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irrationality proof of $\sqrt{n}+\sqrt{n+1}$ for any $n>0$

My attempt:

$\sqrt{n}+\sqrt{n+1}=\frac{p}{q}$

$2n+1+2\sqrt{n(n+1)}=\frac{p^2}{q^2}$

Now we have to show $2n+1+2\sqrt{n(n+1)}$ cannot be a perfect squere but I don't know how.

Martin Sleziak
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Taha Akbari
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4 Answers4

11

Notice that $\sqrt{n}+\sqrt{n+1}$ is a root of $$x^4 -2 (2 n+1) x^2+1.$$ Now use the rational root theorem (assuming that $n$ is an integer) to find that the only possible rational roots are $\pm 1$.

Pjotr5
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$2n + 1 + 2\sqrt{n(n+1)}$ is irrational if and only if $\sqrt{n(n+1)} = \sqrt{n^2 + n}$ is irrational. Then observe that $$n^2 < n^2 + n < n^2 + 2n + 1 = (n+1)^2,$$ which implies that indeed $n(n+1)$ is not a perfect square. But the square roots of positive integers are either positive integers or irrational, by the rational root theorem or Gauss' lemma. So for $x=\sqrt{n} + \sqrt{n+1}$, we see $x^2$ is irrational, and hence $x$ is irrational.

Mr. Chip
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  • Not a valid proof, since it only proves that $\sqrt{n^2+n}$ is not an integer, but we need that it is not rational. – Bill Dubuque Aug 08 '16 at 16:15
  • @BillDubuque (p^2/q^2) with p, q coprime and q not 1 is not an integer. – Jacob Wakem Aug 08 '16 at 16:21
  • @Alephnull It is true (by unique factorization) that sqrt's of integers are either integral or irrational, but the answer neither proves that nor explicitly mentions that it is using that result as a lemma. You need to do one or the other to obtain a complete correct proof. This gap is a common oversight (some students do not even realize that it requires proof). – Bill Dubuque Aug 08 '16 at 16:28
  • @BillDubuque Do you know how special the squareroot function is in this regard? Do you know if there is a good calculus proof of this result that might generalize? – Jacob Wakem Aug 08 '16 at 16:33
  • @Alephnull The proofs of this and related results have nothing to do with calculus. Rather, they depend heavily on the Fundamental Theorem of Arithmetic. A more general result is that any rational root of an integer coeff polynomial with lead coeff = 1 must be an integer (special case of the Rational Root Test). – Bill Dubuque Aug 08 '16 at 16:37
  • @BillDubuque: I don't really agree that there is a definite concept of a "complete and correct proof", but I'll concede that your point deserves mention in the context of this question. – Mr. Chip Aug 08 '16 at 23:46
  • @Mr.Chip The reason why I stressed this point is that it is a common pitfall for beginning students of number theory. Glad to see that you have updated your answer to address that. – Bill Dubuque Aug 09 '16 at 00:10
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If $\sqrt{n+1}+\sqrt{n}$ is rational, so is $$ \frac{1}{\sqrt{n+1}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n} $$ By summing and subtracting, we get that both $\sqrt{n+1}$ and $\sqrt{n}$ are rational.

It's not difficult to show that $\sqrt{n}$ is rational if and only if $n$ is a perfect square, but the only consecutive squares in the natural numbers are $0$ and $1$.

egreg
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    Special case of this answer. – Bill Dubuque Aug 08 '16 at 17:06
  • @BillDubuque I don't think so. The method of course is similar. – egreg Aug 08 '16 at 17:36
  • Really?? Why don't you think so? How is it not a special case? – Bill Dubuque Aug 08 '16 at 17:43
  • @BillDubuque One can derive this result from several other deeper results. Do you really think I don't know algebra? If so, please, leave me alone. Now I'm expecting the usual downvote. – egreg Aug 08 '16 at 17:53
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    Why do you think my comment is meant to imply anything about your knowledge of algebra? Rather, it was meant merely to point readers at more general results. Of course it goes without saying that there may be many ways to generalize it. I highly doubt that my comment will cause your answer to be downvoted. – Bill Dubuque Aug 08 '16 at 18:02
  • @egreg Agreed with Bill: I upvoted both, your answer and his comment. Nice observation btw. – YoTengoUnLCD Aug 09 '16 at 00:33
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Assuming that $\sqrt{n}+\sqrt{n+1}$ is a rational number, it follows by squaring that $\sqrt{n(n+1)}$ is a rational number. Given a positive integer $m$, $\sqrt{m}$ is a rational number only if $m$ is a square, but $n(n+1)$ cannot be a square: if we assume it is a square, it follows that $$4n(n+1) = (2n+1)^2-1$$ is also a square, but the only consecutive squares are $0$ and $1$.

Jack D'Aurizio
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