I am not a mathematician at all and I had a thought about Pi that I can't work out.

Pi is irrational, with an infinite sequence of numbers A recurring number is infinite

Would it be theoretically possible that at one point down Pi's sequence, it just turned into an infinite sequence of one number?

If not, what is stopping this? I recognise through reductio ad absurdum that my proposition is impossible, as it means that PI must at some point end in all of 1-recurring and 2-rcurring...9-recurring simultaneously, but I lack the toolset to understand where my thinking breaks down.


marked as duplicate by Jyrki Lahtonen Jul 22 '15 at 12:48

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  • $\begingroup$ If number have "infinite sequence of one number" then it's rational $\endgroup$ – Michael Galuza Jul 22 '15 at 12:42
  • $\begingroup$ Not sure that the one picked is the best match. It is one of the oldest on our site anyway. $\endgroup$ – Jyrki Lahtonen Jul 22 '15 at 12:49
  • 1
    $\begingroup$ A number is irrational precisely means that it has no recurring sequence. $\endgroup$ – Yves Daoust Jul 22 '15 at 13:19

No -- if all digits of $\pi$ from some point onward are the same digit, that means exactly that the sequence of digits is eventually repeating (namely, that particular digit would eventually repeat), and that would mean it is rational (which is is known not to be).


Note that for $n = 0, 1, 2, 3, 4, 5, 6, 7, 8$ we have

$$\frac n9 = 0.nnnnnnnn....$$

Hence if the decimal expansion for $\pi$ terminated with an infinite string of identical digits after $N$ decimal places, then we could write $\pi$ as a rational number. Namely

$$\sum_{j=0}^N a_j 10^{-j} + \frac n9 10^{-N} = \frac{\sum_{j=0}^N a_n 10^{N-j} }{10^N} + \frac{n}{9 \cdot 10^{N}}$$

($a_0 = 3, a_1 = 1, a_2 = 4, ...$)

  • $\begingroup$ Thanks, though I can't say i follow the proof! It may not be possible to write this in prose, but is there anything that sets a non-infinite limit on the amount of repetition of a digit in Pi? $\endgroup$ – Jon Bates Jul 22 '15 at 12:48
  • $\begingroup$ @JonBates Maybe you got why 0.nnnnn... is rational, for every integer n? And then, why 0.000...0nnnnn... is rational, for every starting block of 0 and every integer n? And then, why A.Bnnnnn... is rational, for every integer A, every starting block B and every integer n? $\endgroup$ – Did Jul 22 '15 at 13:06

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