proof explanation of $\sqrt{\frac{n}{n+1}}\notin \mathbb{Q}$

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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.