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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.

156 Views

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.

A= {x ∈ Z : 0 ≤ x ≤ 12}
= {0, 1, 2, 3, 4, 5, 6, 7, 8,9, 10, 11, 12 } (i) R = {(a, b) : | a – b | is a multiple of 4|
As | a – a | = 0 is divisible by 4 ∴ (a, a) ∈ R ∀ a ∈ A.
∴ R is reflexive.
Next, let (a, b) ∈ R
⇒ | a – b | is divisible by 4
⇒ | – (b – a) | is divisible by 4 ⇒ | b – a | is divisible by 4 ⇒ (b, a) ∈ R
∴ R is symmetric.
Again. (a, b) ∈ R and (b, c) ∈ R
⇒ | a – A | is a multiple of 4 and | b – c | is a multiple of 4 ⇒ a – b is a multiple of 4 and b – c is a multiple of 4 ⇒ (a – b) + (b – c) is a multiple of 4 ⇒ a – c is a multiple of 4 ⇒ | a – c | is a multiple of 4 ⇒ (a, c) ∈ R ∴ R is transitive.
∴ R is an equivalence relation.
Set of elements which are related to | = {a ∈ A : (a, 1) ∈ R}
= {a ∈ A : |a – 1| is a multiple of 4]
= {1, 5, 9}
[ ∵ | 1 – 1 | = 0, | 5 – 1 | = 4 and I 9 – 1 | = 8 are multiples of 4] (ii) R = {(a, b) : a = b} ∴ a = a ∀ a ∈ A,

∴ R is reflexive.
Again, (a, b) ∈ R ⇒ a = b ⇒ b = a ⇒ (b,a) ∈ R ∴ R is symmetric.
Next. (a, b) ∈ R and (b, c) ∈ R
⇒ a = b and b = c
⇒ a = c ⇒ (a, c) ∈ R
∴ R is transitive.
∴ R is an equivalence relation.
Set of elements of A which are related to I = {a ∈ A : (a,1) ∈ R }
= {a ∈ A : a = 1} = {1}.

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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.

111 Views