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Is there a name for the set of real numbers which can be written as the sum of finitely many powers of 2?

Such a set would include all integers of course, but no irrational and not all rational numbers (1/2 is in, but 1/3 is not for example).

Does this set have any interesting properties or uses? Is it well defined?

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    Note that such numbers can be written as a single fraction where the denominator is a power of two. It's a bit easier to phrase, and a lot easier to see which numbers fulfill the requirement. A not very useful fact is that these numbers are exactly the surreal numbers with a finite generation number. – 2017-02-15
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    Note also that all numbers that can be exactly represented as floating point numbers are of that form. – 2017-02-15
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    It is a binary fixed-point arithmetic number. – 2017-02-15

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These are the dyadic rationals (or the positive dyadic rationals, if you don't allow subtractions as well as sums).

This isn't hard to show. First, suppose $q$ is a dyadic rational. let $a$ be odd and $b=2^m$ be a power of $2$ with ${a\over b}=q$; then $q=2^{-m}+2^{-m}+...$ ($a$ times).

In the other direction, suppose I have $q=2^{i_1}+2^{i_2}+...+2^{i_n}$; I want to show that that's a dyadic rational. WLOG, suppose $i_1\le i_2\le . . . \le i_n$. Then we can factor $q$ as $$2^{i_1}(1+2^{j_2}+2^{j_3}+...+2^{j_n}),$$ where $j_k=i_k-i_1$. Now since $i_1\le i_k$ for all $k$, we have that $(1+2^{j_2}+2^{j_3}+...+2^{j_n})$ is an integer; call it "$c$".

So $q=2^{i_1}c$. If $i_1\ge 0$, then $q$ is an integer - certainly a dyadic rational! If $i_1<0$, then $q={c\over 2^{-i_1}}$, so is a dyadic rational.

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    Thank you, thought there would be some straightforward name I didn't know about. – 2017-02-15