Proving/Disproving: Int x, y, z Divisibility Claim

AI Thread Summary
The discussion revolves around proving or disproving the claim that if an integer x does not divide the product yz, then it does not divide either y or z. Participants emphasize the importance of understanding when to prove versus disprove a statement, suggesting that a counterexample suffices for disproving a false claim. An equivalent statement, its contrapositive, is proposed as a method to prove the original claim. Additionally, questions arise about applying contraposition to non-implicative statements, with clarification provided that such statements can still be framed in an implication format. Overall, the conversation highlights the interplay between mathematical logic and proof strategies in computer science contexts.
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I wasn't sure if this went in math, or computer science. I'm posting it here, because it is for a computer science course, although it's technically mathematical proofs...

1. The problem:
Prove or disprove the following claim: For all integers x, y, and z, if x does not divide yz then x does not divide y and x does not divide z.
I wrote this in logic notation first, so I have that down. What I am having difficulty with is the actual prove/disprove part. I understand the steps needed to be taken (1. Assume... etc.), but am having trouble actually applying it. There are other similar questions, but I figured if I could get some help on this question, I could figure out the other ones as well, by applying the same concept...

Also, how do I know what to choose: prove, or disprove? In what circumstances would I attempt to prove, and in what circumstances would I attempt to disprove?

Any help would be greatly appreciated. :)
 
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puddipop said:
I wasn't sure if this went in math, or computer science. I'm posting it here, because it is for a computer science course, although it's technically mathematical proofs...

1. The problem:
Prove or disprove the following claim: For all integers x, y, and z, if x does not divide yz then x does not divide y and x does not divide z.
I wrote this in logic notation first, so I have that down. What I am having difficulty with is the actual prove/disprove part. I understand the steps needed to be taken (1. Assume... etc.), but am having trouble actually applying it. There are other similar questions, but I figured if I could get some help on this question, I could figure out the other ones as well, by applying the same concept...

Also, how do I know what to choose: prove, or disprove? In what circumstances would I attempt to prove, and in what circumstances would I attempt to disprove?
You should attempt to prove a statement only if it is true. If it is false, you should disprove it, which you can do by a single counterexample.
An equivalent statement to the one you are given is its contrapositive: For all integers x, y, and z, if x divides y OR x divides z, then x divides yz. If you can prove that, then you will have proved the statement you're given.
 
Thank you for your help!

I have one more question: I know how to use contrapositive for an implication (P => Q becomes not(Q) => not(P)). But, I'm not sure how to do that for a statement that is not an implication.

For example, how would I apply contraposition to this sort of claim?: All x, y, z belonging to P (prime numbers), x2 + y2 =/= z2 (for all prime numbers x, y, and z, x2 + y2 =/= z2).
 
puddipop said:
Thank you for your help!

I have one more question: I know how to use contrapositive for an implication (P => Q becomes not(Q) => not(P)). But, I'm not sure how to do that for a statement that is not an implication.

For example, how would I apply contraposition to this sort of claim?: All x, y, z belonging to P (prime numbers), x2 + y2 =/= z2 (for all prime numbers x, y, and z, x2 + y2 =/= z2).
I assume you mean x2 and so on. You can write exponents using the X2 button at the top of the input window.

At any rate, that is an implication; namely,
if x, y, and z are prime numbers, then x2 + y2 ≠ z2. Now it should be easy to write the contrapositive.
 
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