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I have to prove that $S=\{a+b\sqrt{2} \mid a,b \in \mathbb Q\}$ is a field and so far everything worked out for me except for finding the multiplicative inverse over $(S\backslash\{0\}, \cdot)$ which is: $$\forall a \in S: \exists (a^{-1}) \in S: a\cdot(a^{-1})=1$$

As a proof I thought of: $(a+b\sqrt{2}) \cdot (a+b\sqrt{2})^{-1} \Rightarrow (a+b\sqrt{2}) \cdot \frac{1}{(a+b\sqrt{2})}$ but I am stuck with the fraction. It is non-trivial for me to compute the inverse of an element $a \in S$.

Any help is highly appreciated! Thank you!

Shaun
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null1
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2 Answers2

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Hint: You want $c,d \in \Bbb{Q}$ such that $c + d \sqrt{2} = \frac{1}{a+b\sqrt{2}}$ for any $a,b \in \Bbb{Q}$. Have you tried rationalizing the denominator on the right (using its conjugate)?

Eric Towers
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    Oh, thank you very much, that is a really good hint! So basically: $$\frac{1}{a+b\sqrt{2}}=\frac{1 \cdot (a-b\sqrt{2})}{(a+b\sqrt{2}) \cdot (a-b\sqrt{2})}=\frac{a}{a^2-2b^2}-(\frac{b}{a^2-2b^2})\sqrt{2}$$ if I made no mistake. – null1 Jan 11 '20 at 19:09
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Hint: from the difference of squares factorization $(a+b\sqrt{2})(a-b\sqrt{2})=a^2-2b^2$.

Oscar Lanzi
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