Discrete Math Exam Proofs: Senioritis & Graduation

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Homework Help Overview

The discussion revolves around potential proofs for a discrete math exam, specifically focusing on properties of functions and their unions. The original poster expresses difficulty in finding proofs and mentions the impact of senioritis as graduation approaches.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster considers using proof by contraposition to demonstrate that if the union of two functions is a function, then both must be functions. They also discuss the necessity of disjoint domains for the union to be well-defined. Participants question the understanding of why the disjoint condition guarantees the truth of the implication.

Discussion Status

Participants are engaging in clarifying definitions and reasoning related to functions and their unions. Some guidance has been provided regarding the definitions needed to approach the proofs, and there is an acknowledgment of the challenges involved in these types of proofs.

Contextual Notes

There is mention of specific requirements from the professor regarding how to demonstrate that a union of functions is a function, including the need to show that it is a relation and well-defined. The discussion also hints at the use of first-order logic and function graphs as potential tools for simplification.

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.

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.
 
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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
 
Office_Shredder said:
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 -.-
 
Last edited:

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