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