# If a number is irrational in base 10, is it irrational in other bases?

If a number is irrational in base 10, is it necessarily irrational in all other bases? Or is it possible for a number to be irrational in only a few bases?

• if you allow irrational bases Jan 3, 2014 at 2:23
• FYI: mathworld.wolfram.com/Base.html irrational bases are weird. Jan 3, 2014 at 2:25
• Strange question, because irrationality is not a property depending on bases. I think you mean something slightly different, no? Jan 3, 2014 at 2:29
• Irrationality is independent on the base in which you write a number. The definition of irrational is being the quotient of two integers.
– OR.
Jan 3, 2014 at 2:29
• @ABC ... is not being ... Jan 3, 2014 at 2:32

You don't understand what "irrational" means. You have probably been told that an irrational number is one whose decimal expansion does not repeat. Although this is the case, it is a secondary property. An irrational number is one that cannot be written in the form $$a\over b$$ where $$a$$ and $$b$$ are integers; a rational number is what that can be written in that form.

Now it is the case that a number has a repeating base-10 representation if, and only if, it is a rational number, that is if it can be written as a fraction $$\frac ab$$. An irrational number always has a non-repeating base-10 representation.

And it is also the case that a number has a repeating base-$$n$$ expansion, for any base $$n$$, if, and only if, it is a rational number; an irrational number has a non-repeating representation in every base. This is probably the question you meant to ask.

Suppose a number $$x$$ has a base-$$n$$ expansion that begins with some sequence of digits $$a_1a_2a_3\ldots a_i = a$$, and then follows with $$b_1b_2b_3\ldots b_j = b$$ repeated forever. Then it turns out that $$x$$ is a rational number, and we can even find a fraction for it; the fraction is $$\frac{a}{n^i} + \frac1{n^i}\frac{b}{n^j-1}.$$

For example suppose we are working in base 8, and we want to find a fraction for the number 0.13456456456… where the digits are understood base 8. Then $$i=2$$, and $$a_1a_2 =$$ 13; and $$j=3$$, and $$b_1b_2b_3 =$$ 456. Then we can calculate that \begin{align}x & = \frac{13_{8}}{8^2} + \frac1{8^2}\frac{456_{8}}{8^3-1} \\ & = \frac{11}{64} + \frac1{64}\frac{302}{511} \\ &=\frac{5621}{32704} + \frac{302}{32704} \\ & = \frac{5923}{32704}\end{align}

and since this is a quotient of two integers, it is rational, because that is what a rational number is.

Its base-8 expansion is of course 0.13456456456456…, because that was how we constructed it, but it also repeats when written in any other base; for example in base 10 it is written $$0.181109\ 344422700587084148727984\ 34442270058708414872798\ \ldots.$$

Similarly, the base-10 decimal 0.13456456456… is equal to the rational number $$\frac{13}{10^2} + \frac1{10^2}\frac{456}{10^3-1} = \frac{13443}{99900} = \frac{4481}{33300}.$$

• an irrational number has a non-repeating expansion in every rational base. Jan 3, 2014 at 2:50
• "base $b$" as normally understood means a certain family of representations of the form $x= \sum_{i=-k}^\infty w_ib^{-i}$ where $b$ is an integer greater than 1 and $0\le w_i \lt b$. There are many other ways to represent numbers, but they are not part of the normal meaning of "any base". Noninteger bases are not part of the normal meaning; factorial bases are not part of the normal meaning; symmetric ternary is not part of the normal meaning, etc.
– MJD
Jan 3, 2014 at 3:07
• @qaphla Do you mean that, e.g. there's no repeating $\pi$ in $\pi$ when expressing $\pi$ in $\pi$ base? May 12, 2020 at 9:54
• @Andrei Is it even possible to have a non integer base? The resulting numbers would look very funny... Jun 18, 2021 at 12:10
• rationality or irrationality is base independent. Sep 25, 2021 at 6:52