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?

Let's generalise this: you want to write $\dfrac{1}{x+y\sqrt{z}}$ in the form $b+c\sqrt{z}$:
$$\dfrac{1}{x+y\sqrt{z}} = \dfrac{(x-y\sqrt{z})}{(x+y\sqrt{z})(x-y\sqrt{z})} = \dfrac{x-y\sqrt{z}}{x^2-y^2z} = \dfrac{x}{x^2-y^2z} + \dfrac{-y}{x^2-y^2z} \sqrt{z}.$$
So $b= \frac{x}{x^2-y^2z}$ and $c= \frac{-y}{x^2-y^2z}$.
It works for complex numbers too.
