hcju/proofwikiqa huggingface dataset:
This is to be done by strong induction on the length of WFFs. By definition of $v$ being a boolean interpretation, $\map v p$ is well-defined for all $p \in \PP_0$, the vocabulary of $\LL_0$. A WFF of length $1$ has (trivially) a unique parsing sequence. Consequently, only a single defining rule for $v$ as a boolean interpretation applies. So the result holds for all WFFs of length $1$. Now, suppose the result is true for all WFFs of length $k$ or less. Let $\mathbf A$ be a WFF of length $k+1$. There are two possibilities: Suppose $\mathbf A = \neg \mathbf B$ for some WFF $\mathbf B$. Then $\mathbf B$ is of length $k$, so by the induction hypothesis has a unique value $v (\mathbf B)$ no matter what parsing sequence is used. So as $\map v {\mathbf A} = \map {f^\neg} {\map v {\mathbf B} }$, it follows that $\mathbf A$ likewise has a unique value. Hence the result holds for $k + 1$ in this situation. Suppose $\mathbf A = \paren {\mathbf B * \mathbf C}$ for some WFFs $\mathbf B$ and $\mathbf C$ and some connective $*$. By Language of Propositional Logic has Unique Parsability, $*$ must be the unique main connective. So $\mathbf B$ and $\mathbf C$ are both WFFs shorter than $k + 1$ and therefore by the induction hypothesis have unique values $\map v {\mathbf B}$ and $\map v {\mathbf C}$. Since $\mathbf A = \paren {\mathbf B * \mathbf C}$ for a unique main connective $*$, it follows that: :$\map v {\mathbf A} = \map {f^*} {\map v {\mathbf B}, \map v {\mathbf C} }$ is well-defined. Hence the result holds for $k + 1$ in this situation. So the result follows by the Second Principle of Mathematical Induction. {{qed}}
It is to be demonstrated that each WFF arises by a unique rule of formation from the bottom-up specification of propositional logic. The rules $\mathbf W : TF$ and $\mathbf W : \mathcal P_0$ need no further treatment. From inspection of the first character it is clear that the remaining $\mathbf W : \neg$ and $\mathbf W : Op$ cannot yield the same WFF. What remains is to establish uniqueness in applying $\mathbf W : \neg$ and $\mathbf W : Op$. For $\mathbf W : \neg$, this means to consider: :$\mathbf A = \neg \mathbf B = \neg \mathbf C$ from which it is immediate that $\mathbf B = \mathbf C$. Lastly, for $\mathbf W : Op$, we have the following lemma: === Lemma === {{:Language of Propositional Logic has Unique Parsability/Lemma}}{{qed|lemma}} Having examined all possible combinations of rules of formation, we conclude that $\mathcal L_0$ has unique parsability. {{qed}}
Let $n$ be an odd number with prime decomposition $\displaystyle n = p_1^{a_1} \cdots p_k^{a_k}$. By Sum of Unitary Divisors of Integer, the sum of its unitary divisors is $\displaystyle \prod_{1 \mathop \le i \mathop \le k} \paren {1 + p_i^{a_i} }$. To be a unitary perfect number, this must be equal to $2 n$. Since each $p$ is odd, each $1 + p_i^{a_i}$ is even. Hence $\displaystyle \prod_{1 \mathop \le i \mathop \le k} \paren {1 + p_i^{a_i} }$ is divisible by $2^k$. Since $n$ is odd, $2 n$ is divisible by $2$ but not $4$. Thus $k = 1$. So $n$ is a prime power. By Sum of Unitary Divisors of Power of Prime, the sum of its unitary divisors is $1 + n$. This cannot be equal to $2 n$, since $n > 1$. Therefore there are no unitary perfect numbers which are odd. {{qed}}
