $X$:
:$V \in X$
Thus by definition of local basis:
:$u \in V$
{{qed|lemma}}
Assume that:
:$\exists Z \in X: u \in Z$
By definition of $X$:
:$Z \subseteq U$
Thus by definition of subset:
:$u \in U$
{{qed|lemma}}
Thus by definition of union:
:$U = \bigcup X$
Hence $P$ is basis of $L$.
{{qed}}
Theorem: Value of Golden Ratio using 666

The Golden Ratio can be given by the following formula:
:$\phi = -2 \sin 666 \degrees = 1.61803 \, 39887 \, 5 \ldots$

Proof:
{{begin-eqn}}
{{eqn | l = -2 \sin 666 \degrees
      | r = -2 \sin 306 \degrees
      | c = Sine of Angle plus Full Angle
}}
{{eqn | r = 2 \sin 54 \degrees
      | c = Sine of Conjugate Angle
}}
{{eqn | r = 2 \cos 36 \degrees
      | c = Sine of Complement equals Cosine
}}
{{eqn | r = 2 \times \frac \phi 2
      | c = Cosine of $36 \degrees$
}}
{{eqn | r = \phi
}}
{{end-eqn}}
{{qed}}
Theorem: Euler Phi Function of 666 equals Product of Digits

The number $666$ has the following interesting property:
:$\map \phi {666} = 6 \times 6 \times 6$
where $\phi$ denotes the Euler $\phi$ function.

Proof:
From Euler Phi Function of Integer:
:$\ds \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$
where $p \divides n$ denotes the primes which divide $n$.
We have that:
:$666 = 2 \times 3^2 \times 37$
Thus:
{{begin-eqn}}
{{eqn | l = \map \phi {666}
      | r = 666 \paren {1 - \dfrac 1 2} \paren {1 - \dfrac 1 3} \paren {1 - \dfrac 1 {37} }
      | c = 
}}
{{eqn | r = 666 \times \frac 1 2 \times \frac 2 3 \times \frac {36} {37}
      | c = 
}}
{{eqn | r = 3 \times 1 \times 2 \times 36
      | c = 
}}
{{eqn | r = 3 \times 2 \times \paren {2^2 \times 3^2}
      | c = 
}}
{{eqn | r = 216
      | c = 
}}
{{eqn | r = 6 \times 6 \times 6
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Pair of Consecutive Powerful Numbers whose First is Odd

The only known pair of consecutive integers which are both powerful numbers such that the first of the pair is odd is:
:$\tuple {675, 676}$

Proof:
By investigation:
{{begin-eqn}}
{{eqn | l = 675
      | r = 3^3 \times 5^2
}}
{{eqn | l = 676
      | r = 2^2 \times 13^2
}}
{{end-eqn}}
That there are no smaller ones can be determined again by investigation.
{{qed}}
Theorem: Interior is Union of Elements of Basis

Let $T = \left({S, \tau}\right)$ be a topological space.
Let $B$ be a basis of $T$.
Let $V$ be a subset of $S$.
Then $V^\circ = \bigcup \left\{ {G \in B: G \subseteq V}\right\}$
where $V^\circ$ denotes the interior of $V$.

Proof:
By definition of interior:
:$\left\{ {G \in B: G \subseteq V}\right\} = \left\{ {G \in B: G \subseteq V^\circ}\right\}$
and
:$V^\circ$ is open.
Thus by Open Set is Union of Elements of Basis:
:$V^\circ = \bigcup \left\{ {G \in B: G \subseteq V}\right\}$
{{qed}}
Theorem: Palindromic Squares with Non-Palindromic Roots

The sequence of palindromic squares with non-palindromic square roots begins:
:$676, 69 \, 696, 94 \, 249, 698 \, 896, 5 \, 221 \, 225, 6 \, 948 \, 496, 522 \, 808 \, 225, \ldots$
This sequence is not explicitly given in {{OEISLink}}.
The sequence of those corresponding non-palindromic square roots begins:
:$26, 264, 307, 836, 2285, 2636, 22 \, 865, 24 \, 846, 30 \, 693, \ldots$
{{OEIS|A251673}}

