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.

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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