Prove that $\sqrt2-\sqrt5$ is irrational I've attempted this proof through contradiction and just wanted to check whether I did this correctly.

Assume that the claim is false, that is 

$\sqrt2 - \sqrt5$ = $\frac {a}{b}$

Now square both sides

$2 - 5$ = $\frac {a^2}{b^2}$

$-3$ = $\frac {a^2}{b^2}$

$-3b^2$ = $a^2$

This is impossible as $a^2$ cannot be negative, so this is a contradiction.

Is this proof valid?
Hint: Your squaring must be: $$(a-b)^2=a^2+b^2-2ab$$
you square it wrong 

if you wish to get 2-5 multiple by square root of 2 + square root of 5

but now be could have written the original question as (sqrt(2)-sqrt(5))/1

now when we multiple we get -3/sqrt(2)+sqrt(5)

now square this equation you get 4.5/sqrt(10) as sqrt(10) is irrational so division is also irrational so the square root of the equation is also irrational