The question is from the book The Foundation of Mathematics by Ian Stewart.

Let $y = 0·1234567891011121314151617181920 . . . $,

whose digits are the natural numbers in decimal form, strung end to end. Prove that y is irrational.

This is how I solved it.

I assumed the number is of the form $a_0.a_1a_2a_3.......$ where $a_n = a + nd$ and $a = 0, d = 1$

So if there is no repetition in the infinite decimal places then the number is irrational.

All I had to prove was that assume to there is a repetition after $a_n$ from $a_{n+1}$ till $a_{n+k}$ and if $a_{n+1} \ne a_{n+k+1}$ then there is no repetition and the series must be irrational.

Is my proof correct?


marked as duplicate by zhoraster, John B, steven gregory, Lord Shark the Unknown, B. Goddard Aug 26 '17 at 17:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


This is Champernowne's number. It is not just irrational, it is transcendental. An easy way to prove that it is irrational is that the number has arbitrarily long strings of $1$s and arbitrarily long strings of $2$s. So given any putative period length, it has infinitely many strings of $1$s and of $2$s of that length. Transcendence is harder....

  • $\begingroup$ Could you explain it in an elaborate way, I am sorry, I am totally lost. $\endgroup$ – Anirudh Murali Aug 26 '17 at 7:00

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