avewright/proofwiki-math huggingface dataset:
Theorem: Product of Number of Edges, Edges per Face and Faces of Tetrahedron

The product of the number of edges, edges per face and faces of a tetrahedron is $72$.

Proof:
A tetrahedron has:
:$6$ edges
and:
:$4$ faces.
Each face is a triangle, and so has $3$ edges.
Hence:
:$6 \times 4 \times 3 = 72$
{{qed}}
Theorem: Product of Number of Edges, Edges per Face and Faces of Cube

The product of the number of edges, edges per face and faces of a cube is $288$.

Proof:
A cube has:
:$12$ edges
and:
:$6$ faces.
Each face is a square, and so has $4$ edges.
Hence:
:$12 \times 6 \times 4 = 288$
{{qed}}
Theorem: Product of Number of Edges, Edges per Face and Faces of Regular Octahedron

The product of the number of edges, edges per face and faces of a regular octahedron is $288$.

Proof:
A regular octahedron has:
:$12$ edges
and:
:$8$ faces.
Each face is a triangle, and so has $3$ edges.
Hence:
:$12 \times 8 \times 3 = 288$
{{qed}}
Theorem: Ordinals are Well-Ordered/Corollary

Let $A$ be a set of ordinals.
Let $\Epsilon {\restriction_A}$ denote the epsilon restriction on $A$.
Then $A$ is strictly well-ordered by $\Epsilon {\restriction_A}$.

Proof:
Let $A$ be a set of ordinals.
Let $\Epsilon {\restriction_A}$ denote the epsilon restriction on $A$.
It is to be shown that $\Epsilon {\restriction_A}$ is antireflexive.
{{AimForCont}} there is a $a \in A$ such that $a \in a$.
Then by {{Corollary|Transitive Set is Proper Subset of Ordinal iff Element of Ordinal}}, we have $a \subsetneq a$.
And so:
:$\exists x \in a: x \notin a$
But this is a contradiction, so $\Epsilon {\restriction_A}$ must be antireflexive.
By Strict Subset Relation is Transitive, it follows that $\Epsilon {\restriction_A}$ is transitive.
By Ordinals are Well-Ordered, it follows that for any non-empty subset $B \subseteq A$ there is a smallest element $x \in B$ such that:
:$\forall y \in B: x \in y \lor x = y$
It has been shown that $\Epsilon {\restriction_A}$ is a strict ordering such that every non-empty subset of $A$ has a smallest element.
Hence the result.
{{qed}}
Category:Ordinals
Category:Ordinals are Well-Ordered
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Theorem: Product of Number of Edges, Edges per Face and Faces of Regular Dodecahedron

The product of the number of edges, edges per face and faces of a regular dodecahedron is $1800$.

Proof:
A regular dodecahedron has:
:$30$ edges
and:
:$12$ faces.
Each face is a pentagon, and so has $5$ edges.
Hence:
:$30 \times 12 \times 5 = 1800$
{{qed}}
Theorem: Product of Number of Edges, Edges per Face and Faces of Regular Icosahedron

The product of the number of edges, edges per face and faces of a regular icosahedron is $1800$.

Proof:
A regular icosahedron has:
:$30$ edges
and:
:$20$ faces.
Each face is a triangle, and so has $3$ edges.
Hence:
:$30 \times 20 \times 3 = 1800$
{{qed}}
Theorem: Smallest 5th Power equal to Sum of 5 other 5th Powers

The smallest positive integer whose fifth power can be expressed as the sum of $5$ other distinct positive  fifth powers is $72$:
:$72^5 = 19^5 + 43^5 + 46^5 + 47^5 + 67^5$

