<s>[INST] √2+√3 is irrational I’m trying to prove that $\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}$ is irrational. 
I have already proved that $\sqrt{2}+\sqrt{3}$ is irrational. Should I use a similar approach as below or is there a different way?
Proof: $\sqrt{2}+\sqrt{3}$ is irrational 
Assume $\sqrt{2}+\sqrt{3}$ is rational so $\sqrt{2}+\sqrt{3}=p/q$ where $p$ and $z$ are integers, $q\neq 0$, $\gcd(p,q)=1$
$r^2=5+2\sqrt{6}$ 
$r^2-5=2\sqrt{6}$ then I squared both sides…
$r^4-10r^2+25=24$ 
$r^4-10r^2+1=0$ 
$r=p/q$ so $p^4/q^4-10p^2/q^2+1=0$ 
 then I got a common denominator 
$p^4-10p^2q^2+q^4=0$
$p^4=10p^2q^2-q^4$
I found that $q=1$ or $-1$ and then I plugged it back into $r^4-10r^2+1=0$ and found that it can’t $=-1$ so there is a contradiction. 
 [/INST] You have a shorter proof: if $\sqrt{2}+\sqrt{3}=\frac{p}{q}$, where $p\in\mathbb  Z$ and $q\in\mathbb N$, $q\neq 0$, then $5+\sqrt{6}=\frac{p^2}{q^2}$. So, $\sqrt{6}=\frac{p^2-5q^2}{q^2}$ is rational, which is known to be false.
For $\sqrt{n}$, with $n\in\mathbb N$, you have the following alternative:
1) $\sqrt{n}$ is an integer (it happens when $n$ is a perfect square)
2) $\sqrt{n}$ is irrational.
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