avewright/proofwiki-math huggingface dataset:
Theorem: Groups of Order 8

Let $G$ be a group of order $8$.
Then $G$ is isomorphic to one of the following:
:$\Z_8$
:$\Z_4 \oplus \Z_2$
:$\Z_2 \oplus \Z_2 \oplus \Z_2$
:$D_4$
:$\Dic 2$
where:
:$\Z_n$ is the cyclic group of order $n$
:$D_4$ is the dihedral group of order $8$
:$\Dic 2$ is the dicyclic group of order $8$, also known as the quaternion group.

Proof:
The abelian cases are handled by {{Corollary|Abelian Group Factored by Prime}}.
{{qed|lemma}}
Let $G$ be non-abelian.
By Lagrange's Theorem the order of non-identity elements in $G$ is either $2$, $4$ or $8$.
{{AimForCont}} that there exists an order $8$ element.
Then $G$ is generated by this element.
So $G$ is by definition cyclic.
But Cyclic Group is Abelian, contradicting the assumption that $G$ is non-abelian.
So there is no order $8$ element.
By Non-Abelian Order 8 Group has Order 4 Element, there exists at least one order $4$ element in $G$.
Let it be denoted by $a$.
Let $A$ denote the subgroup generated by $a$.
By Lagrange's Theorem there are two cosets in $G$: $A$ and $G \setminus A$.
Let $b \in G \setminus A$.
Then $\set {a, b}$ is a generator of $G$.
Now we consider how $a$ and $b$ interact with each other.
Consider the element $x = b a b^{-1}$.
By Subgroup of Index 2 is Normal, $b A b^{-1} = A$.
So $x \in A$.
By Order of Conjugate Element equals Order of Element, the only possible choices are $x = a$ or $x = a^3$.
If $x = a$, then $a$ and $b$ commute.
Since $\set {a, b}$ generates $G$, this makes $G$ an abelian group, which is a contradiction.
So $b a b^{-1} = a^3$.
It suffices to consider the order of $b$:
:If $\order b = 2$, then $G \cong D_4$.
:If $\order b = 4$, then $G \cong \Dic 2$.
{{qed}}
Theorem: Triangular Numbers which are Product of 3 Consecutive Integers

The $6$ triangular numbers which can be expressed as the product of $3$ consecutive integers are:
:$6, 120, 210, 990, 185 \, 836, 258 \, 474 \, 216$
{{OEIS|A001219}}

Proof:
{{begin-eqn}}
{{eqn | l = T_3
      | r = \frac {3 \left({3 + 1}\right)} 2
      | c = Closed Form for Triangular Numbers
}}
{{eqn | r = 6
      | c = 
}}
{{eqn | r = 1 \times 2 \times 3
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = T_{15}
      | r = \frac {15 \left({15 + 1}\right)} 2
      | c = Closed Form for Triangular Numbers
}}
{{eqn | r = 120
      | c = 
}}
{{eqn | r = 4 \times 5 \times 6
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = T_{20}
      | r = \frac {20 \left({20 + 1}\right)} 2
      | c = Closed Form for Triangular Numbers
}}
{{eqn | r = 210
      | c = 
}}
{{eqn | r = 5 \times 6 \times 7
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = T_{44}
      | r = \frac {44 \left({44 + 1}\right)} 2
      | c = Closed Form for Triangular Numbers
}}
{{eqn | r = 990
      | c = 
}}
{{eqn | r = 9 \times 10 \times 11
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = T_{608}
      | r = \frac {608 \left({608 + 1}\right)} 2
      | c = Closed Form for Triangular Numbers
}}
{{eqn | r = 185 \, 136
      | c = 
}}
{{eqn | r = 56 \times 57 \times 58
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = T_{22 \, 736}
      | r = \frac {22 \, 736 \left({22 \, 736 + 1}\right)} 2
      | c = Closed Form for Triangular Numbers
}}
{{eqn | r = 258 \, 474 \, 216
      | c = 
}}
{{eqn | r = 2^3 \times 3 \times 7^2 \times 11 \times 13 \times 29 \times 53
      | c = 
}}
{{eqn | r = \left({2^2 \times 3 \times 53}\right) \times \left({7^2 \times 13}\right) \times \left({2 \times 11 \times 29}\right)
      | c = 
}}
{{eqn | r = 636 \times 637 \times 638
      | c = 
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown these are the only ones}}
Theorem: First Harmonic Number to exceed 20

