If any of you math gurus could help, I would appreciate it. Reps for all answers. I think I have this, but something tells me that I may be off. R= {(a,a),(b,a),(b,b),(c,b),(c,c)} Is this Transitive? I think no, since you need (a,b),(b,c) and (a,c). In this case, the (a,b) is (c,b) and the (b,c) is (b,a), but there is no (a,c). However, in reading some other problems like this, you don't always need (a,b),(b,c) and (a,c)....for example: R= {(1,1),(1,2),(2,1),(2,2),(3,3),(4,4)} According to my answer key, this IS Transitive, since (1,2) and (2,1) AND (1,1) and (2,2) are in the relation. So this confuses me and makes the answer to my original question not so cut-and-dry. So am I over thinking the original question? Thanks.
A discrete question deserves a discreet answer, so I will refrain from offering my controversial opinion on this problem, and simply suggest that both of your ideas have merit. However, since this problem might potentially involve a transitive element, it certainly deserves a tranny type of answer, so I will suggest that the answer in this case could go both ways, and may not be what it seems. I hope you find my input valuable.
^^ Level 12, Wizard. I've done math through grad school, but I don't like this recent trend of math-threads... I don't come here to think! Now, I could have answered the Riemann zeta question a couple years ago, but this one, ehhh, I'm not sure I ever knew the answer. Good luck.
Discrete Math is....it's a requirement for a Computer Science degree. I guess you could call it programming math, or math that's good for designing networks. They suggest that you take it in your senior year of college, after all of the other math requirements are done. I don't pretend to be good at it, but it's kind of fun, like doing puzzles.
For transitivity, (a must be related to b) and (b must be related to a) for all a,b in the set. In your example that showed transitivity, you can see it holds true for all of the equal relations. (1 is related to 1) and (1 is related to 1)[transitivity!] (2 is relat....) etc. You also see the remaining relations (1 is related to 2), and (2 is related to 1) are also transitive so you can conclude transitivity. In your initial example (a,a),(b,a),(b,b),(c,b),(c,c) you see that (b is related to a) but you don't have (a is related to b). The same goes for (c,b) but you need only one not hold for it not to be transitive. Hopefully I helped a little bit.
And my coworker just politely informed me I just described a symmetric relation. Don't listen to me! I'll let him try to explain it. Transitivity is similar to symmetry, however aRb and bRc -> aRc. (1,1), (1,1) -> (1,1) :contained in the set. (1,2), (2,1) -> (1,1) :contained in the set. (2,1), (1,2) -> (2,2) :contained in the set. Therefore, transitive. Similarly, (a,a), (a,a) -> (a,a) (a,a), (b,a) -> (a,a) (b,a), (a,a) -> (b,a) (b,a), (b,a) -> (b,a) (c,c), (c,c) -> (c,c) (c,c), (c,b) -> (c,b) (c,b), (c,c) -> (c,c) (c,b), (c,b) -> (c,b) All are contained in the set, therefore also transitive.
Awesome...ok, ok. Are you saying that R= {(a,a),(b,a),(b,b),(c,b),(c,c)} is transitive due to this? Just double checking....I've been trying to find this for about 8 hours. (I've also proved that it's reflexive and antisymmetric. If it's also transitive, then it's a partially ordered set)
He missed one piece! (c,b) and (b,a) => (c,a) not in the set so NOT transitive. Final answer! Took a group of programmers to come up with that, that's pretty sad.
Ok I thought as much. Thanks for reinforcing what I worked! If you look at the second example in my first post, there is no a,c...not sure how that works.
Why do you say there is no (a,c)? (a,b), (b,c) -> (a,c) (1,1), (1,1) -> (1,1) :contained in the set. (1,2), (2,1) -> (1,1) :contained in the set. (2,1), (1,2) -> (2,2) :contained in the set.
as tamericus already explained, not transtitive R= {(a,a),(b,a),(b,b),(c,b),(c,c)} you have (c,b) and (b,a) therefore you would need (c,a) to be transitive.