Question

The following relation is defined on the set of real numbers.

(i) aRb if a – b > 0

(ii) aRb iff 1 + ab > 0

(iii) aRb if |a| ≤ b. Find whether relation is reflexive, symmetric or transitive.

Answer 1

(i) Consider aRb if a – b > 0

Let us check for this relation whether it is reflexive, transitive and symmetric.

Reflexivity:

Let a be an arbitrary element of R.

Then, a ∈ R

But a − a = 0 ≯ 0

Thus, this relation is not reflexive.

Symmetry:

Let (a, b) ∈ R ⇒ a − b > 0

⇒ − (b − a) > 0 ⇒ b − a < 0

Therefore, the given relation is not symmetric.

Transitivity:

Let (a, b) ∈ R and (b, c) ∈ R.

Then, a − b > 0 and b − c > 0

Adding the two, we get a – b + b − c > 0

⇒ a – c > 0 ⇒ (a, c) ∈ R.

Clearly, the given relation is transitive.

(ii) Consider aRb iff 1 + a b > 0

Now for this relation we have to check whether it is reflexive, transitive and symmetric.

Reflexivity:

Let a be an arbitrary element of R.

Then, a ∈ R

⇒ 1 + a × a > 0

i.e. 1 + a² > 0 [Since, square of any number is positive]

So, the given relation is reflexive.

Symmetry:

Let (a, b) ∈ R ⇒ 1 + ab > 0

⇒ 1 + ba > 0 ⇒ (b, a) ∈ R

So, the given relation is symmetric.

Transitivity:

Let (a, b) ∈ R and (b, c) ∈ R

⇒1 + ab > 0 and 1 + bc >0

But 1 + ac ≯ 0 ⇒ (a, c) ∉ R

Thus, the given relation is not transitive.

(iii) Consider aRb if |a| ≤ b.

Let us check for this relation whether it is reflexive, transitive and symmetric.

Reflexivity:

Let a be an arbitrary element of R.

Then, a ∈ R [Since, |a| = a] ⇒ |a| ≮ a

Clearly, R is not reflexive.

Symmetry:

Let (a, b) ∈ R ⇒ |a| ≤ b ⇒ |b| ≰ a for all a, b ∈ R

⇒ (b, a) ∉ R

Thus, R is not symmetric.

Transitivity:

Let (a, b) ∈ R and (b, c) ∈ R

⇒ |a| ≤ b and |b| ≤ c

Multiplying the corresponding sides, we get

|a| × |b| ≤ b c ⇒ |a| ≤ c ⇒ (a, c) ∈ R

So, R is transitive.