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Relations and Functions

Short Answer Type

351.

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}.

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352.

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.

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353.

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.

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Long Answer Type

354.

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.

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Short Answer Type

355. Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f { (1, 4). (2, 5), (3. 6)} be a function from A to B. Show that f is one-one.

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356. Show that the function f : R →R , defined as f (x) = x² , is neither one-to-one nor onto.

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357. Show that the modulus function f : R → R, given by f (x) = | x |, is neither one-one nor onto, where |x| is x, if x is positive or 0 and | x | is —x, if. x negative.

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358. Prove that the greatest integer function f : R → R, given by f (x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

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359. Let A and B be two sets. Show that f :A x B → B x A such that f (a,b) = (b,a) is a bijective function.

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360.
Consider the identity function 1_N : N → N defined as l_N(x) = x ∀ x ∈ N. Show that although I_N is onto but I_N + I_N : N → N defined as
(I_N + I_N) (x) = I_N(x) + I_N(x) = x + x = 2 x is not onto.

Clearly I_N is onto. But I_N + I_N is not onto, as we can find an element 3 in the codomain N such that there does not exist any x in the domain N with (I_N + I_N) (x) = 2 x = 3.

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