I'm having trouble proving that a multiplicative inverse exists in the following problem:

Prove that the numbers of the form $a+b\sqrt{2}$, where $a$ and $b$ are rational numbers, form a subfield of $\mathbb{C}$.

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Hint: What happens if you multiply $(a + b \sqrt{2})$ by $(a - b \sqrt{2})$?

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  • $\begingroup$ i.e. use that to rationalize the denominator. Note that it cannot be $0\,$ (else $\,\sqrt2\in\Bbb Q)\ \ $ $\endgroup$ – Bill Dubuque Mar 6 '15 at 1:11
  • $\begingroup$ $(a+b\sqrt{2})(a-b\sqrt{2})=a^2-2b^2$. Sorry if I'm slow, but how does that help? What is the next step? $\endgroup$ – clay Mar 6 '15 at 2:50
  • $\begingroup$ Note $a^2 - 2b^2$ is a non-zero rational number. So if you have $a + b \sqrt{2}$ then $(a-b \sqrt{2})/(a^2 - 2b^2)$ is the multiplicative inverse and it is contained in your field. So all you have to show is that your field is closed under multiplication and you will be done. $\endgroup$ – user2566092 Mar 7 '15 at 17:52

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