These are potential proofs for the discrete math exam on Tuesday. I haven't been able to find proofs online. I have senioritis, and I'm graduating in a few weeks.

Is a proof by contraposition the best way to prove this? If you assume h is not a function or g is not a function, then that would imply that h ∪ g is not a function.

I understand why the domains have to be disjoint. You could run into a problem where an element that appears in both domains is not well-defined.

For the first one what you wrote is correct, assuming you understand why the implication is true.

Your remark on the second one is accurate. Are you confused as to why it's always true when the domains are disjoint?

These problems are essentially solved by just writing down the definition of a function, writing down the definition of union and putting the two together

To show something is a function, the professor wants us to show that (in this case) h ∪ g is a relation, Dom(h ∪ g) = A ∪ B, and it is well-defined. I suppose the most direct way is considering the functions as relations and just using the definitions of functions and unions.

What Office Shredder said really, I don't know if you're used to using first order logic but it'll simplify things a whole lot, I'd use some of that. I'd also use the notion of a graph of a function too

I remember my first experience with doing functiony proofs... they were a pain -.-