Prove that $\sqrt{(1+1/n)}$ is irrational for all $n\in \mathbb{N}$ I have trouble solving this probably simple problem:

Prove that $\sqrt{1+1/n}$ is irrational for all $n\in \mathbb{N}$.


Here's another way.
$\sqrt{1 + \frac 1n} = \sqrt{\frac {n+1}{n}} = \sqrt{\frac {n(n+1)}{n^2}}= \frac 1n \sqrt{n(n+1)}$ which rational if and only if $\sqrt{n(n+1)}$ is rational.
Square roots of integers are rational only when that are integer square roots of perfect squares.  So $\sqrt{n(n+1)}$ is rational only if $n(n+1)$ is a perfect square.
But if $n > 0$ (which it must be if $n$ is natural $\frac 1n$ is defined) then:
$n^2 = n*n < n*(n+1) < (n+1)(n+1) = (n+1)^2$.
If $n*(n+1) = k^2$ then $n < k < n+1$ and $k$ can not be an integer so $n(n+1)$ is not a perfect square.
[The exceptions are $n = 0$ or $n = -1$ then $n(n+1) = 0^2$.  There reason the argument hold is if $n= 0$ then $n*n = n*(n+1) < (n+1)(n+1)$ so $n \le k < n+1$ so $k = n = 0$. or if $n = -1 < 0$ and $n+1 = 0$  $n*n > n*(n+1) = (n+1)(n+1)$ and so $k = n+1 = 0$.]
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Ooooh..... this is cute:
Let $(1 +\frac 1n ) = \frac {a^2}{b^2}; a, b\in \mathbb Z; \gcd(a,b) = 1;b \ne 0$
Then $b^2(1 + \frac1n) = a^2$
$(b^2 - a^2) = \frac {b^2}{n}$.
$(b-a)(b+a) = \frac {b^2}{n}\in \mathbb Z$.  
Let $p|\frac {b^2}{n}$ then $p|b$. But as $a,b$ are relatively prime. $p\not |b \pm a$ so $p\not \mid \frac {b^2}{n}$ after all and $\frac {b^2}{n} = 1$
$(b-a)(b+a) = 1$ so $(b-a) = (b+a) = \pm 1$ and $a = 0;(b = \sqrt{n})$ and $1 + \frac 1n = 0$.  Which is not possible.
