Q. 115.0( 4 Votes )

On the set S of all real numbers, define a relation R = {(a, b) : a ≤ b}.

Show that R is (i) reflexive (ii) transitive (iii) not symmetric.

Answer :

Let R = {(a, b) : a ≤ b} be a relation defined on S.


Now,


We observe that any element x S is less than or equal to itself.


(x,x) R x S


R is reflexive.


Let (x,y) R x, y S


x is less than or equal to y


But y cannot be less than or equal to x if x is less than or equal to y.


(y,x) R


For e.g. , we observe that (2,5) R i.e. 2 < 5 but 5 is not less than or equal to 2 (5,2) R


R is not symmetric


Let (x,y) R and (y,z) R x, y ,z S


x ≤ y and y ≤ z


x ≤ z


(x,z) R


For e.g. , we observe that


(4,5) R 4 ≤ 5 and (5,6) R 5 ≤ 6


And we know that 4 ≤ 6 (4,6) R


R is transitive.


Thus, R is reflexive and transitive but not symmetric.


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