We have: :$90 \, 625^2 = 8 \, 212 \, 8 \mathbf {90 \, 625}$ thus demonstrating it is automorphic. By Automorphic Numbers in Base 10, the only other possible candidate is $6^{5^4}$. However: :$6^{5^4} \equiv 09 \, 376 \pmod {10^5}$ begins with a zero. Hence there are no others. {{qed}}
Let $T_{\mathbf H'}$ be the finite propositional tableau obtained by replacing the hypothesis set $\mathbf H$ of $T$ with $\mathbf H \cup \mathbf H'$. By the Tableau Extension Lemma, $T_{\mathbf H'}$ has a finished extension $T'$. By definition of extension, $T_{\mathbf H'}$ is a rooted subtree of $T'$. But $T_{\mathbf H'}$ and $T$ are equal when considered as rooted trees. The result follows. {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^1 10^{5 k}       | r = 100 \, 001       | c =  }} {{eqn | r = 11 \times 9091       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^2 10^{5 k}       | r = 10 \, 000 \, 100 \, 001       | c =  }} {{eqn | r = 3 \times 31 \times 37 \times 2 \, 906 \, 161       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^3 10^{5 k}       | r = 1 \, 000 \, 010 \, 000 \, 100 \, 001       | c =  }} {{eqn | r = 11 \times 101 \times 3541 \times 9091 \times 27961       | c =  }} {{end-eqn}} Notice that: :$\displaystyle \sum_{k \mathop = 0}^m 10^{5 k} \times R_5 = R_{5 \paren {m + 1} }$ where $R_i$ is the $i$th repunit. Suppose $p \divides m + 1$, where $p$ is a prime that is not 5. By Divisors of Repunit with Composite Index: :$R_p \divides R_{5 \paren {m + 1} }$ By Prime not Divisor implies Coprime, $p$ and $5$ are coprime. By Condition for Repunits to be Coprime, $R_p$ and $R_5$ are coprime. By Euclid's Lemma: :$R_p \divides \dfrac {R_{5 \paren {m + 1} } } {R_5} = n$ Suppose $25 \divides m + 1$. By Divisors of Repunit with Composite Index: :$R_5 \divides R_{25}$ and: :$R_{25} \divides R_{5 \paren {m + 1} }$ So we have: :$\dfrac {R_{25}} {R_5} \divides \dfrac {R_{5 \paren {m + 1} } } {R_5} = n$ The final case is $m + 1 = 5$. This is the case $n = 100 \, 001 \, 000 \, 010 \, 000 \, 100 \, 001$. We have: :$n = 21 \, 401 \times 25 \, 601 \times 182 \, 521 \, 213 \, 001$ Thus all cases are covered. {{qed}}
Performing the calculation using long division: <pre>        0.000007000007...       ------------------ 142857)1.000000000000...        1 000000          999999        --------               1000000                999999               ------- </pre>
Let the homomorphism $\phi: G \to \set e$ be defined as: :$\forall g \in G: \map \phi g = e$ Then: :$\map \ker \phi = G$ and: :$\Img \phi = \set e$ By the First Isomorphism Theorem: :$G / \map \ker \phi \cong \Img \phi$ Hence the result: :$G / G \cong \set e$ {{qed}}
The mistake is either ''and conversely'' or ''equal to $999$'', since $999 \, 999$ is an easy counterexample. Here we will show that the result is equal to '''a multiple of''' $999$ {{iff}} $n$ is divisible by $999$. Write $n = \displaystyle \sum_{i \mathop = 0}^k a_i 10^{3 i}$, where $0 \le a_i < 1000$. This divides the digits of $n$ into groups of $3$. Then the statement is equivalent to: :$999 \divides n \iff 999 \divides \displaystyle \sum_{i \mathop = 0}^k a_i$ This statement is true since: {{begin-eqn}} {{eqn | l = n       | r = \sum_{i \mathop = 0}^k a_i 10^{3 i} }} {{eqn | r = \sum_{i \mathop = 0}^k a_i 1000^i }} {{eqn | o = \equiv       | r = \sum_{i \mathop = 0}^k a_i 1^i       | rr = \pmod {999}       | c = Congruence of Powers }} {{eqn | o = \equiv       | r = \sum_{i \mathop = 0}^k a_i       | rr = \pmod {999} }} {{end-eqn}} {{qed}} === Examples === {{:Integer whose Digits when Grouped in 3s add to Multiple of 999 is Divisible by 999/Examples}}
