avewright/proofwiki-math huggingface dataset:
Theorem: Symmetric Bilinear Form can be Diagonalized

Let $\mathbb K$ be a field.
Let $V$ be a vector space over $\mathbb K$ of finite dimension $n>0$.
Let $f$ be a symmetric bilinear form on $V$.
Then there exists an ordered basis for which the relative matrix of $f$ is diagonal.

Proof:
{{ProofWanted}}
Category:Bilinear Forms (Linear Algebra)
e4gaxj5kousfaflt2jcse0v2tz33i05
Theorem: Dimension of Radical of Bilinear Form

Let $\mathbb K$ be a field.
Let $V$ be a vector space over $\mathbb K$ of finite dimension $n > 0$.
Let $f$ be a bilinear form on $V$.
Let $\map \Rad V$ denote the radical of $V$.
Let $\map {\operatorname {rk} } f$ be the rank of $f$.
Then:
:$\map \dim {\map \Rad V} = n - \map {\operatorname {rk} } f$
where $\dim$ denotes dimension.

Proof:
{{ProofWanted}}
Category:Bilinear Forms (Linear Algebra)
2yz8g3246cqb4qistt492aef52neaj3
Theorem: Dimension of Orthogonal Complement With Respect to Bilinear Form

Let $\mathbb K$ be a field.
Let $V$ be a vector space over $\mathbb K$ of finite dimension.
Let $f$ be a nondegenerate bilinear form on $V$.
Let $U\subset V$ be a subspace.
Let $U^\perp$ be its orthogonal complement.
Then:
:$\map \dim U + \map \dim U^\perp = \map \dim V$

Proof:
{{ProofWanted}}
Category:Bilinear Forms (Linear Algebra)
k4ago7gqoii36ny5o63id7iu434c2mv
Theorem: Pluperfect Digital Invariant has less than 61 Digits

Let $n \in \Z_{>0}$ be a pluperfect digital invariant.
Then $n$ has less than $61$ digits.

Proof:
We have that:
:$n \times 9^n < 10^\paren {n - 1}$
when $n > 60$.
So an $n$-digit integer, for $n > 60$, is always greater than the sum of the $n$th powers of its digits.
{{qed}}
Theorem: Anisotropic Vector Gives Composition of Bilinear Space

Let $\mathbb K$ be a field.
Let $\struct {V, f}$ be a bilinear space over $\mathbb K$.
Let $v \in V$ be anisotropic.
Let $\sequence v$ be its span.
Let $v^\perp$ be its orthogonal complement.
Then $\struct {V, f}$ is the internal orthogonal sum of $\sequence v$ and $v^\perp$:
:$V = \sequence v \oplus v^\perp$

Proof:
{{ProofWanted}}
Category:Bilinear Forms (Linear Algebra)
6yx0osn0ure0y4izg9zk4dgq3191k38
Theorem: Basis of Free Module is No Greater than Generator

Let $R$ be a commutative ring with unity.
Let $M$ be a free $R$-module with basis $B$.
Let $S$ be a generating set for $M$.
Then:
:$\size B \le \size S$.
That is, there exists an injection from $B$ to $S$.

Proof:
Because $S$ is a generating set, there is a surjective homomorphism
:$\phi : R^{\paren S} \to M$
where $R^{\paren S}$ is the free $R$-module on $S$.
Because $B$ is a basis, there is an isomorphism
:$\psi : R^{\paren B} \to M$
Thus $f = \psi^{-1} \circ \phi: R^{\paren S} \to R^{\paren B}$ is a surjective module homomorphism.
By Krull's Theorem, there exists a maximal ideal $M \subset R$.
By Maximal Ideal iff Quotient Ring is Field, $R / M$ is a field.
Let $k = R / M$.
Let $\pi: R \to k$ denote the quotient mapping.
By some universal properties, there exists a $k$-module homomorphism $\bar f: k^{\paren S} \to k^{\paren B}$ such that:
:$\pi^B \circ f = \bar f \circ \pi^S$
Where $\pi^B$ denotes direct sum of module homomorphisms.
Because $\pi^B \circ f$ is surjective, so is $\bar f$.
Let:
:$C$ be the canonical basis of $k^{\paren S}$
:$D$ be the canonical basis of $k^{\paren S}$
Because $C$ is a generator of $k^{\paren S}$ and $\bar f$ is surjective, $\map {\bar f} C$ is a generator of $k^{\paren B}$.
We have:
{{begin-eqn}}
{{eqn | l = \size B
      | r = \size D
      | c = Cardinality of Canonical Basis of Free Module on Set
}}
{{eqn | o = \le
      | r = \size {\map {\bar f} C}
      | c = Basis of Vector Space Injects into Generator
}}
{{eqn | o = \le
      | r = \size C
      | c = Cardinality of Image of Mapping not greater than Cardinality of Domain
}}
{{eqn | r = \size S
      | c = Cardinality of Canonical Basis of Free Module on Set
}}
{{end-eqn}}
{{qed}}
{{explain|find those universal properties}}
Theorem: Extended Rolle's Theorem

