avewright/proofwiki-math huggingface dataset:
Theorem: Infinite Number of Even Fermat Pseudoprimes

Despite their relative rarity, there exist an infinite number of even Fermat pseudoprimes.

Proof:
In the context of Wells, Fermat pseudoprime probably means Fermat pseudoprimes to the base $2$.
Consider the equation:
:$2^m - 2 \equiv 0 \pmod m$
Any $m$ satisfying the above equation is a Fermat pseudoprime.
We show that for each even $m$ satisfying the above equation, there exists a prime $p$ such that $m p$ also satisfies the equation.
Write $m = 2 n$.
Then $2^{2 n - 1} - 1 \equiv 0 \pmod n$.
By Zsigmondy's Theorem, there exists a prime $p$ such that:
:$p \divides 2^{2 n - 1} - 1$
:$p \nmid 2^k - 1$ for all $k < 2 n - 1$
By this we have $2^{2 n - 1} \equiv 1 \pmod p$.
By Fermat's Little Theorem $2^{p - 1} \equiv 1 \pmod p$.
Let $d = \gcd \set {2 n - 1, p - 1}$.
We have $d \le 2 n - 1$ and $2^d \equiv 1 \pmod p$.
This gives $d = 2 n - 1$ and thus $2 n - 1 \divides p - 1$.
Thus there exists $x \in \N$ such that:
:$p = \paren {2 n - 1} x + 1 > n$
This shows that $p \perp n$.
From $2 p n - 1 = 2 n - 1 + 2 n \paren {p - 1} = \paren {2 n - 1} \paren {1 + 2 n x}$ we have:
:$2^{m p - 1} = 2^{\paren {2 n - 1} \paren {1 + 2 n x}} \equiv 1 \pmod {n, p}$
By Chinese Remainder Theorem:
:$2^{m p - 1} \equiv 1 \pmod {n p}$
By Congruence by Product of Moduli:
:$2^{m p} \equiv 2 \pmod {m p}$
showing that $m p$ is also an even Fermat pseudoprime.
The first even Fermat pseudoprime is $161 \, 038$.
We can use the above method repeatedly to construct infinitely many even Fermat pseudoprimes, though it is impractical since these numbers grow exponentially.
{{qed}}
Theorem: Smith Numbers are Infinite in Number

There are infinitely many Smith numbers.

Proof:
First we prove the the algorithm above does generate Smith numbers.
Let $n \ge 2$.
We have:
:$m = 10^n - 1 = 3 \times 3 \times R_n$
where $R_n$ is the repunit with $n$ digits.
We apply the Lemma, taking note that $r \ge 3$:
:$\map {S_p} m < 9 \map N m - 0.54 \times 3 = 9 n - 1.62$
Since both $\map {S_p} m$ and $9 n$ are integers, the inequality can be rewritten as:
:$h = 9 n - \map {S_p} m \ge 2$
By the Division Algorithm:
:$\exists! a, b \in \Z: \paren {h - 2} = 7 b + a, 0 \le a < 7$
Since $h - 2 \ge 0$, we must also have $b \ge 0$.
Take $x = a + 2$.
Then $2 \le x \le 8$ and:
:$h - 7 b = a + 2 = x$
Hence both $b, x$ exist and are within the desired range.
Note that:
:$\map {S_p} {\set {2, 3, 4, 5, 8, 7, 15} } = \set {2, 3, 4, 5, 6, 7, 8}$
so we can the corresponding value of $t \in \set {2, 3, 4, 5, 8, 7, 15}$ for each $2 \le x \le 8$ such that $\map {S_p} t = x$.
Since $b \ge 0$, $M = t m \times 10^b$ is an integer.
To show that $M$ is a Smith number, we need to show:
:$\map {S_p} M = \map S M$
We have:
{{begin-eqn}}
{{eqn | l = \map {S_p} M
      | r = \map {S_p} t + \map {S_p} m + \map {S_p} {2^b \times 5^b}
}}
{{eqn | r = x + \paren {9 n - h} + 7 b
}}
{{eqn | r = 9 n
}}
{{end-eqn}}
Note that:
:$t \le 15 < 99 = 10^2 - 1 \le 10^n - 1 = m$
Hence we can apply Generalization of Multiple of Repdigit Base minus $1$:
{{begin-eqn}}
{{eqn | l = \map S M
      | r = \map S {t m \times 10^b}
}}
{{eqn | r = \map S {\sqbrk {\paren {t - 1} \paren {10^n - t} } \times 10^b}
      | c = Generalization of Multiple of Repdigit Base minus $1$
}}
{{eqn | r = \map S {\sqbrk {\paren {t - 1} \paren {m - \paren {t - 1} } } }
      | c = $10^b$ only adds trailing zeros
}}
{{eqn | r = \map S m
      | c = no carries occur in the subtraction $m - \paren {t - 1}$
}}
{{eqn | r = 9 n
}}
{{end-eqn}}
{{explain|Explanation of the meaning of the square brackets}}
and thus $\map {S_p} M = \map S M = 9 n$, showing that $M$ is indeed a Smith number.
{{qed|lemma}}
Note that for each $n$, we can generate a Smith number that is greater than $m = 10^n - 1$.
To generate an infinite sequence of Smith numbers, we choose $n$ equal to the number of digits of $M$ previously generated.
Then the next Smith number will be strictly greater than the previous one, thus forming a strictly increasing infinite sequence of Smith numbers.
{{qed}}
Theorem: Construction of Smith Number from Prime Repunit

