I was trying to prove that every real fraction can be represented by the binary representation.
To be more specific, I want to show that, given any real fraction $\frac{p}{q},\, p\in \mathbb{Z},q\in \mathbb{N}^+$, there exists a series $\sum_{I=-\infty}^\infty a_i2^{i}$ s.t. $\frac{p}{q}=\sum_{I=-\infty}^\infty a_i2^{i}$, where $a_i\in\{0,1\}.$
The integral part is easy, by the successive divisions by $2$. We may obtain the decimal part by successive multiplications by $2$ (subtracting $1$ from the result of multiplication each time if necessary).
But I suddenly realize a problem, what if the decimal part is a repeating decimal, which means it doesn't end. Our step-by-step multiplication method is like an induction, which could not be applied to the infinite case.
Can someone else help me modify the proof? A lot of thanks!
Assume that $q$ is odd. Then $2\in\mathbb{Z}/(q\mathbb{Z})^*$ and by Euler's theorem $$ 2^{\varphi(q)}\equiv 1\pmod{q}, $$ so $q$ is a divisor of $2^{\varphi(q)}-1$ and $$ \frac{1}{q} = \frac{0.11111111\ldots_2}{q}=0.\overline{B} $$ where $B$ is the binary string with $\varphi(q)$ bits representing $\frac{2^{\varphi(q)}-1}{q}$ in base $2$.
Once you have that the reciprocal of any odd natural number has a periodic base-$2$ representation you have very little to fill in.