Proof:
{{begin-eqn}}
{{eqn | l = 19^5 + 43^5 + 46^5 + 47^5 + 67^5
      | r = 2 \, 476 \, 099
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = 147 \, 008 \, 443
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = 205 \, 962 \, 976
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = 229 \, 345 \, 007
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = 1 \, 350 \, 125 \, 107
      | c = 
}}
{{eqn | r = 1 \, 934 \, 917 \, 632
      | c = 
}}
{{eqn | r = 72^5
      | c = 
}}
{{end-eqn}}
Let $\tuple {I, J, K, L, M}$ be the $5$ distinct positive  fifth powers where $I < J < K < L < M$
The following program in R iterates through 97,330,464 calculations and verifies the assertion.
 for (I in 1:19) {
 	for (J in 2:43) {
 		for (K in 3:46) {
 			for (L in 4:47) {
 				for (M in 5:67) {
 					if (abs((I^5+J^5+K^5+L^5+M^5)^0.2 - round((I^5+J^5+K^5+L^5+M^5)^0.2)) < 0.00000001) {
 						print(paste((I^5+J^5+K^5+L^5+M^5)^0.2, I, J, K, L, M)) 
 					}
 				}
 			}
 		}
 	}
 }
Theorem: Smallest Consecutive Even Nontotients

The smallest pair of consecutive even nontotients is $74$ and $76$.

Proof:
From the sequence of nontotients:
{{:Definition:Nontotient/Sequence}}
Hence, by inspection, it can be seen that $74$ and $76$ are the smallest such pair.
{{qed}}
Theorem: Smallest Number not Expressible as Sum of Fewer than 19 Fourth Powers

:$79 = 15 \times 1^4 + 4 \times 2^4$

Proof:
We have $1^4 = 1, 2^4 = 16, 3^4 = 81 > 79$.
Hence for each $n < 79$, we can only use $1^4$ and $2^4$ in our sum.
Write $n = 2^4 a + 1^4 b$.
We can use the greedy algorithm to generate these expressions, since replacing $2^4$ with $16 \times 1^4$ increases the number of fourth powers required.
Suppose $n < 64$.
By Division Theorem, there is a unique way to write $n = 16 q + r$, with $q \in \Z$, $0 \le r < 16$.
{{begin-eqn}}
{{eqn | l = 16 q + r
      | r = n
}}
{{eqn | ll = \leadsto
      | l = 16 q + r
      | o = <
      | r = 64
}}
{{eqn | ll = \leadsto
      | l = 16 q
      | o = <
      | r = 64
}}
{{eqn | ll = \leadsto
      | l = q
      | o = <
      | r = 4
}}
{{end-eqn}}
Thus $q + r \le 3 + 15 = 18$.
It follows that each positive integer less than $64$ can be expressed in not more than $18$ fourth powers.
Suppose $64 \le n \le 78$.
We cannot use more than $4$ $2^4$, since $5 \times 2^4 = 80 > n$.
Thus we write $n = 4 \times 2^4 + \paren {n - 64} \times 1^4$.
Since $n - 64 \le 78 - 64 = 14$, we can use not more than $18$ fourth powers to express $n$.
For $n = 79$, we still cannot use more than $4$ $2^4$, since $5 \times 2^4 = 80 > n$.
Thus $n = 4 \times 2^4 + 15 \times 1^4$ uses the least number of fourth powers.
Hence $79$ is the smallest positive integer which cannot be expressed as the sum of fewer than $19$ fourth powers.
{{qed}}
Theorem: Reciprocal of 81

:$\dfrac 1 {81} = 0 \cdotp \dot 01234 \, 567 \dot 9$

Proof:
Performing the calculation using long division:
<pre>
   0.0123456790...
  ----------------
81)1.0000000000000
     81
     --
     190
     162
     ---
      280
      243
      ---
       370
       324
       ---
        460
        405
        ---
         550
         486
         ---
          640
          567
           --
           730
           729
           ---
             100
              81
             ---
             ...
</pre>
{{qed}}
Theorem: Reciprocal of Square of 1 Less than Number Base

Let $b \in \Z$ be an integer such that $b > 2$.
Let $n = \paren {b - 1}^2$.
The reciprocal of $n$, expressed in base $b$, recurs with period $b - 1$:
:$\dfrac 1 n = \sqbrk {0 \cdotp \dot 012 \ldots c \dot d}_b = \sqbrk {0 \cdotp 012 \ldots cd012 \ldots}_b$
where:
:$c = b - 3$
:$d = b - 1$