Let $R_n$ be a repunit which is prime where $n \ge 3$.
Then $3304 \times R_n$ is a Smith number.
$3304$ is not the only number this works for, neither is it the smallest, but it serves as an example of the technique.

Proof:
Let $\map S n$ denote the sum of the digits of a positive integer $n$.
Let $\map {S_p} n$ denote the sum of the digits of the prime decomposition of $n$.
Then $\map S n = \map {S_p} n$ {{iff}} $n$ is a Smith number.
Let $n \ge 3$.
We have that:
:$3304 = 2 \times 2 \times 2 \times 7 \times 59$
and so for a prime repunit $R_n$:
:$3304 \times R_n = 2 \times 2 \times 2 \times 7 \times 59 \times \underbrace {111 \ldots 11}_{n \text { ones} }$
and so:
:$\map {S_p} {3304 \times R_n} = 2 + 2 + 2 + 7 + 5 + 9 + n \times 1 = n + 27$
Note that:
{{begin-eqn}}
{{eqn | r = 3304 \times R_n
      | o =
}}
{{eqn | r = 3304 \sum_{k = 0}^{n - 1} 10^k
      | c = Basis Representation Theorem
}}
{{eqn | r = 3 \sum_{k = 0}^{n - 1} 10^{k + 3} + 3 \sum_{k = 0}^{n - 1} 10^{k + 2} + 4 \sum_{k = 0}^{n - 1} 10^k
      | c = Basis Representation Theorem
}}
{{eqn | r = 3 \sum_{k = 3}^{n + 2} 10^k + 3 \sum_{k = 2}^{n + 1} 10^k + 4 \sum_{k = 0}^{n - 1} 10^k
      | c = Translation of Index Variable of Summation
}}
{{eqn | r = 3 \times 10^{n + 2} + 6 \times 10^{n + 1} + 6 \times 10^n + 10 \sum_{k = 3}^{n - 1} 10^k + 7 \times 10^2 + 44
}}
{{eqn | r = 3 \times 10^{n + 2} + 6 \times 10^{n + 1} + 6 \times 10^n + \sum_{k = 4}^n 10^k + 7 \times 10^2 + 44
      | c = Translation of Index Variable of Summation
}}
{{eqn | r = 3 \times 10^{n + 2} + 6 \times 10^{n + 1} + 7 \times 10^n + \sum_{k = 4}^{n - 1} 10^k + 7 \times 10^2 + 44
}}
{{eqn | r = [367 \underbrace {11 \dots 1}_{\paren {n - 4} \text { ones} } 0744]
      | c = Basis Representation Theorem
}}
{{end-eqn}}
This gives:
{{begin-eqn}}
{{eqn | l = \map S {3304 \times R_n}
      | r = 3 + 6 + 7 + \paren {n - 4} + 0 + 7 + 4 + 4
}}
{{eqn | r = n + 27
}}
{{eqn | r = \map {S_p} {3304 \times R_n}
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Arithmetic Sequence of 16 Primes

The $16$ integers in arithmetic sequence defined as:
:$2\,236\,133\,941 + 223\,092\,870 n$
are prime for $n = 0, 1, \ldots, 15$.

