# floating point binary arithmetic

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

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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.

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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 $n$, $\,5 \not | \;\,2^n$.

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@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