The first harmonic number that is greater than $20$ is $H_{272 \, 400 \, 600}$.
That is, the number of terms of the harmonic series required for its partial sum to exceed $20$ is $272 \, 400 \, 600$.

Proof:
We have:
:$H_{272 \, 400 \, 599} = \ds \sum_{k \mathop = 1}^{272 \, 400 \, 599} \frac 1 k \approx 19 \cdotp 99999 \, 99979$
and:
:$H_{272 \, 400 \, 600} = \ds \sum_{k \mathop = 1}^{272 \, 400 \, 600} \frac 1 k \approx 20 \cdotp 00000 \, 00016$
Theorem: First Harmonic Number to exceed 10

The first harmonic number that is greater than $10$ is $H_{12 \, 367}$.
That is, the number of terms of the harmonic series required for its partial sum to exceed $10$ is $12 \, 367$.

Proof:
We have:
:$H_{12 \, 366} = \ds \sum_{k \mathop = 1}^{12 \, 366} \frac 1 k \approx 9 \cdotp 99996 \, 214$
and:
:$H_{12 \, 367} = \ds \sum_{k \mathop = 1}^{12 \, 367} \frac 1 k \approx 10 \cdotp 00004 \, 30083$
Theorem: Canonical Homomorphism to Polynomial Ring is Ring Monomorphism

Let $R$ be a commutative ring with unity.
Let $\struct {R \sqbrk X, \iota, X}$ be a polynomial ring over $R$ in one indeterminate $X$.
Then the canonical homomorphism $\iota : R \to R \sqbrk X$ is a ring monomorphism.

Proof:
Let $\operatorname{id} : R \to R$ be the identity mapping.
Let $1$ be the unity of $R$.
By Identity Mapping is Ring Automorphism, $\operatorname{id}$ is a ring homomorphism.
By Universal Property of Polynomial Ring, there exists a ring homomorphism $h : R \sqbrk X \to R$ with $h \circ \iota = \operatorname{id}$.
By Identity Mapping is Injection, $\operatorname{id}$ is an injection.
By Injection if Composite is Injection, $\iota$ is an injection.
Thus $\iota$ is a ring monomorphism.
{{qed}}
Category:Polynomial Theory
mecsg7q3zqi5xrnyyk8bj9zbe1pdjxf
Theorem: Universal Property of Field of Rational Fractions

Let $R$ be an integral domain.
Let $\struct {\map R x, \iota, x}$ be the field of rational fractions over $R$.
Let $\struct {K, f, a}$ be an ordered triple, where:
:$K$ is a field
:$f : R \to K$ is a unital ring homomorphism
:$a$ is a transcendental element of $K$.
Then there exists a unique unital ring homomorphism $\bar f : \map R x \to K$ such that $\bar f \circ \iota = f$ and $\map {\bar f} x = a$:
:$\xymatrix{
R \ar[d]^\iota \ar[r]^{\forall f} & K\\
\map R x \ar[ru]_{\exists ! \bar f} }$

Proof:
Use Universal Property of Polynomial Ring and Universal Poperty of Field of Fractions.
{{ProofWanted}}
Category:Polynomial Theory
Category:Universal Properties
hj182iw7n4d1b25r40gnv4qd4f2848d
Theorem: Equivalence of Definitions of Unital Subalgebra

Let $R$ be a commutative ring.
Let $\struct {A_R, *}$ be an unital algebra over $R$ whose unit is $1_A$.
Let $\struct {B_R, *}$ be a subalgebra of $A_R$.
{{TFAE|def = Unital Subalgebra}}

Proof:
{{ProofWanted}}
Category:Unital Subalgebras
3hdv92dy2uskggb8ofcf5dd6n3aobr7
Theorem: Polydivisible Number/Examples/381,654,729

The integer $381 \, 654 \, 729$ is the only polydivisible number which is penholodigital.

