Are all integer fractions rational?

Question

Any repeating decimal can be written as a fraction $\frac{a}{b}$ where $a$ and $b$ are integers. But is the reverse true. Will any fraction $\frac{a}{b}$ where $a$ and $b$ are integers produce a quotient with a repeating decimal?

Answer 1

The answer is yes. Every rational number has a repeating decimal expansion.

It's an easy exercise to prove this (hint: consider trailing 9s as above).

Answer 2

This question is confused. The DEFINITION of "rational number" is a number that is a quotient of two integers.

It can be shown that every repeating decimal is rational and every rational number is a repeating decimal. But: DO NOT think that "repeating (or terminating) decimal" is the definition of rational number. No mathematician does that.

The answer is "yes" because of the pigeonhole principle. Say, for example you divide $5$ by $82$:

$$ \begin{array}{cccccccccccccc} & & 0 & . & 6 & 0 & 9 & 7 & 5 & \\[10pt] 82 & ) & 5 & . & 0 & 0 & 0 & 0 & 0 & 0 \\ & & 4 & & 9 & 2 \\ \\ & & & & & 8 & 0 \\ & & & & & & 0 \\ \\ & & & & & 8 & 0 & 0 \\ & & & & & 7 & 3 & 8 \\ \\ & & & & & & 6 & 2 & 0 \\ & & & & & & 5 & 7 & 4 \\ \\ & & & & & & & 4 & 6 & 0 \\ & & & & & & & 4 & 1 & 0 \\ \\ & & & & & & & & 5 & 0 & 0 & \longleftarrow\text{ return to starting point}\\ \end{array} $$ (We started by asking how many times $82$ goes into $500$ and we said $6$. Now we are again faced with the question of how many times $82$ goes into $500$, and the answer is still $6$.)

How do you know that you will eventually return to the starting point, and therefore start repeating? The answer is: the pigeonhole principle. Thare are only 82 possible remainders: $0,1,2,3,\ldots, 81$. If you ever get $0$, the decimal expansion terminates. But if you never get $0$, you can't go more than $81$ steps without getting a remainder that you've seen before, since after $81$ steps you've used them all up.