nonzero digits in decimal representation of $\sqrt{2}$

Question

let $1,d_1d_2d_3\dots$ be a decimal representation of $\sqrt{2}$.

Prove that at least one $d_i$ with $10^{1999}<i<10^{2000}$ is nonzero.

I have no idea how to solve it. I think that the given bounds mean the year only, and one can generalize it replacing $1999$ and $2000$ by $n$ and $n+1$ respectively.

Answer 1

One thing to remember to get a grasp on the problem is that the proposed number of consecutive zeroes is nine times the number of digits to that point.

So, assuming that the digits in the stated portion of the decimal expansion are all zero:

Separate $\sqrt 2$ into two parts, $A$ and $B$, where $A$ consists of all the digits with place value greater than $ 10^{-10^{1999}} $ and $B$ is the (irrational) portion less than $10^{-10^{2000}}$. Now $(\sqrt 2)^2 = (A+B)^2 = A^2 + 2AB + B^2$. Now the smallest (non-zero) digit of $A$ has a squared value of no less than $10^{-2*10^{1999}}$ and the largest place value that can be affected by the result of $2AB$ is approximately $2.8\cdot10^{-10^{2000}}$. Therefore these calculations do not interact and we are left with non-zero digits in the result, which contradicts the definition of $\sqrt 2$.

Therefore there must be non-zero digits within that portion of the decimal representation of $\sqrt 2$.

The side observation is that we can't have more consecutive zeros than we have had digits so far in the representation.

Answer 2

Assuming that all those digits are zero, we have that $$m=10^{10^{1999}}\sqrt{2}$$ is almost an integer, but that cannot happen in virtue of Hurwitz' theorem. In particular, for any convergent $\frac{p_n}{q_n}$ of the continued fraction of $\sqrt{2}=[1;2,2,2,\ldots]$ we have: $$\left|\sqrt{2}-\frac{p_n}{q_n}\right|\geq \frac{1}{2}\left|\frac{p_{n+1}}{q_{n+1}}-\frac{p_n}{q_n}\right|=\frac{1}{2q_n q_{n+1}}\approx\frac{1}{2(\sqrt{2}+1)\,q_n^2}. $$

Answer 3

You don't need any fancy theorems. You just need to think about what happens when you square such a number, digit by digit. All those zeroes (followed, sooner or later, by a non-zero digit) will result in a block of zeroes in the product, followed by a non-zero digit. Which is impossible, because the square is $2.0000\ldots$

Edited to add: I have left out the details, but Joffan's answer does an excellent job of filling them in.