Show that the relation R in the set A = { 1, 2, 3, 4, 5 } given by
R = { (a, b) : | a – b | is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.
A = {1, 2, 3, 4, 5}
R = {(a, b) : | a – b | is even}
Now | a – a | = 0 is an even number,
∴ (a, a) ∈ R ∀ a ∈ A
⇒ R is reflexive.
Again (a, b) ∈ R
⇒ | a – b | is even ⇒ | – (b – a) | is even ⇒ | b – a | is even ⇒ (b, a) ∈ R
∴ R is symmetric.
Let (a, b) ∈ R and (b, c) ∈ R
⇒ | a – b | is even and | b – c | is even ⇒ a – b is even and b – c is even ⇒ (a – b) + (b – c) is even ⇒ a – c is even ⇒ | a – c | is even ⇒ (a, c) ∈ R
∴ R is transitive.
∴ R is an equivalence relation.
∵ | 1 – 3 | = 2, | 3 – 5 | = 2 and | 1 – 5 | = 4 are even, all the elements of {1, 3, 5}
are related to each other. ∵ | 2 – 4 | = 2 is even,
all the elements of {2, 4} are related to each other.
Now | 1 – 2 | = 1, | 1 – 4 | = 3, | 3 – 2 | = 1, | 3 – 4 | = 1, | 5 – 2 | = 3 and | 5 – 4 | = 1 are all odd
no element of the set {1, 3, 5} is related to any element of (2, 4}.
Let R be the relation in the set {1. 2, 3, 4} given by R = {(1,2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}.
Choose the correct answer.
(A) R is reflexive and symmetric but not transitive.
(B) R is reflexive and transitive but not symmetric.
(C) R is symmetric and transitive but not reflexive.
(D) R is an equivalence relation.
Show that the relation R in the set A of points in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.
Show that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12 }, given by
(i) R = {(a, b) : | a – b | is a multiple of 4 }
(ii) R = {(a, b) : a = b} is an equivalence relation. Find the set of all elements related to 1 in each case.
Consider the identity function 1N : N → N defined as lN(x) = x ∀ x ∈ N. Show that although IN is onto but IN + IN : N → N defined as
(IN + IN) (x) = IN(x) + IN(x) = x + x = 2 x is not onto.