I got stuck at : $\frac{a^2}{b^2} = 12+2 \sqrt 35$
I understand that $12$ is rational and now I need to prove that $\sqrt{35}$ is irrational.
so I defined $∀c,d∈R$ while $d$ isn't $0$ that: $\frac{c^2}{d^2} = \sqrt 35$ so $- c^2=(d^2)\sqrt{35}$ It means that $c$ divide with $5$ and $7$? Also, how do I prove that if for example $X^2|4$ then $X|4$?
Here's a different approach, assuming you know that $\sqrt5$ and/or $\sqrt7$ are irrational.
If $\sqrt7+\sqrt5$ were rational, then
$${2\over\sqrt7+\sqrt5}=\sqrt7-\sqrt5$$
would also be rational, in which case
$$\sqrt7={\sqrt7+\sqrt5\over2}+{\sqrt7-\sqrt5\over2}$$
and
$$\sqrt5={\sqrt7+\sqrt5\over2}-{\sqrt7-\sqrt5\over2}$$
would be rational as well.
If $a = \sqrt{5} + \sqrt{7}$ is rational, then $b = \sqrt{5} - \sqrt{7} = a - 2 \sqrt{7}$ is irrational. But $ab = -2$, so $a$ and $b$ must be both rational or irrational.
You're on the right track.
Consider the powers of $5$ that divide both sides of $c^2=35d^2$. You have an even number for the RHS but an odd number for the LHS.
Indeed, if $5^m$ is the largest power of $5$ that divides $c$ and $5^n$ is the largest power of $5$ that divides $d$, then we get $2m=2n+1$, a contradiction.
Hint $ $ Subtract $12$ then square it again to show that it's a root of a polynomial with integer coef's that's monic (lead coef $= 1).\,$ Now apply the Rational Root Test (or equivalent).
Assuming √5 + √7 is rational, √5 + √7 = a ∕ b Squaring both sides, 5 + 7 + 2√35 = a²/b² ⇒2√35 = a²/b² - 12 ⇒2√35 = (a²-12b²)/b² ⇒√35 = (a²-12b²)/2b²
According to our assumption, (a²-12b²)/2b² should have been rational but actually it is irrational. Hence our assumption is wrong. Therefore, √5 + √7 is irrational
