# Let A = {1, 2, 3}

We have been given,

A = {1, 2, 3}

Here, R1, R2, and R3 are the binary relations on 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.

Let us take R1.

R1 = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}

(i). Reflexive:

1, 2, 3 A [ A = {1, 2, 3}]

(1, 1) R1

(2, 2) R2

(3, 3) R3

So, for a A, (a, a) R1

R1 is reflexive.

(ii). Symmetric:

1, 2, 3 A

If (1, 3) R1, then (3, 1) R1

[ R1 = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}]

But if (2, 1) R1, then (1, 2) R1

[ R1 = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}]

So, if (a, b) R1, then (b, a) R1

a, b A

R1 is not symmetric.

(iii). Transitivity:

1, 2, 3 A

If (1, 3) R1 and (3, 3) R1

Then, (1, 3) R1

[ R1 = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}]

But, if (2, 1) R1 and (1, 3) R1

Then, (2, 3) R1

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

a, b, c A

R1 is not transitive.

Now, take R2.

R2 = {(2, 2), (3, 1), (1, 3)}

(i). Reflexive:

1, 2, 3 A [ A = {1, 2, 3}]

(1, 1) R2

(2, 2) R2

(3, 3) R2

So, for a A, (a, a) R2

R2 is not reflexive.

(ii). Symmetric:

1, 2, 3 A

If (1, 3) R2, then (3, 1) R2

[ R2 = {(2, 2), (3, 1), (1, 3)}]

If (2, 2) R2, then (2, 2) R2

[ R2 = {(2, 2), (3, 1), (1, 3)}]

So, if (a, b) R2, then (b, a) R2

a, b A

R2 is symmetric.

(iii). Transitivity:

1, 2, 3 A

If (1, 3) R2 and (3, 1) R2

Then, (1, 1) R2

[ R2 = {(2, 2), (3, 1), (1, 3)}]

So, if (a, b) R2 and (b, c) R2, then (a, c) R2

a, b, c A

R2 is not transitive.

Now take R3.

R3 = {(1, 3), (3, 3)}

(i). Reflexive:

1, 3 A [ A = {1, 2, 3}]

(1, 1) R3

(3, 3) R3

So, for a A, (a, a) R3

R3 is not reflexive.

(ii). Symmetric:

1, 3 A

If (1, 3) R3, then (3, 1) R3

[ R3 = {(1, 3), (3, 3)}]

So, if (a, b) R3, then (b, a) R3

a, b A

R3 is not symmetric.

(iii). Transitivity:

1, 3 A

If (1, 3) R3 and (3, 3) R3

Then, (1, 3) R3

[ R3 = {(1, 3), (3, 3)}]

So, if (a, b) R3 and (b, c) R3, then (a, c) R3

a, b, c A

R3 is transitive.

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