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ptex
Apr13-04, 09:55 AM
Question, List the members of the equivalance relation on {1,2,3,4,5} by the given partition. Identify the equivalance classes
A) {(1,2,3),(4,5)}
B) {(1),(2,4),(5,3)}

My solution is;
A) {(1,1),(1,2),(1,3),(2,2),(2,1),(2,3),(3,1),(3,2),( 3,3),(4,4),(4,5),(5,4),(5,5)}

B) {(1),(2,2),(2,4),(4,2),(4,4),(3,3),(3,5),(5,5),(5, 3)}

Then the next qusetion for which I dont know where to begin is;
For the above (A and B) find the matrix of the relation from X to X. Show the ordering that you are using :confused:

Any help would be greatly appreciated.

HallsofIvy
Apr13-04, 10:50 AM
What you have is correct. For the "matrix" question, do this:
List all members of X across the top and vertically on the left on your paper. For each intersection, IF the member of X on the left IS equivalent to the member on the top, write "1", otherwise write "0".

For B (the easier of the two) this is


1 2 3 4 5
1 1 0 0 0 0
2 0 1 0 1 0
3 0 0 1 0 1
4 0 1 0 1 0
5 0 0 1 0 1
The matrix is
[1 0 0 0 0]
[0 1 0 1 0]
[0 0 1 0 1]
[0 1 0 1 0]
[0 0 1 0 1]

The diagonal is all 1s because an equivalence relation is reflexive and the matrix is symmetric because an equivalence relation is symmetric. The number of 1s in each row and column is the number of elements equivalent to that member.

ptex
Apr13-04, 11:05 AM
Thank you but if its not too much trouble why is it that 1 and 2 as in the first "1" is equivalent but not 1 and 3?

philosophking
Apr13-04, 10:20 PM
First of all his horizontal row is just a space off, 1 is not related to 2 in example B. Just like 1 is not related to 3 in example B. He didn't do example A.

ptex
Apr14-04, 11:29 AM
Thank you now I see it.

ptex
Apr14-04, 05:25 PM
I long the same lines could some one check this?

R={(x,y)|x<y};ordering of X:1,2,3,4

My solution is;
? 1 2 3 4
1 0 1 1 1
2 0 0 1 1
3 0 0 0 1
4 0 0 0 0

[0 1 1 1]
[0 0 1 1]
[0 0 0 1]
[0 0 0 0]

OK?

HallsofIvy
Apr15-04, 05:32 AM
Yes (It wont let me post less than 10 letters!)

ptex
Apr15-04, 08:08 AM
Kool I think I get it. Thank you