# Terminating decimal number which is not terminating in binary

I know that when converting a decimal number from base 10 to base 2, the result might be not terminating, even though the number is terminating in base 10.

For instance, 0.2 -> 0.0011 0011 0011 ...

Is the contrary possible ? That means, is there a non-terminating decimal number which, converted to base 2, is terminating ? I can't find any !

If yes, is there a method to find them ?

And by the way, what about irrational numbers ? Could an irrational number in base 10 become rational in base 2 or in another base ?

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What does "periodic" mean here? Why is "0.2" not periodic? – Chris Eagle Feb 29 '12 at 14:17
To answer the last question: absolutely not. The rationality or irrationality of a number has nothing at all to do with the base in which it’s written: it’s an intrinsic property of the number. – Brian M. Scott Feb 29 '12 at 14:18
Note that $10 = 5 \cdot 2$, so that you were able to find a number $p/q$ with $(2,q) = 1$ and $(10,q) > 1$ (e.g. $1/5$). The other way around it does not work, since every divisor of $2$ is a divisor of $10$. More generally, for primes $p$, $1/p$ is periodic in base $b$ iff $p$ does not divide $b$. So if it is periodic in the decimal system, then $p$ does not divide $10$ and hence also does not divide $2$. – TMM Feb 29 '12 at 14:23
@ChrisEagle periodic is not the right english word, I updated the question. – Jérôme Feb 29 '12 at 14:31

Following TMM's comment, for an expansion to terminate in base $10$, you need to be able to write it as $\frac a{10^n}=\frac a{(2\cdot 5)^n}$ for some naturals $a, n$ For it to terminate in base $2$, you need to be able to write it as $\frac b{2^m}$ for some naturals $b,m$ Given that it terminates in base $2$ with some $b,m$, you can take $n=m, a=2^mb$ to show it terminates in base $10$. If it terminates in base $10$, with some $a,n$, you would like to set $m=n, b=\frac a{5^n}$, but the last may not be a natural and these are the ones that will not terminate. Your example of $0.2_{10}$ fits this exactly.

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Continuing and expanding the previous answer. All rational numbers can be considered periodic in decimal or in any other base. For instance, 3/10 + 1/3 = .63333333... which can be considered periodic. However, if you mean that the decimal representation must repeat from the start of the representation, as for instance 1/11 = .09090909... and, assuming also that the number in question is between 0 and 1, then this kind of periodicity is equivalent to the denominator of the reduced fraction (numerator and denominator having no common factors) having no common factor with 10. In other words the denominator must be odd and not divisible by 5. For numbers in the binary system, the denominator must simply be odd. From this it is obvious that that a fraction that is periodic in its decimal representation is also periodic in its binary representation.

As for your second question about irrational numbers, a number is irrational if it is n ot the quotient of two integers. This has nothing to do with the base in which it is written. The representation of such a number in decimal and binary and any other such system must be infinite in length and non-periodic. My favorite representation of irrational numbers is to use a letter such as e, or if you are desparate, a number from another language like pi or gamma from Greek. If necessary create your own symbol or choose your own letter. It has the huge advantage of being a finite representation.

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