# Is sqrt(n+1)+sqrt(n-1) rational or irrational ?

Sep 23, 2016

See below.

#### Explanation:

We will prove that, searching for the rational solutions.

Supposing that $\sqrt{n + 1} + \sqrt{n - 1}$ is rational then

$\sqrt{n + 1} + \sqrt{n - 1} = {p}_{1} / {q}_{1}$ so

$\left(\sqrt{n + 1} - \sqrt{n - 1}\right) \left(\sqrt{n + 1} + \sqrt{n - 1}\right) = n + 1 - \left(n - 1\right) = 2 = \left(\sqrt{n + 1} - \sqrt{n - 1}\right) {p}_{1} / {q}_{2}$ is also rational

So $\sqrt{n + 1} - \sqrt{n - 1} = {p}_{2} / {q}_{2}$ is also rational then

$2 \sqrt{n + 1} = {p}_{1} / {q}_{1} + {p}_{2} / {q}_{2} = \frac{{p}_{1} {q}_{1} + {p}_{2} {q}_{2}}{{q}_{1} {q}_{2}}$

So $\sqrt{n + 1} = \frac{1}{2} \frac{{p}_{1} {q}_{1} + {p}_{2} {q}_{2}}{{q}_{1} {q}_{2}}$

With the same argument we could prove also that

$\sqrt{n - 1}$ is rational.

So $\sqrt{{n}^{2} - 1} = \frac{p}{q}$ is rational and consequently

${n}^{2} = {\left(\frac{p}{q}\right)}^{2} + 1$ Supposing that $\frac{p}{q} = m$ ( remember that $n$ is integer) we will need

${n}^{2} = {m}^{2} + 1$ or

$\left(n + m\right) \left(n - m\right) = 1$ with $n , m$ integers, $m = 0 , n = 1$ being the only rational solution.

Concluding, the only rational solution is for $m = 0 , n = 1$ but then

$\sqrt{2}$ is irrational so for $n \in {\mathbb{N}}^{+}$, $\sqrt{n + 1} + \sqrt{n - 1}$ is irrational.