hcju/proofwikiqa huggingface dataset:
Let $\map F {r, n}$ be the number of ways to distribute $r$ objects into $n$ cells such that there are no more than $2$ objects in each cell. Let there be $d$ cells which are each occupied by $2$ objects. These can be chosen in $\dbinom n d$ ways. There remain $s = r - 2 d$ objects which can then be distributed among $n - d$ cells in $\dbinom {n - d} s$ ways. In each such arrangement, the $r$ objects may be permuted in: :$\dbinom r 2 \dbinom {r - 2} 2 \cdots \dbinom {r - 2 d + 2} 2 \paren {r - 2 d}! = \dfrac {r!} {2^d}$ different ways. Hence: :$\map F {r, n} = \dbinom n d \dbinom {n - d} s \dfrac {r!} {2^d}$ So the probability of exactly $d$ pairs and $s$ singletons, where $d - s \le n$, is given by: :$\dfrac {\map F {r, n} } {n^r}$ If we assume a $365$-day year, we have that the probability that at least $3$ of them have the same birthday is given by: :$\map \Pr r = 1 - \displaystyle \sum_{d \mathop = 0}^{\floor {r / 2} } \dfrac {n! \, r!} {n^r 2^d d! \paren {r - 2 d}! \paren {n + d - r}!}$ where $n = 365$. We require the smallest $r$ for which $\map \Pr r > \dfrac 1 2$. The result yields to calculation. {{qed}}
The factors of $P$ are: :$1, 2, 4, \dots, 2^{n - 1}, 2^n - 1, 2 \paren {2^n - 1}, \dots, 2^{n - 1} \paren {2^n - 1}$ Therefore their product is: {{begin-eqn}} {{eqn | l = \prod_{d \mathop \divides P} d       | r = \paren {\prod_{i \mathop = 0}^{n - 1} 2^i} \paren {\prod_{i \mathop = 0}^{n - 1} 2^i \paren {2^n - 1} } }} {{eqn | r = \paren {\prod_{i \mathop = 0}^{n - 1} 2^i}^2 \paren {2^n - 1}^n }} {{eqn | r = \paren {2^\frac {n \paren {n - 1} } 2}^2 \paren {2^n - 1}^n }} {{eqn | r = \paren {2^{n - 1} }^n \paren {2^n - 1}^n }} {{eqn | r = P^n }} {{end-eqn}} {{qed}}
=== Necessary Condition === {{:Sequence is Bounded in Norm iff Bounded in Metric/Necessary Condition}}{{qed|lemma}} === Sufficent Condition === {{:Sequence is Bounded in Norm iff Bounded in Metric/Sufficient Condition}}{{qed}}
Let $\sequence {x_n} $ be a bounded sequence in $\struct {R, \norm {\,\cdot\,} }$. Then: :$\exists K \in \R_{\gt 0} : \forall n : \norm {x_n} \le K$ Then $\forall n, m \in \N$: {{begin-eqn}} {{eqn | l = \map d { x_n , x_m }       | r = \norm {x_n - x_m}       | c = {{Defof|Metric Induced by Norm on Division Ring}} }} {{eqn | o = \le       | r = \norm {x_n} + \norm {x_m}       | c = Norm of Difference }} {{eqn | o = \le       | r = K + K       | c = {{Defof|Bounded Sequence in Normed Division Ring}} }} {{eqn | r = 2 K }} {{end-eqn}} Hence the sequence $\sequence {x_n} $ is bounded by $2 K$ in the metric space $\struct {R, d}$.
