Consider the number $\alpha$ obtained by writing one after another the decimal representations of $1,1987,1987^2,\ldots$ to the right of the decimal point. Show that $\alpha$ is irrational.

In order to show that $\alpha$ is irrational, we have to show its decimal representation is never periodic from some point forwards. Since $\log_{10}(1987)$ is irrational, it follows from the pigeonhole principle that the fractional part of $n\log_{10}(1987) = \log_{10}(1987^n)$ is dense on $[0,1]$.

Is this way of solving correct or is there a nicer way of solving it?

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    $\begingroup$ I wonder if you could do something by noticing that the sequences $(1 \mod n, 1987 \mod n, 1987^2 \mod n,...)$ with $n=10^k$, $n \in \mathbb{N}$ are periodic but the length between these repetitions in $\alpha$ grows with the number of digits in $1987^m$ $\endgroup$ – n1000 Jul 26 '16 at 3:58
  • $\begingroup$ @user21820 For any periodic sequence, there will be some finite string that is not contained in it (for instance pick any string with the same length as the period that is not equal to a cyclic shift of the repeating part). The OP's argument shows that $\alpha$ does contain every finite string, so it can't be periodic. It's just a slight adjustment to argue that it can't be eventually periodic (which is equivalent to $\alpha$ irrational). $\endgroup$ – Erick Wong Aug 2 '16 at 17:51
  • $\begingroup$ @ErickWong: Ah thanks for your comment! =) I don't recall why I made my earlier comment since I can see how the argument goes now when I read the question again. $\endgroup$ – user21820 Aug 3 '16 at 3:35

Assume, for sake of contradiction, that $\alpha$ is rational. Then, after some amount of digits to the right of the decimal place, $\alpha$'s decimal representation would be periodic. But since $1987$ is relatively prime to $10$, we see that from Euler's Totient Theorem for each $k \geq 1$ we have $1987^{\varphi(10^k)} \equiv 1 \pmod{10^k}$. Thus, $1987^{\varphi(10^k)}$ must contain at least $k-1$ consecutive zeros.

Therefore there exist arbitrarily large positive powers of $1987$ with arbitrarily long blocks of consecutive zeros (since $\varphi(10^k)$ is increasing). But this must mean that the period is a string of $0$s, a contradiction.


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