Let $f: D \to \R$ be differentiable on a closed interval $I \subseteq \R$.
Let $x_0 < x_1 < \dots < x_n \in I$.
Let $\map f {x_i} = 0$ for $i = 0, \ldots, n$.
Then for all $i = 0, \ldots, n-1$:
:$\exists \xi_i \in \openint {x_i} {x_{i + 1} }: \map {f'} {\xi_i} = 0$

Proof:
Since $f$ is differentiable on $I$, f is differentiable on $\closedint {x_i} {x_{i + 1} }$ for $i = 0, \ldots, n - 1$.
Thus a fortiori, $f$ is also continuous on the closed interval $\closedint {x_i} {x_{i + 1} }$ and differentiable on the open interval $\openint {x_i} {x_{i + 1} }$.
For $i = 0, \ldots, n$ we have $\map f {x_i} = 0$.
Hence the conditions of Rolle's Theorem are fulfilled on each interval which yields the result.
Theorem: Pell's Equation/Examples/61

:$x^2 - 61 y^2 = 1$
has the smallest positive integral solution:
:$x = 1 \, 766 \, 319 \, 049$
:$y = 226 \, 153 \, 980$

Proof:
From Continued Fraction Expansion of $\sqrt {61}$:
:$\sqrt {61} = \sqbrk {7, \sequence {1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14} }$
The cycle is of length is $11$.
By Solution of Pell's Equation, the only solutions of $x^2 - 61 y^2 = 1$ are:
:${p_{11 r} }^2 - 61 {q_{11 r} }^2 = \paren {-1}^{11 r}$
for $r = 1, 2, 3, \ldots$
When $r = 1$ this gives:
:${p_{11}}^2 - 61 {q_{11}}^2 = -1$
which is not the solution required.
When $r = 2$ this gives:
:${p_{22} }^2 - 61 {q_{22} }^2 = 1$
From Convergents of Continued Fraction Expansion of $\sqrt {61}$:
:$p_{22} = 1 \, 766 \, 319 \, 049$
:$q_{22} = 226 \, 153 \, 980$
although on that page the numbering goes from $p_0$ to $p_{21}$, and $q_0$ to $q_{21}$.
{{qed}}
Theorem: No Infinitely Descending Membership Chains/Corollary

There cannot exist a sequence $\sequence {x_n}$ whose domain is $\N_{\gt 0}$ such that:
:$\forall n \in \N_{>0}: x_{n + 1} \in x_n$

Proof:
{{AimForCont}} there is a sequence like that.
From the definition of a sequence, let $f$ be the the mapping that is defined by $\sequence {x_n}$.
Let $\omega$ denote the minimally inductive set.
Let $g: \omega \to \N_{>0}$ be defined as:
:$\map g \alpha = \alpha + 1$
Then the composition $f \circ g$ is a mapping whose domain is $\omega$ such that:
:$\forall n \in \omega: \map {\paren {f \circ g} } {n^+} \in \map {\paren {f \circ g} } n$
But this contradicts No Infinitely Descending Membership Chains.
Therefore by Proof by Contradiction there cannot exist such a sequence.
Hence the result.
{{qed}}
Category:Axiom of Foundation
904ut568vdzgrnrefeup55u9l9ceixf
Theorem: Kaprekar's Process for 2-Digit Numbers

Kaprekar's process, when applied to a non-repdigit $2$-digit positive integer leads to the cycle:
:$09 \to 81 \to 63 \to 27 \to 45 \to 09$
Note that it is important to retain the leading zero on the $9$, or the process trivially terminates in $0$.