Let $\sequence {x_n} $ be a bounded sequence in the metric space $\struct {R, d}$. Then: :$\exists K \in \R_{> 0} : \forall n, m : \map d {x_n , x_m} \le K$ By the definition of the metric induced by a norm this is equivalent to: :$\exists K \in \R_{> 0} : \forall n, m : \norm {x_n - x_m} \le K$ Then $\forall n \in \N$: {{begin-eqn}} {{eqn | l = \norm {x_n}       | r = \norm {x_n - x_1 + x_1} }} {{eqn | o = \le       | r = \norm {x_n - x_1} + \norm {x_1}       | c = Norm axiom (N3) (Triangle Inequality) }} {{eqn | o = \le       | r = K + \norm {x_1}       | c = Assumption of bounded sequence }} {{end-eqn}} Hence the sequence $\sequence {x_n}$ is bounded by $K + \norm {x_1}$ in the normed division ring $\struct {R, \norm {\,\cdot\,} }$.
Let $\sequence {x_n} $ be a Cauchy sequence in $R$. Then by definition: :$\forall \epsilon \in \R_{\gt 0}: \exists N \in \N : \forall n, m \ge N: \norm {x_n - x_m} < \epsilon$ Let $n_1$ satisfy: :$\forall n, m \ge n_1: \norm {x_n - x_m} < 1$ Then $\forall n \ge n_1$: {{begin-eqn}} {{eqn | l = \norm {x_n}       | r = \norm {x_n - x_{n_1} + x_{n_1} } }} {{eqn | o = \le       | r = \norm {x_n - x_{n_1} } + \norm {x_{n_1} }       | c = Norm axiom (N3) (Triangle Inequality). }} {{eqn | o = \le       | r = 1 + \norm {x_{n_1} }       | c = as $n, n_1 \ge n_1$ }} {{end-eqn}} Let $K = \max \set {\norm {x_1}, \norm {x_2}, \dots, \norm {x_{n_1 - 1}}, 1 + \norm {x_{n_1} } }$. Then: :$\forall n < n_1: \norm {x_n} \le K$ :$\forall n \ge n_1: \norm {x_n} \le 1 + \norm {x_{n_1} } \le K$ It follows by definition that $\sequence {x_n}$ is bounded. {{qed}}
Let $d$ be the metric induced on $R$ be the norm $\norm {\,\cdot\,}$. Let $\sequence {x_n} $ be a Cauchy sequence in $\struct {R, \norm {\,\cdot\,}}$. By the definition of a Cauchy sequence in a normed division ring then $\sequence {x_n} $ is a Cauchy sequence in $\struct {R, d}$. By Cauchy sequence is bounded in metric space then $\sequence {x_n} $ is a bounded sequence in $\struct {R, d}$. By Sequence is Bounded in Norm iff Bounded in Metric then $\sequence {x_n} $ is a bounded sequence in $\struct {R, \norm {\,\cdot\,} }$. {{qed}}
Let $\sequence {x_n} $ be a Cauchy sequence in $R$. By Norm Sequence of Cauchy Sequence has Limit, $\sequence {\norm {x_n} }$ is a convergent sequence in $\R$. By Convergent Real Sequence is Bounded, $\sequence {\norm {x_n} }$ is bounded. That is: :$\exists M \in \R_{\gt 0}: \forall n \in \N: \norm {x_n} = \size {\norm {x_n} } \le M$ Thus, by definition, $\sequence {x_n}$ is bounded. {{qed}}
Let $\sequence {x_n}$ be a sequence in $R$ that converges to the limit $l \in R$. Let $\epsilon > 0$.  Then also $\dfrac \epsilon 2 > 0$. Because $\sequence {x_n}$ converges to $l$, we have: :$\exists N: \forall n > N: \norm {x_n - l} < \dfrac \epsilon 2$ So if $m > N$ and $n > N$, then: {{begin-eqn}} {{eqn | l = \norm {x_n - x_m}       | r = \norm {x_n - l + l - x_m} }} {{eqn | o = \le       | r = \norm {x_n - l} + \norm {l - x_m}       | c = Triangle Inequality }} {{eqn | o = <       | r = \frac \epsilon 2 + \frac \epsilon 2       | c = by choice of $N$ }} {{eqn | r = \epsilon       | c =  }} {{end-eqn}} Thus $\sequence {x_n}$ is a Cauchy sequence. {{qed}}
