# UGC-NET | UGC NET CS 2018 July – II | Question 89

Which of the following is an equivalence relation on the set of all functions from Z to Z?
(A) {(f, g) | f (x)−g (x) = 1 x ∈ Z}
(B) {(f, g) | f (0) = g (0) or f (1) = g (1)}
(C) {(f, g) | f (0) = g (1) and f (1) = g (0)}
(D) {(f, g) | f (x)−g (x) = k for some k ∈ Z }

Explanation: (A) This relation has none of the three properties. It is not reflexive, since f(x) – f(x) = 0 ≠ 1. It is not symmetric, since if j(x)- g(x) = 1, then g(x)- f(x) = -1 ≠ 1. It is not transitive, since if f(x)- g(x) = 1 and g(x) – h(x) = 1, then f(x) – h(x) = 2 ≠ 1.

(B) This is not an equivalence relation because it is not transitive. Let f(x) = 0, g(x) = x, and h(x) = 1 for all x E Z. Then f is related tog since f(0) = g(0), and g is related to h since g(1) = h(1), but f is not related to h since they have no values in common. By inspection we see that this relation is reflexive and symmetric.

(C) This relation is not reflexive, since there are lots of functions f (for instance, f(x) = x) that do not have the property that f(0) = f(1). It is symmetric by inspection (the roles of f and g are the same). It is not transitive. For instance, let f(0) = g(1) = h(0) = 7, and let f(0) = g(0) = h(1) = 3; fill in the remaining values arbitrarily. Then f and g are related, as are g and h, but f is not related to h since 7 ≠ 3.

(D) This is an equivalence relation. Two functions are related here if they differ by a constant. It is clearly reflexive (the constant is 0). It is symmetric, since if f (x) – g(x) = k, then g(x) – f (x) = -k. It is transitive, since if f(x) – g(x) = k1 and g(x) – h(x) = k2 , then f(x) – h(x) = k3 , where k3 = k1 + k2 (add the first two equations).

So, option (D) is correct.

UGC has taken this question from Kenneth Rosen-7th edition (Page-615, Ques-3, Chapter- 9.5 Equivalence Relations).

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