Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove that the decimal number $\displaystyle \frac{1}{5}$ cannot be represented by a finite expansion in the binary system.

share|cite|improve this question
up vote 4 down vote accepted

Suppose that $\frac{1}{5}$ has a finite binary expansion, then $\frac{1}{5} = \sum_{i = 1}^N \frac{1}{2^i}a_i$, where $a_i \in \{0, 1\}$. This implies $2^N = \sum_{i = 1}^N 5 a_i 2^{N-i}$. Both sides of the equality are integers, but right side is a multiple of $5$ while $2^N$ is not.

share|cite|improve this answer

HINT: From your earlier post, you know that

a real number has a finite representation in the binary system if and only if it is of the form $$ \pm \frac{m}{2^n}$$ where n and m are positive integers.

How can you apply that here to show that $\displaystyle \frac{1}{5}$ cannot be represented by a finite expansion in the binary system?

Are there any $m, n \in \mathbb{Z}$ such that $\displaystyle \frac{m}{2^n} = \frac{1}{5}$? Why not?

Is there any multiple $m$ such that $5m$ is equivalent to $2^n$ for some $n$?

Put differently, note that for any $n$, $\,5 \not | \;\,2^n$.

share|cite|improve this answer
@nour Is there a reason you changed your upvote? I'd really like to know why? That's certainly your choice (upvoting), but I'd like to know why you changed your mind a couple weeks later? – amWhy Dec 9 '12 at 15:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.