Question

Show that the relation R in the set R of real numbers, defined as

R = {(a, b) : a ≤ b2}, is neither reflexive, nor symmetric, nor transitive.

Answer 1

(i) R is not reflexive ∵ a is not less than or equal to a² for all a ∈ R, e.g., \(\frac{1}{2}\) is not less than \(\frac{1}{4}\).
(ii) R is not symmetric, since if a ≤ b², then b is not less than or equal to a², e.g., 2 < 5² but 5 is not less than or equal to 2².
(iii) R is not transitive : Here, also if a ≤ b², b ≤ c², then a is not less than or equal to c², e.g., 2 < (- 2)², – 2 < (-1)². But 2 is not less than (-1)².

Question 3.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric and transitive.
Solution:
(i) R is not reflexive : a ≠ a + 1.
(ii) R is not symmetric : If b = a + 1, then a ≠ b + 1.
(iii) R is not transitive :If b = a + 1, c = b + 1, then c ≠ a + 1.