hcju/proofwikiqa huggingface dataset:
Let $a, b$ be integers with $1 \le b < a \le 8$ and $a + b = 9$. Then: {{begin-eqn}} {{eqn | l = \paren {1 + \sum_{k \mathop = 0}^n a 10^k}^2 - \paren {1 + \sum_{k \mathop = 0}^n b 10^k}^2       | r = \paren {1 + \sum_{k \mathop = 0}^n a 10^k - 1 - \sum_{k \mathop = 0}^n b 10^k} \paren {1 + \sum_{k \mathop = 0}^n a 10^k + 1 + \sum_{k \mathop = 0}^n b 10^k}       | c = Difference of Two Squares }} {{eqn | r = \paren {\sum_{k \mathop = 0}^n \paren {a - b} 10^k} \paren {2 + \sum_{k \mathop = 0}^n 9 \times 10^k}       | c = $a + b = 9$ }} {{eqn | r = \paren {\sum_{k \mathop = 0}^n \paren {a - b} 10^k} \paren {1 + 10^{n + 1} }       | c = Difference of Two Squares }} {{eqn | r = \sum_{k \mathop = 0}^n \paren {a - b} 10^k + \sum_{k \mathop = 0}^n \paren {a - b} 10^{k + n + 1}       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^n \paren {a - b} 10^k + \sum_{k \mathop = n + 1}^{2 n + 1} \paren {a - b} 10^k       | c = Translation of Index Variable of Summation }} {{eqn | r = \sum_{k \mathop = 0}^{2 n + 1} \paren {a - b} 10^k }} {{end-eqn}} which is a repdigit number. The examples above are instances with $\tuple {a, b} = \tuple {5, 4}, \tuple {6, 3}, \tuple {7, 2}, \tuple {8, 1}$. {{qed}}
{{Mistake}} $8042 = 1^3 + 4^3 + 4^3 + 10^3 + 10^3 + 10^3 + 17^3$, among many other expressions. However: $8042$ is conjectured to be the largest positive integer that cannot be expressed as the sum of fewer than $\bf 7$ cubes. $\bf {454}$ is proven to be the largest positive integer that cannot be expressed as the sum of fewer than $8$ cubes.
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}}
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: :$\map {S_p} {3304 \times R_n} = 2 + 2 + 2 + 7 + 5 + 9 + n \times 1 = n + 27$ Then: :$\map S {3304 \times R_n} = n + 27$ because: {{explain|that because}} Hence the result. {{qed}}
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}}
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}}
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}}
$11\,000\,001\,446\,613\,353$ 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}}
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 {{GroupAxiom|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}}
{{WLOG}}, assume that $x \ne 0$. Observe that by 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: :$\left({1 + h}\right)^n = 1 + n h + \cdots + h^n > n h$ 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 to the Squeeze Theorem for Real Sequences: :$\size {x^n} \to 0$ as $n \to \infty$. Hence the result, by the definition of a limit.
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 either 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 Archimedean Principle, we can choose a natural number $M$ such that: :$M > \dfrac 1 {\paren {1 - \size x} \epsilon}$ Then, by Sum of Geometric Sequence: :$\displaystyle M \epsilon \le \sum_{k \mathop = 0}^{M - 1} \size x^k = \frac {1 - \size x^M} {1 - \size x} < M \epsilon$  a contradiction. Hence the result. {{qed}}
Let $\epsilon \in \R_{>0}$ be a strictly positive real number. By the Archimedean Principle, 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, it follows that: :$\forall n \in \N: n \ge N \implies \size {x^n} = \size x^n \le \size x^N < \epsilon$ Hence the result, by the definition of a limit. {{qed}}
Define: :$\displaystyle L = \inf_{n \mathop \in \N} \size x^n$ By the Continuum Property, such an $L$ exists in $\R$. Clearly, $L \ge 0$. Suppose that $L > 0$. 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 either Absolute Value Function is Completely Multiplicative is applied. Hence the result, by the definition of a limit. {{qed}}
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 {{GroupAxiom|0}}. From Integer Multiplication is Associative, we have that $\times$ is associative on $S$. Thus $\struct {S, \times}$ fulfils {{GroupAxiom|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 {{GroupAxiom|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 {{GroupAxiom|3}}. Thus it has been demonstrated that $\struct {S, \times}$ does not satisfy the group axioms. Hence the result. {{qed}}
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.
