Answer:
a=0, b=16, of the pair (0,16)
Step-by-step explanation:
You want to prove the following: There are real numbers a and b such that [tex]\sqrt{a+b}=\sqrt{a}+\sqrt{b}[/tex]
So, to prove this, it is enough to find a pair (a,b) of real numbers for which the equation holds true. There are infinite of such pairs, but one example is (0,16). Indeed:
[tex]\sqrt{a+b}=\sqrt{0+16}=\sqrt{16}=4=0+4=\sqrt{0}+\sqrt{16}=\sqrt{a}+\sqrt{b}[/tex]
In fact, (0,m) or (n,0) ara valid pairs, for nonnegative numbers m,n. Even more is true; these are the only pairs which satisfy the equation.