you can also see this in another way: Anti-symmetric is when all (a,b) are in the relation where a != b we must have (b,a) are not in the relation!. For example: Let R = { (1,3), (3,7), (7,1) } then to check if it's anti-symmetric or not, let's see "a" and "b". Obviously there aren't equal (1 != 3) so flip the order (1,3) -> (3,1), if you look at it there is no (3,1) and it applies same for (3,7) & (7,1) therefore we can say R relation is an anti-symmetric.
@@SawFinMath Here's what I don't understand regarding the example @ 7:45. Antisymmetric states (P^Q)->W. P=(a,b)eR, Q=(b,a)eR, W=(a=b), R={(a,b) | a > b}. (1,2) is not in the relational set because it does satisfy the property a > b. This means the hypothesis (P^Q) will always be false which should mean the implication is always true. Because the implication is ONLY false when the hypothesis is TRUE and the conclusion is FALSE. It seems like this example should be both transitive and anti-symmetric. =====UPDATE===== The answer to R={(a, b) | (a > b)} is both Transitive & Anti-Symmetric. See my above response as to the answer why!
At 9:20 I am still confused about how R_3 is anti-symmetric. I understand that 0 -> 1 is true. But I can't think of an example for R_3 where a = b, which confuses me.
"But I can't think of an example for R_3 where a = b, which confuses me." There is no example, because a > b is never true for a = b. 1 > 1 is false, 2 > 2 is false, pi > pi is false and so on... That's why the first part of the implication (a,b) € R3 ^ (b,a) € R3 is false for any values of a,b. Therefore, the implication is correct (0 -> 1).
Hi, in the last example, when we heck for the transitive property, why don't we test (3,3) and other pairs? Because if we find one counter example, it wouldnt be a transitive. And there is a counter example for (3,3).
Lets say our set was (1,1),(1,2),(2,1),(2,2),(3,1). NOTE when plugging in numbers you have to use existing sets, you cant plug in random values. If you use the transitive property on this then lets say that (a,b) = (3,1). What will be (b,c)? Well b has to be 1 from the (a,b). If b is 1, then the only sets available to test with we have are (1,1) and (1,2) Lets use (b,c) = (1,1). Here you'll see that (a,c) has to be (3,1) which works and thus the property is true. BUT if we use (b,c) to be (1,2), then youll see that (a,c) has to be (3,2), which does not exist in the set. Therefor we would have a contradiction causing this to be false. In the question the prof used, (a,b) = (3,3), and made (b,c) also (3,3), which means (a,c) has to also be (3,3). This causes the property to be true.
Hi@@derp3044 , not gonna lie its been a year since I did this so I have forgotten lol, but I just reviewed it and I think I can help. For R3 the condition is a > b. In your example you use 5 for A and 3 for B. This is valid so lets move on. Now lets check (b, c). This is where I am confused. You made B = 8? That's not possible. B is set to 3 from the first set. There is a error in your logic. That's why you are getting a weird result. I think what you meant was (5, 3) (3, x) where x is a number less than 3. That can be 0, 1 or 2. Lets assume its 2. So we have (5, 3), (3, 2). That means (a,c) is (5, 2), which still satisfies the condition! Hope that helps :)
What happen if the number that we looking for is not in relations? for example suppose my roster method is {(-2,4), (0,0), (2,4)} Basically we're looking for the number that it is end with 4. However there is no anything that start with 4 like{4,2} and so on is it okay if we can say that it is transitive?
irreflexive is the opposite of reflexive. if set A = {a, b, c, d} and B = { {a,a}, {b,b},{c,c}} then A is irreflexive and B is reflexive. however reflexive properties are diagonal and irreflexive are non-diagonal. Moreover, in order to understand asymmetric you first need to understand symmetric and anti-symmetric, so in symmetric if aRb then aRb = bRa. In terms of antisymmetric, It is the opposite of symmetric like if aRb then bRa should not be included in a set but there could be diagonal element (reflexive) like if aRb then there can also be aRa but not bRa In last asymmetric is also same like anti-symmetric but there is only one difference which is you cannot add reflexive (like if aRb then !(bRa | aRa) set in asymmetric set, whereas in anti-symmetric you can include reflexive properties. Note: this is what I understand and I might be wrong because Im still learning, so you can confirm all the points from any good teacher or google.
I was confused with anti-symmetric relations until I looked at the contrapositive. If a relation is antisymmetric, then for all a & b in A: If (a != b) then not ( ((a, b) is in R) and ((b, a) is in R) ). In other words, in an anti-symmetric relation, you can never have a pair of distinct a, b such that both (a, b) and (b, a) are in the relation.
I think the (3, 4) itself is not the element of the relation so you don't need to consider it in the problem because it didn't satisfy the initial requirement of a+b must be equal or less than 3. For example we have a set of numbers R = {0, 1, 2, 3, 4, 5} and we want to find the elements of relation where {(a,b) | a + b ≤ 3}. The elements will only be (0, 0) (0, 1) (1, 0) (1, 1) (0, 2) (2, 0) (1, 2) (2, 1) (0, 3) (3, 0). It shows the symmetric relations within the elements of the relation so the relation itself is symmetrical. The number (3, 4) didn't satisfy the initial requirements to be the elements of the relations where a+b must be equal or less than 3. Sorry if there is any mistakes in the explanation above, I hope it helps!
