1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ Try solving the equation $x-0.1=x^2$ $\endgroup$ – B.Martin Oct 17 '18 at 4:39
4
$\begingroup$

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 $x-x^2=0.1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.