Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$R = \{ (1,3),(2,3), (3,4) \}$ It says this $R$ is anti-symmetric but how, $a$ does not equal $b$

$R = \{ (1,1),(2,2),(3,3),(4,4) \}$ It says this is Reflexive (yup) Symmetric (?), Anti-Symmetric (?), and Transitive (?), and I was wondering how this makes sense.

If you draw a graph to represent these I don't know how its true. Maybe the solutions is messed?

Btw, it's set of relations on the set $A = \{ 1,2,3,4 \}$

share|improve this question
1  
Anti-symmetric just means you can't have both $(a,b)$ and $(b,a)$, right? –  Gerry Myerson Nov 12 '12 at 0:25
1  
No, it's if u have both (a,b)and(b,a) then => a=b –  Aaron Nov 12 '12 at 0:28
1  
OK, then what's the difficulty? There are no $a$ and $b$ such that $R$ has both $(a,b)$ and $(b,a)$, so the "then" part never eventuates, right? –  Gerry Myerson Nov 12 '12 at 1:04
    
Hey you are right! Sorry!! –  Aaron Nov 12 '12 at 1:05
add comment

1 Answer

up vote 2 down vote accepted

Edit: Now that you have changed the first problem, it is actually anti-symmetric. Why? Because being anti-symmetric means obeying an if-statement that says if you have $(x,y)$ and $(y,x)$, then $x = y$. Here, you don't have anything satisfying the if part of this conditional; thus, we say the if-then statement (vacuously) holds.

As for the second example, this is certainly symmetric and transitive: both of those properties are if-then statements, and here the "if" part can only be satisfied by ordered pairs of the form $(n,n)$. In fact, this relation is a formal way of writing what we would normally represent with "$=$", that is, equality in the standard sense of arithmetic, which you probably know is an equivalence relation.

Thinking of the second relation as equality, hopefully you can see why anti-symmetry holds too, though it does so in a somewhat vacuous manner (as in the first example).

share|improve this answer
    
Sorry, check my edit for the first example, I wrote down the wrong question –  Aaron Nov 12 '12 at 0:23
    
Wait I'm not sure I understood that. So if there is no (a,b)and(b,a) in a relation it's always anti symmetric? That is really vague –  Aaron Nov 12 '12 at 0:38
    
In general, given a statement of the form "if P, then Q", we say it's false only when P is satisfied but Q is not. If P is not satisfied, then we don't really care about Q; we just say the statement is true. (In such a scenario, one might say it's "vacuously" true.) That's what's happening here: you have if-then statements where the if part is not satisfied, so the then part isn't important. [Hopefully this all sounds familiar and you aren't seeing it here for the first time!] –  Benjamin Dickman Nov 12 '12 at 0:38
    
Yea I understand because in logic if you have p 0 it's always true regardless the value of Q. My real question is then in every relation there is no (a,b)and(b,a) is it always antisymmetric –  Aaron Nov 12 '12 at 0:40
    
Yes, if for every (a,b) you don't have (b,a), then the relation is antisymmetric. Note: if for every (a,b) you do have (b,a), then the relation is symmetric! Hopefully you can now see how the word "antisymmetric" arises. –  Benjamin Dickman Nov 12 '12 at 0:42
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.