Let $d$ be the metric induced on $R$ be the norm $\norm {\,\cdot\,}$. Let $\sequence {x_n}$ be a sequence in $R$ that converges to the limit $l$ in $\struct {R, \norm {\,\cdot\,}}$. Thus, by definition, $\sequence {x_n} $ converges to the limit $l$ in $\struct {R, d}$. By Convergent Sequence is Cauchy Sequence in metric space, $\sequence {x_n} $ is a Cauchy sequence in $\struct {R, d}$. Thus, by definition, $\sequence {x_n} $ is a Cauchy sequence in $\struct {R, \norm {\,\cdot\,}}$. {{qed}}
Let $M_n$ denote a magic cube of order $n$. By Sum of Terms of Magic Cube, the total of all the entries in $M_n$ is given by: :$T_n = \dfrac {n^3 \paren {n^3 + 1}} 2$ There are $n^2$ rows in $M_n$, each one with the same magic constant. Thus the magic constant $C_n$ of the magic cube $M_n$ is given by: {{begin-eqn}} {{eqn | l = C_n       | r = \dfrac {T_n} {n^2}       | c =  }} {{eqn | r = \dfrac {n^3 \paren {n^3 + 1} } {2 n^2}       | c = Sum of Terms of Magic Cube }} {{eqn | r = \dfrac {n \paren {n^3 + 1} } 2       | c =  }} {{end-eqn}} {{qed}} === Sequence === {{:Magic Constant of Magic Cube/Sequence}} Category:Magic Cubes l460snwa7l9sgr0y71rc3rmnc1ggo1y
Let $M_n$ denote a magic cube of order $n$. $M_n$ is by definition an arrangement of the first $n^3$ (strictly) positive integers into an $n \times n \times n$ cubic array containing the positive integers from $1$ upwards. Thus there are $n^3$ entries in $M_n$, going from $1$ to $n^3$. Thus: {{begin-eqn}} {{eqn | l = T_n       | r = \sum_{k \mathop = 1}^{n^3} k       | c =  }} {{eqn | r = \frac {n^3 \paren {n^3 + 1} } 2       | c = Closed Form for Triangular Numbers }} {{end-eqn}} {{qed}} === Sequence === {{:Sum of Terms of Magic Cube/Sequence}} Category:Magic Cubes f5o21efpxqfe4sl7t92bh1tdmolatpu
Suppose there were an order $2$ magic cube. Take one row of this magic cube. From Magic Constant of Magic Cube, the row and column total is $9$. Any row or column with a $1$ in it must therefore also have an $8$ in it. But there are: :one row :one column both of which have a $1$ in them. Therefore the $8$ would need to go in $2$ distinct cells. But $8$ appears in a magic cube exactly once. Hence there can be no order $2$ magic cube. {{qed}} Category:Magic Cubes ge077rmjlfvg5amhhg2wh5d3hlqzemj