Taking the group axioms in turn: === $G \, 0$: Closure === Let $\mathbf A, \mathbf B \in S$. By definition of matrix product, $\mathbf {A B}$ is a square matrix of order $n$. From Determinant of Matrix Product: :$\det \mathbf A \det \mathbf B = \map \det {\mathbf {A B} }$ Hence the determinant of $\mathbf {A B}$ is either $1$ or $-1$. Thus $\mathbf {A B} \in S$ and so $\struct {S, \times}$ is closed. {{qed|lemma}} === $G \, 1$: Associativity === We have that Matrix Multiplication is Associative. Thus $\times$ is associative on $\struct {S, \times}$. {{qed|lemma}} === $G \, 2$: Identity === From Unit Matrix is Unity of Ring of Square Matrices, the unit matrix $\mathbf I$ serves as the identity element of $\struct {S, \times}$. {{qed|lemma}} === $G \, 3$: Inverses === Because the determinants of the elements of $S$ are $1$ or $-1$, they are by definition invertible. We have that $\mathbf I$ is the identity element of $\struct {\R, \circ}$. From the definition of invertible matrix, the inverse of any invertible matrix $\mathbf A$ is $\mathbf A^{-1}$. {{qed|lemma}} All the group axioms are thus seen to be fulfilled, and so $\struct {S, \times}$ is a group. {{qed}}
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}}
By the definition of convergence: :$\displaystyle \lim_{n \mathop \to \infty} z_n = 0 \iff \lim_{n \mathop \to \infty} \size {z_n} = 0$ By Modulus of Product: :$\forall n \in \N: \size {z_n} = \size {z^n} = \size z^n$ So: :$\displaystyle \lim_{n \mathop \to \infty} \size {z_n} = 0 \iff \lim_{n \mathop \to \infty} \size z^n = 0$ Since $\size z \in \R_{\ge 0}$, by Sequence of Powers of Real Number less than One: :$\displaystyle \lim_{n \mathop \to \infty} \size z^n = 0 \iff \size z < 1$ The result follows. {{qed}}
Let $0_R$ be the zero of $R$.   By the definition of convergence: :$\displaystyle \lim_{n \mathop \to \infty} x_n = 0_R \iff \lim_{n \mathop \to \infty} \norm {x_n} = 0$ By norm axiom (N2) (Multiplicativity) then for each $n \in \N$: :$\norm {x_n} = \norm {x^n} = \norm x^n$. So: :$\displaystyle \lim_{n \mathop \to \infty} \norm {x_n} = 0 \iff \lim_{n \mathop \to \infty} \norm x^n = 0$ Since $\norm x \in \R_{\ge 0}$, by Sequence of Powers of Number less than One: :$\displaystyle \lim_{n \mathop \to \infty} \norm x^n = 0 \iff \norm x < 1$ The result follows. {{qed}}
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
Let $X, Y, Z$ be sets defined as: {{begin-eqn}} {{eqn | l = X       | r = \set {a, b, c}       | c =  }} {{eqn | l = Y       | r = \set {1, 2}       | c =  }} {{eqn | l = Z       | r = \set {z, y, z}       | c =  }} {{end-eqn}} Let $g: X \to Y$ be defined in two-row notation as: :$\dbinom {a \ b \ c } {1 \ 2 \ 2}$ which is seen by inspection to be a surjection. Let $f: Y \to Z$ be defined in two-row notation as: :$\dbinom {1 \ 2} {x \ y}$ which is seen by inspection to be an injection. The composition $f \circ g$ is seen to be: :$\dbinom {a \ b \ c} {x \ y \ y}$ which is: :not an injection (both $b$ and $c$ map to $y$) :not a surjection (nothing maps to $z$). Hence the result. {{qed}}
From Existence of Cyclic Group of Order n we have that one such group of order $3$ is the cyclic group of order $3$. This is exemplified by the additive group of integers modulo $3$, whose Cayley table can be presented as: {{:Modulo Addition/Cayley Table/Modulo 3}}{{qed|lemma}} Consider an arbitrary group $\struct {G, \circ}$ whose identity element is $e$. Let the underlying set of $G$ be: :$G = \set {e, a, b}$ where $a, b \in G$ are arbitrary. Since $e$ is the identity, we can start off the Cayley table for $G$ as: :$\begin{array}{c|ccc} \circ & e & a & b \\ \hline e & e & a & b \\ a & a &   &   \\ b & b &   &   \\ \end{array}$ Consider the element at $a \circ a$. We have that $a \circ a$ must be either $e$ or $b$, as from Group has Latin Square Property