Equivalance Relations Could some one check my work please?

AI Thread Summary
The discussion focuses on identifying equivalence classes and constructing matrices for given equivalence relations based on specified partitions. For partition A, the equivalence classes are correctly identified, while the matrix representation is explained as having "1" for equivalent members and "0" otherwise. In partition B, the matrix is confirmed to be accurate, highlighting the reflexive and symmetric properties of equivalence relations. A follow-up question about a different relation involving ordering is also addressed, with the matrix solution validated. The participants express satisfaction with the clarifications provided.
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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 don't 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.
 
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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

Code:
    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.
 
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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?
 
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.
 
Thank you now I see it.
 
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?
 
Yes (It won't let me post less than 10 letters!)
 
Kool I think I get it. Thank you
 
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