Anomalous Cancellation on 2-Digit Numbers

Theorem
There are exactly four anomalously cancelling vulgar fractions having two-digit numerator and denominator when expressed in base $10$ notation:

Proof
Let $\dfrac {\sqbrk {a x} } {\sqbrk {x b} }$ be an anomalously cancelling vulgar fraction.

Then we have:

From $(3)$:

Therefore $\dfrac {9 a b} {10 a - b}$ must be an integer.

$10 a - b$ 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:

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 $\dfrac b {10 - b} > a$:

Then we eliminate the pairs $\tuple {a, b}$ where $10 a - b$ has a prime factor greater than $9$:

Thus it remains to check whether $\dfrac {9 a b} {10 a - b}$ is an integer for the following pairs $\tuple {a, b}$:


 * $\tuple {1, 2}$, $\tuple {1, 3}$, $\tuple {2, 3}$, $\tuple {1, 4}$, $\tuple {2, 4}$, $\tuple {1, 5}$, $\tuple {2, 5}$, $\tuple {3, 5}$, $\tuple {4, 5}$, $\tuple {2, 6}$, $\tuple {3, 6}$, $\tuple {4, 8}$, $\tuple {5, 8}$

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 $\dfrac {49} {98}$