Prove that there is a digit that appears infinitely often in the decimal expansion of $\sqrt{7}$.

Question

Prove that there is a digit that appears infinitely often in the decimal expansion of $\sqrt{7}$.

Answer 1

I'd prove it by absurdum saying that if there are no digits that appear infinitely often, then each one of the 10 digits appears at most a finite number of times. So the decimal expansion of $\sqrt(7)$ would be finite and this would be a contradiction, because if it was finite, then I can show that there is a rational number that has the same expansion. This leads to say that $\sqrt(7)$ is rational.

Answer 2

Hint: $\sqrt 7$ is irrational. What would happen if all digits appeared only finitely many times? Moreover, this shows that at least $2$ digits must appear infinity often.

Answer 3

Suppose not, that every digit $i \in \{0, \dotsc, 9\}$ appears a finite number of times, say $a_i$. Then The decimal expansion of $\sqrt{7}$ would have exactly $a_0 + \dotsb + a_9$ digits, but it is infinite instead.