We have: {{begin-eqn}} {{eqn | l = 148 \, 349       | r = 0 + 9 + 14 \, 833 + 2 + 9 + 133 \, 496       | c =  }} {{eqn | r = \mathop !1 + \mathop !4 + \mathop !8 + \mathop !3 \mathop + \mathop !4 \mathop + \mathop !9       | c =  }} {{end-eqn}} A computer search can verify solutions under $10^6$ (that is, with no more than $6$ digits) in seconds. Let $n$ be a $k$-digit number, for $k \ge 7$. Then the sum of the subfactorials of its digits is not more than $\mathop !9 \times k$. But we have: {{begin-eqn}} {{eqn | l = n       | o = \ge       | r = 10^{k - 1} }} {{eqn | r = 10^6 \times 10^{k - 7} }} {{eqn | o = \ge       | r = 10^6 \times \paren {1 + 9 \paren {k - 7} }       | c = Bernoulli's Inequality }} {{eqn | o = >       | r = 7 \times \mathop !9 \times \paren {9 k - 62}       | c = $7 \times \mathop !9 = 934472$ }} {{eqn | r = \mathop !9 \paren {63 k - 62 \times 7} }} {{eqn | o = >       | r = \mathop !9 \times k       | c = $k \ge 7$ }} {{eqn | o = \ge       | r = \text{sum of the subfactorials of digits of } n }} {{end-eqn}} So no more numbers have this property. {{qed}}
By long division: <pre>        0.000002997000002...       --------------------- 333667)1.000000000000000000          667334        --------          3326660          3003003          -------           3236570           3003003           -------            2335670            2335669            -------                  1000000                   667334                  -------                   ...... </pre> It remains to be shown that $333 \, 667$ is the only prime number with the required property. We have that: :$333 \, 667 \nmid 10$ From Period of Reciprocal of Prime, the period of such a prime is the order of $10$ modulo $p$. That is, the smallest integer $d$ such that: :$10^d \equiv 1 \pmod p$ From the above long division we know that the period of $\dfrac 1 {333 \, 667}$ is $9$, so $10^9 \equiv 1 \pmod {333 \, 667}$. The only other possible primes $p$ whose reciprocals might have a period of $9$ must also satisfy: :$10^9 \equiv 1 \pmod p$ that is: :$p \divides 10^9 - 1$ Consider: {{begin-eqn}} {{eqn | l = 10^9 - 1       | r = 999 \, 999 \, 999       | c =  }} {{eqn | r = 3^4 \times 37 \times 333 \, 667       | c = prime factorization }} {{end-eqn}} Therefore the only other possible primes whose reciprocals might have a period of $9$ are $3$ and $37$. From Period of Reciprocal of 37 has Length 3: :$\dfrac 1 {37} = 0 \cdotp \dot 02 \dot 7$ and trivially: :$\dfrac 1 3 = 0 \cdotp \dot 3$ which has a period of $1$. As required, the only prime number whose reciprocal has a period of $9$ is $333 \, 667$. {{qed}}
