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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?

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  • $\begingroup$ 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. $\endgroup$
    – jnyan
    Nov 1, 2016 at 9:34
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    $\begingroup$ @jnyan that's not true $\endgroup$
    – mercio
    Nov 1, 2016 at 9:55
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    $\begingroup$ Possibly relevant : 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". $\endgroup$
    – Watson
    Nov 1, 2016 at 17:12
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    $\begingroup$ @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. $\endgroup$
    – Watson
    Nov 1, 2016 at 17:13
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    $\begingroup$ @jnyan : the question is about changing "every $0$ digit", not only one of them. $\endgroup$
    – Watson
    Nov 1, 2016 at 17:20

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