# Does a real number with this decimal expansion for $r$ and $r^2$ exist?

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?

• You are correct that $0.\overline{999}$ does not satisfy the requirement that it be strictly less than $1$ and so is not allowed. – JMoravitz Oct 17 '18 at 4:16
• @JMoravitz that is what I figured. Do you know how I should approach this problem? Is this even possible? – Mohammed Shahid Oct 17 '18 at 4:26
• Try solving the equation $x-0.1=x^2$ – 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 $$0, so $$x^2. 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 $$0 with a terminating decimal expansion such that $$\sqrt{1-4k}$$ does not terminate can be used in place of the 0.1 in $$x-x^2=0.1$$.