<s>[INST] Real number multiplicative inverses expressed in another form I've been asked to express the multiplicative inverse of $3 + \sqrt{5}$ in the form $c + d\sqrt{5}$, where $c,d$ are rational numbers.
I understand that for some rational numbers $c,d$ we must have:
$$1 = (3 + \sqrt{5})(c + d\sqrt{5}).$$
I was able to answer for the multiplicative inverse of $2 +\sqrt{3}$. We find that $1 = (2 +\sqrt{3})(c + d\sqrt{3})$ where $c = 2$ and $d = -1$. However this seems to be related to the original $2 +\sqrt{3}$; and in the problem at hand this is not the case. So I'm rather confused.
How could i go about solving this? I'd like steps without the answer IF possible; moreover, how could i prepare for more general questions in this form?
 [/INST] \begin{align}
Answer &= \frac{1}{3+\sqrt{5}}\\
\mbox{Multiply and Divide by Conjugate}&\\
 &= \frac{1}{3+\sqrt{5}}\cdot \frac{3-\sqrt{5}}{3-\sqrt{5}}\\
 &= \frac{3-\sqrt{5}}{(3)^2-(\sqrt{5})^2}\\
\end{align}
Hidden Answer:


 $$\frac{3}{4}-\frac{\sqrt{5}}{4}$$


</s>