Question

prove that for all a,b in Q, Q(sqrt(a),sqrt(b))=Q(sqrt(a)+sqrt(b)).

Answer 1

Q contained in Q(sqrt(a) contained in Q(sqrt(a),sqrt(b))

So as sqrt a is not in Q, [Q(sqrt(a):Q]=2 and [Q(sqrt(a),sqrt(b)):Q(sqrt(a)]=2.

So [Q(sqrt(a),sqrt(b)):Q]=2 or 4, if b is in Q(sqrt(a)) or not
respectively.

Now we also have,

Q contained in Q(sqrt(a)+sqrt(b)) contained in Q(sqrt(a),sqrt(b))

Depending on whether sqrt(b) is in sqrt(a) or not, extension
Q(sqrt(a)+sqrt(b)):Q has basis {1, sqrt(a)} or
{1,sqrt(a), sqrt(b), sqrt(ab)}. So [Q(sqrt(a)+sqrt(b)):Q]=2 or 4

If sqrt(b) is in sqrt(a), then Q has basis {1, sqrt(a)}and then [Q(sqrt(a)+sqrt(b)):Q]=2 and Q(sqrt(a),sqrt(b)}=2

If sqrt(b) is not in sqrt(a), then Q has basis {1,sqrt(a), sqrt(b), sqrt(ab)} then [Q(sqrt(a)+sqrt(b)):Q]=4 and Q(sqrt(a),sqrt(b)}=4

Now, in either case [Q(sqrt(a)+sqrt(b)):Q(sqrt(a),sqrt(b)}=1 as required.

You acn also argue in the same way like If sqrt(a) is in sqrt(b)...then same will happens.