# Equivalence Relations

1. Nov 1, 2008

### vvvidenov

1. The problem statement, all variables and given/known data
Find relations that satisfying
just Reflexive
just Symmrtic
just Transitive

(R) & (S), but not (T)
(R) & (T), but not (S)
(S) & (T), but not (R)

2. Relevant equations
S=Z

(a,b) $$\in$$R if <=> a>b (T) but, not (S) & (R).

the ex is given in the class, but nothing else was explained. I am confused. Please help if you can.

3. The attempt at a solution

(1,1) (1,2) (1,3)
(2,1) (2,2) (2,3)
(3,1) (3,2) (3,3)

for (1,1) we have equiv. rel (R, S & T)

2. Nov 1, 2008

### vvvidenov

The problem above is first course of Abstarct Algebra. We use textbook (Dan Saracino, second edition).

3. Nov 1, 2008

### Office_Shredder

Staff Emeritus
(a,b) is in R iff a>b. Then (a,a) is not in R as a>a is false. (a,b) in R means (b,a) is not in R as if a>b, b>a is false. And if (a,b) and (b,c) are in R, a>b>c implies a>c so (a,c) is in R. Note in this example (1,1) is NOT in R as 1>1 isn't true.

I think a more intuitive notation is aRb to denote 'a is related to b' where R is a relation. Then aRb iff a>b it's clear that if aRb is true, bRa is false since a>b implies b>a is false. Similarly, aRa is false since a>a is false. But aRb, bRc implies a>b, b>c so a>b>c implies a>b and aRc holds. hence > is a relation which is transitive but not symmetric or reflexive.

Try to think up other relations that satisfy, say, aRa but not aRb -> bRa or aRb, bRc -> aRc

4. Nov 1, 2008

### vvvidenov

aRb iff a-b$$\geq$$0
not (S), but is (T) and (R)

5. Nov 1, 2008

### vvvidenov

I can't think of example that is only (S) , and only (T). I don't think I understand the differences og being only (R,) only (S), only (T), and when we have (T), but not (R) & (S) and so on.

6. Nov 1, 2008

### vvvidenov

Thank you so much. I found similar questions with answers and I get it now.