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?
 A: Hint: Your squaring must be: $$(a-b)^2=a^2+b^2-2ab$$
A: $(\sqrt{2}$ - $\sqrt{5})^2$ is not equal to 2 - 5; but rather, $2 - 2\sqrt{10} + 5$ = $7 - 2\sqrt{10}$
A: If
$\sqrt 2 - \sqrt 5 \in \Bbb Q, \tag 0$
then
$\sqrt 2 - \sqrt 5 = \dfrac{r}{s}, \; r, s \in \Bbb Z;  \tag 1$
we square:
$7 - 2\sqrt{10} = 2 - 2\sqrt{10} + 5 = \dfrac{r^2}{s^2}, \tag{2}$
and re-arrange to find
$\sqrt{10} = \dfrac{1}{2}\left (7 - \dfrac{r^2}{s^2} \right ) \in \Bbb Q; \tag 3$
we may then conclude with the aid of the following well-known
Theorem:  Let $2 \le m, n \in \Bbb N$ be natural numbers.  Then $\sqrt[m] n \in \Bbb Q$ if and only if $n$ is a perfect $m$-th power, in which case $\sqrt[m] n \in \Bbb N$.
Proof of Theorem:  The "if" direction is self evident; as for the "only if" direction,
we may suppose that
$\sqrt[m] n = \dfrac{r}{s}, \; r, s \in \Bbb Z, \tag 4$
with
$\gcd(r, s) = 1; \tag 5$
note we may safely assume
$s \ne \pm 1 \tag{5.5}$
since in the contrary case (4) becomes
$\sqrt[m] n = \pm r \Longrightarrow n = (\pm r)^m, \tag{5.6}$
i.e., $n$ is a perfect $m$-th power.
From (4) we have
$r^m = ns^m, \tag 6$
from which it follows that
$s \mid r^m; \tag 7$
now by virtue of (5) we have, via Bezout's identity, $a, b \in \Bbb Z$ such that
$ar + bs = 1; \tag 8$
multiplying this equation through by $r^{m - 1}$ we obtain
$ar^m + bsr^{m - 1} = r^{m - 1}; \tag 9$
since
$s \mid r^m, \; s \mid bsr^{m - 1}, \tag{10}$
(9) yields
$s \mid r^{m - 1}; \tag{11}$
we may now replace (7) with (11) and continue repeating the process expressed in (8)-(11), each time decrementing $m$ by $1$ until $m = 2$ is reached (the reader may construct a simple inductive argument to formalize this process if he or she so desires); at this point (11) becomes
$s \mid r, \tag{12}$ 
but by virtue of (5.5) this contradicts (5); therefore (4) is impossible and hence
$\sqrt [m]n \notin \Bbb Q. \tag{13}$
End:  Proof of Theorem.
Applying this theorem with $n = 10$, $m = 2$ we see that
$\sqrt{10} \notin \Bbb Q \tag{14}$
in contradiction to (3)and hence 
$\sqrt 2 - \sqrt 5 \notin \Bbb Q \tag{15}$
as well.
A: 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
