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

R₂ on Z defined by (a, b) ∈ R₂⇔ |a – b| ≤ 5

Check for Reflexivity:

∀ a ∈ Z,

(a, a) ∈ R₂ needs to be proved for reflexivity.

If (a, b) ∈ R₂

Then, |a – b| ≤ 5 …(1)

So, for (a, a) ∈ R₁

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

|a – a| ≤ 5

⇒ 0 ≤ 5

⇒ (a, a) ∈ R₂

So, ∀ a ∈ Z, then (a, a) ∈ R₂

∴ R₂ is reflexive.

Check for Symmetry:

∀ a, b ∈ Z

If (a, b) ∈ R₂

We have, |a – b| ≤ 5 …(2)

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

|b – a| ≤ 5

Since, the value is in mod, |b – a| = |a – b|

⇒ The statement |b – a| ≤ 5 is true.

⇒ (b, a) ∈ R₂

So, if (a, b) ∈ R₂, then (b, a) ∈ R₂

∀ a, b ∈ Q₀

∴ R₁ is symmetric.

Check for Transitivity:

∀ a, b, c ∈ Z

If (a, b) ∈ R₂ and (b, c) ∈ R₂

⇒ |a – b| ≤ 5 and |b – c| ≤ 5

Since, inequalities cannot be added or subtract. We need to take example to check for,

|a – c| ≤ 5

Take values a = 18, b = 14 and c = 10

Check: |a – b| ≤ 5

⇒ |18 – 14| ≤ 5

⇒ |4| ≤ 5 is true.

Check: |b – c| ≤ 5

⇒ |14 – 10| ≤ 5

⇒ |4| ≤ 5

Check: |a – c| ≤ 5

⇒ |18 – 10| ≤ 5

⇒ |8| ≤ 5 is not true.

⇒ (a, c) ∉ R₂

So, if (a, b) ∈ R₂ and (b, c) ∈ R₂, then (a, c) ∉ R₁

∀ a, b, c ∈ Z

∴ R₂ is not transitive.