For each node $v \in T$, let $p (v)$ be the path from $v$ to $r_T$, the root of $T$. This path is unique by Path in Tree is Unique. Let $\mathcal V$ be the subtree of $T$ consisting those nodes $v$ of $T$ such that $p (v)$ is not contradictory. Suppose that $\mathcal V$ were infinite. Then by König's Tree Lemma, $\mathcal V$ has an infinite branch $\Gamma$. Since $\mathcal V \subseteq T$, it follows that $\Gamma$ is also a branch of $T$. However, by construction, it is impossible that $\Gamma$ is contradictory. This contradicts that $T$ is a tableau confutation. Hence $\mathcal V$ is finite. Next, define a finite propositional tableau $T'$ by: :$v \in T' \iff \pi (v) \in \mathcal V$ that is, the rooted tree formed by $\mathcal V$ and all its children. Then by construction, for each leaf node $v$ of $T'$, we have that $v \notin \mathcal V$. That is, that $p (v)$ is a contradictory branch of $T'$. By Leaf of Rooted Tree is on One Branch, every branch of $T'$ is contradictory. Hence $T'$ is a tableau confutation of $\mathbf H'$, as desired. {{qed}}
{{begin-eqn}} {{eqn | l = \frac 1 {\order G} \sum_{g \mathop \in G} \size {X^g}       | r = \frac 1 {\order G} \sum_{g \mathop \in G} \size {\set {x \in X: g x = x} }       | c = by definition }} {{eqn | r = \frac 1 {\order G} \sum_{x \mathop \in X} \size {\set {g \in G: g x = x} }       | c = Same summation, different indexing }} {{eqn | r = \frac 1 {\order G} \sum_{x \mathop \in X} \order {\Stab x}       | c = {{Defof|Stabilizer}} }} {{eqn | r = \frac 1 {\order G} \sum_{x \mathop \in X} \frac {\order G} {\order {\Orb x} }       | c = Orbit-Stabilizer Theorem }} {{eqn | r = \sum_{x \mathop \in X} \frac 1 {\order {\Orb x} }       | c =  }} {{eqn | r = \sum_{\Orb x \mathop \in X / G} \paren {\sum_{x \mathop \in \Orb x} \frac 1 {\order {\Orb x} } }       | c =  }} {{eqn | r = \sum_{\Orb x \mathop \in X / G} 1       | c =  }} {{eqn | r = \order {X / G}       | c =  }} {{end-eqn}} {{qed}}
Let $C$ denote the sequence in question. We have that: :$\dfrac {1 \, 122 \, 659 - 1} 2 = 561 \, 329 = 83 \times 6763$ and so is not prime. Thus $1 \, 122 \, 659$ is not a safe prime, as is required for $C$ to be a Cunningham chain of the first kind. Then: {{begin-eqn}} {{eqn | l = 2 \times 561 \, 329 + 1       | r = 1 \, 122 \, 659       | c = which is the $87 \, 359$th prime }} {{eqn | l = 2 \times 1 \, 122 \, 659 + 1       | r = 2 \, 245 \, 319       | c = which is the $165 \, 760$th prime }} {{eqn | l = 2 \times 2 \, 245 \, 319 + 1       | r = 4 \, 490 \, 639       | c = which is the $315 \, 347$th prime }} {{eqn | l = 2 \times 4 \, 490 \, 639 + 1       | r = 8 \, 981 \, 279       | c = which is the $601 \, 286$th prime }} {{eqn | l = 2 \times 8 \, 981 \, 279 + 1       | r = 17 \, 962 \, 559       | c = which is the $1 \, 149 \, 096$th prime }} {{eqn | l = 2 \times 17 \, 962 \, 559 + 1       | r = 35 \, 925 \, 119       | c = which is the $2 \, 199 \, 933$rd prime }} {{eqn | l = 2 \times 35 \, 925 \, 119 + 1       | r = 71 \, 850 \, 239       | c = which is the $4 \, 220 \, 407$th prime }} {{eqn | l = 2 \times 71 \, 850 \, 239 + 1       | r = 143 \, 700 \, 479       | c =  }} {{eqn | r = 13 \times 47 \times 479 \times 491       | c = and so is not prime }} {{end-eqn}} Establishing that this is indeed the smallest such Cunningham chain of the first kind of length $7$ can be done by a computer search. {{qed}}