Proof:
By Basis Representation Theorem, the number $\sqbrk {12 \ldots cd}_b$ can be written as:
{{begin-eqn}}
{{eqn | l = \sqbrk {12 \ldots cd}_b
      | r = 1 \times b^{b - 3} + 2 \times b^{b - 4} + \dots + \paren {b - 3} \times b + \paren {b - 1}
}}
{{eqn | r = 1 + \sum_{k \mathop = 0}^{b - 3} \paren {b - 2 - k} b^k
}}
{{eqn | r = 1 + \frac {\paren{b - 2} \paren {1 - b^{b - 2} } } {1 - b} + \frac {b \paren {1 - \paren {b - 2} b^{b - 3} + \paren {b - 3} b^{b - 2} } } {\paren {1 - b}^2}
      | c = Sum of Arithmetic-Geometric Sequence, taking $a = n = b - 2, d = -1, r = b$
}}
{{eqn | r = \frac {1 - 2 b + b^2 + b - 2 - b^{b - 1} + 2 b^{b - 2} - b^2 + 2 b + b^b - 2 b^{b - 1} - b + b^{b - 1} - 2 b^{b - 2} - b^b + 3 b^{b - 1} } {\paren {1 - b}^2}
}}
{{eqn | r = \frac {b^{b - 1} - 1} {\paren {1 - b}^2}
}}
{{end-eqn}}
Thus the recurring basis expansion $\sqbrk {0 \cdotp \dot 012 \ldots c \dot d}_b$ can be written as:
{{begin-eqn}}
{{eqn | l = \sqbrk {0 \cdotp \dot 012 \ldots c \dot d}_b
      | r = \sqbrk {12 \ldots cd}_b \sum_{n \mathop = 1}^\infty \paren {b^{1 - b} }^n
}}
{{eqn | r = \frac {b^{b - 1} - 1} {\paren {1 - b}^2} \times \frac {b^{1 - b} } {1 - b^{1 - b} }
      | c = Sum of Infinite Geometric Sequence
}}
{{eqn | r = \frac {b^{b - 1} - 1} {\paren {1 - b}^2} \times \frac 1 {b^{b - 1} - 1}
      | c = multiplying top and bottom by $b^{b - 1}$
}}
{{eqn | r = \frac 1 {\paren {1 - b}^2}
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Square of 1 Less than Number Base

Let $b \in \Z$ be an integer such that $b > 2$.
Let $n = b - 1$.
The square of $n$ is expressed in base $b$ as:
:$n^2 = \left[{c1}\right]_b$
where $c = b - 2$.

Proof:
{{begin-eqn}}
{{eqn | l = n^2
      | r = \left({b - 1}\right)^2
      | c = 
}}
{{eqn | r = b^2 - 2 b + 1
      | c = 
}}
{{eqn | r = b \left({b - 2}\right) + 1
      | c = 
}}
{{end-eqn}}
The result follows by definition of number base.
{{qed}}
Theorem: Positive Integers whose Square Root equals Sum of Digits

The following positive integers have a square root that equals the sum of their digits:
:$0, 1, 81$
and there are no more.

Proof:
{{begin-eqn}}
{{eqn | l = \sqrt 0
      | r = 0
      | c = 
}}
{{eqn | l = \sqrt 1
      | r = 1
      | c = 
}}
{{eqn | l = \sqrt {81}
      | r = 9
      | c = 
}}
{{eqn | r = 8 + 1
      | c = 
}}
{{end-eqn}}
By considering the square roots, we are looking for positive integers with a square for which its sum of digits is equal to the original number.
It is easy to verify the result up to $36$.
We prove that for $n > 36$, the sum of digits of $n^2$ cannot equal $n$.
Let $n$ be a $d$-digit number.
Then $n < 10^d$.
Thus $n^2 < 10^{2 d}$.
Therefore $n^2$ has at most $2 d$ digits.
Hence the sum of digits of $n^2$ is at most $18 d$.
If $n$ is a $2$-digit number, we have $n > 36 = 18 d$.
Now suppose $d \ge 3$.
Then:
{{begin-eqn}}
{{eqn | l = n
      | o = \ge
      | r = 10^{d - 1}
}}
{{eqn | r = 10 \times 10^{d - 2}
}}
{{eqn | o = \ge
      | r = 10 \paren {1 + 9 \paren {d - 2} }
      | c = Bernoulli's Inequality
}}
{{eqn | r = 90 d - 170
}}
{{eqn | o = \ge
      | r = 18 d + 216 - 170
      | c = as $d \ge 3$
}}
{{eqn | o = >
      | r = 18 d
}}
{{end-eqn}}
Therefore for $n > 36$, the sum of digits of $n^2$ cannot equal $n$.
{{qed}}
Theorem: Closed Form for Heptagonal Numbers