Let $S_m$ be the $m$th set of $m$ consecutive integers. Let $S_k$ be the $k$th set of $m$ consecutive integers after the sets with an even number of elements have been removed. Then $S_k = S_m$ where $m = 2 k - 1$. By the method of construction: :the largest integer in $S_m$ is $T_m$, the $m$th triangular number :there are $m$ integers in $S_m$. We also have that the middle integer in $S_m$ is $T_m - \dfrac {m - 1} 2$ (by inspection). Thus the sum of the elements of $S_k$ is: {{begin-eqn}} {{eqn | l = \sum S_k       | r = m \times \paren {T_m - \dfrac {m - 1} 2}       | c =  }} {{eqn | r = m \times \paren {\dfrac {m \paren {m + 1} } 2 - \dfrac {m - 1} 2}       | c =  }} {{eqn | r = m \times \paren {\dfrac {m \paren {m + 1} } 2 - \dfrac {m - 1} 2}       | c =  }} {{eqn | r = m \times \paren {\dfrac {m^2 + 1} 2}       | c =  }} {{eqn | r = \dfrac {m^3 + m} 2       | c =  }} {{eqn | r = \dfrac {\paren {2 k - 1}^3 + \paren {2 k - 1} } 2       | c =  }} {{eqn | r = \dfrac {\paren {8 k^3 - 12 k^2 + 6 k - 1} + \paren {2 k - 1} } 2       | c =  }} {{eqn | r = 4 k^3 - 6 k^2 + 4 k - 1       | c =  }} {{end-eqn}} We need to calculate the sum of all $S_k$ from $1$ to $n$. Hence we have: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n S_k       | r = \sum_{k \mathop = 1}^n \paren {4 k^3 - 6 k^2 + 4 k - 1}       | c =  }} {{eqn | r = 4 \sum_{k \mathop = 1}^n k^3 - 6 \sum_{k \mathop = 1}^n k^2 + 4 \sum_{k \mathop = 1}^n k - \sum_{k \mathop = 1}^n 1       | c =  }} {{eqn | r = 4 \paren {\frac {n^2 \paren {n + 1}^2} 4} - 6 \sum_{k \mathop = 1}^n k^2 + 4 \sum_{k \mathop = 1}^n k - \sum_{k \mathop = 1}^n 1       | c = Sum of Sequence of Cubes }} {{eqn | r = n^2 \paren {n + 1}^2 - 6 \paren {\frac {n \paren {n + 1} \paren {2 n + 1} } 6} + 4 \sum_{k \mathop = 1}^n k - \sum_{k \mathop = 1}^n 1       | c = Sum of Sequence of Squares }} {{eqn | r = n^2 \paren {n + 1}^2 - n \paren {n + 1} \paren {2 n + 1} + 4 \paren {\frac {n \paren {n + 1} } 2} - \sum_{k \mathop = 1}^n 1       | c = Closed Form for Triangular Numbers }} {{eqn | r = n^2 \paren {n + 1}^2 - n \paren {n + 1} \paren {2 n + 1} + 2 n \paren {n + 1} - n       | c = simplification }} {{eqn | r = n^4 + 2 n^3 + n^2 - 2 n^3 - 3 n^2 - n + 2 n^2 + 2 n - n       | c = multiplying everything out }} {{eqn | r = n^4       | c = simplifying }} {{end-eqn}} {{qed}}
The total number of elements in $S_0, S_1, \ldots, S_r$ is: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 0}^r \card {S_j}       | r = \sum_{j \mathop = 0}^r 3^j       | c =  }} {{eqn | r = \dfrac {3^{r + 1} - 1} {3 - 1}       | c = Sum of Geometric Sequence }} {{eqn | r = \dfrac {3^{r + 1} - 1} 2       | c = simplifying }} {{end-eqn}} Thus for any given $S_n$: :$\displaystyle \sum S_n = \sum k \sqbrk {\dfrac {3^n - 1} 2 < k \le \dfrac {3^{n + 1} - 1} 2}$ using Iverson's convention. So $\displaystyle \sum S_n$ can be evaluated as the difference between two triangular numbers: {{begin-eqn}} {{eqn | l = \sum S_n       | r = \sum_{k \mathop = 1}^{\paren {3^{n + 1} - 1} / 2} k - \sum_{k \mathop = 1}^{\paren {3^n - 1} / 2} k       | c =  }} {{eqn | r = \dfrac 1 2 \paren {\dfrac {3^{n + 1} - 1} 2 \paren {\dfrac {3^{n + 1} - 1} 2 + 1} - \dfrac {3^n - 1} 2 \paren {\dfrac {3^n - 1} 2 + 1} }       | c = Closed Form for Triangular Numbers }} {{eqn | r = \dfrac 1 2 \paren {\dfrac {3^{n + 1} - 1} 2 \paren {\dfrac {3^{n + 1} - 1 + 2} 2} - \dfrac {3^n - 1} 2 \paren {\dfrac {3^n - 1 + 2} 2} }       | c = simplifying }} {{eqn | r = \dfrac 1 8 \paren {\paren {3^{n + 1} - 1} \paren {3^{n + 1} + 1} - \paren {3^n - 1} \paren {3^n + 1} }       | c = simplifying }} {{eqn | r = \dfrac 1 8 \paren {\paren {3^{2 n + 2} - 1} - \paren {3^{2 n} - 1} }       | c = Difference of Two Squares }} {{eqn | r = \dfrac 1 8 \paren {3^{2 n + 2} - 3^{2 n} }       | c = simplifying }} {{eqn | r = \dfrac 1 8 \paren {3^{2 n} \paren {3^2 - 1} }       | c = extracting $3^{2 n}$ as a factor }} {{eqn | r = 3^{2 n}       | c = Oh look, that second factor magically equals $8$ }} {{end-eqn}} Hence the result. {{qed}}
Let $d$ be the metric induced on $R$ be the norm $\norm {\,\cdot\,}$. By the definition of a convergent sequence in a normed division ring then: :$\sequence {x_n}$ converges to $x$ in $\struct {R, \norm {\,\cdot\,} }$ {{iff}} $\sequence {x_n}$ converges to $x$ in $\struct {R, d}$. By Convergent Subsequence of Cauchy Sequence in Metric Space: :$\sequence {x_n}$ converges to $x$ in $\struct {R, d}$ {{iff}} $\sequence {x_n}$ has a subsequence that converges to $x$ In $\struct {R, d}$. By the definition of a convergent sequence in a normed division ring: :$\sequence {x_n}$ has a subsequence that converges to $x$ In $\struct {R, d}$ {{iff}} $\sequence {x_n}$ has a subsequence that converges to $x$ in $\struct {R, \norm {\,\cdot\,} }$. The result follows. {{qed}}
By every convergent sequence is a Cauchy sequence then $\NN \subseteq \CC$. The proof is completed in these steps: :$(1): \quad \NN$ is an ideal of $\CC$. :$(2): \quad \NN$ is a maximal left ideal. :$(3): \quad \NN$ is a maximal right ideal. === Lemma 1 === {{:Null Sequences form Maximal Left and Right Ideal/Lemma 1}}{{qed|lemma}} === Lemma 2 === {{:Null Sequences form Maximal Left and Right Ideal/Lemma 2}}{{qed|lemma}} === Lemma 3 === {{:Null Sequences form Maximal Left and Right Ideal/Lemma 3}}{{qed}}
The Test for Ideal is applied to prove the result. === Lemma 1.1 === {{:Null Sequences form Maximal Left and Right Ideal/Lemma 1/Lemma 1.1|Lemma 1}}{{qed|lemma}} === Lemma 1.2 === {{:Null Sequences form Maximal Left and Right Ideal/Lemma 1/Lemma 1.2}}{{qed|lemma}} === Lemma 1.3 === {{:Null Sequences form Maximal Left and Right Ideal/Lemma 1/Lemma 1.3}}{{qed|lemma}} By Test for Ideal then the result follows. {{qed}}
By Lemma 1 of Null Sequences form Maximal Left and Right Ideal: :$\NN$ is an ideal of $\CC$. Hence $\NN$ is a left ideal of $\CC$. It remains to show that $\NN$ is maximal. === Lemma 2.1 === {{:Null Sequences form Maximal Left and Right Ideal/Lemma 2/Lemma 2.1}}{{qed|lemma}} === Lemma 2.2 === {{:Null Sequences form Maximal Left and Right Ideal/Lemma 2/Lemma 2.2}}{{qed|lemma}} The result follows by definition of maximal left ideal. {{qed}}
