I've heard the proof that this number is irrational is accessible to even a novice to number theory:

$\alpha = 0.2 \ 3 \ 5 \ 7 \ 11 \ 13 \ 17 \ldots$

The proof may utilize that a number is irrational iff its decimal expansion either terminates or is periodic, but then I have to show that the set of primes doesn't eventually look something like this:

$\mathbb P = \{\ldots 171, 17171, 171717171, \ldots \}$

I was also told that there is a proof not dependent on that theorem. edit: by "that theorem" the theorem that characterizes irrationals as non-repeating decimals.

Does anyone know simple proofs of this? Thanks.

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    $\begingroup$ possible duplicate of Is $0.23571113171923293137\dots$ transcendental? $\endgroup$ – user171358 Oct 29 '14 at 14:10
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    $\begingroup$ Copeland–Erdős constant $\endgroup$ – user171358 Oct 29 '14 at 14:12
  • $\begingroup$ Hint: between any number and its double exists a prime (I don't know how to prove this). Using this, can you see why $\mathbb P$ can't end the way you wrote? $\endgroup$ – Akiva Weinberger Oct 29 '14 at 14:21
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    $\begingroup$ @DigitalBrain "$x$ is irrational" is not a duplicate of "$x$ is transcendental". The method of proof would be different for irrationality. $\endgroup$ – user147263 Oct 29 '14 at 14:34

Hint: If $ \gcd (a, d) = 1$, then there exists infinitely many primes in the arithmetic progression $ a + n d$.

Note: This is a high power theorem whose proof is (arguably) inaccessible to a novice in number theory, but the statement is easy to understand and to accept as fact.

Hint: A number is rational if and only if its decimal representation finally repeats/terminates in 0.

Note: This is a simple fact. I'm not sure if this is the theorem you are referencing in the statement, but it doesn't require "primes eventually look something like this".

Hint: Prove that for any $k$, there must exist a sequence of $k$ 0's in the number.

Hence, this number is irrational.

  • $\begingroup$ A slightly more elementary approach is to use the fact that for every $n\in\mathbb N$ there is a prime of the form $kn+1$. (This can be proven elementary, for example with cyclotomic polynomials.) $\endgroup$ – punctured dusk Oct 29 '14 at 15:10
  • $\begingroup$ Re: the comment at the second hint, what I'm saying is that if the list of all primes eventually starts looking like the set I gave, then the number would be rational. $\endgroup$ – GFauxPas Oct 29 '14 at 15:40

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