We have set,

A = {a, b, c}

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

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 = {(a, a) (a, b) (a, c) (b, b) (b, c) (c, a) (c, b) (c, c)}

(i). Reflexive:

For all a, b, c ∈ A. [∵ A = {a, b, c}]

Then, (a, a) ∈ R₁

(b, b) ∈ A

(c, c) ∈ A

[∵ R₁ = {(a, a) (a, b) (a, c) (b, b) (b, c) (c, a) (c, b) (c, c)}]

So, ∀ a, b, c ∈ A, then (a, a), (b, b), (c, c) ∈ R.

∴ R₁ is reflexive.

(ii). Symmetric:

If (a, a), (b, b), (c, c), (a, c), (b, c) ∈ R₁

Then, clearly (a, a), (b, b), (c, c), (c, a), (c, b) ∈ R₁

∀ a, b, c ∈ A

[∵ R₁ = {(a, a) (a, b) (a, c) (b, b) (b, c) (c, a) (c, b) (c, c)}]

But, we need to try to show a contradiction to be able to determine the symmetry.

So, we know (a, b) ∈ R₁

But, (b, a) ∉ R₁

So, if (a, b) ∈ R₁, then (b, a) ∉ R₁.

∀ a, b ∈ A

∴ R₁ is not symmetric.

(iii). Transitive:

If (b, c) ∈ R₁ and (c, a) ∈ R₁

But, (b, a) ∉ R₁ [Check the Relation R₁ that does not contain (b, a)]

∀ a, b ∈ A

[∵ R₁ = {(a, a) (a, b) (a, c) (b, b) (b, c) (c, a) (c, b) (c, c)}]

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

∀ a, b, c ∈ A

∴ R₁ is not transitive.

Now, we have

R₂ = {(a, a)}

(i). Reflexive:

Here, only (a, a) ∈ R₂

for a ∈ A. [∵ A = {a, b, c}]

[∵ R₂ = {(a, a)}]

So, for a ∈ A, then (a, a) ∈ R₂.

∴ R₂ is reflexive.

(ii). Symmetric:

For symmetry,

If (x, y) ∈ R, then (y, x) ∈ R

∀ x, y ∈ A.

Notice, in R₂ we have

R₂ = {(a, a)}

So, if (a, a) ∈ R₂, then (a, a) ∈ R₂.

Where a ∈ A.

∴ R₂ is symmetric.

(iii). Transitive:

Here,

(a, a) ∈ R₂ and (a, a) ∈ R₂

Then, obviously (a, a) ∈ R₂

Where a ∈ A.

[∵ R₂ = {(a, a)}]

So, if (a, a) ∈ R₂ and (a, a) ∈ R₂, then (a, a) ∈ R₂, where a ∈ A.

∴ R₂ is transitive.

Now, we have

R₃ = {(b, a)}

(i). Reflexive:

∀ a, b ∈ A [∵ A = {a, b, c}]

But, (a, a) ∉ R₃

Also, (b, b) ∉ R₃

[∵ R₃ = {(b, a)}]

So, ∀ a, b ∈ A, then (a, a), (b, b) ∉ R₃

∴ R₃ is not reflexive.

(ii). Symmetric:

If (b, a) ∈ R₃

Then, (a, b) should belong to R₃.

∀ a, b ∈ A. [∵ A = {a, b, c}]

But, (a, b) ∉ R₃

[∵ R₃ = {(b, a)}]

So, if (a, b) ∈ R₃, then (b, a) ∉ R₃

∀ a, b ∈ A

∴ R₃ is not symmetric.

(iii). Transitive:

We have (b, a) ∈ R₃ but do not contain any other element in R₃.

Transitivity can’t be proved in R₃.

[∵ R₃ = {(b, a)}]

So, if (b, a) ∈ R₃ but since there is no other element.

∴ R₃ is not transitive.

Now, we have

R₄ = {(a, b) (b, c) (c, a)}

(i). Reflexive:

∀ a, b, c ∈ A [∵ A = {a, b, c}]

But, (a, a) ∉ R₄

Also, (b, b) ∉ R₄ and (c, c) ∉ R₄

[∵ R₄ = {(a, b) (b, c) (c, a)}]

So, ∀ a, b, c ∈ A, then (a, a), (b, b), (c, c) ∉ R₄

∴ R₄ is not reflexive.

(ii). Symmetric:

If (a, b) ∈ R₄, then (b, a) ∈ R₄

But (b, a) ∉ R₄

[∵ R₄ = {(a, b) (b, c) (c, a)}]

So, ∀ a, b ∈ A, if (a, b) ∈ R₄, then (b, a) ∉ R₄.

⇒ R₄ is not symmetric.

It is sufficient to show only one case of ordered pairs violating the definition.

∴ R₄ is not symmetric.

(iii). Transitivity:

We have,

(a, b) ∈ R₄ and (b, c) ∈ R₄

⇒ (a, c) ∈ R₄

But, is it so?

No, (a, c) ∉ R₄

So, it is enough to determine that R₄ is not transitive.

∀ a, b, c ∈ A, if (a, b) ∈ R₄ and (b, c) ∈ R₄, then (a, c) ∉ R₄.

∴ R₄ is not transitive.