1. Irrational Roots

Find a polynomial f(x) in Z[x] such that r is a root and use f(x) to show that r is irrational.

r=sqrt(3-sqrt(5))

The hint says r^2=3-sqrt(5), so (r^2-3)^2=?

So I suppose they don't just want (x-sqrt(3-sqrt(5)), but I don't know how to construct a polynomial so that I can show the irrationality.

Thanks.

You're right that $x-\sqrt{3-\sqrt{5}}$ won't do because is is not a member of $\mathbb{Z}[x]$. Just working out the first few powers of $r$, I get:

$r=\sqrt{3-\sqrt{5}}$,
$r^{2} = 3-\sqrt{5}$,
$r^{3} = (3-\sqrt{5})\sqrt{3-\sqrt{5}}$,
$r^{4} = 14+6\sqrt{5}$.

Now just try and spot linear multiples of these which sum to an integer, which you are free to minus away at the end using the constant term in the polynomial you are constructing. We need to eliminate the $\sqrt{5}$ term. Try $r^{4}-6r^{2}$ and go from there.

3. (r^2-3)^2=(-sqrt(5))^2
Then we have a polynomial $f(x)\in\mathbb{Z}[x]$ (actually $\in\mathbb{Q}[z]$ suffices but they're equivalent) of which r is a root. If f(x) can be shown irreducible and having degree >=2, then r is irrational

How about answering that question? If $r= \sqrt{3- \sqrt{5}}$ then $r^2= 3- \sqrt{5}$ and $r^2- 3= -\sqrt{5}$. Now, what is $(r^2- 3)^2$?