From Square of Repunit: {{begin-eqn}} {{eqn | l = 121       | r = 11^2 }} {{eqn | l = 12 \, 321       | r = 111^2 }} {{eqn | l = 1 \, 234 \, 321       | r = 1111^2 }} {{end-eqn}} and so on. Then from 1+2+...+n+(n-1)+...+1 = n^2: {{begin-eqn}} {{eqn | l = 1 + 2 + 1       | r = 2^2 }} {{eqn | l = 1 + 2 + 3 + 2 + 1       | r = 3^2 }} {{eqn | l = 1 + 2 + 3 + 4 + 3 + 2 + 1       | r = 4^2 }} {{end-eqn}} and so on. Then: {{begin-eqn}} {{eqn | l = 11^2 \times 2^2       | r = 22^2 }} {{eqn | l = 111^2 \times 3^2       | r = 333^2 }} {{eqn | l = 1111^2 \times 4^2       | r = 4444^2 }} {{end-eqn}} The pattern breaks down after $9$: :$1 \, 111 \, 111 \, 111^2 = 1 \, 234 \, 567 \, 900 \, 987 \, 654 \, 321$ {{qed}}
Suppose $m, n \in \N$ and $m < n$. Write $\map F {n, m} = n! \paren {n + 1}! \cdots m!$. Suppose we have $\map F {n, m} > r!$ for some $r \in \N$. Suppose further that there is a prime $p$ where $m < p \le r$. We claim that $\map F {n, m}$ cannot be a factorial of any number. {{AimForCont}} $\map F {n, m} = s!$ for some $s \in \N$. Since $s! > r!$, we must have $r! \divides s!$. Since $p \le r$, we must have $p \divides r!$. Thus we have $p \divides s!$. However, since $n, n + 1, \dots, m < p$, we must have $p \nmid k!$ for each $n \le k \le m$. Thus $p \nmid \map F {n, m} = s!$, which is a contradiction. Hence $\map F {n, m}$ cannot be a factorial of some number. We have the following lemmata: === Lemma 1 === {{:Factorial as Product of Consecutive Factorials/Lemma 1}}{{qed|lemma}} === Lemma 2 === {{:Factorial as Product of Consecutive Factorials/Lemma 2}}{{qed|lemma}} We also have: :$\forall n \in \N: n > 1 \implies \exists p: 2 n < p < 3 n$ {{finish|The result above is shown by M. El Bachraoui in their paper Primes in the Interval [2n, 3n]. I do not have the time to comprehend their proof yet}} === Case $1$: $m$ is an even number larger than $2$ === Write $m = 2 k$, where $k > 1$. Then: {{begin-eqn}} {{eqn | l = \map F {n, m}       | o = \ge       | r = \paren {m - 1}! \, m! }} {{eqn | r = \paren {2 k - 1}! \, \paren {2 k}! }} {{eqn | o = >       | r = \paren {3 k - 1}!       | c = Lemma 1 }} {{end-eqn}} There is a prime $p$ where $m = 2 k < p \le 3 k - 1$. Therefore $\map F {n, m}$ is not a factorial of some number. === Case $2$: $m$ is an odd number larger than $11$ === Write $m = 2 k - 1$, where $k \ge 7$. Then: {{begin-eqn}} {{eqn | l = \map F {n, m}       | o = \ge       | r = \paren {m - 1}! \, m! }} {{eqn | r = \paren {2 k - 2}! \, \paren {2 k - 1}! }} {{eqn | o = >       | r = \paren {3 k - 1}!       | c = Lemma 2 }} {{end-eqn}} There is a prime $p$ where $m = 2 k - 1 < 2 k < p \le 3 k - 1$. Therefore $\map F {n, m}$ is not a factorial of some number. === Case $3$: Particular values of $m$ === The cases above leaves us with $m = 1, 2, 3, 5, 7, 9, 11$ to check. We have: {{begin-eqn}} {{eqn | l = 10! \times 11!       | o = >       | r = 13!       | c = and $13$ is a prime }} {{eqn | l = 8! \times 9!       | o = >       | r = 11!       | c = and $11$ is a prime }} {{eqn | l = 6! \times 7!       | r = 10! }} {{eqn | l = 5! \times 6! \times 7!       | o = >       | r = 11!       | c = and $11$ is a prime }} {{eqn | l = 4! \times 5!       | r = 2880       | c = is not a factorial of some number }} {{eqn | l = 3! \times 4! \times 5!       | o = >       | r = 7!       | c = and $7$ is a prime }} {{eqn | l = \map F {n, 3}       | r = 12       | c = is not a factorial of some number }} {{eqn | l = \map F {n, 2}       | r = 2! }} {{eqn | l = \map F {0, 1}       | r = 1! }} {{eqn | r = 0! }} {{end-eqn}} Thus there are no more. {{qed}}
