Notational ambiguity in "$\sqrt 2$ = a rational number"

Question

I was reading chapter 29 page 296 of the book A Book Of Abstract Algebra (second edition) written by Charles C. Pinter, and the equation below was given: $$\sqrt{3} = a + b\sqrt{2}$$ where $a, b \in \mathbb {Q}$.

They were solving for $\sqrt{2} $ in terms of $a$ and $b$. The instruction given was to square both sides and solve for $\sqrt{2}$ What I did was $$\big( \sqrt{3} \ \big)^2 = \big(a + b\sqrt{2}\ \big )^2 \implies 3 = a^2 + 2b^2 + 2ab\sqrt{2}.$$ Looking at this, I don't find anyway to get to their final answer of $\sqrt{2} = a$ when solving for $\sqrt{2}$.

Answer 1

Quoting from page 296 of Pinter:

Observe that $\sqrt3$ cannot be in $\mathbb{Q}(\sqrt2)$; for if it were, we would have $\sqrt3=a+b\sqrt2$ for rational $a$ and $b$; squaring both sides and solving for $\sqrt2$ would give us $\sqrt2=$ a rational number, which is impossible.

Note, the "a" in "$\sqrt2=$ a rational number" is the article a, not the variable $a$, i.e., "the square root of $2$ would be a rational number."