Proof:
First it is demonstrated that indeed $381 \, 654 \, 729$ has this property:
{{begin-eqn}}
{{eqn | l = 3
      | r = 1 \times 3
}}
{{eqn | l = 38
      | r = 2 \times 19
}}
{{eqn | l = 381
      | r = 3 \times 127
}}
{{eqn | l = 3816
      | r = 4 \times 954
}}
{{eqn | l = 38 \, 165
      | r = 5 \times 7633
}}
{{eqn | l = 381 \, 654
      | r = 6 \times 63 \, 609
}}
{{eqn | l = 3 \, 816 \, 547
      | r = 7 \times 545 \, 221
}}
{{eqn | l = 38 \, 165 \, 472
      | r = 8 \times 4 \, 770 \, 684
}}
{{eqn | l = 381 \, 654 \, 729
      | r = 9 \times 42 \, 406 \, 081
}}
{{end-eqn}}
{{qed|lemma}}
It remains to be demonstrated that it is the only such number.
Let $\sqbrk {abcdefghi}$ be a polydivisible number which is penholodigital.
By Divisibility by 5:
:$e = 5$
By Divisibility by 2:
:$b, d, f, h$ are even.
Hence:
:$a, c, g, i$ are odd.
By Divisibility by 3:
:$a + b + c$, $d + 5 + f$, $g + h + i$ are divisible by $3$.
By Divisibility by 8:
:$\sqbrk {fgh}$ is divisible by $8$.
Because $f$ is even:
:$\sqbrk {f00}$ is divisible by $8$.
We must therefore have:
:$\sqbrk {gh}$ divisible by $8$.
Thus:
:$\sqbrk {gh}$ must be one of $16, 32, 72, 96$.
Because $i$ is odd and $g + h + i$ are divisible by $3$:
:$\sqbrk {ghi}$ must be one of $321, 327, 723, 729, 963$.
By Divisibility by 4:
:$\sqbrk {cd}$ is divisible by $4$.
Because $c$ is odd:
:$d$ must be $2$ or $6$.
Suppose $h = 6$.
Then $\sqbrk {ghi} = 963$ and $d = 2$.
Since $d + 5 + f$ is divisible by $3$ and $f$ is even:
:$f = 8$
The remaining even number $b$ must be $4$.
So:
:$\sqbrk {abcdefghi} = \sqbrk {a4c258963}$
Thus the only remaining options for $a$ and $c$ are $1$ and $7$.
However, we find that neither $1472589$ nor $7412689$ is divisible by $7$.
Therefore we must have:
:$h = 2$
It follow therefore that:
:$d = 6$
Because $d + 5 + f$ is divisible by $3$, and $f$ is even:
:$f = 4$
The remaining even number $b$ must be $8$.
So:
:$\sqbrk {abcdefghi} = \sqbrk {a8c654g2i}$
Since $a + 8 + c$ is divisible by $3$:
:$\sqbrk {a8c}$ must be one of $183, 189, 381, 387, 783, 789, 981, 987$.
Suppose $g = 3$.
Then:
:$\sqbrk {a8c}$ must be one of $189, 789, 981, 987$.
Inspection of the possible numbers reveals that none of $1896543$, $7896543$, $9816543$, $9876543$ is divisible by $7$.
Suppose $g = 7$.
Then:
:$\sqbrk {a8c}$ must be one of $183, 189, 381, 981$.
Among $1836547$, $1896547$, $3816547$, $9816547$, only $3816547$ is divisible by $7$.
The results follow.
{{qed}}
Theorem: Pandigital Product of Pandigital Pairs in 3 Ways

