2
$\begingroup$

How to prove that 0.101001000100001..... is irrational? There's the fact that it is non recurring but is there any mathematical proof like that we give for square root of two?

$\endgroup$
3
  • 2
    $\begingroup$ There is a proof in the sense that you prove that $p/q$ is always has a recurring decimal expansion, and your number does not, but that does not seem to be what you're looking for. Can you be more precise? $\endgroup$ Commented Jan 12, 2017 at 12:48
  • $\begingroup$ I've been told this can be proved by using binomials, though I can't see how. How can we prove your first statement? Thanks. $\endgroup$
    – Petra
    Commented Jan 12, 2017 at 12:52
  • $\begingroup$ @user406333 it suffices to show that every $q$ divides a number of the form $99\cdots900\cdots0$, and that any fraction with such a number in the denominator will eventually repeat. $\endgroup$ Commented Jan 12, 2017 at 12:55

1 Answer 1

4
$\begingroup$

Note that your number is nothing but $$0.1+0.001+0.000001+\cdots$$ Assume that the number is rational. Then it must be of the form $$p/q=0.1+0.001+0.000001+\cdots$$ for some integers $p,q$. Then it should hold that $$p.0000\cdots=p=q\cdot(0.1+0.001+0.000001+\cdots),$$ But if $q$ has $n$ digits, then the non zero digits of the summands after the n'th, on the right hand side, occupy different digits of the number, and so they cannot cancel out.

Example: if $q$ were 576 then we would sum $$5.76.+0.576+0.000576+0.0000000576+0.000000000000576\cdots)$$ and you see that from the third summand on, the non zero digits are non-intersecting.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .