Anning's Theorem

Theorem
In any base greater than $1$, the fraction:
 * $\dfrac {101 \, 010 \, 101} {110 \, 010 \, 011}$

has the property that if the two $1$'s in the center of the numerator and the denominator are replaced by the same odd number of $1$'s, the value of the fraction remains the same.

For example:
 * $\dfrac {101 \, 010 \, 101} {110 \, 010 \, 011} = \dfrac {1 \, 010 \, 111 \, 110 \, 101} {1 \, 100 \, 111 \, 110 \, 011} = \dfrac {9091} {9901}$ (in base $10$).

Proof
Let $b$ be the base in question.

Let $F = \dfrac {101 \, 010 \, 101} {110 \, 010 \, 011}$.

Then:
 * $F = \dfrac {1 + b^2 + b^4 + b^6 + b^8} {1 + b + b^4 + b^7 + b^8}$

It is necessary to prove that for all $k \in \Z_{>0}$:
 * $F = \dfrac {1 + b^2 + b^4 + b^5 + \cdots + b^{2 k + 2} + b^{2 k + 4} + b^{2 k + 6} } {1 + b + b^4 + b^5 + \cdots + b^{2 k + 2} + b^{2 k + 5} + b^{2 k + 6} }$

This is done by:
 * multiplying the numerator of one by the denominator of the other

and then:
 * multiplying the denominator of one by the numerator of the other

and checking that they are equal.

Thus we proceed:

and:

Equality can be demonstrated.