The pandigital integer $0 \, 429 \, 315 \, 678$ can be expressed as the product of a pandigital doubleton in $3$ different ways:
{{begin-eqn}}
{{eqn | l = 0 \, 429 \, 315 \, 678
      | r = 04 \, 926 \times 87 \, 153
}}
{{eqn | r = 07 \, 923 \times 54 \, 186}}
{{eqn | r = 15 \, 846 \times 27 \, 093}}
{{end-eqn}}

Proof:
We have that:
:$0 \, 429 \, 315 \, 678 = 2 \times 3^2 \times 11 \times 19 \times 139 \times 821$
Then:
{{begin-eqn}}
{{eqn | l = 04 \, 926 \times 87 \, 153
      | r = \paren {2 \times 3 \times 821} \times \paren {3 \times 11 \times 19 \times 139}
}}
{{eqn | l = 07 \, 923 \times 54 \, 186
      | r = \paren {3 \times 19 \times 139} \times \paren {2 \times 3 \times 11 \times 821}
}}
{{eqn | l = 15 \, 846 \times 27 \, 093
      | r = \paren {2 \times 3 \times 19 \times 139} \times \paren {3 \times 11 \times 821}
}}
{{end-eqn}}
{{qed}}
Theorem: Smallest Positive Integer which is Sum of 2 Fourth Powers in 2 Ways

The smallest positive integer which can be expressed as the sum of $2$ fourth powers in $2$ different ways is:
{{begin-eqn}}
{{eqn | l = 635 \, 318 \, 657
      | r = 59^4 + 158^4
      | c = 
}}
{{eqn | r = 133^4 + 134^4
      | c = 
}}
{{end-eqn}}

Proof:
The fact that these are the smallest can be demonstrated by calculation.
{{qed}}
Theorem: Largest Penholodigital Square

The largest penholodigital square is $923 \, 187 \, 456$:
:$923 \, 187 \, 456 = 30 \, 384^2$

Proof:
{{ProofWanted|Needs to be demonstrated that there are none higher. Could be done by checking all the squares from $30 \, 385^2$ up to $31 \, 426$ but that's too boring for now.}}
Theorem: Largest 9-Digit Prime Number

The largest prime number with $9$ digits is $999 \, 999 \, 937$.

Proof:
Consider the numbers $\sqbrk {999 \, 999 \, 9ab}$.
Since $999 \, 999 \, 000$ is divisible by $2, 3, 5, 7, 11, 13$,
if $\sqbrk {9ab}$ is divisible by these primes, so is $\sqbrk {999 \, 999 \, 9ab}$.
After this elimination the only $\sqbrk {ab} > 37$ that remains are:
:$41, 43, 47, 53, 61, 67, 71, 77, 83, 89, 91, 97$
We have:
{{begin-eqn}}
{{eqn | l = 999 \, 999 \, 941
      | r = 113 \times 149 \times 59 \, 393
}}
{{eqn | l = 999 \, 999 \, 943
      | r = 5 \, 623 \times 177 \, 841
}}
{{eqn | l = 999 \, 999 \, 947
      | r = 163 \times 6 \, 134 \, 969
}}
{{eqn | l = 999 \, 999 \, 953
      | r = 29 \times 31 \times 773 \times 1 \, 439
}}
{{eqn | l = 999 \, 999 \, 961
      | r = 3 \, 673 \times 272 \, 257
}}
{{eqn | l = 999 \, 999 \, 967
      | r = 3 \, 257 \times 307 \, 031
}}
{{eqn | l = 999 \, 999 \, 971
      | r = 193 \times 5 \, 181 \, 347
}}
{{eqn | l = 999 \, 999 \, 977
      | r = 2 \, 971 \times 336 \, 587
}}
{{eqn | l = 999 \, 999 \, 983
      | r = 337 \times 2 \, 967 \, 359
}}
{{eqn | l = 999 \, 999 \, 989
      | r = 4 \, 327 \times 231 \, 107
}}
{{eqn | l = 999 \, 999 \, 991
      | r = 67 \times 14 \, 925 \, 373
}}
{{eqn | l = 999 \, 999 \, 997
      | r = 71 \times 2 \, 251 \times 6 \, 257
}}
{{end-eqn}}
And we do have $999 \, 999 \, 937$ is prime.
{{qed}}
Theorem: Smallest Pandigital Square

