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I'm solving some exercises about fields and am trying to find the inverse for $a_1 + \sqrt{2}b_1$, i.e. $\frac{1}{a_1 + \sqrt{2}b_1}$. This means I need to split the fraction into something of the form $x_1 + \sqrt{2}x_2$ but I can't seem to remember how to do such a basic thing! Can anyone help me out?

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  • $\begingroup$ You have to say in what field you are working, it's probably $\mathbb{Q}(\sqrt{2})$ but if for exmaple it is $\mathbb{R}$ then the inverse is exactly what you wrote $\endgroup$ – Belgi Jul 26 '12 at 9:19
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Hint: $\displaystyle \frac{1}{a_1+\sqrt{2} b_1}= \frac{a_1- \sqrt{2}b_1}{(a_1+\sqrt{2}b_1)(a_1- \sqrt{2}b_1)}$.

It is the same idea that for complex numbers.

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  • $\begingroup$ Thanks! The solution is $\tfrac{a}{a^2 - 2 b^2} + \sqrt{2} \tfrac{-b}{a^2-2b^2}$. $\endgroup$ – fabian789 Jul 26 '12 at 8:07

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