# Reflexive = transistive relation?

1. May 9, 2010

### jwxie

Given A = {1,2,3}
R1 = {1,1 2,2 3,3}

I know it is reflexive, and I know it is symmetric. But what about its transitivity?

Def of transitive: a,b in R, b,c in R, then a,c is also in R

let a = 1
let b = 1
let c = 1

(1,1) and (1,1)
So yes, the book says it is an equivalence relation, so its transitivity is also valid.

therefore, are all reflexive relations transitive?

but what if the questions asks: constructs a reflexive and sysmmetric but not transitive?

I can do 11 22 33 and add 12 21 to make them both reflexive and symmetric. since i added 12 and 21, thus these 2 ordered pairs destroyed the transitivity?

2. May 9, 2010

### CRGreathouse

No. You need to construct a relation where (a, b) and (b, c) are in the relation, but (a, c) is not.

3. May 9, 2010

### jwxie

Well can't I make a = 1 b = 1 and c =1?

The book said 11 22 33 is an equivalence relations on A, so I am guessing that's how he did it.

4. May 10, 2010

### Landau

This is wrong. You can't just "take" a,b,c=1. You need to check that FOR ALL a,b,c in R the implication "if (a,b) and (b,c) are in R, then (a,c) is in R" is valid.

Btw, you need to be careful with dropping brackets.
R1 = {(1,1), (2,2), (3,3)} is correct,
R1 = {1,1 2,2 3,3} is nonsensical.

5. May 10, 2010

### Redbelly98

Staff Emeritus
No. See the second example here:
http://en.wikipedia.org/wiki/Equivalence_relation#Relations_that_are_not_equivalences
This is not even a relation! A relation is a collection of order pairs from A, a subset of $A\times A$. Did you mean {(1,1), (2,2), (3,3)}? If that is the case, then, yes it is transitive. Transitive means "if (a, b) and (b, c) are in the relation, then (a, c) is also in the relation. Here, in order that (a, b) and (b, c) be in the relation, we must have a= b= c. so that (a, b)= (a, c) is in the set.