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Let S be the set of all people in the world, a and b be people in S, and R be the relation given in each item below. Is (S,R) a partially ordered set? i) a = b or a is a descendant of b ii) a and be do NOT have a common friend

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I answered Yes for i) and No for ii) by the way
Did you get this question from a textbook?
i rephrased it a little to shorten the question.. is there something wrong with the way I rephrased it?

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Post the question exactly as it is from the book, then post the name, author, and edition of the book.
Um, okay, this is from Kenneth Rosen's Discrete Mathematics and Its Applications, 7th ed. Here's a screenshot from the book too:
I only need parts c and d from the problem in the book by the way
Why do you need the author, title, ed, etc of the book?
To see if I had the book. Hang on
Oh okay..
Which chapter are you working on?
9.6, Partial Orderings
Recall the following: A relation, R, on a set A is (i) Reflexive of (a,a) ∈ R for all a ∈ A (ii) Anti-symmetric if (a,b),(b,a)∈R implies that a = b (iii) Transitive if (a,b),(b,c) ∈ R implies that (a,c)∈ R (iv) A partial ordering if it is reflexive, anti-symmetric and transitive. Suppose S is the set of all people in the world: (a) R = {(a,b) ∈ R | a is no shorter than b} is a relation on S. Then, R is not anti-symmetric because there may be two different persons with the same height. Hence, (S,R) is not a proset. (b) R = {(a,b) ∈ R | a weighs more than b} is a relation on S. Then R is not reflexive because no other person weighs more than him or herself.
Hence, (b) is not a proset either
Thank you, but I only need parts c and d, as I've said above :D
You are correct then, lol
But you should post a more complete answer than just yes or no
Haha thanks, just want to make sure. My teacher said she doesn't require the actual explanation.
Yes, but that is the proper way to answer the questions. Yes or No are technically not mathematical responses.
If I were the teacher, I would tell you that these are not "Yes" or "No" questions.
Haha that would be my inclination as well. I'm going to ask her again.

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