Answer to Question #311158 in Discrete Mathematics for Gurleen

Question

Give an example of relation which is Reflexive, anti symmetric & transitive but not symmetric

Answer 1

Let $\N$ be the set of all natural numbers. Let R be a relation defined on $\N$ as $a \text{R} b \iff a \text{ divides } b$.

Since a divides itself for all $a \in \N$, R is reflexive.

For if $a\text{R}b \text{ and } b\text{R}c$, then $a\text{R}c$.

To prove the above statement, we have$a\text{R}b \iff a \text{ divides } b \implies b = ma \text{ for some positive integer m}\\ b\text{R}c \iff b \text{ divides } c \implies c = nb \text{ for some positive integer n}\\ \text{Using } b = ma, \text{ we get } c = mna \implies a \text{ divides }c \implies a\text{R}c$

Hence R is transitive.

Now, if $a\text{R}b \text{ and } b\text{R}a$ then $b= ma \text{ and }a = nb \implies a = mna \iff mn = 1\iff m=n,$ since m and n are integers. This proves R is antisymmetric.

R is not symmetric, because $4\text{R}8$, but $8\not\text{R}4.$

Thus we have shown a relation R which is reflexive, transitive, antisymmetric but not symmetric.