{{begin-eqn}} {{eqn | l = \map \tau 1       | r = 1       | c = {{TauLink|1}} }} {{eqn | l = \map \tau {625}       | r = 5       | c = {{TauLink|625}} }} {{eqn | l = \map \tau {6561}       | r = 9       | c = {{TauLink|6561}} }} {{eqn | l = \map \tau {4 \, 100 \, 625}       | r = 45       | c = {{TauLink|4,100,625|4 \, 100 \, 625}} }} {{end-eqn}} Suppose $N = \map \tau {N^4}$. By Tau Function Odd Iff Argument is Square, $N$ must be odd. The case $N = 1$ is trivial. Suppose $N$ is a prime power. Write $N = p^n$. By Tau of Power of Prime: :$N = \map \tau {p^{4 n} } = 4 n + 1$ By Bernoulli's Inequality: :$N = p^n \ge 1 + n \paren {p - 1}$ This gives us the inequality: :$4 n + 1 \ge 1 + n \paren {p - 1}$ which can be simplified to: :$4 \ge p - 1$ The only odd primes satisfying the inequality are $3$ and $5$. We have: :$\map \tau {3^4} = 5 > 3^1$ :$\map \tau {3^8} = 9 = 3^2$ :$\map \tau {3^{4 n} } = 4 n + 1 < 3^n$ for $n > 2$ :$\map \tau {5^4} = 5 = 5^1$ :$\map \tau {5^{4 n} } = 4 n + 1 < 5^n$ for $n > 1$ :$\map \tau {p^{4 n} } = 4 n + 1 < p^n$ for any $p > 5$ Hence $625$ and $6561$ are the only prime powers satisfying the property. Note that Tau Function is Multiplicative. To form an integer $N$ with our property, we must choose and multiply prime powers from the list above. The product $625 \times 6561 = 4 \, 100 \, 625$ gives equality. If we chose any $\tuple {p, n}$ with $\map \tau {p^{4 n} } < p^n$, we must choose $3^4$ in order for equality to possibly hold. Then $\map \tau {3^4} = 5 \divides N$, so a tuple $\tuple {5, n}$ must be chosen. If $\tuple {5, 1}$ was chosen, $5^2 \nmid N$. But $\map \tau {3^4 \times 5^4} = 25 \divides N$, which is a contradiction. Suppose $\tuple {5, n}$ with $n \ge 2$ was chosen. Then $\map \tau {3^4 \times 5^{4 n} } = 20 n + 1 < 3 \times 5^n$, a contradiction. Thus we have exhausted all cases. {{qed}}
Define: :$\ds S = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n \paren {T - I}^n$ $S$ converges since $\norm {T - I} < 1$. We have that $\ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n \norm {T - I}^n$ is the Newton-Mercator Series. This converges since $\norm {T - I} < 1$. Hence the series for $S$ converges absolutely, and so $S$ is well defined. Using the series definition for the matrix exponential:  {{begin-eqn}}  {{eqn | l = e^S       | r = I + S + \frac 1 {2!} S^2 + \frac 1 {3!} S^3 + \cdots }} {{eqn | r = I + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n \paren {T - I}^n + \frac 1 {2!} \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n \paren {T - I}^n}^2 + \frac 1 {3!} \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n \paren {T - I}^n}^3 + \cdots }} {{eqn | ll= \leadsto       | l = e^S       | r = I + \paren {T - I} + c_2 \paren {T - I}^2 + c_3 \paren {T - I}^3 + \cdots       | c = grouping terms by powers of $T - I$ }} {{eqn | r = T + c_2 \paren {T - I}^2 + c_3 \paren {T - I}^3 + \cdots }} {{end-eqn}} If $c_i = 0$ for $i \ge 2$, then $e^S = T$, and the result is shown. The Newton-Mercator Series is a Taylor expansion for $\map \ln {1 + x}$. When combined with the Power Series Expansion for Exponential Function, it gives:  {{begin-eqn}} {{eqn | l = e^{\map \ln {1 + x} }       | r = 1 + \map \ln {1 + x} + \frac 1 {2!