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

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

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  • $\begingroup$ Still need to prove that $\sqrt{7}$ is irrational $\endgroup$
    – Surb
    Oct 19 '18 at 20:51
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    $\begingroup$ @Surd: My answer is a hint, not a complete solution. It also depends on the starting assumptions of the problem. $\endgroup$
    – JavaMan
    Oct 19 '18 at 20:52
  • $\begingroup$ @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. $\endgroup$
    – LuxGiammi
    Oct 19 '18 at 20:54
  • $\begingroup$ 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. $\endgroup$
    – Surb
    Oct 19 '18 at 20:58
  • $\begingroup$ 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 $\endgroup$
    – LuxGiammi
    Oct 20 '18 at 6:19
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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.

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  • $\begingroup$ Come on...... :) $\endgroup$
    – Hugo
    Oct 19 '18 at 20:51
  • $\begingroup$ Sorry I am new at these topics, I don't understand exactly, can you explain more specific ? Thank you. $\endgroup$
    – user606126
    Oct 19 '18 at 21:14
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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.

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  • $\begingroup$ 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”? $\endgroup$
    – JavaMan
    Oct 20 '18 at 11:29
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    $\begingroup$ 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.. $\endgroup$
    – LuxGiammi
    Oct 20 '18 at 12:59

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