The smallest pandigital square is $1 \, 026 \, 753 \, 849$:
:$1 \, 026 \, 753 \, 849 = 32 \, 043^2$

Proof:
We check all the squares of numbers from $\ceiling {\sqrt {1 \, 023 \, 456 \, 789} } = 31 \, 992$ up to $32 \, 042$, with the following constraints:
Since all these squares has $10$ as its two leftmost digits, the number cannot end with $0$, $1$ or $9$.
A pandigital number is divisible by $9$, so our number must be divisible by $3$.
These constraints leaves us with the following $12$ candidates:
{{begin-eqn}}
{{eqn | l = 31 \, 992^2
      | r = 1 \, 023 \, 488 \, 064
}}
{{eqn | l = 31 \, 995^2
      | r = 1 \, 023 \, 680 \, 025
}}
{{eqn | l = 31 \, 998^2
      | r = 1 \, 023 \, 872 \, 004
}}
{{eqn | l = 32 \, 004^2
      | r = 1 \, 024 \, 256 \, 016
}}
{{eqn | l = 32 \, 007^2
      | r = 1 \, 024 \, 448 \, 049
}}
{{eqn | l = 32 \, 013^2
      | r = 1 \, 024 \, 832 \, 169
}}
{{eqn | l = 32 \, 016^2
      | r = 1 \, 025 \, 024 \, 256
}}
{{eqn | l = 32 \, 022^2
      | r = 1 \, 025 \, 408 \, 484
}}
{{eqn | l = 32 \, 025^2
      | r = 1 \, 025 \, 600 \, 625
}}
{{eqn | l = 32 \, 028^2
      | r = 1 \, 025 \, 792 \, 784
}}
{{eqn | l = 32 \, 034^2
      | r = 1 \, 026 \, 177 \, 156
}}
{{eqn | l = 32 \, 037^2
      | r = 1 \, 026 \, 369 \, 369
}}
{{end-eqn}}
By inspection, none of these numbers are pandigital.
{{qed}}
Theorem: Sound Proof System is Consistent

Let $\LL$ be a logical language.
Let $\mathscr M$ be a formal semantics for $\LL$.
Let $\mathscr P$ be a proof system for $\LL$.
Suppose that $\mathscr P$ is sound for $\mathscr M$.
Then $\mathscr P$ is consistent.

Proof:
By assumption, some logical formula $\phi$ is not an $\mathscr M$-tautology.
Since $\mathscr P$ is sound for $\mathscr M$, $\phi$ is also not a $\mathscr P$-theorem.
But then by definition $\mathscr P$ is consistent.
{{qed}}
Theorem: Equivalence of Definitions of Consistent Proof System

{{TFAE|def = Consistent (Logic)/Proof System/Propositional Logic|view = Consistent Proof System for Propositional Logic}}
Let $\LL_0$ be the language of propositional logic.
Let $\mathscr P$ be a proof system for $\LL_0$.

Proof:
=== Definition 1 implies Definition 2 ===
Suppose that $\neg \vdash_{\mathscr P} \phi$.
Suppose additionally that there is some logical formula $\psi$ such that:
:$\vdash_{\mathscr P} \psi, \neg \psi$
By the Rule of Explosion:
:$\psi, \neg \psi \vdash_{\mathscr P} \phi$
By Provable Consequence of Theorems is Theorem, we conclude:
:$\vdash_{\mathscr P} \phi$
in contradiction to our assumption.
{{qed|lemma}}
Theorem: Rule of Explosion/Variant 3

