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Is 0.123456789101112... rational or irrational? How do you prove it? I think it is irrational but I'm not sure how I can arrive at a contradiction. How do you even represent this number in series?

Xiao Lei Zhang
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    It's Champernowne's number, q.v., and it's transcendental. – Gerry Myerson Aug 20 '17 at 22:50
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    The decimal expansion has arbitrary long streaks of $000\cdots000$ and $111\cdots 111$ etc. so it cannot be periodic. – Winther Aug 20 '17 at 22:51
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    Previous discussions: https://math.stackexchange.com/questions/2281342/simplified-formula-for-champernownes-constant, https://math.stackexchange.com/questions/1329899/is-the-champernowne-constant-an-automatic-number, https://math.stackexchange.com/questions/703883/champernowne-constant-summation-and-behavior-of-terms-in-continued-fraction-ex, https://math.stackexchange.com/questions/2222659/does-champernownes-constant-converge-to-the-digits-of-pi, https://math.stackexchange.com/questions/1707374/why-do-we-care-about-the-champernowne-constant and many more. – Gerry Myerson Aug 20 '17 at 22:54

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The decimal expansion has arbitrary long streaks of 000⋯000 and 111⋯111 etc. so it cannot be periodic. – Winther 1 hour ago

And numbers with non-periodic decimal expansions are irrational. So that number is irrational.

Chris Culter
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COMMENT.- The inequality $$\left|\pi-\frac pq\right|\gt \frac{1}{q^{42}}$$ valid for all rational $\dfrac pq$ with $q\ge2$ rated by A. Baker as “striking inequality " was published by Mahler in 1953. Mahler also proved that the number $0.1234567891011 ....$ , whose decimal part follows the sequence of natural integers is transcendental. What is irrational is almost obvious because obviously lacks any period.

Piquito
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