Discrete Mathematics Questions and Answers – Relations – Equivalence Classes and Partitions

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This set of Discrete Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Relations – Equivalence Classes and Partitions”.

1. Suppose a relation R = {(3, 3), (5, 5), (5, 3), (5, 5), (6, 6)} on S = {3, 5, 6}. Here R is known as _________
a) equivalence relation
b) reflexive relation
c) symmetric relation
d) transitive relation
View Answer

Answer: a
Explanation: Here, [3] = {3, 5}, [5] = {3, 5}, [5] = {5}. We can see that [3] = [5] and that S/R will be {[3], [6]} which is a partition of S. Thus, we can choose either {3, 6} or {5, 6} as a set of representatives of the equivalence classes.
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2. Consider the congruence 45≡3(mod 7). Find the set of equivalence class representatives.
a) {…, 0, 7, 14, 28, …}
b) {…, -3, 0, 6, 21, …}
c) {…, 0, 4, 8, 16, …}
d) {…, 3, 8, 15, 21, …}
View Answer

Answer: a
Explanation: Note that a set of class representatives is the subset of a set which contains exactly one element from each equivalence class. Now, for integers n, a and b, we have congruence a≡b(mod n), then the set of equivalence classes are {…, -2n, -n, 0, n, 2n,…}, {…, 1-2n, 1-n, 1, 1+n, 1+2n,…}. The required answer is {…, 0, 7, 14, 28, …}.

3. Which of the following relations is the reflexive relation over the set {1, 2, 3, 4}?
a) {(0,0), (1,1), (2,2), (2,3)}
b) {(1,1), (1,2), (2,2), (3,3), (4,3), (4,4)}
c) {,(1,1), (1,2), (2,1), (2,3), (3,4)}
d) {(0,1), (1,1), (2,3), (2,2), (3,4), (3,1)
View Answer

Answer: b
Explanation: {(1,1), (1,2), (2,2), (3,3), (4,3), (4,4)} is a reflexive relation because it contains set = {(1,1), (2,2), (3,3), (4,4)}.

4. Determine the partitions of the set {3, 4, 5, 6, 7} from the following subsets.
a) {3,5}, {3,6,7}, {4,5,6}
b) {3}, {4,6}, {5}, {7}
c) {3,4,6}, {7}
d) {5,6}, {5,7}
View Answer

Answer: b
Explanation: {3,5}, {3,6,7}, {4,5,6}. It is not a partition because these sets are not pairwise disjoint. The elements 3, 5 and 6 appear repeatedly these sets. {1}, {2,3,6}, {4}, {5} – this is a partition as they are pairwise disjoint. {3,4,6}, {7} – this is not a partition as element 5 is missing.
{5,6}, {5,7} – this is not a partition because it is missing the elements 3, 4 in any of the sets.

5. Determine the number of equivalence classes that can be described by the set {2, 4, 5}.
a) 125
b) 5
c) 16
d) 72
View Answer

Answer: b
Explanation: Suppose B={2, 4, 5} and B×B = (2,2), (4,4), (5,5), (2,4), (4,2), (4,5), (5,4), (2,5), (5,2). A relation R on set B is said to be equivalence relation if R is reflexive, Symmetric, transitive. Hence, total number of equivalence relation=5 out of 23=8 relations.
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6. Determine the number of possible relations in an antisymmetric set with 19 elements.
a) 23585
b) 2.02 * 1087
c) 9.34 * 791
d) 35893
View Answer

Answer: b
Explanation: Number of antisymmetric relation is given:-|A|=n, |AxA|=n xn. Then, N=total number of diagonal will n and we know that N = 2n * 3(n2-n)/2. So, the number of relations should be = 2.02 * 1087.

7. For a, b ∈ Z define a | b to mean that a divides b is a relation which does not satisfy ___________
a) irreflexive and symmetric relation
b) reflexive relation and symmetric relation
c) transitive relation
d) symmetric relation
View Answer

Answer: b
Explanation: Suppose, a=0, then we know that 0 does not divide 0, 0 ∤ 0 and it is not reflexive. Again, 2 | 4 but 4 does not 2 and so it is not a symmetric relation.

8. Which of the following is an equivalence relation on R, for a, b ∈ Z?
a) (a-b) ∈ Z
b) (a2+c) ∈ Z
c) (ab+cd)/2 ∈ Z
d) (2c3)/3 ∈ Z
View Answer

Answer: b
Explanation: Let a ∈ R, then a−a = 0 and 0 ∈ Z, so it is reflexive. To see that a-b ∈ Z is symmetric, then a−b ∈ Z -&gt say, a−b = m, where m ∈ Z ⇒ b−a = −(a−b)=−m and −m ∈ Z. Thus, a-b is symmetric. To see that a-b is transitive, let a, b, c ∈ R. Thus, a−b ∈ Z; b−c ∈ Z. Let a−b = i and b−c = j, for integers i,j ∈ Z. Then a−c ='(a−b)+(b−c)=i + j. So, a−c ∈ Z. Therefore a – c is transitive. Hence, (a-b) is an equivalence relation on the set R. Rest of the options are not equivalence relations.

9. Determine the set of all integers a such that a ≡ 3 (mod 7) such that −21 ≤ x ≤ 21.
a) {−21, −18, −11, −4, 3, 10, 16}
b) {−21, −18, −11, −4, 3, 10, 17, 24}
c) {−24, -19, -15, 5, 0, 6, 10}
d) {−23, −17, −11, 0, 2, 8, 16}
View Answer

Answer: b
Explanation: For an integer a we have x ≡ 3 (mod 7) if and only if a = 7m + 3. Thus, by calculating multiples of 7, add 3 and restrict the value of a, so that −21 ≤ x ≤ 21. The set for a = {−21, −18, −11, −4, 3, 10, 17, 24}.
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10. For a, b ∈ R define a = b to mean that |x| = |y|. If [x] is an equivalence relation in R. Find the equivalence relation for [17].
a) {,…,-11, -7, 0, 7, 11,…}
b) {2, 4, 9, 11, 15,…}
c) {-17, 17}
d) {5, 25, 125,…}
View Answer

Answer: c
Explanation: We can find that [17] = {a ∈ R|a = 17} = {a ∈ R||a| = |17|} = {-17, 17} and [−17] = {a ∈ R|a = −17} = {a ∈ R||a| = |−17|}= {−17, 17}. Hence, the required equivalence relation is {-17, 17}.

Sanfoundry Global Education & Learning Series – Discrete Mathematics.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn