Consider a pair of integer-valued sequences $(a(n), b(n))$ such that the continued fraction
$$a(0) + \cfrac{b(1)}{a(1) + \cfrac{b(2)}{a(2) + \cdots}},$$
exists. Given a strictly-nonzero sequence $c(n)$, we have the identity
$$a(0) + \cfrac{b(1)}{a(1) + \cfrac{b(2)}{a(2) + \cdots}} = a(0) + \cfrac{c(1) b(1)}{c(1) a(1) + \cfrac{c(1) c(2) b(2)}{c(2) a(2) + \cdots}}. $$
More explicitly, we replace $a(n)$ with $c(n) a(n)$ for $n \geq 0$ and $b(n)$ with $c(n) c(n - 1) b(n)$ for $n \geq 1$, taking $c(0) = 1$.
Define the relation on pairs of sequences $(a, b) \sim (a', b')$ to mean that there exists such a transforming sequence $c$; namely, \begin{align*} a(n) &= c(n) a'(n), \quad n \geq 0 \\ b(n) &= c(n - 1) c(n) b'(n), \quad n \geq 1, \end{align*} where $c(n) \neq 0$ for $n \geq 1$ and $c(0) = 1$.
If we allow the transforming sequence $c(n)$ to be rational-valued, then $\sim$ is an equivalence relation, and "generates the same continued fraction" is an invariant under its equivalence classes.
Is $(a, b) \sim (a', b')$ equivalent to "$(a, b)$ and $(a', b')$ generate the same continued fraction"?
In other words, if $(a, b)$ and $(a', b')$ generate the same continued fraction, does there exist a "transforming sequence" $c(n)$ which takes $(a', b')$ to $(a, b)$?
A counterexample: $$\sqrt{3}-1=\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{2+\ddots}}}}=\cfrac{2}{1+\cfrac{3}{1+\cfrac{2}{1+\cfrac{3}{1+\ddots}}}}$$ Here $$\begin{aligned} a(0)&=0&b(n)&=1&a(2n-1)&=1&a(2n)&=2\\ a'(0)&=0&a'(n)&=1&b'(2n-1)&=2&b'(2n)&=3 \end{aligned}\qquad(n>0)$$ and clearly $\color{blue}{(a,b)\nsim(a',b')}$, since otherwise we would have $c(n)=a(n)/a'(n)=a(n)$ for $n>0$, hence $c(n-1)c(n)=2$ for $n>1$, hence $b(n)/b'(n)=2$ for $n>1$, which is a contradiction.