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I would like to understand the following proof of @Hw Chu

in this question Prove that $\sqrt{\frac{n}{n+1}}\notin \mathbb{Q}$

If $\displaystyle\sqrt{\frac{n}{n+1}} \in \mathbb Q$, then $\displaystyle \sqrt{n(n+1)} = (n+1)\sqrt{\frac{n}{n+1}} \in \mathbb Q$, and since $n \in \mathbb N_{>0}$, $\sqrt{n(n+1)} \in \mathbb Z.$

Would someone elaborate this proof I mean proof step by step

Michael Hardy
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3 Answers3

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And then, as has been shown here many times, if $m$ is not the square of an integer, then $\sqrt{m}$ is irrational.

But $n^2 < n(n+1) < (n+1)^2$, so $n(n+1)$ is not the square of an integer, so $\sqrt{n(n+1)}$ is irrational.

marty cohen
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If $\sqrt{\frac{n}{n+1}}\in\Bbb{Q}$ then there exists $q\in\Bbb{Q}$ such that $\frac{n}{n+1}=q^2$. Then $$n(n+1)=\frac{n}{n+1}\cdot(n+1)^2=q^2(n+1)^2,$$ so $\sqrt{n(n+1)}=q(n+1)\in\Bbb{Q}$. To see that this implies that $\sqrt{n(n+1)}\in\Bbb{Z}$, see the excellent answers to this question.

Servaes
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We have

If $\displaystyle\sqrt{\frac{n}{n+1}} \in \mathbb Q \implies \sqrt{n(n+1)} = (n+1)\sqrt{\frac{n}{n+1}} \in \mathbb Q$

that is a simple step by rationalization, then

and since $n \in \mathbb N_{>0}$, $\sqrt{n(n+1)} \in \mathbb Z$

here since $\sqrt{n(n+1)}$ is a product of an integer and a rational number it must be an integer, indeed assume $\sqrt{n(n+1)}=\frac p q$ with $\gcd(p,q)=1$ then $n(n+1)=\frac{p^2}{q^2}$ and $\gcd(p^2,q^2)=1$ by FTA

$n^2 < n(n+1) < (n+1)^2$.

this step prove that $\sqrt{n(n+1)}$ can't be an integer.

user
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  • A product of an integer and a rational number need not be an integer. – Servaes Apr 19 '18 at 20:31
  • @Servaes even if it is in the form $\sqrt{n(n+1)}$? – user Apr 19 '18 at 20:33
  • I do not understand. Do you mean your rational number is $\sqrt{\frac{n}{n+1}}$ and your integer is $n+1$? Why would that imply that the product is an integer? – Servaes Apr 19 '18 at 20:35
  • @Servaes the product is or rational or an integer but since it is in the form $\sqrt{n(n+1)}$ it is an integer that is a rational $\frac a b$ with b=1. – user Apr 19 '18 at 20:37
  • Indeed the product is rational, possibly an integer. How the form $\sqrt{n(n+1)}$ imply that it is an integer? This requires an argument, and likely a somewhat involved one. – Servaes Apr 19 '18 at 20:45
  • @Servaes $\sqrt{n(n+1)}$ rational $\implies n(n+1)$ is rational but $n(n+1)$ is integer thus $\sqrt{n(n+1)}$ is (supposed to be) an integer – user Apr 19 '18 at 20:49
  • The last implication is not easy, see this question. – Servaes Apr 19 '18 at 20:50
  • @Servaes but here we are working by the initial assumption that is $\sqrt{\frac{n}{n+1}} \in \mathbb Q $ which implies that $\sqrt{n(n+1)} = (n+1)\sqrt{\frac{n}{n+1}} \in \mathbb Q$. – user Apr 19 '18 at 20:53
  • I agree; the assumption that $\sqrt{\frac{n}{n+1}}\in\Bbb{Q}$ implies that also $\sqrt{n(n+1)}\in\Bbb{Q}$. But to show that this implies $\sqrt{n(n+1)}\in\Bbb{Z}$ is not easy, see the question I linked before. – Servaes Apr 19 '18 at 20:56
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    @Servaes It seems trivial to me, indeed assume $\sqrt{n(n+1)}=\frac p q$ with $\gcd(p,q)=1$ then $n(n+1)=\frac{p^2}{q^2}$ and $\gcd(p^2,q^2)=1$ by FTA. – user Apr 19 '18 at 20:59
  • That is indeed very straightforward; I had a far more roundabout argument in mind for some reason, which blinded me. Unfortunately I can't retract my downvote unless there is an edit. Do you mind if I summarize this argument in your answer, or would you prefer to do that yourself? – Servaes Apr 19 '18 at 21:03
  • @Servaes Finally we have clarified that point! I will add that to my answer! – user Apr 19 '18 at 21:05