Math, asked by prasanna264, 1 month ago

Let A = { 1,2,3 }. Define the relation R1, R2, R3 and R4 on A as
(i) R1 = { (1,1) ,(1,2) ,(2,1), (2,2) }
(ii) R2 = { (1,1) ,(2,2) , (3,3) ,(1,2) ,(2,3) ,(1,3) }
(iii) R3 = { (1,1) ,(2,2) ,(3,3) ,(1,2) , (2,1) ,(1,3) ,(3,1) }
(iv) R4 = { (1,1) ,(2,2) ,(3,3) ,(1,2) ,(2,1) }
Check whether R1, R2, R3 and R4 are equivalence relations or not. If yes, then
find the equivalence classes of all the elements of A

Answers

Answered by satishkumarpura222

Answer:

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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