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 ?
For the sake of Contradiction let us assume that  $2-\sqrt{2} \in \Bbb Q$ , say $2-\sqrt{2}=r \in \Bbb Q \implies \sqrt{2}= 2-r$ but since both $2,r \in \Bbb Q$ and $(\Bbb Q,+)$ is a subgroup of $(\Bbb R,+)$ (i.e. basically addition of two rational numbers is rational) it follows that $\sqrt{2} \in \Bbb Q$ , a contradiction!
Yet another proof: $0&lt;2-\sqrt{2}&lt;1$, so $(2-\sqrt{2})^n\to 0$.  Note that $(2-\sqrt{2})^n=a_n-b_n\sqrt{2}$ for some integers $a_n,b_n$ by binomial theorem, so $$0&lt;(2-\sqrt{2})^n=(a_n-2b_n)+b_n(2-\sqrt{2}).\tag{1}$$
If $2-\sqrt{2}$ were rational, say $=p/q$, $p,q\in\mathbb{N}$, then by equation (1) $(2-\sqrt{2})^n$ must be at least $1/q$, contradicting $(2-\sqrt{2})^n\to 0$.