Proof:
First we note that:
:$2\,236\,133\,941 - 223\,092\,870 = 2\,013\,041\,071 = 53 \times 89 \times 426\,763$
and so this arithmetic sequence of primes does not extend to $n < 0$.
{{begin-eqn}}
{{eqn | l = 2\,236\,133\,941 + 0 \times 223\,092\,870
      | r = 2\,236\,133\,941
      | c = which is prime
}}
{{eqn | l = 2\,236\,133\,941 + 1 \times 223\,092\,870
      | r = 2\,459\,226\,811
      | c = which is prime
}}
{{eqn | l = 2\,236\,133\,941 + 2 \times 223\,092\,870
      | r = 2\,682\,319\,681
      | c = which is prime
}}
{{eqn | l = 2\,236\,133\,941 + 3 \times 223\,092\,870
      | r = 2\,905\,412\,551
      | c = which is prime
}}
{{eqn | l = 2\,236\,133\,941 + 4 \times 223\,092\,870
      | r = 3\,128\,505\,421
      | c = which is prime
}}
{{eqn | l = 2\,236\,133\,941 + 5 \times 223\,092\,870
      | r = 3\,351\,598\,291
      | c = which is prime
}}
{{eqn | l = 2\,236\,133\,941 + 6 \times 223\,092\,870
      | r = 3\,574\,691\,161
      | c = which is prime
}}
{{eqn | l = 2\,236\,133\,941 + 7 \times 223\,092\,870
      | r = 3\,797\,784\,031
      | c = which is prime
}}
{{eqn | l = 2\,236\,133\,941 + 8 \times 223\,092\,870
      | r = 4\,020\,876\,901
      | c = which is prime
}}
{{eqn | l = 2\,236\,133\,941 + 9 \times 223\,092\,870
      | r = 4\,243\,969\,771
      | c = which is prime
}}
{{eqn | l = 2\,236\,133\,941 + 10 \times 223\,092\,870
      | r = 4\,467\,062\,641
      | c = which is prime
}}
{{eqn | l = 2\,236\,133\,941 + 11 \times 223\,092\,870
      | r = 4\,690\,155\,511
      | c = which is prime
}}
{{eqn | l = 2\,236\,133\,941 + 12 \times 223\,092\,870
      | r = 4\,913\,248\,381
      | c = which is prime
}}
{{eqn | l = 2\,236\,133\,941 + 13 \times 223\,092\,870
      | r = 5\,136\,341\,251
      | c = which is prime
}}
{{eqn | l = 2\,236\,133\,941 + 14 \times 223\,092\,870
      | r = 5\,359\,434\,121
      | c = which is prime
}}
{{eqn | l = 2\,236\,133\,941 + 15 \times 223\,092\,870
      | r = 5\,582\,526\,991
      | c = which is prime
}}
{{end-eqn}}
But note that $2\,236\,133\,941 + 16 \times 223\,092\,870 = 5\,805\,619\,861 = 79 \times 73\,488\,859$ and so is not prime.
{{qed}}
Theorem: Smallest 17 Primes in Arithmetic Sequence

The smallest $17$ primes in arithmetic sequence are:
:$3\,430\,751\,869 + 87\,297\,210 n$
for $n = 0, 1, \ldots, 16$.

