Is there more to explain why a hypothesis doesn't hold, rather than that it arrives at a contradiction? Yesterday, I had the pleasure of teaching some maths to a high-school student. She wondered why the following doesn't work:
$\sqrt{a+b}=\sqrt{a}+\sqrt{b}$.
I explained it as follows (slightly less formal)


*

*For your hypothesis to hold, it should hold given an arbitrary set of operations performed on your equation.

*For example, it should hold if we square the equation, and after that take the square root, i.e. (note that I applied her logic in the second line; I know it's not OK to do maths like that)
$\sqrt{a+b}=\sqrt{a}+\sqrt{b}\\
a+b=a+b+2\sqrt{ab}\\
\sqrt{a+b}=\sqrt{a}+\sqrt{b}+\sqrt{2\sqrt{ab}}$

*We now arrive at a contradiction, which means that your hypothesis is false.
However, she then went on to ask 'But why then is it false? You only proved that it's false!'. As far as I'm concerned, my little proof is a perfect why explanation as far as mathematicians are concerned, but I had a hard time convincing her - the only thing I could think of is to say that the square root operator is not a linear operator, but I don't really think that adds much (besides, I really don't want to be explaining and proving linearity to a high school student).
So, my question: is there anything 'more' as to why the above doesn't work, or was I justified in trying to convince her that this is really all there is to it?

Here's another explanation approach:

Suppose that $$\sqrt{a+b}=\sqrt a + \sqrt b$$

If this is true, then we can square both sides of the equation:
$$(\sqrt {a+b})^2=a+b=(\sqrt a+\sqrt b)^2=a+b+2\sqrt {ab}$$
This resulting equality can then be manipulated as follows:
$$a+b=a+b+2\sqrt{ab}\\
0=2\sqrt {ab}\\
0=ab$$
By the rules of real numbers, which do not have zero-divisors, if $ab=0$ then either $a=0$ or $b=0$.  Looking back at the original supposition, it will hold given either condition.  So the answer is that the supposition is correct as long as at least one of $a,b$ is $0$, or in more explanatory language, "you can do that, as long as either $a$ or $b$ is zero."
