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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?

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    $\begingroup$ if you allow irrational bases $\endgroup$
    – Memming
    Jan 3 '14 at 2:23
  • $\begingroup$ FYI: mathworld.wolfram.com/Base.html irrational bases are weird. $\endgroup$
    – Memming
    Jan 3 '14 at 2:25
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    $\begingroup$ Strange question, because irrationality is not a property depending on bases. I think you mean something slightly different, no? $\endgroup$
    – Ian Mateus
    Jan 3 '14 at 2:29
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    $\begingroup$ Irrationality is independent on the base in which you write a number. The definition of irrational is being the quotient of two integers. $\endgroup$
    – OR.
    Jan 3 '14 at 2:29
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    $\begingroup$ @ABC ... is not being ... $\endgroup$ Jan 3 '14 at 2:32
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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 $$\frac{13}{10^2} + \frac1{10^2}\frac{456}{10^3-1} = \frac{13443}{99900} = \frac{4481}{33300}.$$

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    $\begingroup$ an irrational number has a non-repeating expansion in every rational base. $\endgroup$
    – qaphla
    Jan 3 '14 at 2:50
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    $\begingroup$ "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. $\endgroup$
    – MJD
    Jan 3 '14 at 3:07
  • $\begingroup$ @qaphla Do you mean that, e.g. there's no repeating $\pi$ in $\pi$ when expressing $\pi$ in $\pi$ base? $\endgroup$
    – Andrei
    May 12 '20 at 9:54
  • $\begingroup$ @Andrei Is it even possible to have a non integer base? The resulting numbers would look very funny... $\endgroup$
    – Paul Uszak
    Jun 18 '21 at 12:10
  • $\begingroup$ rationality or irrationality is base independent. $\endgroup$ Sep 25 '21 at 6:52

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