714 views
1. Mr. X claims the following:

If a relation R is both symmetric and transitive, then R is reflexive. For this, Mr. X offers the following proof:

“From xRy, using symmetry we get yRx. Now because R is transitive xRy and yRx together imply xRx. Therefore, R is reflexive”.

2. Give an example of a relation R which is symmetric and transitive but not reflexive.
edited | 714 views
0

Ex: empty relation(no ordered pairs).

https://gateoverflow.in/847/gate2002-2-17

Let set $A$ be ${1,2,3}$, and let a relation $R$ on $A$ be

$\left \{ (1,1),(1,2),(2,1),(2,2) \right \}$

$R$ is both symmetric and transitive, but not reflexive. The key point here is that there may be some element in set A which is not related to any of the element in $R$, but to be reflexive, all elements must be related to themselves.
edited
+1
R={(1,1),(1,2),(2,1),(2,2)}
0
0
A={1,2,3,4}

R={(1,2),(2,1),(2,3),(3,2)} is this example correct?
0

@Anshul Shankar  your example is not correct because it is not transitive (1,2),(2,1) are in R but (1,1)  is not in R, also (2,3),(3,2) are in R but (2,2) is not in R.

Another Example A = {1},

Relation on A , R = {}

This relation R is symmetric & Transitive but not Reflexive.
edited
0
Did you mean empty relation?
0
Yes! he
Reflexivity of a relation R is to ensure each element of the set on which a relation is defined must be related to itself. Since the relation R is not known to be reflexive, then for some element z in A we can't be sure of whether (z,z)∊R even if (x.x) ∊R as implied by (x,y) and (y,x). So this is the flaw in Mr. X's proof.

Quintessential example:

Let R be a relation defined on a set A={1,2,3} as follows

R={(1,1)(1,2)(2,1)(2,2)}.This relation is both symmetric as well as transitive but not reflexive.Why? This is simply because (3,3) does not belong to the relation R.

1
2