Proof:
First we note that:
:$3\,430\,751\,869 - 87\,297\,210 = 3\,343\,454\,659 = 17\,203 \times 194\,353$
and so this arithmetic sequence of primes does not extend to $n < 0$.
{{begin-eqn}}
{{eqn | l = 3\,430\,751\,869 + 0 \times 87\,297\,210
      | r = 3\,430\,751\,869
      | c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 1 \times 87\,297\,210
      | r = 3\,518\,049\,079
      | c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 2 \times 87\,297\,210
      | r = 3\,605\,346\,289
      | c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 3 \times 87\,297\,210
      | r = 3\,692\,643\,499
      | c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 4 \times 87\,297\,210
      | r = 3\,779\,940\,709
      | c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 5 \times 87\,297\,210
      | r = 3\,867\,237\,919
      | c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 6 \times 87\,297\,210
      | r = 3\,954\,535\,129
      | c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 7 \times 87\,297\,210
      | r = 4\,041\,832\,339
      | c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 8 \times 87\,297\,210
      | r = 4\,129\,129\,549
      | c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 9 \times 87\,297\,210
      | r = 4\,216\,426\,759
      | c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 10 \times 87\,297\,210
      | r = 4\,303\,723\,969
      | c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 11 \times 87\,297\,210
      | r = 4\,391\,021\,179
      | c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 12 \times 87\,297\,210
      | r = 4\,478\,318\,389
      | c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 13 \times 87\,297\,210
      | r = 4\,565\,615\,599
      | c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 14 \times 87\,297\,210
      | r = 4\,652\,912\,809
      | c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 15 \times 87\,297\,210
      | r = 4\,740\,210\,019
      | c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 16 \times 87\,297\,210
      | r = 4\,827\,507\,229
      | c = which is prime
}}
{{end-eqn}}
But note that $3\,430\,751\,869 + 17 \times 87\,297\,210 = 4\,914\,804\,439 = 41 \times 97 \times 1 235807$ and so is not prime.
{{ProofWanted|It remains to be shown that there are no smaller such APs}}
Theorem: Smallest 18 Primes in Arithmetic Sequence

The smallest $18$ primes in arithmetic sequence are:
:$107\,928\,278\,317 + 9\,922\,782\,870 n$
for $n = 0, 1, \ldots, 16$.

Proof:
First we note that:
:$107\,928\,278\,317 - 9\,922\,782\,870 = 98\,005\,495\,447 = 29 \times 149 \times 22\,681\,207$
and so this arithmetic sequence of primes does not extend to $n < 0$.
{{begin-eqn}}
{{eqn | l = 107\,928\,278\,317 + 0 \times 9\,922\,782\,870
      | r = 107\,928\,278\,317
      | c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 1 \times 9\,922\,782\,870
      | r = 117\,851\,061\,187
      | c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 2 \times 9\,922\,782\,870
      | r = 127\,773\,844\,057
      | c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 3 \times 9\,922\,782\,870
      | r = 137\,696\,626\,927
      | c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 4 \times 9\,922\,782\,870
      | r = 147\,619\,409\,797
      | c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 5 \times 9\,922\,782\,870
      | r = 157\,542\,192\,667
      | c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 6 \times 9\,922\,782\,870
      | r = 167\,464\,975\,537
      | c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 7 \times 9\,922\,782\,870
      | r = 177\,387\,758\,407
      | c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 8 \times 9\,922\,782\,870
      | r = 187\,310\,541\,277
      | c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 9 \times 9\,922\,782\,870
      | r = 197\,233\,324\,147
      | c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 10 \times 9\,922\,782\,870
      | r = 207\,156\,107\,017
      | c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 11 \times 9\,922\,782\,870
      | r = 217\,078\,889\,887
      | c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 12 \times 9\,922\,782\,870
      | r = 227\,001\,672\,757
      | c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 13 \times 9\,922\,782\,870
      | r = 236\,924\,455\,627
      | c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 14 \times 9\,922\,782\,870
      | r = 246\,847\,238\,497
      | c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 15 \times 9\,922\,782\,870
      | r = 256\,770\,021\,367
      | c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 16 \times 9\,922\,782\,870
      | r = 266\,692\,804\,237
      | c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 17 \times 9\,922\,782\,870
      | r = 276\,615\,587\,107
      | c = which is prime
}}
{{end-eqn}}
But note that $107\,928\,278\,317 + 18 \times 9\,922\,782\,870 = 286\,538\,369\,977 = 23 \times 181 \times 68\,829\,779$ and so is not prime.
{{ProofWanted|It remains to be shown that there are no smaller such APs}}
Theorem: Prime Gap of 654

There exists a prime gap of $654$ between $11\,000\,001\,446\,613\,353$ and $11\,000\,001\,446\,614\,007$.

