# Discrete Mathematics Questions and Answers – Types of Relations

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This set of Discrete Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Types of Relations”.

1. The binary relation {(1,1), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2)} on the set {1, 2, 3} is __________
a) reflective, symmetric and transitive
b) irreflexive, symmetric and transitive
c) neither reflective, nor irreflexive but transitive
d) irreflexive and antisymmetric

Explanation: Not reflexive -> (3,3) not present; not irreflexive -> (1, 1) is present; not symmetric -> (2, 1) is present but not (1, 2); not antisymmetric – (2, 3) and (3, 2) are present; not asymmetric -> asymmetry requires both antisymmetry and irreflexivity. So, it is transitive closure of relation.

2. Consider the relation: R’ (x, y) if and only if x, y>0 over the set of non-zero rational numbers,then R’ is _________
a) not equivalence relation
b) an equivalence relation
c) transitive and asymmetry relation
d) reflexive and antisymmetric relation

Explanation: Reflexive: a, a>0
Symmetric: if a, b>0 then both must be +ve or -ve, which means b, a > 0 also exists
Transitive: if a, b>0 and b, c>0 then to have b as same number, both pairs must be +ve or -ve which implies a, c>0. Hence, R’ is an equivalence relation.

3. Let S be a set of n>0 elements. Let be the number Br of binary relations on S and let Bf be the number of functions from S to S. The expression for Br and Bf, in terms of n should be ____________
a) n2 and 2(n+1)2
b) n3 and n(n+1)
c) n and n(n+6)
d) 2(n*n) and nn

Explanation: For a set with n elements the number of binary relations should be 2(n*n) and the number of functions should be nn. Hence Br = 2(n*n) and Bf = nn.
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4. Let A be a set of k (k>0) elements. Which is larger between the number of binary relations (say, Nr) on A and the number of functions (say, Nf) from A to A?
a) number of relations
b) number of functions
c) the element set
d) number of subsets of the relation

Explanation: For a set with k elements the number of binary relations should be 2(n*n) and the number of functions should be nn. Now, 2(n*n) => n2log (2) [taking log] and nn => nlog (n) [taking log]. It is known that n2log (2) > nlog (n). Hence, the number of binary relations > the number of functions i.e, Nr > Nf.

5. Consider the binary relation, A = {(a,b) | b = a – 1 and a, b belong to {1, 2, 3}}. The reflexive transitive closure of A is?
a) {(a,b) | a >= b and a, b belong to {1, 2, 3}}
b) {(a,b) | a > b and a, b belong to {1, 2, 3}}
c) {(a,b) | a <= b and a, b belong to {1, 2, 3}}
d) {(a,b) | a = b and a, b belong to {1, 2, 3}}

Explanation: By definition of Transitive closure we have that a is related to all smaller b (as every a is related to b – 1) and from the reflexive property a is related to a.
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6. Let R1 be a relation from A = {1, 3, 5, 7} to B = {2, 4, 6, 8} and R2 be another relation from B to C = {1, 2, 3, 4} as defined below:
i. An element a in A is related to an element b in B (under R1) if a * b is divisible by 3.
ii. An element a in B is related to an element b in C (under R2) if a * b is even but not divisible by 3. Which is the composite relation R1R2 from A to C?
a) R1R2 = {(1, 2), (1, 4), (3, 3), (5, 4), (5,6), (7, 3)}
b) Φ
c) R1R2 = {(1, 2), (1,6), (3, 2), (3, 4), (5, 4), (7, 2)}
d) R1R2 = {(2,2), (3, 2), (3, 4), (5, 1), (5, 3), (7, 1)}

Explanation: By definition, i) R1 = {(1,6), (3,2), (3,4), (3,6), (3,8), (5,6), (7,6)} and ii) R2 = {(1,2), (1,4), (1,8), (5,2), (5,4), (5,8), (7,2), (7,4), (7,8)}. So, R1R2 = Φ.

7. The time complexity of computing the transitive closure of a binary relation on a set of n elements should be ________
a) O(n)
b) O(logn)
c) O(n(n+(3/2)))
d) O(n3)

Explanation: Calculation of transitive closure results into matrix multiplication. We can do matrix multiplication in O(n3) time. There are better algorithms that do less than cubic time.

8. Let A and B be two non-empty relations on a set S. Which of the following statements is false?
a) A and B are transitive ⇒ A∩B is transitive
b) A and B are symmetric ⇒ A∪B is symmetric
c) A and B are transitive ⇒ A∪B is not transitive
d) A and B are reflexive ⇒ A∩B is reflexive

Explanation: In terms of set theory, the binary relation R defined on the set X is a transitive relation if, for all a, b, c ∈ X, if aRb and bRc, then aRc. If there are two relations on a set satisfying transitive property then there union must satisfy transitive property.

9. Determine the characteristics of the relation aRb if a2 = b2.
a) Transitive and symmetric
b) Reflexive and asymmetry
c) Trichotomy, antisymmetry, and irreflexive
d) Symmetric, Reflexive, and transitive

Explanation: Since, x2 = y2 is just a special case of equality, so all properties that apply to x = y also apply to this case. Hence, the relation satisfies symmetric, reflexive and transitive closure.

10. Let R be a relation between A and B. R is asymmetric if and only if ________
a) Intersection of D(A) and R is empty, where D(A) represents diagonal of set
b) R-1 is a subset of R, where R-1 represents inverse of R
c) Intersection of R and R-1 is D(A)
d) D(A) is a subset of R, where D(A) represents diagonal of set

Explanation: A relation is asymmetric if and only if it is both antisymmetric and irreflexive. As a consequence, a relation is transitive and asymmetric if and only if it is a strict partial order. If D(A) is a diagonal of A set and intersection of D(A) and R is empty, then R is asymmetric.

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