Theorem: Anomalous Cancellation on 2-Digit Numbers

There are exactly four anomalously cancelling proper fractions having two-digit numerator and denominator when expressed in base $10$ notation:
{{:Anomalous Cancellation on 2-Digit Numbers/Examples}}

Proof:
Let $\dfrac {\sqbrk {a x} } {\sqbrk {x b} }$ be an anomalously cancelling proper fraction.
Then we have:
{{begin-eqn}}
{{eqn | n = 1
      | l = a, b, x \le 9
      | o =
      | c = as they are digits in base $10$
}}
{{eqn | n = 2
      | l = a < b
      | o =
      | c = this fraction is a proper fraction
}}
{{eqn | n = 3
      | l = \dfrac a b = \dfrac {\sqbrk {a x} } {\sqbrk {x b} }
      | o =
      | c = this fraction is anomalously cancelling
}}
{{end-eqn}}
From $(3)$:
{{begin-eqn}}
{{eqn | l = \dfrac a b
      | r = \dfrac {10 a + x} {10 x + b}
      | c = 
}}
{{eqn | ll = \leadsto
      | l = a \paren {10 x + b}
      | r = b \paren {10 a + x}
      | c = 
}}
{{eqn | ll = \leadsto
      | l = 10 a x + a b
      | r = 10 a b + b x
      | c = 
}}
{{eqn | ll = \leadsto
      | l = 10 a x - b x
      | r = 9 a b
      | c = 
}}
{{eqn | ll = \leadsto
      | l = x
      | r = \dfrac {9 a b} {10 a - b}
      | c = 
}}
{{end-eqn}}
Therefore $\dfrac {9 a b} {10 a - b}$ must be an integer.
{{AimForCont}} $10 a - b$ is divisible by a prime $p$ such that $p > 9$.
We have that:
:$\paren {10 a - b} \divides 9 a b$
and so:
:$p \divides 9 a b$
where $\divides$ denotes divisibility.
By Euclid's Lemma for Prime Divisors, that means one of the elements of $\set {9, a, b}$ must be divisible by $p$.
Hence from Absolute Value of Integer is not less than Divisors, either $a$ or $b$ must be greater than $9$.
This contradicts $(1)$ above.
Hence by Proof by Contradiction it cannot be the case that $10 a - b$ is divisible by a prime greater than $9$.
We continue:
{{begin-eqn}}
{{eqn | l = \dfrac {9 a b} {10 a - b}
      | o = \le
      | r = 9
      | c = from $(1)$: $x \le 9$
}}
{{eqn | ll = \leadsto
      | l = \dfrac {a b} {10 a - b}
      | o = \le
      | r = 1
      | c =
}}
{{eqn | ll = \leadsto
      | l = a b
      | o = \le
      | r = 10 a - b
      | c = as $10 a - b \ge 10 - b > 0$
}}
{{eqn | ll = \leadsto
      | l = b
      | o = \le
      | r = 10 a - a b
      | c = adding $b - a b$ to both sides
}}
{{eqn | ll = \leadsto
      | l = a
      | o = \ge
      | r = \dfrac b {10 - b} 
      | c = as $10 - b > 0$
}}
{{end-eqn}}
It remains for us to check whether $x = \dfrac {9 a b} {10 a - b}$ is an integer for $\dfrac b {10 - b} \le a < b$, when $b$ ranges from $2$ to $9$.