Proof:
$11\,000\,001\,446\,613\,353$ is a prime number.
$11\,000\,001\,446\,614\,007$ is a prime number.
It can be checked that all numbers between these two are composite.
{{qed}}
Theorem: Pair of Large Twin Primes

The integers defined as:
:$1\,159\,142\,985 \times 2^{2304} \pm 1$
are a pair of twin primes each with $703$ digits.

Proof:
$1\,159\,142\,985 \times 2^{2304} - 1$:
{{Alpertron-factorizer|date = $6$th March $2022$|time = $0.2$ seconds}}
$1\,159\,142\,985 \times 2^{2304} + 1$:
{{Alpertron-factorizer|date = $6$th March $2022$|time = $0.8$ seconds}}
{{qed}}
Theorem: Integers under Subtraction do not form Group

Let $\struct {\Z, -}$ denote the algebraic structure formed by the set of integers under the operation of subtraction.
Then $\struct {\Z, -}$ is not a group.

Proof:
It is to be demonstrated that $\struct {\Z, -}$ does not satisfy the group axioms.
First it is noted that Integer Subtraction is Closed.
Thus $\struct {\Z, -}$ fulfils {{Group-axiom|0}}.
However, we then have Subtraction on Numbers is Not Associative.
So, for example:
:$3 - \paren {2 - 1} = 2 \ne \paren {3 - 2} - 1 = 0$
Thus it has been demonstrated that $\struct {\Z, -}$ does not satisfy the group axioms.
Hence the result.
{{qed}}
Theorem: Sequence of Powers of Number less than One/Necessary Condition

Let $x \in \R$ be such that $\size{x} < 1$.
Let $\sequence {x_n}$ be the sequence in $\R$ defined as $x_n = x^n$.
Then $\sequence {x_n}$ is a null sequence.

Proof:
{{WLOG}}, assume that $x \ne 0$.
Observe that {{hypothesis}}:
:$0 < \size x < 1$
Thus by Ordering of Reciprocals:
:$\size x^{-1} > 1$
Define:
:$h = \size x^{-1} - 1 > 0$
Then:
:$x = \dfrac 1 {1 + h}$
By the binomial theorem, we have that:
:$\paren {1 + h}^n = 1 + n h + \cdots + h^n > n h$
because $h > 0$.
By Absolute Value Function is Completely Multiplicative, it follows that:
:$0 \le \size {x^n} = \size x^n = \dfrac 1 {\paren {1 + h}^n} < \dfrac 1 {n h}$
From Sequence of Reciprocals is Null Sequence:
:$\dfrac 1 n \to 0$ as $n \to \infty$
By the Multiple Rule for Real Sequences:
:$\dfrac 1 {n h} \to 0$ as $n \to \infty$
By the {{Corollary|Squeeze Theorem for Real Sequences}}:
:$\size {x^n} \to 0$
as $n \to \infty$.
Hence the result, by the definition of a limit.
Theorem: Sequence of Powers of Number less than One/Necessary Condition

Let $x \in \R$ be such that $\size{x} < 1$.
Let $\sequence {x_n}$ be the sequence in $\R$ defined as $x_n = x^n$.
Then $\sequence {x_n}$ is a null sequence.

Proof:
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Suppose that:
:$\exists N \in \N: \size x^N < \epsilon$
Then the result follows by the definition of a limit, because:
:$\forall n \in \N: n \ge N \implies \size {x^n} = \size x^n \le \size x^N < \epsilon$
where Absolute Value Function is Completely Multiplicative is applied.
It remains to consider the case that:
:$\forall n \in \N: \size x^n \ge \epsilon$
By the Axiom of Archimedes, we can choose a natural number $M$ such that:
:$M > \dfrac 1 {\paren {1 - \size x} \epsilon}$
Then, by Sum of Geometric Sequence:
:$\ds M \epsilon \le \sum_{k \mathop = 0}^{M - 1} \size x^k = \frac {1 - \size x^M} {1 - \size x} < M \epsilon$ 
a contradiction.
Hence the result.
{{qed}}
Theorem: Sequence of Powers of Number less than One/Necessary Condition

Let $x \in \R$ be such that $\size{x} < 1$.
Let $\sequence {x_n}$ be the sequence in $\R$ defined as $x_n = x^n$.
Then $\sequence {x_n}$ is a null sequence.