By Lemma 1 of Null Sequences form Maximal Left and Right Ideal then $\NN$ is an ideal of $\CC$. Hence $\NN$ is a right ideal of $\CC$. It remains to show that $\NN$ is maximal. By Lemma 2.1 of Null Sequences form Maximal Left and Right Ideal then $\NN \subsetneq \CC$. By maximal right ideal it needs to be shown that: :There is no right ideal $\JJ$ of $\CC$ such that $\NN \subsetneq \JJ \subsetneq \CC$ Let $\JJ$ be a Right ideal of $\CC$ such that $\NN \subsetneq \JJ \subseteq \CC$. It will be shown that $\JJ$ = $\CC$, from which the result will follow. Let $\sequence {x_n} \in \JJ \setminus \NN$ By Inverse Rule for Cauchy sequences then  :$\exists K \in \N: \sequence {\paren {x_{K + n} }^{-1} }_{n \mathop \in \N}$ is a Cauchy sequence. Let $\sequence {y_n}$ be the sequence defined by: :$y_n = \begin{cases} 0 & : n \le K \\ \paren {x_n}^{-1} & : n > K \end{cases}$ By Cauchy Sequence with Finite Elements Prepended is Cauchy Sequence then $\sequence {y_n} \in \CC$ By the definition of a right ideal the product $\sequence {x_n} \sequence {y_n} = \sequence {x_n y_n} \in \JJ$ By the definition of $\sequence {y_n}$ then: :$x_n y_n = \begin{cases} 0 & : n \le K \\ 1 & : n > K \end{cases}$ Let $\mathcal 1 = \tuple {1, 1, 1, \dots}$ be the unity of $\CC$ Then $\mathcal 1 - \sequence {x_n} \sequence {y_n}$ is the sequence $\sequence {w_n}$ defined by: :$w_n = \begin {cases} 1 & : n \le K \\ 0 & : n > K \end {cases}$ By Convergent Sequence with Finite Elements Prepended is Convergent Sequence then $\sequence {w_n}$ is convergent to 0. So $\sequence {w_n} \in \NN \subsetneq \JJ$ Since \sequence {x_n} $\sequence {y_n}, \sequence {w_n} \in \JJ$ by the definition of a ring ideal then: :$\sequence {w_n} + \sequence {x_n} \sequence {y_n} = \mathcal 1 \in \JJ$ By the definition of a right ideal then: :$\forall \sequence {a_n} \in \CC, \mathcal 1 \circ \sequence {a_n} = \sequence {a_n} \in \JJ$ Hence $\JJ = \CC$ {{qed}}
The proof proceeds by induction. For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition: :$F_n = \displaystyle \prod_{j \mathop = 0}^{n - 1} F_j + 2$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = F_1       | r = 2^{\paren {2^1} } + 1       | c = {{Defof|Fermat Number}} }} {{eqn | r = 5       | c =  }} {{eqn | r = 3 + 2       | c =  }} {{eqn | r = \paren {2^{\paren {2^0} } + 1} + 2       | c =  }} {{eqn | r = F_0 + 2       | c = {{Defof|Fermat Number}} }} {{eqn | r = \displaystyle \prod_{j \mathop = 0}^{1 - 1} F_j + 2       | c =  }} {{end-eqn}} Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$F_k = \displaystyle \prod_{j \mathop = 0}^{k - 1} F_j + 2$ from which it is to be shown that: :$F_{k + 1} = \displaystyle \prod_{j \mathop = 0}^k F_j + 2$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = F_{k + 1}       | r = 2^{\paren {2^{k + 1} } } + 1       | c = {{Defof|Fermat Number}} }} {{eqn | r = 2^{\paren {2 \times 2^k} } + 1       | c =  }} {{eqn | r = 2^{\paren {2^k} } \times 2^{\paren {2^k} } + 1       | c =  }} {{eqn | r = \paren {F_k - 1} \times \paren {F_k - 1} + 1       | c = {{Defof|Fermat