The closed-form expression for the $n$th heptagonal number is:
:$H_n = \dfrac {n \paren {5 n - 3} } 2$

Proof:
Heptagonal numbers are $k$-gonal numbers where $k = 7$.
From Closed Form for Polygonal Numbers we have that:
:$\map P {k, n} = \dfrac n 2 \paren {\paren {k - 2} n - k + 4}$
Hence:
{{begin-eqn}}
{{eqn | l = H_n
      | r = \frac n 2 \paren {\paren {7 - 2} n - 7 + 4}
      | c = Closed Form for Polygonal Numbers
}}
{{eqn | r = \dfrac {n \paren {5 n - 3} } 2
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Integers for which Divisor Sum of Phi equals Divisor Sum

The following positive integers have the property that the divisor sum of their Euler $\phi$ value equals their divisor sum:
:$\map {\sigma_1} {\map \phi n} = \map {\sigma_1} n$
:$1, 87, 362, 1257, 1798, 5002, 9374, \ldots$
{{OEIS|A033631}}

Proof:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {\map \phi 1}
      | r = \map {\sigma_1} 1
      | c = {{EulerPhiLink|1}}
}}
{{eqn | r = 1
      | c = {{DSFLink|1}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {\map \phi {87} }
      | r = \map {\sigma_1} {56}
      | c = {{EulerPhiLink|87}}
}}
{{eqn | r = 120
      | c = {{DSFLink|56}}
}}
{{eqn | r = \map {\sigma_1} {87}
      | c = {{DSFLink|87}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {\map \phi {362} }
      | r = \map {\sigma_1} {180}
      | c = {{EulerPhiLink|362}}
}}
{{eqn | r = 546
      | c = {{DSFLink|180}}
}}
{{eqn | r = \map {\sigma_1} {362}
      | c = {{DSFLink|362}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {\map \phi {1257} }
      | r = \map {\sigma_1} {836}
      | c = {{EulerPhiLink|1257}}
}}
{{eqn | r = 1680
      | c = {{DSFLink|836}}
}}
{{eqn | r = \map {\sigma_1} {1257}
      | c = {{DSFLink|1257}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {\map \phi {1798} }
      | r = \map {\sigma_1} {840}
      | c = {{EulerPhiLink|1798}}
}}
{{eqn | r = 2880
      | c = {{DSFLink|840}}
}}
{{eqn | r = \map {\sigma_1} {1798}
      | c = {{DSFLink|1798}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {\map \phi {5002} }
      | r = \map {\sigma_1} {2400}
      | c = {{EulerPhiLink|5002}}
}}
{{eqn | r = 7812
      | c = {{DSFLink|2400}}
}}
{{eqn | r = \map {\sigma_1} {5002}
      | c = {{DSFLink|5002}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {\map \phi {9374} }
      | r = \map {\sigma_1} {4536}
      | c = {{EulerPhiLink|9374}}
}}
{{eqn | r = 14 \, 520
      | c = {{DSFLink|4536}}
}}
{{eqn | r = \map {\sigma_1} {9374}
      | c = {{DSFLink|9374}}
}}
{{end-eqn}}
Theorem: Prime Gaps of 8

The following pairs of consecutive prime numbers are those whose difference is $8$:
:$\tuple {89, 97}, \tuple {359, 367}, \tuple {389, 397}, \tuple {401, 409}, \ldots$
{{OEIS|A031926}}

Proof:
Demonstrated by listing the prime gaps.
Theorem: Smallest Cunningham Chain of the First Kind of Length 6

The smallest Cunningham chain of the first kind of length $6$ is:
:$\tuple {89, 179, 359, 719, 1439, 2879}$

