avewright/proofwiki-math huggingface dataset:
Theorem: Sequence is Bounded in Norm iff Bounded in Metric

Let $\struct {R, \norm {\,\cdot\,} } $ be a normed division ring.
Let $d$ be the metric induced on $R$ be the norm $\norm {\,\cdot\,}$.
Let $\sequence {x_n}$ be a sequence in $R$.
Then:
:$\sequence {x_n} $ is a bounded sequence in the normed division ring $\struct {R, \norm {\,\cdot\,} }$ {{iff}} $\sequence {x_n} $ is a bounded sequence in the metric space $\struct {R, d}$.

Proof:
=== Necessary Condition ===
{{:Sequence is Bounded in Norm iff Bounded in Metric/Necessary Condition}}{{qed|lemma}}
Theorem: Sequence is Bounded in Norm iff Bounded in Metric/Necessary Condition

Let $\struct {R, \norm {\,\cdot\,}} $ be a normed division ring.
Let $d$ be the metric induced on $R$ be the norm $\norm {\,\cdot\,}$.
Let $\sequence {x_n}$ be a sequence in $R$.
Let $\sequence {x_n} $ be a bounded sequence in the normed division ring $\struct {R, \norm {\,\cdot\,}}$ 
Then:
:$\sequence {x_n} $ is a bounded sequence in the metric space $\struct {R, d}$

Proof:
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 $\struct {R, d}$.
Theorem: Sequence is Bounded in Norm iff Bounded in Metric/Sufficient Condition

Let $\struct {R, \norm {\,\cdot\,}} $ be a normed division ring.
Let $d$ be the metric induced on $R$ be the norm $\norm {\,\cdot\,}$.
Let $\sequence {x_n}$ be a sequence in $R$.
Let $\sequence {x_n} $ be a bounded sequence in the metric space $\struct {R, d}$
Then:
:$\sequence {x_n} $ is a bounded sequence in the normed division ring $\struct {R, \norm {\,\cdot\,} }$

Proof:
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-mult|3}}
}}
{{eqn | o = \le
      | r = K + \norm {x_1}
      | c = {{hypothesis}}: {{Defof|Bounded Sequence in Metric Space}}
}}
{{end-eqn}}
Hence the sequence $\sequence {x_n}$ is bounded by $K + \norm {x_1}$ in the normed division ring $\struct {R, \norm {\,\cdot\,} }$.
Theorem: Magic Constant of Magic Cube

The magic constant of a magic cube of order $n$ is given by:
:$C_n = \dfrac {n \paren {n^3 + 1} } 2$

Proof:
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}}
Theorem: Sum of Terms of Magic Cube

The total of all the entries in a magic cube of order $n$ is given by:
:$T_n = \dfrac {n^3 \paren {n^3 + 1} } 2$

Proof:
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}}
Theorem: Smallest Magic Cube is of Order 3

Apart from the trivial order $1$ magic cube:
{{:Magic Cube/Examples/Order 1}}
the smallest magic cube is the order $3$ magic cube:
{{:Magic Cube/Examples/Order 3}}

Proof:
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
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Theorem: Fourth Power as Summation of Groups of Consecutive Integers

Take the positive integers and group them in sets such that the $m$th set contains the next $m$ positive integers:
:$\set 1, \set {2, 3}, \set {4, 5, 6}, \set {7, 8, 9, 10}, \set {11, 12, 13, 14, 15}, \ldots$
Remove all the sets with an even number of elements.
Then the sum of all the integers in the first $n$ sets remaining equals $n^4$.

Proof:
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}}
Theorem: Even Power of 3 as Sum of Consecutive Positive Integers

Take the positive integers and group them in sets such that the $n$th set contains the next $3^n$ positive integers:
:$\set 1, \set {2, 3, 4}, \set {5, 6, \ldots, 13}, \set {14, 15, \cdots, 40}, \ldots$
Let the $n$th such set be denoted $S_{n - 1}$, that is, letting $S_0 := \set 1$ be considered as the zeroth.
Then the sum of all the elements of $S_n$ is $3^{2 n}$.

Proof:
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 $S_n$:
:$\ds \sum S_n = \sum k \sqbrk {\dfrac {3^n - 1} 2 < k \le \dfrac {3^{n + 1} - 1} 2}$
using Iverson's convention.
So $\ds \sum S_n$ 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}}
Theorem: Convergent Subsequence of Cauchy Sequence/Normed Division Ring

Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $\sequence{x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {R, \norm {\,\cdot\,} }$. 
Let $x \in R$.
Then $\sequence {x_n}$ converges to $x$ {{iff}} $\sequence {x_n}$ has a subsequence that converges to $x$.

Proof:
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}}
Theorem: Null Sequences form Maximal Left and Right Ideal

Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$.
Let $\NN$ be the set of null sequences.
That is:
:$\ds \NN = \set {\sequence {x_n}: \lim_{n \mathop \to \infty} x_n = 0 }$
Then $\NN$ is a ring ideal of $\CC$ that is a maximal left ideal and a maximal right ideal.

Proof:
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.
Theorem: Null Sequences form Maximal Left and Right Ideal/Lemma 1

:$\NN$ is an ideal of $\CC$.

Proof:
The Test for Ideal is applied to prove the result.
Theorem: Null Sequences form Maximal Left and Right Ideal/Lemma 2

:$\NN$ is a maximal left ideal.

Proof:
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.
Theorem: Null Sequences form Maximal Left and Right Ideal/Lemma 3

:$\NN$ is a maximal right ideal.

