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$R = \{ (1,3),(2,3), (3,4) \}$ 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 $A = \{ 1,2,3,4 \}$

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Anti-symmetric just means you can't have both $(a,b)$ and $(b,a)$, right? – Gerry Myerson Nov 12 '12 at 0:25
No, it's if u have both (a,b)and(b,a) then => a=b – Aaron Nov 12 '12 at 0:28
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
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).

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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

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