Proof:
By definition, a Cunningham chain of the first kind is a sequence of prime numbers $\tuple {p_1, p_2, \ldots, p_n}$ such that:
: $p_{k + 1} = 2 p_k + 1$
: $\dfrac {p_1 - 1} 2$ is not prime
: $2 p_n + 1$ is not prime.
Thus each term except the last is a Sophie Germain prime.
{{:Definition:Sophie Germain Prime/Sequence}}
Let $P: \Z \to \Z$ be the mapping defined as:
:$\map P n = 2 n + 1$
Applying $P$ iteratively to each of the smallest Sophie Germain primes in turn:
{{begin-eqn}}
{{eqn | l = \map P 2
      | r = 5
      | c = 
}}
{{eqn | l = \map P 5
      | r = 11
      | c = 
}}
{{eqn | l = \map P {11}
      | r = 23
      | c = 
}}
{{eqn | l = \map P {23}
      | r = 47
      | c = 
}}
{{eqn | l = \map P {47}
      | r = 95
      | c = which is not prime
}}
{{end-eqn}}
Thus $\tuple {2, 5, 11, 23, 47}$ is a Cunningham chain of the first kind of length $5$.
{{begin-eqn}}
{{eqn | l = \map P 3
      | r = 7
      | c = 
}}
{{eqn | l = \map P 7
      | r = 15
      | c = which is not prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map P {29}
      | r = 59
      | c = 
}}
{{eqn | l = \map P {59}
      | r = 119
      | c = which is not prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map P {41}
      | r = 83
      | c = 
}}
{{eqn | l = \map P {83}
      | r = 167
      | c = 
}}
{{eqn | l = \map P {167}
      | r = 335
      | c = which is not prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map P {53}
      | r = 107
      | c = 
}}
{{eqn | l = \map P {107}
      | r = 215
      | c = which is not prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map P {83}
      | r = 167
      | c = 
}}
{{eqn | l = \map P {167}
      | r = 335
      | c = which is not prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map P {89}
      | r = 179
      | c = 
}}
{{eqn | l = \map P {179}
      | r = 359
      | c = 
}}
{{eqn | l = \map P {359}
      | r = 719
      | c = 
}}
{{eqn | l = \map P {719}
      | r = 1439
      | c = 
}}
{{eqn | l = \map P {1439}
      | r = 2879
      | c = 
}}
{{eqn | l = \map P {2879}
      | r = 5759
      | c = which is not prime.
}}
{{end-eqn}}
It is noted that $\dfrac {89 - 1} 2 = 44$ which is not prime.
Hence the sequence of $6$:
:$\tuple {89, 179, 359, 719, 1439, 2879}$
{{qed}}
Theorem: Necessary and Sufficient Condition for First Order System to be Mutually Consistent

Let $\mathbf y$, $\boldsymbol \psi$ be $n$-dimensional vectors.
Let $g$ be a twice differentiable mapping.
Let
:$(1): \quad \map {\boldsymbol \psi} {x, \mathbf y} = \map {\mathbf y'} {x, \mathbf y}$
:$(2): \quad \mathbf p \sqbrk {x, \mathbf y, \map {\boldsymbol \psi} {x, \mathbf y} } = \map {g_{\mathbf y} } {x, \mathbf y}$
where $\mathbf p$ is a momentum.
Then the boundary conditions defined by $(1)$ are mutually consistent {{iff}} the mapping $\map g {x, \mathbf y}$ satisfies the Hamilton-Jacobi equation:
:$(3): \quad \dfrac {\partial g} {\partial x} + \map H {x, \mathbf y, \dfrac {\partial g} {\partial \mathbf y} } = 0$

Proof:
=== Necessary condition ===
Take the partial derivative of $(3)$ {{WRT|Differentiation}} $x$:
:$(4): \quad \frac {\partial^2 \map g {x, \mathbf y} } {\partial \mathbf y \partial x} = -\dfrac \partial {\partial \mathbf y} \map H {x, \mathbf y, \dfrac {\partial g} {\partial \mathbf y} }$
By the Schwarz-Clairaut Theorem, the order of partial derivatives of $g$ can be exchanged:
:$(5): \quad \dfrac {\partial^2 \map g {x, \mathbf y} } {\partial \mathbf y \partial x} = \dfrac {\partial \mathbf p \sqbrk {x, \mathbf y, \map {\boldsymbol \psi} {x, \mathbf y} } } {\partial x}$
Since $\map g {x, \mathbf y}$ depends on $\mathbf y$ only in a direct way, it follows that:
:$\dfrac {\partial g} {\partial \mathbf y} = g_{\mathbf y}$
Hence, from $(2)$, $(4)$ and $(5)$ it follows that:
:$\dfrac {\partial \mathbf p \sqbrk {x, \mathbf y, \map {\boldsymbol \psi} {x, \mathbf y} } } {\partial x} = -\dfrac \partial {\partial \mathbf y} \map H {x, \mathbf y, \mathbf p \sqbrk {x, \mathbf y, \map {\boldsymbol \psi} {x, \mathbf y} } }$
These are the consistency equations.
{{qed|lemma}}
Theorem: 2-Digit Numbers forming Longest Reverse-and-Add Sequence

