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

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

• Hi and welcome to MSE. What are your thoughts on the problem? What have you tried?
– Surb
Oct 19 '18 at 20:49

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.

• Still need to prove that $\sqrt{7}$ is irrational
– Surb
Oct 19 '18 at 20:51
• @Surd: My answer is a hint, not a complete solution. It also depends on the starting assumptions of the problem. Oct 19 '18 at 20:52
• @surb in my answer I've taken it for known, because not only the proof is simple, but also because the problem does not ask to prove it. Oct 19 '18 at 20:54
• 1) I upvoted Javaman's answer because his justification is fair. 2) "the problem does not ask to prove it" this is certainly a very (very very) bad reason. The problem doesn't ask to prove that there is a digit appearing infinitely many times in the expansion of $\sqrt{p}$ for prime $p$ as well... But I guess you'd agree this fact can't be used here.
– Surb
Oct 19 '18 at 20:58
• Of course I agree on the last point, but when you prove a theorem you don't prove all the theorems or lemmas you need to prove that theorem and take them for true (or already proved elsewhere). I haven' seen a single answer on MSE about, for instance, limits evaluation, that proves that the limit of a sum, under certain hypothesis, is the sum of limits. That's what I meant in saying that "the problem does not ask to prove it". The proof of irrationality of $sqrt(7)$ in my opinion was out of scope of the question. Of course, I absolutely agree on the fact that if the question does not ask Oct 20 '18 at 6:19

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.

• Come on...... :)
– Hugo
Oct 19 '18 at 20:51
• Sorry I am new at these topics, I don't understand exactly, can you explain more specific ? Thank you. Oct 19 '18 at 21:14

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.

• What do you mean by “if there are no digits that appear infinitely often, then each one of the 10 digits appears at most once”? Oct 20 '18 at 11:29
• Sorry, my mistake. Thanks for pointing it out. Edit: I've corrected it. I'm sorry, but when I wrote the answer the time for me was 11.40 PM, and I was getting pretty tired.. Oct 20 '18 at 12:59