Proof:
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 7 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}}
Theorem: Product of Sequence of Fermat Numbers plus 2

Let $F_n$ denote the $n$th Fermat number.
Then:
{{begin-eqn}}
{{eqn | q = \forall n \in \Z_{>0}
      | l = F_n
      | r = \prod_{j \mathop = 0}^{n - 1} F_j + 2
      | c = 
}}
{{eqn | r = F_0 F_1 \dotsm F_{n - 1} + 2
      | c = 
}}
{{end-eqn}}

Proof:
The proof proceeds by induction.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$F_n = \ds \prod_{j \mathop = 0}^{n - 1} F_j + 2$
Theorem: 492 is Sum of 3 Cubes in 3 Ways

$492$ can be expressed as the sum of $3$ cubes, either positive or negative in $3$ known ways.
{{begin-eqn}}
{{eqn | l = 492
      | r = 50^3 + \paren {-19}^3 + \paren {-49}^3
}}
{{eqn | r = 123 \, 134^3 + 9179^3 + \paren {-123 \, 151}^3
}}
{{eqn | r = 1 \, 793 \, 337 \, 644^3 + \paren {-81 \, 3701 \, 167}^3 + \paren {-1 \, 735 \, 662 \, 109}^3
}}
{{end-eqn}}

Proof:
Brute force.
Theorem: Smallest n needing 6 Numbers less than n so that Product of Factorials is Square

Let $n \in \Z_{>0}$ be a positive integer.
Then it is possible to choose at most $6$ positive integers less than $n$ such that the product of their factorials is square.
The smallest $n$ that actually requires $6$ numbers to be chosen is $527$.

Proof:
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.}}
Theorem: Null Sequences form Maximal Left and Right Ideal/Lemma 4

:$\NN \ne \O$

Proof:
From Constant Sequence Converges to Constant in Normed Division Ring, the zero $\tuple {0, 0, 0, \dots}$ of $\CC$ to converges $0 \in R$.
Therefore $\tuple {0, 0, 0, \dots} \in \NN$.
{{qed}}
Theorem: Null Sequences form Maximal Left and Right Ideal/Lemma 5

:$\forall \sequence {x_n}, \sequence {y_n} \in \NN: \sequence {x_n} + \paren {-\sequence {y_n} } \in \NN$

Proof:
Let $\ds \lim_{n \mathop \to \infty} x_n = 0$ and $\ds \lim_{n \mathop \to \infty} y_n = 0$.
The sequence $\sequence {x_n} + \paren {-\sequence {y_n} } = \sequence {x_n - y_n}$.
From Difference Rule for Sequences in Normed Division Ring:
:$\ds \lim_{n \mathop \to \infty} x_n - y_n = 0 - 0 = 0$
The result follows.
{{qed}}
Theorem: Null Sequences form Maximal Left and Right Ideal/Lemma 6

:$\quad \forall \sequence {x_n} \in \NN, \sequence {y_n} \in \CC: \sequence {x_n} \sequence {y_n} \in \NN, \sequence {y_n} \sequence {x_n} \in \NN$

Proof:
Let $\ds \lim_{n \mathop \to \infty} x_n = 0$.
By the definition of the product on the ring of Cauchy sequences then:
{{begin-eqn}}
{{eqn | l = \sequence {x_n} \sequence {y_n}
      | r = \sequence {x_n y_n}
}}
{{eqn | l = \sequence {y_n} \sequence {x_n}
      | r = \sequence {y_n x_n}
}}
{{end-eqn}}
By Product of Sequence Converges to Zero with Cauchy Sequence Converges to Zero:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} x_n y_n
      | r = 0
}}
{{eqn | l = \lim_{n \mathop \to \infty} y_n x_n
      | r = 0
}}
{{end-eqn}}
The result follows.
{{qed}}
Theorem: Null Sequences form Maximal Left and Right Ideal/Lemma 7

:$\NN \subsetneq \CC$.

Proof:
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}}
Theorem: Null Sequences form Maximal Left and Right Ideal/Lemma 8

:There is no left ideal $\JJ$ of $\CC$ such that $\NN \subsetneq \JJ \subsetneq \CC$

Proof:
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}}
Theorem: Difference between Two Squares equal to Repdigit

{{begin-eqn}}
{{eqn | l = 6^2 - 5^2
      | r = 11
      | c = 
}}
{{eqn | l = 56^2 - 45^2
      | r = 1111
      | c = 
}}
{{eqn | l = 556^2 - 445^2
      | r = 111 \, 111
      | c = 
}}
{{eqn | o = :
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 7^2 - 4^2
      | r = 33
      | c = 
}}
{{eqn | l = 67^2 - 34^2
      | r = 3333
      | c = 
}}
{{eqn | l = 667^2 - 334^2
      | r = 333 \, 333
      | c = 
}}
{{eqn | o = :
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 8^2 - 3^2
      | r = 55
      | c = 
}}
{{eqn | l = 78^2 - 23^2
      | r = 5555
      | c = 
}}
{{eqn | l = 778^2 - 223^2
      | r = 555 \, 555
      | c = 
}}
{{eqn | o = :
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 9^2 - 2^2
      | r = 77
      | c = 
}}
{{eqn | l = 89^2 - 12^2
      | r = 7777
      | c = 
}}
{{eqn | l = 889^2 - 112^2
      | r = 777 \, 777
      | c = 
}}
{{eqn | o = :
      | c = 
}}
{{end-eqn}}

Proof:
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}}
Theorem: Largest Number Not Expressible as Sum of Fewer than 8 Cubes

$8042$ is (probably) the largest positive integer that cannot be expressed as the sum of fewer than $8$ cubes.

Proof:
It is believed that this entry is a 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.