Let $m \in \Z_{>0}$ be a positive integer expressed in decimal notation.
Let $\map r m$ be the reverse-and-add process on $m$.
Let $r$ be applied iteratively to $m$.
The $2$-digit integers $m$ which need the largest number of iterations before reaching a palindromic number are $89$ and $98$, both needing $24$ iterations.

Proof:
The sequence obtained by iterating $r$ on $89$ is:
:$89, 187, 968, 1837, 9218, 17347, 91718, 173437, 907808, 1716517, 8872688,$
:$17735476, 85189247, 159487405, 664272356, 1317544822, 3602001953, 7193004016, 13297007933,$
:$47267087164, 93445163438, 176881317877, 955594506548, 170120002107, 8713200023178$
{{OEIS|A033670}}
Note that $\map r {89} = \map r {98} = 187$, so the sequence obtained by iterating $r$ on $98$ is the same.
To show that no other $2$-digit integers have longer sequences, we define an equivalence relation on the $2$-digit integers as follows:
:$\sqbrk {a_1 a_2} \sim \sqbrk {b_1 b_2} \iff a_1 + a_2 = b_1 + b_2$
For each $2$-digit integer $\sqbrk {a_1 a_2}$:
:$\map r {\sqbrk {a_1 a_2} } = \sqbrk {a_1 a_2} + \sqbrk {a_2 a_1} = 11 \paren {a_1 + a_2}$
Notice that if $\sqbrk {a_1 a_2} \sim \sqbrk {b_1 b_2}$:
:$\map r {\sqbrk {a_1 a_2} } = 11 \paren {a_1 + a_2} = 11 \paren {b_1 + b_2} = \map r {\sqbrk {b_1 b_2} }$
So the value of $r$ is equal in each equivalence class.
For each $2$-digit integer $\sqbrk {a_1 a_2}$, we have:
:$18 = 9 + 9 \ge a_1 + a_2 \ge 1 + 0 = 1$
For $2$-digit integers where $a_1 + a_2 \le 9$:
:$\map r {\sqbrk {a_1 a_2} } = 11 \paren {a_1 + a_2}$ is a $2$-digit multiple of $11$, which is a palindrome.
So we consider each equivalence class of $2$-digit integers with $a_1 + a_2 \ge 10$.
We have the sequences:
:$19, 110, 121$
:$29, 121$
:$39, 132, 363$
:$49, 143, 484$
:$59, 154, 605, 1111$
:$69, 165, 726, 1353, 4884$
:$79, 176, 847, 1595, 7546, 14003, 44044$
:$89, \cdots, 8713200023178$, with $24$ iterations
:$99$
So:
:every $2$-digit integer not in the equivalence class of $89$ has a shorter iteration
and:
:$\eqclass {89} \sim = \set {89, 98}$
{{qed}}
Theorem: Reciprocal of 89

:$\dfrac 1 {89} = 0 \cdotp \dot 01123 \, 59550 \, 56179 \, 77528 \, 08988 \, 76404 \, 49438 \, 20224 \, 719 \dot 1$

Proof:
Performing the calculation using long division:
<pre>
   0.0112359550561797752808988764044943820224719101...
  ------------------------------------------------------
89)1.0000000000000000000000000000000000000000000000000
     89       534       712       801        89
     --       ---       ---       ---       ---  
     110       160       880       390       810
      89        89       801       356       801
     ---       ---       ---       ---       ---
      210       710       790       340        90
      178       623       712       267        89
      ---       ---       ---       ---        --
       320       870       780       730        100
       267       801       712       712         89
       ---       ---       ---       ---        --- 
        530       690       680       180       ...
        445       623       623       178
        ---       ---       ---       ---
         850       670       570        200
         801       623       534        178
         ---       ---       ---        ---
          490       470       360        220
          445       445       356        178
           --       ---       ---        ---
           450       250        400       420
           445       178        356       356
           ---       ---        ---       ---
             500      720        440       640
             445      712        356       623
             ---      ---        ---       ---
              550       800       840       170
              534       712       801        89
</pre>
{{qed}}
Theorem: Cauchy Sequence is Bounded/Real Numbers

Every Cauchy sequence in $\R$ is bounded.

