Prove that an infinite decimal whose digits are natural numbers strung end to end is irrational [duplicate]

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. GoddardAug 26 '17 at 17:04

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....