Proof:
Let $n \in \Z_{>0}$ be a $2$-digit positive integer.
{{WLOG}}, let $n$ be expressed in decimal notation as:
:$n = 10 x + y$
where:
: $x > y$
:$0 \le x \le 9, 0 \le y \le 9$
The reversal of $n$ is $10 y + x$.
We have:
{{begin-eqn}}
{{eqn | l = 10x + y - 10 y - x
      | r = 9 \left({x - y}\right)
      | c = where $1 \le x - y \le 9$
}}
{{end-eqn}}
Thus after the first iteration of Kaprekar's process, $n$ is one of:
:$09$
:$18$
:$27$
:$36$
:$45$
or their reversals.
Applying Kaprekar's process to $09$ gives:
{{begin-eqn}}
{{eqn | l = 09
      | r = 90 - 09
      | c =
}}
{{eqn | r = 81
      | c =
}}
{{eqn | o = \to
      | r = 81 - 18
      | c =
}}
{{eqn | r = 63
      | c =
}}
{{eqn | o = \to
      | r = 63 - 36
      | c =
}}
{{eqn | r = 27
      | c =
}}
{{eqn | o = \to
      | r = 72 - 27
      | c =
}}
{{eqn | r = 45
      | c =
}}
{{eqn | o = \to
      | r = 45 - 54
      | c =
}}
{{eqn | r = 09
      | c =
}}
{{end-eqn}}
All those multiples of $9$ can be seen to be in (or end up in) this loop.
Hence the result.
{{qed}}
Theorem: Numbers with 6 or more Prime Factors

The sequence of positive integers with $6$ or more prime factors (not necessarily distinct) begins:
:$64, 96, 128, 144, 160, 192, 216, 224, 240, 256, \ldots$
{{OEIS|A046305}}

Proof:
{{begin-eqn}}
{{eqn | l = 64
      | r = 2 \times 2 \times 2 \times 2 \times 2 \times 2
      | c = 
}}
{{eqn | l = 96
      | r = 2 \times 2 \times 2 \times 2 \times 2 \times 3
      | c = 
}}
{{eqn | l = 128
      | r = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \paren {\times 2}
      | c = 
}}
{{eqn | l = 144
      | r = 2 \times 2 \times 2 \times 2 \times 3 \times 3
      | c = 
}}
{{eqn | l = 160
      | r = 2 \times 2 \times 2 \times 2 \times 2 \times 5
      | c = 
}}
{{eqn | l = 192
      | r = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \paren {\times 3}
      | c = 
}}
{{eqn | l = 216
      | r = 2 \times 2 \times 2 \times 3 \times 3 \times 3
      | c = 
}}
{{eqn | l = 224
      | r = 2 \times 2 \times 2 \times 2 \times 2 \times 7
      | c = 
}}
{{eqn | l = 240
      | r = 2 \times 2 \times 2 \times 2 \times 3 \times 5
      | c = 
}}
{{eqn | l = 256
      | r = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \paren {\times 2 \times 2}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Cube as Sum of Sequence of Centered Hexagonal Numbers

:$C_n = \ds \sum_{i \mathop = 1}^n H_i$
where:
:$C_n$ denotes the $n$th cube number
:$H_i$ denotes the $i$th centered hexagonal number.

Proof:
From Closed Form for Centered Hexagonal Numbers:
:$H_n = 3 n \paren {n - 1} + 1$
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n H_i
      | r = \sum_{i \mathop = 1}^n \paren {3 i \paren {i - 1} + 1}
      | c = 
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {3 i^2 - 3 i + 1}
      | c = 
}}
{{eqn | r = 3 \sum_{i \mathop = 1}^n i^2 - 3 \sum_{i \mathop = 1}^n i + \sum_{i \mathop = 1}^n 1
      | c = 
}}
{{eqn | r = 3 \frac {n \paren {n + 1} \paren {2 n + 1} } 6 - 3 \sum_{i \mathop = 1}^n i + \sum_{i \mathop = 1}^n 1
      | c = Sum of Sequence of Squares
}}
{{eqn | r = 3 \frac {n \paren {n + 1} \paren {2 n + 1} } 6 - 3 \frac {n \paren {n + 1} } 2 + \sum_{i \mathop = 1}^n 1
      | c = Closed Form for Triangular Numbers
}}
{{eqn | r = \frac {n \paren {\paren {n + 1} \paren {2 n + 1} - 3 \paren {n + 1} + 2} } 2
      | c = simplification
}}
{{eqn | r = \frac {n \paren {2 n^2 + 3 n + 1 - 3 n - 3 + 2} } 2
      | c = multiplying out
}}
{{eqn | r = \frac {n \paren {2 n^2} } 2
      | c = simplification
}}
{{eqn | r = n^3
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Existence of Number to Power of Prime Minus 1 less 1 divisible by Prime Squared