Proof:
By investigating all square numbers which are palindromic.
{{begin-eqn}}
{{eqn | l = 676
      | r = 26^2
}}
{{eqn | l = 69 \, 696
      | r = 264^2
}}
{{eqn | l = 94 \, 249
      | r = 307^2
}}
{{eqn | l = 698 \, 896
      | r = 836^2
}}
{{eqn | l = 5 \, 221 \, 225
      | r = 2285^2
}}
{{eqn | l = 6 \, 948 \, 496
      | r = 2636^2
}}
{{eqn | l = 522 \, 808 \, 225
      | r = 22 \, 865^2
}}
{{end-eqn}}
and so on.
{{qed}}
Theorem: Tetrahedral Numbers which are Sum of 2 Tetrahedral Numbers

The sequence of tetrahedral numbers which are the sum of two other tetrahedral numbers begins:
:$20, 680, 29260, 34220, 70300, \dots$
{{OEIS|A034404}}

Proof:
{{begin-eqn}}
{{eqn | o =
      | r = 20
      | c = the $4$th tetrahedral number
}}
{{eqn | r = 10
      | c = the $3$rd tetrahedral number
}}
{{eqn | o =
      | ro= +
      | r = 10
      | c = the $3$rd tetrahedral number
}}
{{eqn | o =
      | r = 680
      | c = the $15$th tetrahedral number
}}
{{eqn | r = 120
      | c = the $8$th tetrahedral number
}}
{{eqn | o =
      | ro= +
      | r = 560
      | c = the $14$th tetrahedral number
}}
{{eqn | o =
      | r = 29 \, 260
      | c = the $55$th tetrahedral number
}}
{{eqn | r = 27 \, 720
      | c = the $54$th tetrahedral number
}}
{{eqn | o =
      | ro= +
      | r = 1540
      | c = the $20$th tetrahedral number
}}
{{eqn | o =
      | r = 34 \, 220
      | c = the $58$th tetrahedral number
}}
{{eqn | r = 29 \, 260
      | c = the $55$th tetrahedral number
}}
{{eqn | o =
      | ro= +
      | r = 4960
      | c = the $30$th tetrahedral number
}}
{{eqn | o =
      | r = 70 \, 300
      | c = the $74$th tetrahedral number
}}
{{eqn | r = 59 \, 640
      | c = the $70$th tetrahedral number
}}
{{eqn | o =
      | ro= +
      | r = 10 \, 660
      | c = the $39$th tetrahedral number
}}
{{end-eqn}}
{{qed}}
Theorem: Open Set is Union of Elements of Basis

Let $T = \struct {S, \tau}$ be a topological space.
Let $B$ be a basis of $T$.
Let $V$ be an open subset of $S$.
Then $V = \bigcup \set {G \in B: G \subseteq V}$

Proof:
Let $x$ be arbitrary.
We will prove that:
:$x \in V \implies \exists Y \in \set {G \in B: G \subseteq V}: x \in Y$
Assume that:
:$x \in V$
By definition of basis:
:$\exists F \subseteq B: V = \bigcup F$
By definition of union:
:$\exists Y \in F: x \in Y$
By Set is Subset of Union/General Result:
:$Y \subseteq V$
Thus by definition of subset:
:$Y \in \set {G \in B: G \subseteq V}$
Thus:
:$x \in Y$
{{qed|lemma}}
Assume that:
:$\exists Y \in \set {G \in B: G \subseteq V}: x \in Y$
By assumption:
:$Y \subseteq V$
Thus by definition of subset:
:$x \in V$
Hence by definition of union:
:$V = \bigcup \set {G \in B: G \subseteq V}$
{{qed}}