Let N be the set of all natural numbers. Let R be a relation defined on N as aRb⟺a divides b.
Since a divides itself for all a∈N, R is reflexive.
For if aRb and bRc, then aRc.
To prove the above statement, we haveaRb⟺a divides b⟹b=ma for some positive integer mbRc⟺b divides c⟹c=nb for some positive integer nUsing b=ma, we get c=mna⟹a divides c⟹aRc
Hence R is transitive.
Now, if aRb and bRa then b=ma and a=nb⟹a=mna⟺mn=1⟺m=n, since m and n are integers. This proves R is antisymmetric.
R is not symmetric, because 4R8, but 8R4.
Thus we have shown a relation R which is reflexive, transitive, antisymmetric but not symmetric.
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