In $\sqrt 2 =\frac ab$. Why must gdc(a, b) = 1 for $\sqrt 2$ $\in\mathbb{Q}$?

Question

In the proof by contradiction of square root of two being irrational it is implied that if both a and b are even or odd then they cannot be on the lowest terms( $gdc(a, b) = 1$ ). Why must they be on the lowest terms for √2 to be rational?

√2 = a/b
2 = a²/b²
2b² = a² => 2 | a² => 2 | a 
4 | a²
4 | 2b²
2 | b² => 2 | b

This proves that both a and b are even and thus have common factor of 2. Why does that contradict that √2 is rational?

One proof Why is it obvious

We additionally assume that this a/b is simplified to lowest terms, since that can obviously be done with any fraction.

Given the answers the question becomes why can we presume in the theorem that gdc(a, b) = 1 ?

Answer 1

A different way to look at this is: if you think there is a pair of positive integers $a,~ b$ such that $\frac{a}{b} = \sqrt{2}$ then you can generate a new pair of integers, $a',~ b'$ where $a' < a$ and $b' < b$. But then the same argument applies again, giving you a descending chain of $a$ and $b$ values. But there is no infinite descent of positive integers. So we couldn't have had the original pair $a,~ b$ after all, and thus there is no rational number $\sqrt{2}$.

Answer 2

Because in the statement of the theorem, it is required that $gcd(a,b)=1$.

You have found a contradiction to this, that $gcd(a,b)=2$.

Hence $\sqrt 2$ is irrational, as you assumed it to be rational.