Help with a proof for the irrationality of $\sqrt 2$

Question

The following proof comes from Ian Stewart's Galois Theory.

Assume for a contradiction that there exists integers $a$, $b$ with $b\neq0$ such that $(\frac{a}{b})^2=2$.

1. We may assume that $a, b>0$.
2. Observe that if such an expression exists, then there must be one in which $b$ is as small as possible.
3. Show that $(\frac{2b-a}{a-b})^2=2$.
4. Show that $2b-a>0$, $a-b>0$.
5. Show that $a-b<b$, a contradiction.

I do not follow step 4 & 5. Any hint would be appreciated!

Answer 1

In step 1 and 2 it's already assumed that $a,b$ is a pair of positive integers that satisfy $\displaystyle \left(\dfrac{a}{b}\right)^2=2$, and among all the possible pairs of $(a,b)$, this pair is where $b$ is the smallest. In other words, any smaller pair with a smaller $b$ would lead to contradiction. That's what is explained in step 4 and 5.