How would I find the degrees of the following field extensions

i) Q(sqrt(11))/Q

ii) Q(sqrt(11),sqrt(5))/Q

iii) Q(sqrt(11),sqrt(5),sqrt(55))/Q

i) Since the minimal polynomial of √11 (which is x^2 - 11) is irreducible over Q and is of degree 2, we see that [Q(√11) : Q] = 2.

ii) Similarly, note that the minimal polynomial of √5 (which is x^2 - 5) is irreducible over Q(√11) and is also of degree 2. So, [Q(√11, √5) : Q(√11)] = 2.

(This comes from the fact that we can't write √5 = a + b√11 for any a,b in Q. This is easy to prove algebraically by contraction.)

Therefore, [Q(√11, √5) : Q] = [Q(√11, √5) : Q(√11)] * [Q(√11) : Q] = 2 * 2 = 4.

iii) This time, note that √55 = √5 * √11.
So, √55 is in Q(√11, √5).
==> Q(√11, √5, √55) = Q(√11, √5) [the other containment is trivially true].

So, [Q(√11, √5, √55) : Q] = 4.

I hope this helps!