We note first that the following pairs $\tuple {a, b}$ are such that $\dfrac b {10 - b} > a$:
{{begin-eqn}}
{{eqn | lo= b = 6:
      | l = \dfrac b {10 - b}
      | r = \dfrac 3 2 > 1
      | c = which eliminates $\tuple {1, 6}$
}}
{{eqn | lo= b = 7:
      | l = \dfrac b {10 - b}
      | r = \dfrac 7 3 > 2
      | c = which eliminates $\tuple {1, 7}$ and $\tuple {2, 7}$
}}
{{eqn | lo= b = 8:
      | l = \dfrac b {10 - b}
      | r = 4 > 3
      | c = which eliminates $\tuple {1, 8}$, $\tuple {2, 8}$ and $\tuple {3, 8}$
}}
{{eqn | lo= b = 9:
      | l = \dfrac b {10 - b}
      | r = 9
      | c = which eliminates all possible such solutions such that $b = 9$
}}
{{end-eqn}}
Then we eliminate the pairs $\tuple {a, b}$ where $10 a - b$ has a prime factor greater than $9$:
{{begin-eqn}}
{{eqn | lo= \tuple {3, 4}:
      | l = 10 a - b
      | r = 26 = 13 \times 2
      | c = 
}}
{{eqn | lo= \tuple {4, 6}:
      | l = 10 a - b
      | r = 34 = 17 \times 2
      | c = 
}}
{{eqn | lo= \tuple {5, 6}:
      | l = 10 a - b
      | r = 44 = 11 \times 4
      | c = 
}}
{{eqn | lo= \tuple {3, 7}:
      | l = 10 a - b
      | r = 23
      | c = 
}}
{{eqn | lo= \tuple {4, 7}:
      | l = 10 a - b
      | r = 33 = 11 \times 3
      | c = 
}}
{{eqn | lo= \tuple {5, 7}:
      | l = 10 a - b
      | r = 43
      | c = 
}}
{{eqn | lo= \tuple {6, 7}:
      | l = 10 a - b
      | r = 53
      | c = 
}}
{{eqn | lo= \tuple {6, 8}:
      | l = 10 a - b
      | r = 52 = 13 \times 4
      | c = 
}}
{{eqn | lo= \tuple {7, 8}:
      | l = 10 a - b
      | r = 62 = 31 \times 2
      | c = 
}}
{{end-eqn}}
Thus it remains to check whether $\dfrac {9 a b} {10 a - b}$ is an integer for the following pairs $\tuple {a, b}$:
:$\tuple {1, 2}$, $\tuple {1, 3}$, $\tuple {2, 3}$, $\tuple {1, 4}$, $\tuple {2, 4}$, $\tuple {1, 5}$, $\tuple {2, 5}$, $\tuple {3, 5}$, $\tuple {4, 5}$, $\tuple {2, 6}$, $\tuple {3, 6}$, $\tuple {4, 8}$, $\tuple {5, 8}$
{{begin-eqn}}
{{eqn | lo= \tuple {1, 2}:
      | l = \dfrac {9 a b} {10 a - b}
      | r = \dfrac {18} 8
      | c = not an integer
}}
{{eqn | lo= \tuple {1, 3}:
      | l = \dfrac {9 a b} {10 a - b}
      | r = \dfrac {27} 7
      | c = not an integer
}}
{{eqn | lo= \tuple {2, 3}:
      | l = \dfrac {9 a b} {10 a - b}
      | r = \dfrac {54} {16}
      | c = not an integer
}}
{{eqn | lo= \tuple {1, 4}:
      | l = \dfrac {9 a b} {10 a - b}
      | r = \dfrac {36} 6 = 6
      | c = which leads to the solution $\dfrac {16} {64}$
}}
{{eqn | lo= \tuple {2, 4}:
      | l = \dfrac {9 a b} {10 a - b}
      | r = \dfrac {72} {16}
      | c = not an integer
}}
{{eqn | lo= \tuple {1, 5}:
      | l = \dfrac {9 a b} {10 a - b}
      | r = \dfrac {45} 5 = 9
      | c = which leads to the solution $\dfrac {19} {95}$
}}
{{eqn | lo= \tuple {2, 5}:
      | l = \dfrac {9 a b} {10 a - b}
      | r = \dfrac {90} {15} = 6
      | c = which leads to the solution $\dfrac {26} {65}$
}}
{{eqn | lo= \tuple {3, 5}:
      | l = \dfrac {9 a b} {10 a - b}
      | r = \dfrac {135} {25}
      | c = not an integer
}}
{{eqn | lo= \tuple {4, 5}:
      | l = \dfrac {9 a b} {10 a - b}
      | r = \dfrac {180} {35}
      | c = not an integer
}}
{{eqn | lo= \tuple {2, 6}:
      | l = \dfrac {9 a b} {10 a - b}
      | r = \dfrac {108} {14}
      | c = not an integer
}}
{{eqn | lo= \tuple {3, 6}:
      | l = \dfrac {9 a b} {10 a - b}
      | r = \dfrac {108} {14}
      | c = not an integer
}}
{{eqn | lo= \tuple {4, 8}:
      | l = \dfrac {9 a b} {10 a - b}
      | r = \dfrac {288} {32} = 9
      | c = which leads to the solution $\dfrac {49} {98}$
}}
{{eqn | lo= \tuple {5, 8}:
      | l = \dfrac {9 a b} {10 a - b}
      | r = \dfrac {360} {42}
      | c = not an integer
}}
{{end-eqn}}
So the only solutions for $\tuple {a, x, b}$ are:
:$\tuple {1, 6, 4}$, $\tuple {1, 9, 5}$, $\tuple {2, 6, 5}$ and $\tuple {4, 9, 8}$
These correspond to the fractions:
:$\dfrac {16} {64}$, $\dfrac {19} {95}$, $\dfrac {26} {65}$ and $\dfrac {49} {98}$
{{qed}}