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?

  • 1
    $\begingroup$ 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
  • 4
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 4
    $\begingroup$ @ABC ... is not being ... $\endgroup$ – Andrés E. Caicedo Jan 3 '14 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. This definition has nothing to do with the base in which the numbers are written, and this is why your question, as you phrased it, does not really make sense. It's like asking if a verse of a song will still rhyme even if it is printed in colored ink.

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 $$\frac{13}{10^2} + \frac1{10^2}\frac{456}{10^3-1} = \frac{13443}{99900} = \frac{4481}{33300}.$$

  • 13
    $\begingroup$ an irrational number has a non-repeating expansion in every rational base. $\endgroup$ – qaphla Jan 3 '14 at 2:50
  • 4
    $\begingroup$ "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 at 9:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.