# If a relation is reflexive is it symmetric and transitive?

If a relation is reflexive is it symmetric and transitive ?

let ~ means " in relation with "

if A is a set , ~ is a relation on $A$, prove that:

if $a$~$a$ for any $a$ $\in$ A then

1- $x$~$y$ $\rightarrow$ $y$~$x$

2- $x$ ~$y$ , $y$ ~ $z$ $\rightarrow$ x~z

if this is wrong , give an example to a reflexive relation which is not transitive or symmetric

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No, you prove it – Ethan May 1 '13 at 10:40
@Ethan: That is nasty. But I approve. To elaborate a bit, reflexivity, symmetry, and transitivity are the three axioms which define an equivalence relation. Why are there three and not just one? – Harald Hanche-Olsen May 1 '13 at 10:41
@Ethan , if it not right , can you give plz an example ? – Maths Lover May 1 '13 at 10:43
Try it with a set of just two elements. There are only four possible reflexive relations on it, so that's not a lot of work. – Harald Hanche-Olsen May 1 '13 at 10:44
@Maths Lover : the 3 conditions are indendendent of each other. There are eight possibilities (whether a relation is/isn't reflexive, transitive, symmmetric), and there are simple examples of all 8. – Stefan Smith May 1 '13 at 10:50

Let $A=\{a,b,c,d,e\}$ and $$R=\{(a,a),(b,b),(c,c),(d,d),(e,e),(c,e),(e,b)\}$$

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Or somewhat more economically: Let $A=\{a,b,c\}$ and let $R = \{(a,a),(b,b),(c,c),(a,b),(b,c)\}$ – Abel May 1 '13 at 10:51
@Abel: Oh yes! These days around the world, we should be so care about even an alphabet. It is very valuable. Thanks :D – Babak S. May 1 '13 at 10:56
@Abel : I am stealing your answer and posting it as my own if you dont, before Babak modifies his answer to yours! :) – Arjang May 1 '13 at 11:01
@Arjang Why? It's Babak's answer, just condensed a bit. – Abel May 1 '13 at 11:02
@Arjang: Do you wanna to post an answer like Abel pointed here? I can remove mine and then you can do that. :-) I's absolutely OK. – Babak S. May 1 '13 at 11:05

The relation of divisibility, in any ring with $1$, is an example of reflexive, transitive, but non symmetric relation

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can you plz define the relation of divisiblity ? you mean ,if R is a ring , $x$~$y$ iff $xy = 1$ ? where x,y $\in$ R – Maths Lover May 1 '13 at 13:20
@MathsLover $x$~$y$ iff $\exists a$ such that $y=xa$ – Federica Maggioni May 1 '13 at 13:23
$\le$ is a simple example of a relation that is reflexive and transitive, but not symmetric. – Scott Jul 1 '14 at 18:03

For this one time, I'll help you out... but indeed, next time maybe study the problem a bit more before asking for an answer ;-)

Take for A the set $A = \{ 1,2 \}$ and consider the relation ~ defined by 1 ~ 1, 2 ~ 2, and 1 ~ 2.

This relation satisfies a ~ a for any $a \in A$ (please check that!), but it does not satisfy 1) because 1 ~ 2 holds while 2~1 does not (this is because you should fill in the same element for `a' on both sides)...

Similarly, consider the same set but now with a relation ~ definded by 1 ~ 1, 2 ~ 2, 1~2 and 2~1. This relation does satisfy 1) but it does not satisfy 2) since 1~2 and 2~1 but 1 $\neq$ 2 (can you see why this is so?).

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Your second example is transitive, since $x\sim z$ for all $x,z\in\{1,2\}$. In fact, there are no non-transitive reflexive relations on $\{1,2\}$. – Abel May 1 '13 at 11:00