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 ?
 A: 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!
A: Step 1 of your proof is incorrect:
From: 
$$2- \sqrt{2} = {a/b} $$
it does not follow that:
$$2 = (a/b)^2$$
but rather that:
$$(a/b)^2=(2- \sqrt{2})^2=4-4\sqrt{2}+2=6-4\sqrt{2}$$
Now, I am sure you want to know: so how do you prove it then?! Well, likely you will get many Answers that will show you how ... but your question was whether your proof was correct ... it is not.
A: 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.
A: That proof is not correct.
You start with $2-\sqrt{2}=\frac{a}{b}$ but on your next line you jump to $2=\left(\frac{a}{b}\right)^2$.
What you instead want to do is show that $\sqrt{2}$ is irrational (which seems like what you've done) then prove the following property: a rational minus an irrational number yields an irrational number. This can be done by considering the contrapositive, which states that if $p,q$ are rationals, then $p+q$ is rational. Here's a hint on how to prove this:

 Let $p=\frac{a}{b},q=\frac{c}{d}$. What is $p+q$ in terms of $a,b,c,d$?

A: Yet another proof: $0<2-\sqrt{2}<1$, 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 $$0<(2-\sqrt{2})^n=(a_n-2b_n)+b_n(2-\sqrt{2}).\tag{1}$$
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 $(2-\sqrt{2})^n\to 0$.
