- #1
Shackleford
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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.
If h ∪ g is a function, then h and g are functions.
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.
Let h and g be functions. If Dom(h) = A, Dom(g) = B, and A ∩B = ∅, then h ∪ g is 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.