Question

Test whether the following relations R₁, R₂ and R₃ are (i) reflexive (ii) symmetric and (iii) transitive :

R₁ on Q₀ defined by (a, b) ϵ R₁⇔ a = 1/b

Answer 1

Here, R₁, R₂, R_3, and R₄ are the binary relations.

So, recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

So, using these results let us start determining given relations.

We have

R₁ on Q₀ defined by (a, b) ∈ R₁⇔ a = 1/b

Check for Reflexivity:

∀ a, b ∈ Q₀,

(a, a), (b, b) ∈ R₁ needs to be proved for reflexivity.

If (a, b) ∈ R₁

Then, a = 1/b …(1)

So, for (a, a) ∈ R₁

Replace b by a in equation (1), we get

a = 1/a

But, we know

a ≠ 1/a

⇒ (a, a) ∉ R₁

So, ∀ a ∈ Q₀, then (a, a) ∉ R₁

∴ R₁ is not reflexive.

Check for Symmetry:

If (a, b) ∈ R₁

Then, (b, a) ∈ R₁

∀ a, b ∈ Q₀

If (a, b) ∈ R₁

We have, a = 1/b …(2)

Now, for (b, a) ∈ R₁

Replace a by b & b by a in equation (2), we get

b = 1/a

⇒ (b, a) ∈ R₂

So, if (a, b) ∈ R₁, then (b, a) ∈ R₁

∀ a, b ∈ Q₀

∴ R₁ is symmetric.

Check for Transitivity:

If (a, b) ∈ R₁ and (b, c) ∈ R₁

a = 1/b and b = 1/c

We need to eliminate b.

We have

a = 1/b

b = 1/a

Putting b = 1/a in b = 1/c, we get

1/a = 1/c

a = c

But, a ≠ 1/c

⇒ (a, c) ∉ R₁

So, if (a, b) ∈ R₁ and (b, c) ∈ R₁, then (a, c) ∉ R₁

∀ a, b, c ∈ Q₀

∴ R₁ is not transitive.