Number}} }} {{eqn | r = \paren {\paren {\prod_{j \mathop = 0}^{k - 1} F_j + 2} - 1} \paren {F_k - 1} + 1       | c = Induction Hypothesis }} {{eqn | r = \paren {\prod_{j \mathop = 0}^{k - 1} F_j} F_k + F_k - \prod_{j \mathop = 0}^{k - 1} F_j - 1 + 1       | c = multiplying out }} {{eqn | r = \prod_{j \mathop = 0}^k F_j + F_k - \paren {F_k - 2}       | c = Induction Hypothesis, and simplifying }} {{eqn | r = \prod_{j \mathop = 0}^k F_j + 2       | c = simplifying }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{>0}: F_n = \displaystyle \prod_{j \mathop = 0}^{n - 1} F_j + 2$ {{qed}}
Obviously the product cannot be a square if $n$ is a prime. For $n$ composite, we can write: :$n = a b$ where $a, b \in \Z_{>1}$. Then: {{begin-eqn}} {{eqn | o =       | r = n! \paren {n - 1}! \paren {a!} \paren {a - 1}! \paren {b!} \paren {b - 1}! }} {{eqn | r = n a b \paren {\paren {n - 1}! \paren {a - 1}! \paren {b - 1}!}^2 }} {{eqn | r = \paren {n! \paren {a - 1}! \paren {b - 1}!}^2 }} {{end-eqn}} which is a square. Hence no more than $6$ factorials is required. To show that $527$ is the smallest that actually requires $6$, observe that: {{tidy}} {{explain|It might be worth extracting some of the below statements into lemmata, for example: "If $n$ is itself square, then so is $n! \paren {n - 1}!$" and "... Then $n! \paren {n - 1}! b! \paren {b - 1}!$ is square" -- they're really easy to prove, even I can do them :-) but it takes more than a glance to recognise that they are true.}} If $n$ is itself square, then so is $n! \paren {n - 1}!$. If $n$ is not square-free, write $n = a^2 b$, where $b$ is square-free. Then $n! \paren {n - 1}! b! \paren {b - 1}!$ is square. If $n$ is divisible by $2$, write $n = 2 m$. Then $\paren {2 m}! \paren {2 m - 1}! \paren {m!} \paren {m - 1}! \paren {2!}$ is square. If $n$ is divisible by $3$, write $n = 3 m$. Then $\paren {3 m}! \paren {3 m - 1}! \paren {2 m}! \paren {2 m - 1}! \paren {3!}$ is square. If $n$ is divisible by $5$, write $n = 5 m$. Then $\paren {5 m}! \paren {5 m - 1}! \paren {m!} \paren {m - 1}! \paren {6!}$ is square. If $n$ is divisible by $7$, write $n = 7 m$. Then $\paren {7 m}! \paren {7 m - 1}! \paren {5 m}! \paren {5 m - 1}! \paren {7!}$ is square. If $n$ is divisible by $11$, write $n = 11 m$. Then $\paren {11 m}! \paren {11 m - 1}! \paren {7 m}! \paren {7 m - 1}! \paren {11!}$ is square. The remaining numbers less than $527$ that are not of the above forms are: :$221, 247, 299, 323, 377, 391, 403, 437, 481, 493$ Each of the following is a product of $5$ factorials which is square: :$221! \, 220! \, 18! \, 11! \, 7!$ :$247! \, 246! \, 187! \, 186! \, 20!$ :$299! \, 298! \, 27! \, 22!$ :$323! \, 322! \, 20! \, 14! \, 6!$ :$377! \, 376! \, 29! \, 23! \, 10!$ :$391! \, 389! \, 24! \, 21! \, 17!$ :$403! \, 402! \, 33! \, 30! \, 14!$ :$437! \, 436! \, 51! \, 49! \, 28!$ :$481! \, 479! \, 38! \, 33! \, 22!$ :$493! \, 491! \, 205! \, 202! \, 7!$ {{finish|The fact that $527$ has no such representation can be verified by a direct (but lengthy) computation.}}
