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
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.
PREVIOUS
NEXT
Try our Mini Course
Master Important Topics in 7 Days
Learn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
VIEW ALL COURSES
RELATED QUESTIONS :
Fill in theMathematics – Exemplar
State True Mathematics – Exemplar
State True Mathematics – Exemplar
State True Mathematics – Exemplar
Let A = {1, 2, 3}Mathematics – Exemplar
Show that the relMathematics – Board Papers
Let N denote the Mathematics – Board Papers
POPULAR QUESTIONS :
Q
Which of the prodChemistry – Exemplar 816 views
Q
Do you think apomBiology – Board Papers 863 views
Q
“The Volga is oneGeography – Board Papers 182 views
Goprep is better on the App
Never miss a class, open this in Goprep App to get full experience
OPEN IN APP