Discrete Math Problem - Properties of Relations

PTthenMath

New member
Good Morning!

I'm new to this forum, and wasn't sure where to post a Discrete Math problem, but as the class has probability topics; I figured this was appropriate.

Quick background: Taking discrete math online, and it's very difficult to follow. The text is beyond my comprehension. I got the below problem incorrect on an assignment, and when I asked the professor for help, I was told he/she needed to check the text and would get back to me. Great class, huh? So, I'll show my thought process, and hopefully someone can steer me in the right direction. Also, searched all over YouTube for videos, and didn't find any that clarified relations (at least for my brain). Please do not tell me the analogies such as "the father of". That's only confusing me more! I learn best by seeing numerous examples, so I understand the pattern, and then I start understanding through reading. Thanks for any help/clarification, I really appreciate it.

"Let A={1,2,3,4}. Determine whether the relation is reflexive, irreflexive, symmetric, asymmetric, antisymmetric, or transitive.
R={(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)}"

1. Ok, I know (haha, think) this isn't reflexive. For that to be true, I'd need to see (1,1), (2,2), (3,3), or (4,4).

2. I know this is irreflexive as there are no identical pairs (e.g. no (2,2))

3. It is not symmetric from the get-go as the first ordered pair is (1,2) and there’s no (2,1)

4. It is not antisymmetric as there are no values (x,y),(y,x) where can draw the conclusion x=y (note: this one confuses me slightly, so I may be misinterpreting it)

5. It is asymmetric as there is neither symmetry nor asymmetry. (note: again, not sure I'm interpreting this one correctly)

6. It is transitive (the most confusing one for me). Okay, so if (x,y) and (y,z) we can conclude the relation (x,z). I see (1,2) and (2,3) as well as (1,3). That's transitive, right? Does that alone make it transitive, or does this have to hold true for other values as well?

Anyway, that's it. I got it wrong with no explanation why, so any clarification on relations would really help me out. Also, I've tried drawing this as a Digraph, and that doesn't help either. Hopefully someone can find a way to make me grasp this concept!!!!

stapel

Super Moderator
Staff member
First off, let me thank you for showing your work and reasoning so nicely!

"Let A={1,2,3,4}. Determine whether the relation is reflexive, irreflexive, symmetric, asymmetric, antisymmetric, or transitive.
R={(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)}"

1. Ok, I know (haha, think) this isn't reflexive. For that to be true, I'd need to see (1,1), (2,2), (3,3), or (4,4).
Correct.

2. I know this is irreflexive as there are no identical pairs (e.g. no (2,2))
Correct.

3. It is not symmetric from the get-go as the first ordered pair is (1,2) and there’s no (2,1)
Correct.

4. It is not antisymmetric as there are no values (x,y),(y,x) where can draw the conclusion x=y (note: this one confuses me slightly, so I may be misinterpreting it)
Actually, antisymmetry is the condition that, if you have (x, y) and (y, x) (that is, if you appear to have symmetry), it must be that x = y (so the symmetry is trivial). This would be like having the point (1, 1) in the set, with x = 1 and y = 1. Or, put a bit differently, antisymmetry is the property that, if you have (x, y) with x not equal to y, then you cannot have (y, x). (For asymmetry, you can't even have (y, x) if x = y.)

5. It is asymmetric as there is neither symmetry nor asymmetry. (note: again, not sure I'm interpreting this one correctly)
You say that it "is asymmetric as there is [not] asymmetry." I'm not sure what you're meaning here...?

To determine asymmetry, check for reverses of points. For instance, for the point (2, 1), there would have to not be (1, 2) for this to be asymmetric.

6. It is transitive (the most confusing one for me). Okay, so if (x,y) and (y,z) we can conclude the relation (x,z). I see (1,2) and (2,3) as well as (1,3). That's transitive, right? Does that alone make it transitive, or does this have to hold true for other values as well?
It has to hold for all elements for which you have (a, b) and (b, c). In your case, the pairs would be (1, 2) and (2, 3), (1, 2) and (2, 4), (1, 3) and (3, 4), and (2, 3) and (3, 4). Do all the pairs work? PTthenMath

New member
Whew...thanks for some help. Let's see where I'm at...

Antisymmetry: Ok, so it must appear symmetric before we can even consider antisymmetric. I think that clears it up.

Asymmetry: I meant it is asymmetric as there is neither symmetry nor ANTIsymmetry; typed too fast! Regardless, your explanation fixed my brain here.

Transitive....moment of truth? Yes, all the pairs work. (1,2) and (2,3) gives me (1,3) - that checks. (1,2) and (2,4) gives me (1,4) - that checks. (1,3) and (3,4) gives me (1,4) - that checks. And lastly, (2,3) and (3,4) gives me (2,4) - that checks. So, yes, transitive. P.S. If you say I'm wrong, I'll cry.

PTthenMath

New member
1. OK, so for antisymmetric, we must see symmetric first. As it's clearly not symmetric, we don't even need to go down that road, and can conclude antisymmetric.

2. Ah, I had a typo there, didn't I. Ok, so for asymmetry, there must be no instance of symmetry. Got it.

3. For the transitive, I see:
- (1,2) and (2,3), we also have (1,3) - that checks
- (1,2) and (2,4), we also have (1,4) - that checks
- (1,3) and (3,4), we also have (1,4) - that checks
- (2,3) and (3,4), we also have (2,4) - that checks

So, we can conclude antisymmetric, asymmetric, and transitive...right?

stapel

Super Moderator
Staff member
Antisymmetry: Ok, so it must appear symmetric before we can even consider antisymmetric. I think that clears it up.
No. To be antisymmetric, it must be that any symmetry is trivial. If you have (x, y) and (y, x), then x = y. If you don't have any apparent symmetry, then the set is trivially antisymmetric.

Transitive....moment of truth? Yes, all the pairs work. (1,2) and (2,3) gives me (1,3) - that checks. (1,2) and (2,4) gives me (1,4) - that checks. (1,3) and (3,4) gives me (1,4) - that checks. And lastly, (2,3) and (3,4) gives me (2,4) - that checks. So, yes, transitive.
Yes. PTthenMath

New member
Awesome. I think I got it.

I'll get back to some practice problems, and come crawling back if necessary.

Thanks!