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

1.	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» We have R = {(a,b) :a ≤ b²} , where a, b ∈ R For reflexivity, we observe that 1/2 ≤ (1/2)² is not true. So, R is not reflexive as (1/2, 1/2) ∉ R For symmetry, we observe that − 1 ≤ 3² but 3 > (−1)² ∴ (−1, 3) ∈ R but (3, −1) ∉R. So, R is not symmetric. For transitivity, we observe that 2 ≤ (−3)² and −3 ≤ (1)² but 2 > (1)² ∴ (2, − 3) ∈ R and (−3,1) ∈ R but (2, 1) ∉ R. So, R is not transitive. Hence, R is neither reflexive, nor symmetric and nor transitive.

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»

We have R = {(a,b) :a ≤ b²} , where a, b ∈ R

For reflexivity, we observe that 1/2 ≤ (1/2)² is not true.

So, R is not reflexive as (1/2, 1/2) ∉ R

For symmetry, we observe that − 1 ≤ 3² but 3 > (−1)²

∴ (−1, 3) ∈ R but (3, −1) ∉R.

So, R is not symmetric.

For transitivity, we observe that 2 ≤ (−3)² and −3 ≤ (1)² but 2 > (1)²

∴ (2, − 3) ∈ R and (−3,1) ∈ R but (2, 1) ∉ R. So, R is not transitive.

Hence, R is neither reflexive, nor symmetric and nor transitive.