Uniqueness of Continued Fraction Expansion for Rationals

Recall how we found the continued fraction for a rational.

We used the division algorithm to determine the terms in the continued fraction. Here's a recap:

By the division algorithm, we have these unique equations: $$a=bq_1+r_1$$ $$b=r_1q_2+r_2$$ $$r_1=r_2q_3+r_3$$ $$\vdots$$ $$r_{n-2}=r_{n-1}q_n+0$$ Then, we can rearrange each of them so that we have what we want: $$a=bq_1+r_1\implies \frac{a}{b}=q_1+\frac{r_1}{b}$$ $$b=r_1q_2+r_2\implies \frac{b}{r_1}=q_2+\frac{r_2}{r_1}$$ $$r_1=r_2q_3+r_3\implies \frac{r_1}{r_2}=q_3+\frac{r_3}{r_2}$$ $$\vdots\qquad\qquad\qquad\qquad\vdots$$ $$r_{n-2}=r_{n-1}q_n+0\implies \frac{r_{n-2}}{r_{n-1}}=q_n+0.$$

Now, it's time to substitute repeatedly. $$\begin{align} \dfrac{a}{b}&=q_1+\dfrac{1}{q_2+\frac{r_2}{r_1}}\\ &=q_1+\dfrac{1}{q_2+\dfrac{1}{q_3+\frac{r_3}{r_2}}}\\ &\qquad\qquad \vdots\\ &=q_1+\dfrac{1}{q_2+\dfrac{1}{\ddots+\dfrac{1}{q_n}}}\\ &=[q_1,q_2,\cdots,q_n]. \end{align}$$

Since all the quotients when we used the division algorithm were unique, we know the continued fraction expansion is as well, because we used those unique values.

Note: If we have a fraction of the form $[a_1,a_2,\cdots,a_{n-1},2]$, then note that this is equal to $[a_1,a_2,\cdots,a_{n-1},1,1]$. We are not able to get the second representation from the division algorithm though.