Let's assume that $\pi = \sum_{i=0}^{\infty}\frac{a_i}{10^i}.$ Also, let $n$ be an arbitrary natural number.

Is there $j,k,m \in \mathbb{N}, (j < k < m)$ such that $|b_1 - b_{2}| = n$ when $b_1 = \sum_{i=j}^{k}a_i10^{k-i}, b_2 = \sum_{i=k+1}^{m}a_i10^{m-i}$?

for example: 0.14(15)(92)... so for |92-15|=77 there are j,k,m as wanted.

the original question was:

can a number $x = \sum_{i=0}^{\infty}\frac{x_i}{10^i}$ such that $\forall n \in \Bbb N\ \exists j,k,m \in \mathbb{N}, (j < k < m)$ such that $|b_1 - b_{2}| = n$ when $b_1 = \sum_{i=j}^{k}a_i10^{k-i}, b_2 = \sum_{i=k+1}^{m}a_i10^{m-i}$ is a rational number?

i would like for now a solution only for the first part but any solution will be welcomed.

Austin Weaver
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ned grekerzberg
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  • To clarify, $3.14(15)(92)$ would constitute a positive answer if it were true that $|92-15|=4$? – Hagen von Eitzen Oct 12 '17 at 16:49
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    Doesn't this boil down to asking whether or not $\pi$ is normal (which is, AFAIK, unknown)? or have I misunderstood something? – Xander Henderson Oct 12 '17 at 16:50
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    That may not be less difficult than Riemann Hypothesis, say, but it's far less interesting. The decimal system if just one of countably infinite ones, so who really cares? –  Oct 12 '17 at 16:52
  • respond to Hagen von Eitzen : why?, i said for any arbitrary n natural number so for 77 (which is natural) the statement is true. – ned grekerzberg Oct 12 '17 at 16:54

1 Answers1


It is widely believed that the decimal expansion of $\pi$ contains every string of digits, see for example Does Pi contain all possible number combinations?

If one accepts this as true, your statement follows readily. However, to show this is considered as very difficult.

While your problem is somewhat simpler, I am afraid it is sufficiently similar that it is too difficult to be answered definitely at this point in time (and likely quite a bit of time in the future).

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