:$p, \neg p \vdash q$

Proof:
{{BeginTableau|p, \neg p \vdash q|Instance 2 of the Hilbert-style systems}}
{{Assumption|1|p}}
{{Assumption|2|\neg p}}
{{TableauLine
 |n = 3
 |f = q \implies (p \lor q)
 |rlnk = Definition:Hilbert Proof System/Instance 2
 |rtxt = Axiom $A2$
}}
{{TableauLine
 |n = 4
 |f = \neg p \implies (q \lor \neg p)
 |rlnk = Definition:Hilbert Proof System/Instance 2
 |rtxt = Rule $RST \, 1$
 |dep = 3
 |c = $\neg p \, / \, q$, $q \, / \, p$
}}
{{TableauLine
 |n = 5
 |f = q \lor \neg p
 |rlnk = Definition:Hilbert Proof System/Instance 2
 |rtxt = Rule $RST \, 3$
 |dep = 2, 4
}}
{{TableauLine
 |n = 6
 |f = (q \lor \neg p) \implies (\neg p \lor q)
 |rlnk = Definition:Hilbert Proof System/Instance 2
 |rtxt = Axiom $A3$, Rule $RST \, 1$
 |c = $\neg p \, / \, q$, $q \, / \, p$
}}
{{TableauLine
 |n = 7
 |f = \neg p \lor q
 |rlnk = Definition:Hilbert Proof System/Instance 2
 |rtxt = Rule $RST \, 3$
 |dep = 5, 6
}}
{{TableauLine
 |n = 8
 |f = p \implies q
 |rlnk = Definition:Hilbert Proof System/Instance 2
 |rtxt = Rule $RST \, 2 \, (2)$
}}
{{TableauLine
 |n = 9
 |f = q
 |rlnk = Definition:Hilbert Proof System/Instance 2
 |rtxt = Rule $RST \, 3$
 |dep = 1, 8
}}
{{EndTableau}}
{{qed}}
Theorem: Smallest Integer which is Sum of 3 Fifth Powers in 2 Ways

The smallest positive integer which can be expressed as the sum of $3$ fifth powers in $2$ different ways:
The positive integer $1 \, 375 \, 298 \, 099$ can be expressed as the sum of $3$ fifth powers in $2$ different ways:
{{begin-eqn}}
{{eqn | l = 1 \, 375 \, 298 \, 099
      | r = 24^5 + 28^5 + 67^5
      | c = 
}}
{{eqn | r = 3^5 + 54^5 + 62^5
      | c = 
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = 1 \, 375 \, 298 \, 099
      | r = 7 \, 962 \, 624 + 17 \, 210 \, 368 + 1 \, 350 \, 125 \, 107
      | c = 
}}
{{eqn | r = 24^5 + 28^5 + 67^5
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 1 \, 375 \, 298 \, 099
      | r = 243 + 459 \, 165 \, 024 + 916 \, 132 \, 832
      | c = 
}}
{{eqn | r = 3^5 + 54^5 + 62^5
      | c = 
}}
{{end-eqn}}
{{ProofWanted|that it is the smallest}}
Theorem: Automorphic Numbers with 10 Digits

The only $10$-digit automorphic numbers are:
:$1 \, 787 \, 109 \, 376$
:$8 \, 212 \, 890 \, 625$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = 1 \, 787 \, 109 \, 376^2
      | r = \enspace 3 \, 193 \, 759 \, 92 \mathbf {1 \, 787 \, 109 \, 376}
}}
{{eqn | l = 8 \, 212 \, 890 \, 625^2
      | r = 67 \, 451 \, 572 \, 41 \mathbf {8 \, 212 \, 890 \, 625}
}}
{{end-eqn}}
thus demonstrating they are automorphic.
By Automorphic Numbers in Base 10, there are no others.
{{qed}}
Theorem: Left-Truncated Automorphic Number is Automorphic

Let $n$ be an automorphic number, expressed in some conventional number base.
Let any number of digits be removed from the left-hand end of $n$.
Then what remains is also an automorphic number.

Proof:
Let $n$ be an automorphic number of $d$ digits, expressed in base $b$.
By {{Defof|Automorphic Number}}, we have:
:$n^2 \equiv n \pmod {b^d}$
Let some digits be removed from the left-hand end of $n$, so that only $d'$ digits remain.
This only makes sense when $d' < d$.
Define this new number as $n'$.
Then we have:
:$n \equiv n' \pmod {b^{d'} }$
Thus we have:
{{begin-eqn}}
{{eqn | l = n^2
      | o = \equiv
      | r = n
      | rr = \pmod {b^{d'} }
      | c = Congruence by Divisor of Modulus: from $n^2 \equiv n \pmod {b^d}$ and $b^{d'} \divides b^d$
}}
{{eqn | l = n^2
      | o = \equiv
      | r = \paren {n'}^2
      | rr = \pmod {b^{d'} }
      | c = Congruence of Powers: from $n \equiv n' \pmod {b^{d'} }$
}}
{{eqn | ll = \leadsto
      | l = \paren {n'}^2
      | o = \equiv
      | r = n'
      | rr = \pmod {b^{d'} }
}}
{{end-eqn}}
Hence $n'$ is an automorphic number of $d'$ digits in base $b$.
{{qed}}
Theorem: Square whose Divisor Sum is Cubic

The number $1 \, 857 \, 437 \, 604$ is a square number whose divisor sum is a cube.

Proof:
{{begin-eqn}}
{{eqn | l = 1 \, 857 \, 437 \, 604
      | r = 43 \, 098^2
      | c = 
}}
{{eqn | l = \map {\sigma_1} {1 \, 857 \, 437 \, 604}
      | r = 5 \, 168 \, 743 \, 489
      | c = {{DSFLink|1,857,437,604|1 \, 857 \, 437 \, 604}}
}}
{{eqn | r = 1729^3
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Largest Right-Truncatable Primes allowing 1

Let $1$ be temporarily considered to be a prime number.
Under that consideration, the largest right-truncatable prime numbers are:
:$1 \, 979 \, 339 \, 333$
:$1 \, 979 \, 339 \, 339$

Proof:
We have that:
{{begin-eqn}}
{{eqn | o = 
      | r = 1 \, 979 \, 339 \, 333
      | c = is prime
}}
{{eqn | o = 
      | r = 1 \, 979 \, 339 \, 339
      | c = is prime
}}
{{end-eqn}}
For both, the truncation process is the same:
{{begin-eqn}}
{{eqn | o = 
      | r = 197 \, 933 \, 933
      | c = is the $10 \, 970 \, 817$th prime
}}
{{eqn | o = 
      | r = 19 \, 793 \, 393
      | c = is the $1 \, 252 \, 285$th prime
}}
{{eqn | o = 
      | r = 1 \, 979 \, 339
      | c = is the $147 \, 488$th prime
}}
{{eqn | o = 
      | r = 197 \, 933
      | c = is the $17 \, 815$th prime
}}
{{eqn | o = 
      | r = 19 \, 793
      | c = is the $2240$th prime
}}
{{eqn | o = 
      | r = 1979
      | c = is the $299$th prime
}}
{{eqn | o = 
      | r = 197
      | c = is the $45$th prime
}}
{{eqn | o = 
      | r = 19
      | c = is the $8$th prime
}}
{{eqn | o = 
      | r = 1
      | c = has been defined temporarily to be prime
}}
{{end-eqn}}
{{finish|It remains to be demonstrated that there are no such primes which are larger}}
Theorem: Completely Multiplicative Function is Multiplicative

Let $f: \Z \to \Z$ be a function on the integers $\Z$.
Let $f$ be completely multiplicative.
{{Questionable|Complete multiplicativity is defined for fields, but $\Z$ is not a field.}}
Then $f$ is multiplicative.
{{Questionable|Multiplicativity is defined for $f : \N \to \N$, undefined for $f : \Z \to \Z$}}

Proof:
By definition of complete multiplicativity:
:$\forall m, n \in \Z: \map f {m n} = \map f m \map f n$
Hence by True Statement is implied by Every Statement:
:$\forall m, n \in \Z: m \perp n \implies \map f {m n} = \map f m \map f n$
So $f$ is multiplicative.
{{qed}}
Category:Number Theory
Category:Completely Multiplicative Functions
Category:Multiplicative Functions
o0un74g2nun14pdzgs9fdjn2ytf6qzr
Theorem: Smallest n for which 2^n-3 is Divisible by n

The smallest positive integer $n$ such that $2^n - 3$ is divisible by $n$ is $4 \, 700 \, 063 \, 497$.