} \paren {\map \ln {1 + x} }^2 + \frac 1 {3!} \paren {\map \ln {1 + x} }^3 + \cdots }} {{eqn | r = 1 + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n x^n + \frac 1 {2!} \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n x^n}^2 + \frac 1 {3!} \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n x^n}^3 + \cdots }} {{eqn | r = 1 + x + c_2 x^2 + c_3 x^3 + \cdots       | c = grouping terms by powers of $x$ }} {{end-eqn}} But $e^{\map \ln {1 + x} } = 1 + x$. Thus: :$1 + x = 1 + x + c_2 x^2 + c_3 x^3 + \cdots \implies c_i = 0$ for $i \ge 2$. {{qed}}
Let $x_n \to x$ in $V$. We have: :$x_n \to x \implies \norm {x_n - x} \to 0$ By the Reverse Triangle Inequality: :$\size {\norm {x_n} - \norm x} \le \norm {x_n - x}$ Hence: :$\size {\norm {x_n} - \norm {x_n} } \to 0$ {{mistake|Should the above read $\size {\norm {x_n} - \norm x} \to 0$?}} Thus: :$\norm {x_n} \to \norm x$ {{finish|Explain why, with use of links, this shows continuity.}} Category:Norm Theory qnjnc1afwm0bbaokk4abjx9wel2yoyw
{{begin-eqn}} {{eqn | l = 9 \, 843 \, 019 + 0 \times 30       | r = 9 \, 843 \, 019       | c = which is the $654 \, 926$th prime }} {{eqn | l = 9 \, 843 \, 019 + 1 \times 30       | r = 9 \, 843 \, 049       | c = which is the $654 \, 927$th prime }} {{eqn | l = 9 \, 843 \, 019 + 2 \times 30       | r = 9 \, 843 \, 079       | c = which is the $654 \, 928$th prime }} {{eqn | l = 9 \, 843 \, 019 + 3 \times 30       | r = 9 \, 843 \, 109       | c = which is the $654 \, 929$th prime }} {{eqn | l = 9 \, 843 \, 019 + 4 \times 30       | r = 9 \, 843 \, 139       | c = which is the $654 \, 930$th prime }} {{end-eqn}} But note that $9 \, 843 \, 019 + 5 \times 30 = 9 \, 843 \, 169 = 7^2 \times 200 \, 881$ and so is not prime. Inspection of tables of primes (or a computer search) will reveal that this is the smallest such sequence. {{qed}}
We have that: :the $4485-5852-7373$ triangle $T_1$ is Pythagorean :the $3059-8580-9109$ triangle $T_2$ is Pythagorean :the $1380-19 \, 019-19 \, 069$ triangle $T_3$ is Pythagorean. Then from Area of Triangle, their areas $A_1$, $A_2$ and $A_3$ respectively are given by: {{begin-eqn}} {{eqn | l = A_1       | r = \dfrac {4485 \times 5852} 2       | c =  }} {{eqn | r = \dfrac {\paren {3 \times 5 \times 13 \times 23} \times \paren {2^2 \times 7 \times 11 \times 19} } 2       | c =  }} {{eqn | r = 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 19 \times 23       | c =  }} {{eqn | r = 13 \, 123 \, 110       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = A_2       | r = \dfrac {3059 \times 8580} 2       | c =  }} {{eqn | r = \dfrac {\paren {7 \times 19 \times 23} \times \paren {2^2 \times 3 \times 5 \times 11 \times 13} } 2       | c =  }} {{eqn | r = 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 19 \times 23       | c =  }} {{eqn | r = 13 \, 123 \, 110       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = A_3       | r = \dfrac {1380 \times 19 \, 019} 2       | c =  }} {{eqn | r = \dfrac {\paren {2^2 \times 3 \times 5 \times 23} \times \paren {7 \times 11 \times 13 \times 19} } 2       | c =  }} {{eqn | r = 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 19 \times 23       | c =  }} {{eqn | r = 13 \, 123 \, 110       | c =  }} {{end-eqn}} {{ProofWanted|It remains to be shown that this is the smallest such triple.}}