Let $p$ be a prime number.
Then there exists at least one positive integer $n$ greater than $1$ such that:
:$n^{p - 1} \equiv 1 \pmod {p^2}$

Proof:
{{begin-eqn}}
{{eqn | l = p^2
      | o = \equiv
      | r = 0
      | rr = \pmod {p^2}
}}
{{eqn | l = 1
      | o = \equiv
      | r = 1
      | rr = \pmod {p^2}
}}
{{eqn | ll= \leadsto
      | l = p^2 + 1
      | o = \equiv
      | r = 0 + 1
      | rr= \pmod {p^2}
      | c = Modulo Addition is Well-Defined
}}
{{eqn | o = \equiv
      | r = 1
      | rr = \pmod {p^2}
}}
{{eqn | ll= \leadsto
      | l = \paren {p^2 + 1}^{p - 1}
      | o = \equiv
      | r = 1^{p - 1}
      | rr= \pmod {p^2}
      | c = Congruence of Powers
}}
{{eqn | o = \equiv
      | r = 1
      | rr= \pmod {p^2}
}}
{{end-eqn}}
Hence $p^2 + 1$ fulfils the conditions for the value of $n$ whose existence was required to be demonstrated.
{{qed}}
Theorem: Sum of 2 Squares in 2 Distinct Ways which is also Sum of Cubes

The smallest positive integer which is both the sum of $2$ square numbers in two distinct ways and also the sum of $2$ cube numbers is $65$:
{{begin-eqn}}
{{eqn | l = 65
      | m = 16 + 49
      | mo= =
      | r = 4^2 + 7^2
      | c = 
}}
{{eqn | m = 1 + 64
      | mo= =
      | r = 1^2 + 8^2
      | c = 
}}
{{eqn | o = 
      | mo= =
      | r = 1^3 + 4^3
      | c = 
}}
{{end-eqn}}

Proof:
From Sum of 2 Squares in 2 Distinct Ways, the smallest $2$ positive integer which are the sum of $2$ square numbers in two distinct ways are $50$ and $65$.
But $50$ cannot be expressed as the sum of $2$ cube numbers:
{{begin-eqn}}
{{eqn | l = 50 - 1^3
      | r = 49
      | c = which is not cubic
}}
{{eqn | l = 50 - 2^3
      | r = 42
      | c = which is not cubic
}}
{{eqn | l = 50 - 3^3
      | r = 23
      | c = which is not cubic
}}
{{eqn | l = 50 - 4^3
      | r = -14
      | c = and we have fallen off the end
}}
{{end-eqn}}
Hence $65$ is that smallest number.
{{qed}}
Theorem: Magic Constant of Order 5 Magic Square

The magic constant of the order $5$ magic square is $65$.

Proof:
Let $M_5$ denote an order $5$ magic square
By Sum of Terms of Magic Square, the total of all the entries in $M_5$ is given by:
:$T_5 = \dfrac {5^2 \paren {5^2 + 1} } 2 = \dfrac {25 \times 26} 2 = 325$
As there are $5$ rows of $M_5$, the magic constant of $M_5$ is given by:
:$S_5 = \dfrac {325} 5 = 65$
{{qed}}
Theorem: Triple of Triangular Numbers whose Pairwise Sums are Triangular

The following triplet of triangular numbers has the property that the sum of each pair of them, and their total, are all triangular numbers:
:$66, 105, 105$

Proof:
Throughout we use Closed Form for Triangular Numbers, which gives that the $n$th triangular number can be expressed as:
:$T_n = \dfrac {11 \times 12} 2$
We have:
{{begin-eqn}}
{{eqn | l = 66
      | r = \frac {11 \times 12} 2
      | c = and so is triangular
}}
{{eqn | l = 105
      | r = \frac {14 \times 15} 2
      | c = and so is triangular
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = 66 + 105
      | r = 171
      | c = 
}}
{{eqn | r = \frac {18 \times 19} 2
      | c = and so is triangular
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 105 + 105
      | r = 210
      | c = 
}}
{{eqn | r = \frac {20 \times 21} 2
      | c = and so is triangular
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 66 + 105 + 105
      | r = 210
      | c = 
}}
{{eqn | r = \frac {23 \times 24} 2
      | c = and so is triangular
}}
{{end-eqn}}
{{qed}}
Theorem: Largest Even Integer not expressible as Sum of 2 k Odd Composite Integers

