Question -
Answer -
Given R = {(a, b): a ≤ b3}
It is observed that(1/2, 1/2) in R as 1/2 > (1/2)3 = 1/8
∴ R is not reflexive.
Now,
(1, 2) ∈ R (as 1 < 23 = 8)
But,
(2, 1) ∉ R (as 2 > 13 = 1)
∴ R is not symmetric.
We have (3, 3/2),(3/2, 6/5) in “R as” 3 < (3/2)3 and 3/2 < (6/5)3
But (3,6/5) ∉ R as 3 > (6/5)3
∴ R is not transitive.
Hence, Ris neither reflexive, nor symmetric, nor transitive.