If $a,b,c$ are rational numbers and $a\sqrt2+b\sqrt3+c\sqrt5=0$, show that $a=b=c=0$ If $a,b,c$ are rational numbers and $a\sqrt2+b\sqrt3+c\sqrt5=0$, show that $a=b=c=0$


I can solve for two terms only but for additional $c\sqrt5$ term, I got stuck. I also want to know how to solve this for general $n$ terms i.e. $a\sqrt2+b\sqrt3+c\sqrt5+\ldots+n \text{ terms}=0$  will imply $a=b=c=\ldots=0$
Assume $a\sqrt 2+b\sqrt3+c\sqrt 5=0$. Then $5c^2=2a^2+2ab\sqrt 6+3b^2$. From the irrationality of $\sqrt 6$, we conclude that $ab=0$. If $b\ne 0$, this makes $a= 0$ and $5c^2=3b^2$, so $\sqrt{15}=\frac{5c}b$ rational, contradiction. 
If $a\ne 0$, this makes $b=0$ and $5c^2=2a^2$, ao $\sqrt{10}=\frac{5c}{a}$ rational, contradiction. We conclude that $a=b=0$, but then also $c=0$.



The general statement is:


  Let $a_1,\ldots, a_n$ be different square-free positive integers. Then $\sqrt{a_1},\ldots, \sqrt{a_n}$ are linearly independent over the rationals.


Its proof involves but a bit deeper techniques. Edit: Found it, see my answer here
Multiply your equation by $\sqrt{2}$ to get,

$2a + \sqrt{6}b + \sqrt{10}c = 0 \implies a = 0,  \sqrt{6}b + \sqrt{10}c = 0$ since the rational and irrational parts both need to be zero.

Now multiply $\sqrt{6}b + \sqrt{10}c = 0$ by $\sqrt{6}$ and apply the same logic.