Proof:
{{ProofWanted|Brute force by the Lehmers, probably}}
Theorem: Smallest Odd Abundant Number not Divisible by 3

The smallest odd abundant number not divisible by $3$ is $5 \, 391 \, 411 \, 025$.

Proof:
We have:
:$5 \, 391 \, 411 \, 025 = 5^2 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23 \times 29$
showing it is not divisible by $3$.
Then from {{DSFLink|5,391,411,025|5 \, 391 \, 411 \, 025}} we have:
:$\map {\sigma_1} {5 \, 391 \, 411 \, 025} = 10 \, 799 \, 308 \, 800 = 2 \times 5 \, 391 \, 411 \, 025 + 16 \, 486 \, 750$
demonstrating that $5 \, 391 \, 411 \, 025$ is abundant.
{{ProofWanted|It is still to be shown it is the smallest such.}}
Theorem: Closure of Irreducible Subspace is Irreducible

Let $T = \struct {S, \tau}$ be a topological space.
Let $Y \subseteq S$ be a subset of $S$ which is irreducible in $T$.
Then its closure $Y^-$ in $T$ is also irreducible in $T$.

Proof:
By definition, $Y$ is an irreducible subset of $S$ in $T$ {{iff}} the subspace $\struct {Y, \tau_Y}$ is an irreducible topological space.
That is, such that two arbitrary non-empty open sets of $\struct {Y, \tau_Y}$ are not disjoint.
The open sets of $T$ in $Y^-$ are the same as the open sets of $\struct {Y, \tau_Y}$.
{{finish|Prove the above statement}}
Hence by definition of irreducible space, any two open sets in $Y^-$ are not disjoint in $Y^-$.
<!--{{explain|Have we proved that while two non-empty open sets in $Y$ have an intersection in $S$, they also have an intersection in $Y$?}}-->
That is $Y^-$ is also irreducible.
More generally, we can also show that if $Y^-$ is irreducible for a subset $Y \subseteq S$, then $Y$ is also irreducible in $T$. 
{{AimForCont}} $Y$ is not irreducible.
Then there exist two proper subsets $Y_1$, $Y_2$ of $Y$ which are closed in $Y$ such that $Y = Y_1 \cup Y_2$.
{{explain|How do we know that?}}
Then:
:$Y^- = {Y_1}^- \cup {Y_2}^-$
which contradicts the assumption.
{{qed}}
Theorem: Closure of Irreducible Subspace is Irreducible

Let $T = \struct {S, \tau}$ be a topological space.
Let $Y \subseteq S$ be a subset of $S$ which is irreducible in $T$.
Then its closure $Y^-$ in $T$ is also irreducible in $T$.

Proof:
In view of {{Defof|Irreducible Subset}}, it suffices to show that for all closed sets $A_1$ and $A_2$ in $T$:
:$Y^- \subseteq A_1 \cup A_2 \implies \exists i_0 \in \set {1, 2} : Y^- \subseteq A_{i_0}$
To this end, let $A_1$ and $A_2$ be closed sets in $T$ such that:
:$Y^- \subseteq A_1 \cup A_2$
Then, in particular:
:$Y \subseteq A_1 \cup A_2$
Since $Y \subseteq S$ is an irreducible subset:
:$\exists i_0 \in \set {1, 2} : Y \subseteq A_{i_0}$
By Closure of Subset of Closed Set of Topological Space is Subset:
:$Y^- \subseteq A_{i_0}$
{{qed}}
Theorem: Point is Contained in Irreducible Component

Let $X$ be a topological space.
Let $x\in X$ be a point.
Then $x$ is contained in some irreducible component of $X$.

Proof:
Because Trivial Topological Space is Irreducible, this is a special case of Irreducible Subspace is Contained in Irreducible Component.
{{qed}}
Category:Irreducible Spaces
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