Prove that there are no two integers $a$ and $b$, such that $\frac{a}{b}$ is non-terminating and non-recurring number

Question

I understand that all non-terminating and recurring real numbers can be expressed as $\frac{a}{b}$ where $a, b$ are integers. But I am unable to prove converse of this statement.

Answer 1

Suppose there exist two integers $p$ and $q$ where $q\neq0$

When you divide $\frac{p}{q}$ using long division, there are only $q$ remainders that are possible. If $0$ appears as a remainder, the decimal expansion terminates. If $0$ never occurs, then the algorithm can run at most $q − 1$ steps without using any remainder more than once. After that, a remainder must recur, and then the decimal expansion repeats.

For more, read about irrational numbers

Answer 2

Factor $b$ as $b=2^h5^kc$, where $c$ is divisible neither for $2$ nor for $5$. Then $$ \frac{a}{b}=\frac{2^k5^ha}{10^{h+k}c} $$ and we can reduce to the case of $a/b$ where $b$ is divisible neither for $2$ nor for $5$ (because the original number is the same with the decimal point suitably shifted) and $b>1$. It is also not restrictive to assume $0<a<b$.

We then know that the decimal number we get is not terminating. Can you prove it?

The process of long division (adding zeros as we need them) will never leave a zero remainder, so the remainder $r$ at each stage satisfies $0<r<b$.

Since only $b-1$ remainders are available, at some stage we must get a remainder we already got before. At that point, the divisions will repeat.