Let $k \in \Z_{>0}$ be a (strictly) positive integer.
The largest even integer which cannot be expressed as the sum of $2 k$ odd positive composite integers is $18 k + 20$.

Proof:
Let $n$ be an even integer greater than $18 k + 20$.
Then $n - 9 \paren {2 k - 2}$ is an even integer greater than $18 k + 20 - 9 \paren {2 k - 2} = 38$.
By Positive Even Integers not Expressible as Sum of 2 Composite Odd Numbers, every even integer greater than $38$ can be expressed as the sum of $2$ odd positive composite integers.
Thus $n - 9 \paren {2 k - 2}$ can be expressed as the sum of $2$ odd positive composite integers.
So, let $a$ and $b$ be odd composite integers such that $a + b = n - 9 \paren {2 k - 2}$.
Then:
:$9 \paren {2 k - 2} + a + b = n$
This is an expression for $n$ as the sum of $2 k$ odd positive composite integers, in which $2 k - 2$ of them are occurrences of $9$.
Thus such an expression can always be found for $n > 18 k + 20$.
{{qed|lemma}}
It remains to be shown that $18 k + 20$ is not expressible as the sum of $2 k$ odd positive composite integers.
The $k = 1$ case is demonstrated in Positive Even Integers not Expressible as Sum of 2 Composite Odd Numbers.
Now suppose $k \ge 2$. 
The two smallest odd positive composite integers are $9$ and $15$.
Suppose $18 k + 20$ is expressible as the sum of $2 k$ odd positive composite integers.
Then at least $2 k - 3$ of them are $9$'s, because:
:$9 \paren {2 k - 4} + 4 \times 15 = 18 k + 24 > 18 k + 20$
Then the problem reduces to finding an expression of $18 k + 20 - 9 \paren {2 k - 3} = 47$ as the sum of $3$ odd positive composite integers.
The first few odd positive composite integers are:
:$9, 15, 21, 25, 27, 33, 35, 39, 45$
Their differences with $47$ are:
:$38, 32, 26, 22, 20, 14, 12, 8, 2$
The integers above are in the set of integers not expressible as a sum of $2$ odd positive composite integers.
The full set of these can be found in Positive Even Integers not Expressible as Sum of 2 Composite Odd Numbers.
Thus $18 k + 20$ is not expressible as the sum of $2 k$ odd positive composite integers.
This proves the result.
{{qed}}
Theorem: Cube of 71 is Odd Integers in Sequence

The cube of $71$, when expressed in decimal notation, is the odd integers from $3$ to $11$ written in sequence:
::$71^3 = 357 \, 911$

Proof:
Verified by direct calculation.
Theorem: Prime Numbers which Divide Sum of All Lesser Primes

The following sequence of prime numbers has the property that each is a divisor of the sum of all primes equal to or smaller than them:
:$2$, $5$, $71$, $369 \, 119$, $415 \, 074 \, 643$, $55 \, 691 \, 042 \, 365 \, 834 \, 801$
{{OEIS|A007506}}
As of the time of the last edit to this page (April $2025$), no others are known.

Proof:
Verified by calculation.
Theorem: 4 Positive Integers in Arithmetic Sequence which have Same Euler Phi Value

The following sets of $4$ positive integers which form an arithmetic sequence are the smallest which all have the same Euler $\phi$ value:
:$72, 78, 84, 90$
:$216, 222, 228, 234$
:$76 \, 326, 76 \, 332, 76 \, 338, 76 \, 344$

