Prove that quadratic field Q(√2) is closed

Question

For the quadratic field Q(√2), how can I prove that it is closed? This means that any addition, subtraction, multiplication, or division between any a+b√2 and c+d√2 will still be from the set Q(√2)?

Addition and subtraction are pretty trivial, so I won't include it here. For multiplication, I did the following:

$$(a+b√2)(c+d√2)$$ $$=ac+bc√2+ad√2+2bd$$ $$=(ac+2bd)+(bc+ad)√2$$

Therefore Q(√2) is closed for multiplication.

For division, I did the following:

$$\frac{a+b√2}{c+d√2}$$ $$=\frac{(a+b√2)(c-d√2)}{(c+d√2)(c-d√2)}$$ $$=\frac{ac-ad√2+bc√2-2bd}{c^2-2d^2}$$ $$=\frac{(ac-2bd)+(bc-ad)√2}{c^2-2d^2}$$

How do I prove that this value is of the set Q(√2)?

Answer 1

$$\frac{(ac-2bd) + (bc-ad)\sqrt{2}}{c^2 - 2d^2} = \frac{ac-2bd}{c^2 - 2d^2} + \frac{bc-ad}{c^2 - 2d^2}\sqrt{2}$$

and $a,b,c,d \in \Bbb Q.$

Answer 2

You're almost there.

Is $\dfrac{ac-2bd}{c^2-2d^2}\in\mathbb Q$?

Is $\dfrac{bc-ad}{c^2-2d^2}$ ?

Then $\dfrac{(ac-2bd)+(bc-ad)\sqrt2}{c^2-2d^2}\in\mathbb Q(\sqrt2)$.

(Note: you should exclude $c=d=0$.)