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I'm trying to understand the cardinality between the set of all infinite binary (base-2) fractions and the set of all infinite decimal (base-10) fractions.

I can easily think of infinite binary fractions that do not match up with infinite decimal fractions, such as $\frac{1}{1010} = 0.00011001100110011\ldots$ (in binary) is not infinite in decimal $\frac{1}{2} = 0.5$ but does every infinite decimal fraction pair up with an infinite binary one (ex: $\frac{1}{3} = 0.333333\ldots$ maps to $\frac{1}{11} = 0.010101\ldots$)? Or does the infinite binary set have greater cardinality?


By "infinite" fraction I mean a fraction that only repeats non-zero values.

Example: $\frac{1}{2} = 0.5000\ldots$ (in decimal) and $\frac{1}{10} = 0.1000\ldots$ (in binary) would not be considered infinite fractions.

For context: I'm trying to determine if programming using decimal floating points is better than binary floating points because of less rounding issues due to these infinite fractions. I already know they are better in regard to having the same rounding issues as is experienced in the real world, which primarily uses a base-10 numeric system.

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the cardinality is just the same, it seems... but may i ask for one detail, so by infinite you mean, for instance, 0.100000000...000... does not belong to any of the sets you consider? – W_D Jul 25 '13 at 18:31
$\frac1n$ in base $b$ will be non-terminating if at least one of the prime divisors of $n$ does not divide $b$. So, any $\frac1n$ in base $10$ will be non-terminating, must be non-terminating in base $2$. But what will happen, if $n=2^a\cdot5^b$ where $a\ge0,b>0$ ? – lab bhattacharjee Jul 25 '13 at 18:31
@AlexWhite Yes, that's what I'm referring to. – Nick Gotch Jul 25 '13 at 18:39
All decimals are infinite repeating, really. In base ten, even $0.5=0.4\overline{9}$. – alex.jordan Aug 1 '13 at 18:58
@alex.jordan Do you have a source/proof for $0.5=4\overline{9}$? It just blows my mind. – Nick Gotch Aug 1 '13 at 19:12

2 Answers 2

up vote 2 down vote accepted

Fractions have terminating decimal representations exactly when the (reduced) denominator factors into powers of $2$ and $5$ only, while they have terminating binary representations exactly when the denominator is a power of $2$.

So yes (if I understand the question), if a fraction does not terminate in decimal, its (reduced) denominator has a prime factor other that $2$ or $5$, and so it will repeat in binary too.

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Thank you, this makes sense. – Nick Gotch Aug 1 '13 at 19:31
@NickGotch Note that you can only find a terminating decimal representation of a fraction that has a repeating binary representation if the denominator only has powers of $2$ and $5$ in its prime factorization, and at least one power of $5$. So one-fifth would be such a number. (And the examples in your posted problem are not equivalent to each other - the binary example one-tenth, but the decimal one is one-half.) – alex.jordan Aug 1 '13 at 22:01
Thanks. I did realize those weren't equivalent, I was just trying to think up quick examples of what I didn't mean by "infinite" fractions. – Nick Gotch Aug 1 '13 at 22:11

Let $B,D$ denote the sets of all infinite binary fractions and infinite decimal fractions respec. Since all the infinite binary fractions of the form $$0.b_0b_1b_2\cdots$$have their $b$'s from $\{0,1\}$, they are just a subset of all the infinite decimal fractions of the form $$0.d_0d_1d_2\cdots$$ where the $d$'s are from $\{0,1,\cdots\ ,9\}$. So there exist a surjection from $B$ to some subset of $D$, namely $B$ itself. Now, clearly, if $0.d_1d_2d_3\cdots $ is any element from $D$, one can find a unique representation of that in $B$. So there exists a surjection from $D$ to some subset of $B$. Then, by Schroeder-Bernstein theorem $B$ and $D$ have the same cardinality.

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I don't think this answer correctly answers the question based on how I'm defining "infinite fractions." See my clarification above. I still haven't been able to come up with a single infinite decimal fraction that maps to a finite binary fraction but I can easily find the reverse. If the cardinality is the same, I'd expect there to be some. – Nick Gotch Aug 1 '13 at 16:36
@NickGotch, look at alex.jordan's comment. – Samrat Mukhopadhyay Aug 1 '13 at 19:28

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