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

Question

I would like to show that $$\forall n\in\mathbb{N}^*, \quad \sqrt{\frac{n}{n+1}}\notin \mathbb{Q}$$

I'm interested in more ways of proofing this.

My method :

suppose that $\sqrt{\frac{n}{n+1}}\in \mathbb{Q}$ then there exist $(p,q)\in\mathbb{Z}\times \mathbb{N}^*$ such that $\sqrt{\frac{n}{n+1}}=\frac{p}{q}$ thus

$$\dfrac{n}{n+1}=\dfrac{p^2}{q^2} \implies nq^2=(n+1)p^2 \implies n(q^2-p^2)=p^2$$ since $p\neq q\implies p^2\neq q^2$ then

$$n=\dfrac{p^2}{(q^2-p^2)}$$

since $n\in \mathbb{N}^*$ then $n\in \mathbb{Q}$

• I'm stuck here and I would like to see different ways to prove $\sqrt{\frac{n}{n+1}}\notin \mathbb{Q}$

Answer 1

Finishing your proof

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

All prime or unitary factors of $q-p$ and of $q+p$ must be prime or unitary factors of $p$ so then must be common factors of both $p$, and $q$. But $p$ and $q$ are presumed to be relatively prime. So the only factors of $q-p$ and $q+p$ are $1$ so $q^2 - p^2 = 1$ which... is not possible.

We can assume $q > p$ (else $\frac pq \ge 1 >\sqrt{\frac n{n+1}}$) so let $q = p + k; k \ge 1$ so $q^2 = p^2 + 2k + 1 > p^2 + 1$.

....

Basically it is well known that $\sqrt n$ is rational for integer $n$ only if $n$ is a perfect square. It can be verified that if $\gcd(a,b)= 1$ then $\sqrt{\frac ab} = \frac pq$ where $\gcd(p,q) = 1$ then $\frac ab = \frac {p^2}{q^2}$ and both $\frac ab$ and $\frac {p^2}{q^2}$ are in lowest terms so $a = p^2$ and $b = q^2$. As $\gcd(n,n+1) = 1$ and then for $\sqrt{\frac n{n+1}}$ to be rational then but $n$ and $n+1$ are perfect squares.

But there are many ways to show that's impossible, but mainly if $(n+k)^2 - n^2 = 2k + k^2$ which if $k \ge 1$ is greater than $1$. Or if $m > n$ then $m^2 - n^2 = (m-n)(m+n) \ge m+n> 1$.

Answer 2

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

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

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Remark: Actually the proof in your post is still in the right track. You actually can assume $\gcd(p,q) = 1$. Factorize

$$ n = \frac{p^2}{(q-p)(q+p)}. $$

Since $\gcd(p, q-p) = \gcd(p, q+p) = 1$, and $n$ is an integer, it is enforced that $q-p, q+p \in \{0,1\}$, which is impossible if $n \in \mathbb N_{>0}$.

Answer 3

From RRT for $P(x)=(n+1)x^2-n=0$, if $x=\frac{p}{q}, \gcd(p,q)=1$ is a solution for P(x), then $\color{green}{p\mid n}$ and $\color{red}{q\mid n+1}$. In this particular case it's even $\color{green}{p^2\mid n}$ and $\color{red}{q^2\mid n+1}$ from $$(n+1)p^2-nq^2=0 \tag{1}$$

Aslo, from $\gcd(n,n+1)=1$ and Bezout $\exists a,b\in \mathbb{Z}$ s.t. $$an+b(n+1)=1 \tag{2}$$ Combining $(1)$ and $(2)$ $$aq^2n+bq^2(n+1)=q^2 \Rightarrow a(n+1)p^2+bq^2(n+1)=q^2\Rightarrow\\ \color{red}{n+1\mid q^2} \Rightarrow q^2=t(n+1)$$ and back to $(1)$ $$p^2-tn=0 \Rightarrow \color{green}{n \mid p^2}$$ as a result $n+1=q^2$ and $n=p^2$ or we have 2 consecutive integers which are also perfect squares, which is not possible, since the next perfect square is $(p+1)^2=p^2+2p+1 > n+1$.

Answer 4

Another approach: If $n/(n+1)$ is a square, then both $n$ and $n+1$ are squares, since they are relatively prime. Therefore, there would be two squares, which difference is $1$:

$$(p+q)(p-q)=p^2-q^2=1$$

Therefore, $p+q=1$ and $p-q=1$. It follows that $p=1$ and $q=0$. Therefore, $n=0$.