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If every $0$ digit in the expansion of $\sqrt{2}$ in base $10$ is replaced with $1$, is the resulting sequence eventually periodic?

Watson
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WangAtChicago
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  • Since the sequence is not periodic, if you change zero to any number, it will still be non periodic since other numbers at other places also decide periodicity. – jnyan Nov 01 '16 at 09:34
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    @jnyan that's not true – mercio Nov 01 '16 at 09:55
  • A periodic sequence would have a finite numbers repeating, say 95467,95467,95467..... Say this was obtained by changing some digit, say changing 4 to 5, then original sequence would have been 94467,94467,94467, a repeating sequence. I can't think of any counterexamples – jnyan Nov 01 '16 at 10:45
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    Possibly relevant : https://math.stackexchange.com/questions/971137. I quote Milo Brandt : "Strictly speaking, it is unknown whether $\pi$ with every $4$ changed to $6$ would be irrational (since we don't know even that every digit of $\pi$ occurs infinitely often), but if it's a normal number, then it would still be irrational". – Watson Nov 01 '16 at 17:12
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    @jnyan : I think you actually proved the converse : if $x$ is rational, then replacing every $0$ digit by $1$ yields a rational number $y(x)$. But here it is a different question: $x_0=\sqrt 2$ is irrational, but does it follow that $y(x_0)$ is irrational? For instance, $x=0.10100100010000…$ is irrational but $y(x)=0.1111111…$ is rational. – Watson Nov 01 '16 at 17:13
  • @watson, no I meant to get a rational by changing only one digit, the original number should also be rational. And the question is to get rational from irrational – jnyan Nov 01 '16 at 17:15
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    @jnyan : the question is about changing "every $0$ digit", not only one of them. – Watson Nov 01 '16 at 17:20
  • It has not been answered by anybody, since I had posted the question here. – WangAtChicago Sep 21 '22 at 03:59

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