Does there exist a real number $0< x <1$, such that the decimal expansions of $x$ and $x^2$ are the same, starting from the millionth term, and neither expansion has an infinite tail of zeroes?

I was thinking $x=0.\overline{999}$, but does that work? Isn't that just equal to 1 which is not allowed.? If this works, how would I prove it?

  • $\begingroup$ You are correct that $0.\overline{999}$ does not satisfy the requirement that it be strictly less than $1$ and so is not allowed. $\endgroup$ – JMoravitz Oct 17 '18 at 4:16
  • $\begingroup$ @JMoravitz that is what I figured. Do you know how I should approach this problem? Is this even possible? $\endgroup$ – Mohammed Shahid Oct 17 '18 at 4:26
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    $\begingroup$ Try solving the equation $x-0.1=x^2$ $\endgroup$ – B.Martin Oct 17 '18 at 4:39

We can concoct an example quite easily. Suppose we want the difference between $x$ and $x^2$ to be 0.1: $$x-x^2=0.1$$ where the order $x-x^2$ is mandated by $0<x<1$, so $x^2<x$. Solving this, we get two admissible values $x=\frac{1\pm\sqrt{0.6}}2$.

Thus (taking $x=\frac{1+\sqrt{0.6}}2$) we have $$x=0.88729833\dots$$ $$x^2=0.78729833\dots$$ so their decimal expansions agree after the first place, and indeed after the millionth place.

Any number $0<k<0.25$ with a terminating decimal expansion such that $\sqrt{1-4k}$ does not terminate can be used in place of the 0.1 in $x-x^2=0.1$.


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