Proof:
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
By the Axiom of Archimedes, there exists a natural number $M$ such that:
:$M > \dfrac 1 {\paren {1 - \size x} \epsilon}$
By the Well-Ordering Principle, there exists a smallest natural number $m$ such that:
:$\exists N \in \N: m > M \size x^N$
Note that:
:$m - 1 \le M \size x^{N + 1}$
By elementary algebra, it follows that:
:$1 - \dfrac 1 m = \dfrac {m - 1} m \le \size x < 1 - \dfrac 1 {M \epsilon}$
Hence, $m < M \epsilon$.
Therefore:
:$\size x^N < \epsilon$
By Absolute Value Function is Completely Multiplicative:
:$\forall n \in \N: n \ge N \implies \size {x^n} = \size x^n \le \size x^N < \epsilon$
Hence the result, by the definition of a limit.
{{qed}}
Theorem: Sequence of Powers of Number less than One/Necessary Condition

Let $x \in \R$ be such that $\size{x} < 1$.
Let $\sequence {x_n}$ be the sequence in $\R$ defined as $x_n = x^n$.
Then $\sequence {x_n}$ is a null sequence.

Proof:
Define:
:$\ds L = \inf_{n \mathop \in \N} \size x^n$
By the Continuum Property, such an $L$ exists in $\R$.
Clearly, $L \ge 0$.
{{AimForCont}} $L > 0$.
Then, by the definition of the infimum, we can choose $n \in \N$ such that $\size x^n < L \size x^{-1}$.
But then $\size x^{n + 1} < L$, which contradicts the definition of $L$.
Therefore, $L = 0$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
By the definition of the infimum, there exists an $N \in \N$ such that $\size x^N < \epsilon$.
It follows that:
:$\forall n \in \N: n \ge N \implies \size {x^n} = \size x^n \le \size x^N < \epsilon$
where Absolute Value Function is Completely Multiplicative is applied.
Hence the result, by the definition of a limit.
{{qed}}
Theorem: Odd Integers under Multiplication do not form Group

Let $S$ be the set of odd integers:
:$S = \set {x \in \Z: \exists n \in \Z: x = 2 n + 1}$
Let $\struct {S, \times}$ denote the algebraic structure formed by $S$ under the operation of multiplication.
Then $\struct {S, \times}$ is not a group.

Proof:
It is to be demonstrated that $\struct {S, \times}$ does not satisfy the group axioms.
First it is noted that Integer Multiplication is Closed.
Then from Odd Number multiplied by Odd Number is Odd, $S$ is closed under $\times$.
Thus $\struct {S, \times}$ fulfils {{Group-axiom|0}}.
From Integer Multiplication is Associative, we have that $\times$ is associative on $S$.
Thus $\struct {S, \times}$ fulfils {{Group-axiom|1}}.
Then we have that:
:$\forall x \in S: 1 \times x = x = x \times 1$
and as $1 \in S$ it follows that $1$ is the identity element of $\struct {S, \times}$
Thus $\struct {S, \times}$ fulfils {{Group-axiom|2}}.
Now consider $3 \in S$.
There exists no $x \in S$ such that $3 \times x = 1$.
Thus $x$ has no inverse element in $S$.
Thus $\struct {S, \times}$ does not fulfil {{Group-axiom|3}}.
Thus it has been demonstrated that $\struct {S, \times}$ does not satisfy the group axioms.
Hence the result.
{{qed}}
Theorem: Sequence of Powers of Number less than One/Sufficient Condition

Let $x \in \R$.
Let $\sequence {x_n}$ be the sequence in $\R$ defined as $x_n = x^n$.
Let $\sequence {x_n}$ be a null sequence.
Then $\size x < 1$.

