# Are irrational numbers irrational by nature? [duplicate]

I remember hearing an interesting theory once, I don't know the source. Since there are some numbers that are precisely expressible in decimal notation that repeat in a binary base, and vice versa, perhaps there exists a base in which irrational numbers are rational.

The more I think about it, the less likely this seems, and my guess is that whatever proof that $\pi$ or $e$ are irrational doesn't involve the decimal base notation. But perhaps there's some research into this?

## marked as duplicate by Rob Arthan, Community♦Oct 9 '15 at 23:00

• There is no integer base in which an irrational number has a finite expansion. The definition of rational and irrational is base independent. That being said, if $x > 0$ is an irrational number (and hence not an integer), then $x$ written in base $x$ is $10$. However that says nothing particularly interesting. – Simon S Oct 9 '15 at 22:09
As long as we are restricting ourselves to integer bases, a rational number will always have a repeating pattern in any base, and an irrational number will not repeat in any base. If $x$ has a repeating pattern to base $b$, then there are exponents $n$, $m$ such that $b^nx - b^mx = y$, where $m$ is chosen to leave nothing but the repeating pattern to the right of the radix, and $n$ does the same, but with one repetition also to the left, which means that $y$ is an integer. But then $x = {y\over b^n-b^m}$, which is rational.
• Just a footnote: "a rational number will always [end with] a repeating pattern in any base" -- the repeating pattern may be $0$. – BrianO Oct 9 '15 at 22:20
• A quibble/complication: $x$ may not be "all repeating pattern" -- there can be "other stuff" to the left of the final infinite string of $y$s. Working out the details gets a little gnarly with indices and exponents. But you got the crucial fact across. – BrianO Oct 9 '15 at 22:27
• @BrianO That is exactly why I have two powers of $b$. $b^m$ is what is needed to bring the first repetition up to the decimal point, $b^n$ brings the 2nd repetition to the decimal point, so the difference is an integer. Of course, I was wrong to say that $y$ is the repeating digits. I'll correct that. – Paul Sinclair Oct 9 '15 at 23:01