The continued fraction of a rational number terminates

Statement

Let $x$ be a real number. The continued fraction expansion of $x$ is finite if and only if $x$ is rational.

Remarks

If $x = a/b$ with $b > 0$, the expansion follows the Euclidean algorithm on $a$ and $b$. So it stops after finitely many steps. Conversely, a finite simple continued fraction is a rational number. For example, $\frac{43}{19}$: $43 = 2 \cdot 19 + 5$, $19 = 3 \cdot 5 + 4$, $5 = 1 \cdot 4 + 1$, $4 = 4 \cdot 1$. So $\frac{43}{19} = [2; 3, 1, 4]$. A rational number has exactly two finite expansions, since $[a_0; \dots, a_n] = [a_0; \dots, a_n - 1, 1]$ when $a_n > 1$.

Explanations

L1 human

For a fraction, the continued fraction algorithm is the Euclidean algorithm in another form. The remainders get smaller, so the process must end. An irrational number gives an infinite continued fraction.