# Relations R1, R2, R3 and R4 are defined on a set A = {a, b, c} as follows :R1 = {(a, a) (a, b) (a, c) (b, b) (b, c), (c, a) (c, b) (c, c)}R2 = {(a, a)}R3 = {(b, a)}R4 = {(a, b) (b, c) (c, a)}Find whether or not each of the relations R1, R2, R3, R4 on A is (i) reflexive (iii) symmetric (iii) transitive.

We have set,

A = {a, b, c}

Here, R1, R2, R3, 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.

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