Relations and Functions Test

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Relations and Functions Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

Let A = {1, 2, 3} and consider the relation R = {1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1,3)}. Then R is

A
neither symmetric, nor transitive

B
reflexive but not transitive

C
reflexive but not symmetric

D
symmetric and transitive

Solution

As (1, 1), (2, 2), (3, 3 ) ∈∈ R, therefore R is reflexive. Since (1,2) ∈ R, but (2,1) ∉ R. Therefore,R is not symmetric.
Question 2
1 / -0

Given an arbitrary equivalence relation R in an arbitrary set X, R divides X into

A
three sets

B
intersecting sets

C
mutually disjoint subsets

D
two sets

Solution

An equivalence relation R gives a partitioning of the set A into mutually disjoint equivalence classes,i.e. union of equivalence classes is the set A itself.
Question 3
1 / -0

A binary operation ∗ : A × A → A for any non-empty set A,is said to be associative if

A
a * b) * c = a *(b * c), ∀ a, b, c ∈ A

B
(a * b) * c ≠ a *(b * c), ∀ a, b ∈ A

C
(a * b) * c = a *(b * c), ∀ a, c ∈ A

D
(a * b) * c ≠ a *(b * c), ∀ a, b, c ∈ A

Solution

A binary operation ∗: A × A → A is said to be associative binary operation, if (a ∗ b) ∗ c = a ∗ (b ∗ c), ∀ a, b,c ∈ A.
Question 4
1 / -0

Equivalence classes are

A
mutually disjoint subsets

B
intersecting sets

C
trivial sets

D
power sets

Solution

An equivalence relation R gives a partitioning of the set A into mutually disjoint equivalence classes,i.e. union of equivalence classes is the set A itself. Any two equivalence classes i.e. subsets are either equal or disjoint.
Question 5
1 / -0

Pick the true statement from the following

A
The composition of functions is commutative.

B
Every function is invertible

C
A binary operation on a set has always the identity element

D
The composition of functions is associative

Solution

Composite functions are always associative i.e. for three functions f,g,h. We have (fog)oh = fo(goh)
Question 6
1 / -0

Let A = { 2 , 3 , 6 }. Which of the following relations on A are reflexive ?

A
R₁ = {( 2,2 ) , ( 3,3 ) , ( 6,6 )}

B
R₂ = {( 2,2 ) , ( 3,3 ) , ( 3,6 ) , ( 6,3 )}

C
R₃ = {( 2,2 ) , ( 3,6 ) , ( 2,6 )}

D
None of these

Solution

R₁ is a reflexive on A, because ( a,a ) ∈ R₁for each a ∈ A

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Relations and Functions Test - 20

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Question 1

neither symmetric, nor transitive

reflexive but not transitive

reflexive but not symmetric

symmetric and transitive

Solution

Question 2

three sets

intersecting sets

mutually disjoint subsets

two sets

Solution

Question 3

a * b) * c = a *(b * c), ∀ a, b, c ∈ A

(a * b) * c ≠ a *(b * c), ∀ a, b ∈ A

(a * b) * c = a *(b * c), ∀ a, c ∈ A

(a * b) * c ≠ a *(b * c), ∀ a, b, c ∈ A

Solution

Question 4

mutually disjoint subsets

intersecting sets

trivial sets

power sets

Solution

Question 5

The composition of functions is commutative.

Every function is invertible

A binary operation on a set has always the identity element

The composition of functions is associative

Solution

Question 6

R1 = {( 2,2 ) , ( 3,3 ) , ( 6,6 )}

R2 = {( 2,2 ) , ( 3,3 ) , ( 3,6 ) , ( 6,3 )}

R3 = {( 2,2 ) , ( 3,6 ) , ( 2,6 )}

None of these

Solution

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a * b) * c = a (b c), ∀ a, b, c ∈ A

(a * b) * c ≠ a (b c), ∀ a, b ∈ A

(a * b) * c = a (b c), ∀ a, c ∈ A

(a * b) * c ≠ a (b c), ∀ a, b, c ∈ A

R₁ = {( 2,2 ) , ( 3,3 ) , ( 6,6 )}

R₂ = {( 2,2 ) , ( 3,3 ) , ( 3,6 ) , ( 6,3 )}

R₃ = {( 2,2 ) , ( 3,6 ) , ( 2,6 )}