Proof:
Let $\sequence {a_n}$ be a Cauchy sequence in $\R$.
Then there exists $N \in \N$ such that:
:$\size {a_m - a_n} < 1$
for all $m, n \ge N$.
So for all $m \ge N$, we have:
{{begin-eqn}}
{{eqn | l = \size {a_m}
      | r = \size {a_N + a_m - a_N}
}}
{{eqn | o = \le 
      | r = \size {a_N} + \size {a_m - a_N}
      | c = Triangle Inequality
}}
{{eqn | o = \le 
      | r = \size {a_N} + 1
}}
{{end-eqn}}
So $\sequence {a_n}$ is bounded, as claimed.
{{qed}}
Theorem: Cauchy Sequence is Bounded/Real Numbers

Every Cauchy sequence in $\R$ is bounded.

Proof:
Let $\sequence {a_n}$ be a Cauchy sequence in $\R$.
Then there exists $N \in \N$ such that:
:$\size {a_m - a_n} < 1$
for all $m, n \ge N$.
Note that for $m \le N$:
{{begin-eqn}}
{{eqn | l = \size {a_m}
      | o = \le
      | r = \max \set {\size {a_1}, \size {a_2}, \dotsc, \size {a_N} }
}}
{{eqn |o = < 
      | r = \max \set {\size {a_1}, \size {a_2}, \dotsc, \size {a_N} } + 1
}}
{{eqn-intertext|and for $m > N$:}}
{{eqn | l = \size {a_m} 
      | r = \size {a_N + a_m - a_N} 
}}
{{eqn | o = \le
      | r = \size {a_N} + \size {a_m - a_N}
}}
{{eqn | o = < 
      | r = \size {a_N} + 1
}}
{{eqn | o = \le 
      | r = \max \set {\size {a_1}, \size {a_2}, \dotsc, \size {a_N} } + 1
}}
{{end-eqn}}
Hence for all $m \in \N$:
:$\size {a_m} < \max \set {\size {a_1}, \size {a_2}, \dotsc, \size {a_N} } + 1$
Therefore $\sequence {a_n}$ is bounded, as claimed.
{{qed}}
Theorem: Reciprocal of 89 as Sum of Fibonacci Numbers by Negative Powers of 10

:$\ds \sum_{k \mathop \ge 0} \dfrac {F_k} {10^{k + 1} } = \dfrac 1 {89}$
where $F_k$ is the $k$th Fibonacci number:
:$F_0 = 0, F_1 = 1, F_k = F_{k - 1} + F_{k - 2}$
That is:
<pre>
1 / 89 = 0.0
       + 0.01
       + 0.001
       + 0.0002
       + 0.00003
       + 0.000005
       + 0.0000008
       + 0.00000013
       + 0.000000021
       + 0.0000000034
       + 0.00000000055
       + ..............
</pre>

Proof:
First we note that from Reciprocal of $89$:
:$\dfrac 1 {89} = 0 \cdotp \dot 01123 \, 59550 \, 56179 \, 77528 \, 08988 \, 76404 \, 49438 \, 20224 \, 719 \dot 1$
We have that:
:$89 = 10^2 - 10 - 1$
So:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \dfrac {F_k} {10^{k + 1} }
      | r = \dfrac 1 {10} \sum_{k \mathop \ge 0} F_k \paren {\dfrac 1 {10} }^k
}}
{{eqn | r = \dfrac 1 {10} \cdot \dfrac {\dfrac 1 {10} } {1 - \dfrac 1 {10} - \paren {\dfrac 1 {10} }^2 }
      | c = Generating Function for Fibonacci Numbers
}}
{{eqn | r = \dfrac 1 {10^2 - 10 - 1}
      | c = 
}}
{{eqn | r = \dfrac 1 {89}
      | c = 
}}
{{end-eqn}}
{{qed}}