it cannot be $a$. If $a \circ a = e$, then that leaves only $b$ to complete the middle row. But $b$ already exists in the final column. So $a \circ a$ cannot be $e$, so must be $b$. Hence the Cayley table for $G$ so far is: :$\begin{array}{c|ccc} \circ  & e & a & b \\ \hline e & e & a & b \\ a & a & b &   \\ b & b &   &   \\ \end{array}$ The rest of the table is completed by following the result that Group has Latin Square Property: :$\begin{array}{c|ccc} \circ  & e & a & b \\ \hline e & e & a & b \\ a & a & b & e \\ b & b & e & a \\ \end{array}$ and it is seen by inspection that $G$ is indeed the cyclic group of order $3$. {{qed}}
=== Necessary Condition === {{refactor}} [For other proofs of the Necessary Condition visit here.] {{:Sequence of Powers of Number less than One/Necessary Condition/Proof 1}}{{qed|lemma}} === Sufficient Condition === {{:Sequence of Powers of Number less than One/Sufficient Condition}}{{qed}} Category:Limits of Sequences Category:Sequence of Powers of Number less than One 5faaam982gem8e9cdufdfzgmq3qwztx
=== Topologically Equivalent implies Convergently Equivalent === {{:Equivalence of Definitions of Equivalent Division Ring Norms/Topologically Equivalent implies Convergently Equivalent}}{{qed|lemma}} === Convergently Equivalent implies Null Sequence Equivalent === {{:Equivalence of Definitions of Equivalent Division Ring Norms/Convergently Equivalent implies Null Sequence Equivalent}}{{qed|lemma}} === Null Sequence Equivalent implies Open Unit Ball Equivalent === {{:Equivalence of Definitions of Equivalent Division Ring Norms/Null Sequence Equivalent implies Open Unit Ball Equivalent}}{{qed|lemma}} === Open Unit Ball Equivalent implies Norm is Power of Other Norm === {{:Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm}}{{qed|lemma}} === Norm is Power of Other Norm implies Topologically Equivalent === {{:Equivalence of Definitions of Equivalent Division Ring Norms/Norm is Power of Other Norm implies Topologically Equivalent}}{{qed|lemma}} === Norm is Power of Other Norm implies Cauchy Sequence Equivalent === {{:Equivalence of Definitions of Equivalent Division Ring Norms/Norm is Power of Other Norm implies Cauchy Sequence Equivalent}}{{qed|lemma}} === Cauchy Sequence Equivalent implies Open Unit Ball Equivalent === {{:Equivalence of Definitions of Equivalent Division Ring Norms/Cauchy Sequence Equivalent implies Open Unit Ball Equivalent}}{{qed}}
Let $\sequence {x_n}$ converge to $l$ in $\norm {\, \cdot \,}_1$. Let $\epsilon \in \R_{> 0}$ be given. Let $\map {B_\epsilon^2} i$ denote the open ball centered on $l$ of radius $\epsilon$ in $\struct {R, \norm {\, \cdot \,}_2}$. By Open Ball of Metric Space is Open Set then $\map {B_\epsilon^2} l$ is open set in $\struct {R, d_2}$. Since $d_1$ and $d_2$ are topologically equivalent metrics then $\map {B_\epsilon^2} l$ is open set in $\struct {R, d_1}$. By the definition of an open set in a metric space then: :$\exists \delta \in \R_{> 0}: \map {B_\delta^1} l \subseteq \map {B_\epsilon^2} l$ Hence: :$\forall x \in R: \norm {x - l}_1 < \delta \implies \norm {x - l}_2 < \epsilon$ Since $\sequence {x_n}$ converges to $l$ in $\norm {\, \cdot \,}_1$ then: :$\exists N \in \N: \forall n \ge N: \norm {x_n - l}_1 < \delta$ Hence: :$\exists N \in \N: \forall n \ge N: \norm {x_n - l}_2 < \epsilon$ Since $\sequence {x_n}$ and $\epsilon > 0$ were arbitrary then it has been shown that for all sequences $\sequence {x_n}$ in $R$:  :$\sequence {x_n}$ converges to $l$ in $\norm {\, \cdot \,}_1 \implies \sequence {x_n}$ converges to $l$ in $\norm {\, \cdot \,}_2$.  By a similar argument it is shown that for all sequences $\sequence {x_n}$ in $R$: :$\sequence {x_n}$ converges to $l$ in $\norm {\, \cdot \,}_2 \implies \sequence {x_n}$ converges to $l$ in $\norm {\, \cdot \,}_1$.  The result follows.  {{qed}}