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I was working through some basic Number Theory Problems in Rosen and came across the following problem :

Show that the real number $0.1248163264...$ represented in base 10 is an irrational number

I am slightly stumped. Can someone help me out? A hint would be great.

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    $\begingroup$ Does it repeat? $\endgroup$
    – bonsoon
    Apr 25, 2015 at 4:30
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    $\begingroup$ @bonsoon , how can I check whether it repeats or not ? $\endgroup$
    – pranav
    Apr 25, 2015 at 4:34
  • $\begingroup$ What is the generator for the sequence? I know this question is a bit old but it's completely unintelligible to me. $\endgroup$
    – Joshua
    Sep 26, 2016 at 4:01

3 Answers 3

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HINT: I expect that you know that a number is rational if and only if its decimal expansion is eventually periodic. Suppose that the expansion eventually repeats with period $p$.

  • Show that there are two consecutive powers of $2$ whose after the initial aperiodic segment whose lengths (when written in the usual way in base ten) are the same multiple of $p$.

I’ve left a further hint in the spoiler-protected block below; mouse-over to see it.

Show that on the one hand these two powers of $2$ must end in the same digit, and on the other hand that they cannot end in the same digit.

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A number is rational if and only if it has an eventually repeating decimal expansion.

So you should show that this decimal expansion does not have an eventually repeating expansion.

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    $\begingroup$ I like that phrasing. I was thinking about how to phrase it. Eventually repeating is nice. $\endgroup$
    – davidlowryduda ♦
    Apr 25, 2015 at 4:34
  • $\begingroup$ @mixedmath , how can show that this does not have a repeating expansion ? Would be grateful for some pointers :) $\endgroup$
    – pranav
    Apr 25, 2015 at 4:34
  • $\begingroup$ @mixedmath , please can you help me out ? $\endgroup$
    – pranav
    Apr 25, 2015 at 4:47
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    $\begingroup$ You asked for a hint. I think this is a great hint. If you want another, you might notice that the last digits of $2^n$ repeat both mod $10$ and mod $100$ (and mod anything, for that matter). But there are $n$ such that $2^n$ has $k$ digits for all $k \geq 1$. Or you might show it in some other way. It's time to start experimenting! $\endgroup$
    – davidlowryduda ♦
    Apr 25, 2015 at 4:49
  • $\begingroup$ Thanks @mixedmath , I will give it a shot and get back to you with my solution :) $\endgroup$
    – pranav
    Apr 25, 2015 at 4:57
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Starting from the first digit after the decimal point, the digit gets multiplied by two every single time without stopping, also expanding the numbers of digits going on without an end. This has no repetition, proving it's irrational.

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