Proof:
{{begin-eqn}}
{{eqn | l = 78 - 72
      | r = 6
      | c = 
}}
{{eqn | l = 84 - 78
      | r = 6
      | c = 
}}
{{eqn | l = 90 - 84
      | r = 6
      | c = 
}}
{{end-eqn}}
demonstrating that this is indeed an arithmetic sequence, with a common difference of $6$.
Now we show:
{{begin-eqn}}
{{eqn | l = \map \phi {72}
      | r = 24
      | c = {{EulerPhiLink|72}}
}}
{{eqn | l = \map \phi {78}
      | r = 24
      | c = {{EulerPhiLink|78}}
}}
{{eqn | l = \map \phi {84}
      | r = 24
      | c = {{EulerPhiLink|84}}
}}
{{eqn | l = \map \phi {90}
      | r = 24
      | c = {{EulerPhiLink|90}}
}}
{{end-eqn}}
{{qed|lemma}}
{{ProofWanted|It remains to be shown that this is the smallest such arithmetic sequence of $4$ integers.}}
{{begin-eqn}}
{{eqn | l = 222 - 216
      | r = 6
      | c = 
}}
{{eqn | l = 228 - 222
      | r = 6
      | c = 
}}
{{eqn | l = 234 - 228
      | r = 6
      | c = 
}}
{{end-eqn}}
demonstrating that this is indeed an arithmetic sequence, with a common difference of $6$.
Now we show:
{{begin-eqn}}
{{eqn | l = \map \phi {216}
      | r = 72
      | c = {{EulerPhiLink|216}}
}}
{{eqn | l = \map \phi {222}
      | r = 72
      | c = {{EulerPhiLink|222}}
}}
{{eqn | l = \map \phi {228}
      | r = 72
      | c = {{EulerPhiLink|228}}
}}
{{eqn | l = \map \phi {234}
      | r = 72
      | c = {{EulerPhiLink|234}}
}}
{{end-eqn}}
{{qed|lemma}}
{{begin-eqn}}
{{eqn | l = 76 \, 332 - 76 \, 326
      | r = 6
      | c = 
}}
{{eqn | l = 76 \, 338 - 76 \, 332
      | r = 6
      | c = 
}}
{{eqn | l = 76 \, 344 - 76 \, 338
      | r = 6
      | c = 
}}
{{end-eqn}}
demonstrating that this is indeed an arithmetic sequence, with a common difference of $6$.
Now we show:
{{begin-eqn}}
{{eqn | l = \map \phi {76 \, 326}
      | r = 25 \, 440
      | c = {{EulerPhiLink|76,326|76 \, 326}}
}}
{{eqn | l = \map \phi {76 \, 332}
      | r = 25 \, 440
      | c = {{EulerPhiLink|76,332|76 \, 332}}
}}
{{eqn | l = \map \phi {76 \, 338}
      | r = 25 \, 440
      | c = {{EulerPhiLink|76,338|76 \, 338}}
}}
{{eqn | l = \map \phi {76 \, 344}
      | r = 25 \, 440
      | c = {{EulerPhiLink|76,344|76 \, 344}}
}}
{{end-eqn}}
{{qed}}
Theorem: Positive Integers which are Euler Phi Value for 17 Integers

There are $17$ positive integers which have an Euler $\phi$ value of the following:
:$72, 96, 120, \ldots$

Proof:
From Numbers with Euler Phi Value of 72, it is seen that $17$ positive integers have an Euler $\phi$ value of $72$:
:$73, 91, 95, 111, 117, 135, 146, 148, 152, 182, 190, 216, 222, 228, 234, 252, 270$
{{proof wanted|boring}}
Theorem: Euler Phi Function of Non-Square Semiprime

Let $n \in \Z_{>0}$ be a semiprime with distinct prime factors $p$ and $q$.
Let $\map \phi n$ denote the Euler $\phi$ function.
Then:
:$\map \phi n = \paren {p - 1} \paren {q - 1}$

Proof:
As $p$ and $q$ are distinct prime numbers, it follows that $p$ and $q$ are coprime.
Thus by Euler Phi Function is Multiplicative:
:$\map \phi n = \map \phi p \, \map \phi q$
From Euler Phi Function of Prime:
:$\map \phi p = p - 1$
:$\map \phi q = q - 1$
Hence the result.
{{qed}}
Theorem: Euler Phi Function of Non-Square Semiprime

Let $n \in \Z_{>0}$ be a semiprime with distinct prime factors $p$ and $q$.
Let $\map \phi n$ denote the Euler $\phi$ function.
Then:
:$\map \phi n = \paren {p - 1} \paren {q - 1}$

