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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?

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

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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....

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

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