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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)?

John Liu
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  • It is already in the correct form! That is, you have shown it is $\frac {ac-2bd}{c^2-2d^2}+ \frac {bc-ad}{c^2-2d^2},\sqrt 2$ which is clearly in the form $A+B\sqrt 2$ with $A,B\in \mathbb Q$. – lulu Nov 01 '20 at 22:41
  • Closed in which space? – Bernard Nov 01 '20 at 23:18

2 Answers2

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

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

Edward Evans
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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 $c=d=0$.)

J. W. Tanner
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