# matrix relation of sets. symmetric, antisymmetric,reflexive,transitive

by sapiental
Tags: antisymmetric, matrix, reflexive, relation, sets, symmetric, transitive
 P: 120 1. The problem statement, all variables and given/known data relation A = {a,b,c} for the following matrix [1,0,0;1,1,0;0,1,1] is it reflexive, transitive, symmetric, antisymmetric 2. Relevant equations ordered pairs. 3. The attempt at a solution i wrote the ordered pairs as (a,a),(b,a),(b,b),(c,b),(c,c) I only that it is reflexive for a,a b,b and c,c also it is antisymmetric because there are no edges in opposite directions between distinct verticies. am I missing anything. thanks!
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P: 38,412
 Quote by sapiental 1. The problem statement, all variables and given/known data relation A = {a,b,c} for the following matrix [1,0,0;1,1,0;0,1,1] is it reflexive, transitive, symmetric, antisymmetric 2. Relevant equations ordered pairs. 3. The attempt at a solution i wrote the ordered pairs as (a,a),(b,a),(b,b),(c,b),(c,c) I only that it is reflexive for a,a b,b and c,c also it is antisymmetric because there are no edges in opposite directions between distinct verticies. am I missing anything. thanks!
I don't know what you mean by "reflexive for a,a b,b and c,c. A relation is reflexive if and only if it contains (x,x) for all x in the base set. Since only a, b, and c are in the base set, and the relation contains (a,a), (b,b), and (c,c), yes, it is reflexive.

To be symmetric, since it contains (b,a) it would have to contain (a,b) and it doesn't: not symmetric. Since it does NOT contain (a,b) or (b,c), yes, it is anti-symmetric.

What about transitive? A relation is transitive if and only if, whenever (x,y) and (y,z) are in the relation, so is (x,z). Can you find pairs so that is NOT true?
 P: 120 Hey, thanks for the reply! I didn't put parenthesis around the ordered pairs (a,a),(b,b),(c,c) for the first problem, sorry. I don't think it's transitive since we have (c,b) and (b,a), and it doesn't contain (c,a). How does that sound? Thanks
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