Proof:
A semiprime with distinct prime factors is a square-free integer.
:$\map \phi n = \ds \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$
where $p \divides n$ denotes the primes which divide $n$.
As there are $2$ prime factors: $p$ and $q$, this devolves to:
:$\map \phi n = \paren {p - 1} \paren {q - 1}$
except when $p = 2$, in which case:
:$\map \phi n = q - 1$
But when $p = 2$, we have that $p - 1 = 1$ and so:
:$\paren {p - 1} \paren {q - 1} = q - 1$
Hence the result.
{{qed}}
Theorem: Euler Phi Function of Square-Free Integer

Let $n$ be an integer such that $n \ge 2$.
Let $n$ be square-free.
Let $\map \phi n$ be the Euler $\phi$ function of $n$.
That is, let $\map \phi n$ be the count of strictly positive integers less than or equal to $n$ which are prime to $n$.
Then:
:$\map \phi n = \ds \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$
where $p \divides n$ denotes the primes which divide $n$.

Proof:
We have that the Euler Phi Function is Multiplicative.
Let the prime decomposition of $n$ be:
:$\ds n = \prod_{1 \mathop \le i \mathop \le r} p_i =  p_1 p_2 \cdots p_r$
From the definition of prime number, each of the prime factors of $n$ is coprime to all other divisors of $n$.
From Euler Phi Function of Prime, we have:
:$\map \phi {p_i} = \paren {p_i - 1}$
Thus:
:$\map \phi n = \ds \prod_{1 \mathop \le i \mathop \le r} \paren {p_i - 1}$
or:
:$\map \phi n = \ds \prod_{p \mathop \divides n} \paren {p - 1}$
where $p$ ranges over all prime numbers
When $p = 2$ we have that:
:$p - 1 = 1$
and so:
:$\ds \prod_{p \mathop \divides n} \paren {p - 1} = \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$
Hence the result.
{{qed}}
Theorem: Euler Phi Function of Square-Free Integer

Let $n$ be an integer such that $n \ge 2$.
Let $n$ be square-free.
Let $\map \phi n$ be the Euler $\phi$ function of $n$.
That is, let $\map \phi n$ be the count of strictly positive integers less than or equal to $n$ which are prime to $n$.
Then:
:$\map \phi n = \ds \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$
where $p \divides n$ denotes the primes which divide $n$.

Proof:
From Euler Phi Function of Integer:
:$\ds \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$
As $n$ is square-free:
:$\ds n = \prod_{p \mathop \divides n} p$
Hence:
:$\ds \map \phi n = \prod_{p \mathop \divides n} p \paren {1 - \frac 1 p}$
and so:
:$\ds \map \phi n = \prod_{p \mathop \divides n} \paren {p - 1}$
When $p = 2$ we have that:
:$p - 1 = 1$
and so:
:$\ds \prod_{p \mathop \divides n} \paren {p - 1} = \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$
Hence the result.
{{qed}}
Theorem: Euler Phi Function of 2 times Odd Prime

Let $n \in \Z_{>0}$ be a semiprime of the form $2 p$, where $p$ is an odd prime.
Let $\map \phi n$ denote the Euler $\phi$ function.
Then:
:$\map \phi n = p - 1$

Proof:
By definition $n$ is a semiprime.
As $p$ is an odd prime, $n$ is not square.
Thus from Euler Phi Function of Non-Square Semiprime:
:$\map \phi n = \paren {2 - 1} \paren {p - 1}$
Hence the result.
{{qed}}
Theorem: Set is Not Element of Itself

There cannot exist a set which is an element of itself.
That is:
:$\neg \exists a: a \in a$

Proof:
{{AimForCont}} $a$ is such a set.
Then $a \in a$ and $a \in \set a$.
$a \ne \O$ because the empty set has no elements by definition.
It is also seen that:
{{begin-eqn}}
{{eqn | l = a \cap \set a
      | r = \set {x: x \in a \land x \in \set a}
      | c = {{Defof|Set Intersection}}
}}
{{eqn | r = \set {x: x \in a \land x = a}
      | c = {{Defof|Singleton}}
}}
{{eqn | r = a
}}
{{end-eqn}}
By the Axiom of Foundation:
:$a \cap \set a = \O$
Thus $a = \O$.
But it was previously established that $a \ne \O$.
From this contradiction it follows that there cannot exist such a set.
Hence the result.
{{qed}}
{{AoF}}
Category:Axiom of Foundation
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