- #1
mr_coffee
- 1,629
- 1
Binary Relations--Transitivity
Hello everyone I'm not sure if I'm hitting all the ordered pairs on this because i can't seem to find a good method to figure it out. I also think the relation to itself is confusing me.
here is the question:
S is a binary relation defined on A = {0,1,2,3}
to speed up this message, I'm going to represent an ordered pair like 00 instead of (0,0).
Let S = {00, 03, 10, 12, 20, 32}
Find S^t, the transititve closure of S.
note: ... 32 .. stands for (3,2), etc
Okay i know for somthing to be transitive, if you have an arrow from point 0 to 0, then it has a relation to itself..
Also
If i had say a relation like (0,1) and (1,2) then I can make an ordered pair from (0,2).
Thats what I'm doing in this problem
So i got the following but I'm not sure if I'm right.
S^t = S UNION { 11, 33, 22, 30, 01, 02 13, 31}
Thanks.
Hello everyone I'm not sure if I'm hitting all the ordered pairs on this because i can't seem to find a good method to figure it out. I also think the relation to itself is confusing me.
here is the question:
S is a binary relation defined on A = {0,1,2,3}
to speed up this message, I'm going to represent an ordered pair like 00 instead of (0,0).
Let S = {00, 03, 10, 12, 20, 32}
Find S^t, the transititve closure of S.
note: ... 32 .. stands for (3,2), etc
Okay i know for somthing to be transitive, if you have an arrow from point 0 to 0, then it has a relation to itself..
Also
If i had say a relation like (0,1) and (1,2) then I can make an ordered pair from (0,2).
Thats what I'm doing in this problem
So i got the following but I'm not sure if I'm right.
S^t = S UNION { 11, 33, 22, 30, 01, 02 13, 31}
Thanks.