Proof:
By Reciprocal of Null Sequence:
:$\sequence {x_n}$ converges to $0$ {{iff}} $\sequence {\dfrac 1 {x_n} }$ diverges to $\infty$.
By the definition of divergence to $\infty$:
:$\exists N \in \N: \forall n \ge N: \size {\dfrac 1 {x_n} } > 1$
In particular:
:$\size {\dfrac 1 {x_N} } > 1$
By Ordering of Reciprocals:
:$\size {x_N} < 1$
That is:
:$\size {x_N} = \size {x^N} = \size x^N < 1$
{{AimForCont}} $\size x \ge 1$.
By Inequality of Product of Unequal Numbers:
:$\size x^N \ge 1^N = 1$
This is a contradiction.
So $\size x < 1$ as required.
Theorem: Square Matrices with +1 or -1 Determinant under Multiplication forms Group

Let $n \in \Z_{>0}$ be a strictly positive integer.
Let $S$ be the set of square matrices of order $n$ of real numbers whose determinant is either $1$ or $-1$.
Let $\struct {S, \times}$ denote the algebraic structure formed by $S$ whose operation is (conventional) matrix multiplication.
Then $\struct {S, \times}$ is a group.

Proof:
Taking the group axioms in turn:
Theorem: Real Sine Function is neither Injective nor Surjective

The real sine function is neither an injection nor a surjection.

Proof:
This is immediately apparent from the graph of the sine function:
{{:Graph of Sine Function}}
For example:
:$\map \sin 0 = \map \sin \pi = 0$
and so the real sine function is not an injection.
Then, for example:
:$\nexists x \in \R: \map \sin x = 2$
and so the real sine function is not a surjection.
{{qed}}
Theorem: Sequence of Powers of Number less than One/Complex Numbers

Let $z \in \C$.
Let $\sequence {z_n}$ be the sequence in $\C$ defined as $z_n = z^n$.
Then:
:$\size z < 1$ {{iff}} $\sequence {z_n}$ is a null sequence.

Proof:
By the definition of convergence:
:$\ds \lim_{n \mathop \to \infty} z_n = 0 \iff \lim_{n \mathop \to \infty} \size {z_n} = 0$
By Modulus of Product:
:$\forall n \in \N: \size {z_n} = \size {z^n} = \size z^n$
So:
:$\ds \lim_{n \mathop \to \infty} \size {z_n} = 0 \iff \lim_{n \mathop \to \infty} \size z^n = 0$
Since $\size z \in \R_{\ge 0}$, by Sequence of Powers of Real Number less than One:
:$\ds \lim_{n \mathop \to \infty} \size z^n = 0 \iff \size z < 1$
The result follows.
{{qed}}
Theorem: Sequence of Powers of Number less than One/Normed Division Ring

Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring
Let $x \in R$.
Let $\sequence {x_n}$ be the sequence in $R$ defined as $x_n = x^n$.
Then:
:$\norm x < 1$ {{iff}} $\sequence {x_n}$ is a null sequence.

Proof:
Let $0_R$ be the zero of $R$.
By the definition of convergence:
:$\ds \lim_{n \mathop \to \infty} x_n = 0_R \iff \lim_{n \mathop \to \infty} \norm {x_n} = 0$
By {{Norm-axiom-mult|2}} then for each $n \in \N$:
:$\norm {x_n} = \norm {x^n} = \norm x^n$.
So:
:$\ds \lim_{n \mathop \to \infty} \norm {x_n} = 0 \iff \lim_{n \mathop \to \infty} \norm x^n = 0$
Since $\norm x \in \R_{\ge 0}$, by Sequence of Powers of Number less than One:
:$\ds \lim_{n \mathop \to \infty} \norm x^n = 0 \iff \norm x < 1$
The result follows.
{{qed}}
Theorem: Sequence of Powers of Number less than One/Rational Numbers

Let $x \in \Q$.
Let $\sequence {x_n}$ be the sequence in $\Q$ defined as $x_n = x^n$.
Then:
:$\size x < 1$ {{iff}} $\sequence {x_n}$ is a null sequence.

Proof:
By the definition of convergence of a rational sequence:
:$\sequence {x_n}$ is a null sequence in the rational numbers {{iff}} $\sequence {x_n}$ is a null sequence in the real numbers
By Sequence of Powers of Real Number less than One:
:$\sequence {x_n}$ is a null sequence in the real numbers {{iff}} $\size x < 1$
{{qed}}
Category:Limits of Sequences
Category:Sequence of Powers of Number less than One
hzx5m2gf5fkreya5guv3y0jmzbwmq3f