Q. 105.0( 5 Votes )

Let R = {(a, b) : a = b2} for all a, b N.

Show that R satisfies none of reflexivity, symmetry and transitivity.

Answer :

We have, R = {(a, b) : a = b2} relation defined on N.


Now,


We observe that, any element a N cannot be equal to its square except 1.


(a,a) R a N


For e.g. (2,2) R 2 ≠ 22


R is not reflexive.


Let (a,b) R a, b N


a = b2


But b cannot be equal to square of a if a is equal to square of b.


(b,a) R


For e.g., we observe that (4,2) R i.e 4 = 22 but 2 ≠ 42 (2,4) R


R is not symmetric


Let (a,b) R and (b,c) R a, b,c N


a = b2 and b = c2


a ≠ c2


(a,c) R


For e.g., we observe that


(16,4) R 16 = 42 and (4,2) R 4 = 22


But 16 ≠ 22


(16,2) R


R is not transitive.


Thus, R is neither reflexive nor symmetric nor transitive.


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