# Prove or disprove $aB_m-bA_m=[q_n,\cdots,q_m]$

Some definitions and notation:

Let the natural numbers $$q_0,q_1,\cdots ,q_n$$ be the $$n$$ terms in the continued fraction expansion of the rational number $$\frac{a}{b}$$, that is $$\frac{a}{b}=q_0 +\frac1{q_1+}\cdots\frac1{q_n}=\frac{[q_0,q_1,\cdots, q_n]}{[q_1,q_2,\cdots,q_n]}.$$

Thus, the convergents to $$\frac{a}{b}$$ are the numbers $$\frac{A_m}{B_m} =\frac{[q_0,\cdots,q_m]}{[q_1,\cdots,q_m]},\quad 0\leq m\leq n.$$

Prove or disprove that $$\lvert aB_{m}-bA_{m}\rvert=[q_n,\cdots,q_m],$$ for $$1\leq m\leq n-1$$.

This question grew out of my efforts to find an alternative proof of the fact that the sequence of convergents gets successively closer and closer to $$\frac{a}{b}$$.

In trying to prove that theorem by contradiction, I derived the following inequality $$\tag{1}\left\lvert \frac{aB_{m-1}-bA_{m-1}}{bB_{m-1}}\right\rvert\lt \left\lvert\frac{aB_m-bA_m}{bB_m}\right\rvert.$$ Then, using the identity $$\tag{2} A_m=q_mA_{m-1}+A_{m-2}$$ my thought was that it might be possible to work recursively back from the initial case $$m=n-1$$.

Thus, in the case $$m=n-1$$ we have $$aB_{n-2}-bA_{n-2}=(q_nA_{n-1}+A_{n-2})B_{n-2}-(q_nB_{n-1}+B_{n-2})A_{n-2},$$ and so $$\lvert aB_{n-2}-bA_{n-2} \rvert=\lvert q_n(A_{n-1}B_{n-2}-B_{n-1}A_{n-2})\rvert =q_n.$$

In a similar vein, I worked out that $$\lvert aB_{n-3}-bA_{n-3}\rvert=\lvert q_nq_{n-1}(-1)^{n-3}+(-1)^{n-3}\rvert$$ and $$\lvert aB_{n-4}-bA_{n-4}\rvert=\lvert q_nq_{n-1}q_{n-2}(-1)^{n-4}+q_n(-1)^{n-4} +q_{n-2}(-1)^{n-4}\rvert.$$ In general, this seems to imply $$\tag{3}\left\lvert aB_{m}-bA_{m}\right\rvert=[q_n,\cdots,q_{m}],$$ but I am unsure how I might prove this.