From Constant Sequence Converges to Constant in Normed Division Ring, the zero $\tuple {0, 0, 0, \dots}$ of $\mathcal C$ to converges $0 \in R$. Therefore $\tuple {0, 0, 0, \dots} \in \mathcal N$. {{qed}}
Let $\displaystyle \lim_{n \mathop \to \infty} x_n = 0$ and $\displaystyle \lim_{n \mathop \to \infty} y_n = 0$. The sequence $\sequence {x_n} + \paren {-\sequence {y_n} } = \sequence {x_n - y_n}$. By Difference Rule for Sequences, $\displaystyle \lim_{n \mathop \to \infty} x_n - y_n = 0 - 0 = 0.$ The result follows. {{qed}}
Let $\displaystyle \lim_{n \mathop \to \infty} x_n = 0$ By the definition of the product on the ring of Cauchy sequences then: :$\sequence {x_n} \sequence {y_n} = \sequence {x_n y_n}$ :$\sequence {y_n} \sequence {x_n} = \sequence {y_n x_n}$ By product of sequence converges to zero with Cauchy sequence then: :$\displaystyle \lim_{n \mathop \to \infty} x_n y_n = 0$ :$\displaystyle \lim_{n \mathop \to \infty} y_n x_n = 0$ The result follows. {{qed}}
By every convergent sequence is a Cauchy sequence then $\NN \subseteq \CC$. From Constant Sequence Converges to Constant in Normed Division Ring, the unity $\tuple {1, 1, 1, \dotsC}$ of $\CC$ converges to $1 \in R$, and therefore $\tuple {1, 1, 1, \dotsc} \in \CC \setminus \NN$ So $\NN \subsetneq \CC$. {{qed}}
Let $\JJ$ be a left ideal of $\CC$ such that $\NN \subsetneq \JJ \subseteq \\CC$. It will be shown that $\JJ$ = $\CC$, from which the result will follow. Let $\sequence {x_n} \in \JJ \setminus \NN$ By Inverse Rule for Cauchy sequences then  :$\exists K \in \N: \sequence { \paren {x_{K + n} }^{-1} }_{n \mathop \in \N}$ is a Cauchy sequence. Let $\sequence {y_n}$ be the sequence defined by: :$y_n = \begin {cases} 0 & : n \le K \\ \paren {x_n}^{-1} & : n > K \end {cases}$ By Cauchy Sequence with Finite Elements Prepended is Cauchy Sequence then $\sequence {y_n} \in \CC$ By the definition of a left ideal the product $\sequence {y_n} \sequence {x_n} = \sequence {y_n x_n} \in \JJ$ By the definition of $\sequence {y_n}$ then: :$y_n x_n = \begin {cases} 0 & : n \le K \\ 1 & : n > K \end {cases}$ Let $\mathcal {1} = \tuple {1, 1, 1, \dotsc}$ be the unity of $\CC$ Then $\mathcal {1} - \sequence {y_n} \sequence {x_n}$ is the sequence $\sequence {w_n}$ defined by :$w_n = \begin{cases} 1 & : n \le K \\ 0 & : n > K \end {cases}$ By Convergent Sequence with Finite Elements Prepended is Convergent Sequence then $\sequence {w_n}$ is convergent to 0. So $\sequence {w_n} \in \NN \subsetneq \JJ$ Since $\sequence {y_n} \sequence {x_n}, \sequence {w_n} \in \JJ$ by the definition of a ring ideal then: :$\sequence {w_n} + \sequence {y_n} \sequence {x_n} = \mathcal {1} \in \JJ$ By the definition of a left ideal then: :$\forall \sequence {a_n} \in \CC, \sequence {a_n} \circ \mathcal {1} = \sequence {a_n} \in \JJ$ Hence $\JJ = \CC$ {{qed}}