<s>[INST] Show that $2 - \sqrt{2}$ is irrational I suppose $2 - \sqrt{2} $ is rational.
so $$2- \sqrt{2} = {a/b} $$
where a,b are integers and gcd(a,b) = 1.
$$\text{Step 1. } 2 = (a/b)^2 \text{ //squared both sides }$$
$$\text{Step 2. } 2b^2 = a^2 \text{ //We see $a^2$ is even }$$
$$\text{Step 3. }2b^2 = (2k)^2$$
Step 3 since $a^2$ is even and so $a$ is even too. Also, k is an integer.
$$\text{Step 4. }b^2 = (2k)^2 \text{ //We see $b^2$ is even }$$
Step 4 since $b^2$ is even and so $b$ is even too.
We see $ a$ and $b$ are even. We see $gcd(a,b) \neq 1$
Is this proof correct ?
 [/INST] As an alternate approach, note that $2-\sqrt 2$ is a root of $x^2-4x+2$ (the conjugate root is $2+\sqrt 2$ so just compute $(x-(2-\sqrt 2))(x-(2+\sqrt 2))$. 
By the Rational Root Theorem, this quadratic has no rational roots.
</s>