Mathematics Questions and Answers – Types of Relations

»

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

1. Which of these is not a type of relation?
a) Reflexive
b) Surjective
c) Symmetric
d) Transitive
View Answer

Answer: b
Explanation: Surjective is not a type of relation. It is a type of function. Reflexive, Symmetric and Transitive are type of relations.

2. An Equivalence relation is always symmetric.
a) True
b) False
View Answer

Answer: a
Explanation: The given statement is true. A relation R in a set A is said to be an equivalence relation if R is reflexive, symmetric and transitive. Hence, an equivalence relation is always symmetric.

3. Which of the following relations is symmetric but neither reflexive nor transitive for a set A = {1, 2, 3}.
a) R = {(1, 2), (1, 3), (1, 4)}
b) R = {(1, 2), (2, 1)}
c) R = {(1, 1), (2, 2), (3, 3)}
d) R = {(1, 1), (1, 2), (2, 3)}
View Answer

Answer: b
Explanation: A relation in a set A is said to be symmetric if (a1, a2)∈R implies that (a1, a2)∈R,for every a1, a2∈R.
Hence, for the given set A={1, 2, 3}, R={(1, 2), (2, 1)} is symmetric. It is not reflexive since every element is not related to itself and neither transitive as it does not satisfy the condition that for a given relation R in a set A if (a1, a2)∈R and (a2, a3)∈R implies that (a1, a3)∈ R for every a1, a2, a3∈R.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!
advertisement
advertisement

4. Which of the following relations is transitive but not reflexive for the set S={3, 4, 6}?
a) R = {(3, 4), (4, 6), (3, 6)}
b) R = {(1, 2), (1, 3), (1, 4)}
c) R = {(3, 3), (4, 4), (6, 6)}
d) R = {(3, 4), (4, 3)}
View Answer

Answer: a
Explanation: For the above given set S = {3, 4, 6}, R = {(3, 4), (4, 6), (3, 6)} is transitive as (3,4)∈R and (4,6) ∈R and (3,6) also belongs to R . It is not a reflexive relation as it does not satisfy the condition (a,a)∈R, for every a∈A for a relation R in the set A.

5. Let R be a relation in the set N given by R={(a,b): a+b=5, b>1}. Which of the following will satisfy the given relation?
a) (2,3) ∈ R
b) (4,2) ∈ R
c) (2,1) ∈ R
d) (5,0) ∈ R
View Answer

Answer: a
Explanation: (2,3) ∈ R as 2+3 = 5, 3>1, thus satisfying the given condition.
(4,2) doesn’t belong to R as 4+2 ≠ 5.
(2,1) doesn’t belong to R as 2+1 ≠ 5.
(5,0) doesn’tbelong to R as 0⊁1

6. Which of the following relations is reflexive but not transitive for the set T = {7, 8, 9}?
a) R = {(7, 7), (8, 8), (9, 9)}
b) R = {(7, 8), (8, 7), (8, 9)}
c) R = {0}
d) R = {(7, 8), (8, 8), (8, 9)}
View Answer

Answer: a
Explanation: The relation R= {(7, 7), (8, 8), (9, 9)} is reflexive as every element is related to itself i.e. (a,a) ∈ R, for every a∈A. and it is not transitive as it does not satisfy the condition that for a relation R in a set A if (a1, a2)∈R and (a2, a3)∈R implies that (a1, a3) ∈ R for every a1, a2, a3 ∈ R.

7. Let I be a set of all lines in a XY plane and R be a relation in I defined as R = {(I1, I2):I1 is parallel to I2}. What is the type of given relation?
a) Reflexive relation
b) Transitive relation
c) Symmetric relation
d) Equivalence relation
View Answer

Answer: d
Explanation: This is an equivalence relation. A relation R is said to be an equivalence relation when it is reflexive, transitive and symmetric.
Reflexive: We know that a line is always parallel to itself. This implies that I1 is parallel to I1 i.e. (I1, I2)∈R. Hence, it is a reflexive relation.
Symmetric: Now if a line I1 || I2 then the line I2 || I1. Therefore, (I1, I2)∈R implies that (I2, I1)∈R. Hence, it is a symmetric relation.
Transitive: If two lines (I1, I3) are parallel to a third line (I2) then they will be parallel to each other i.e. if (I1, I2) ∈R and (I2, I3) ∈R implies that (I1, I3) ∈R.
advertisement

8. Which of the following relations is symmetric and transitive but not reflexive for the set I = {4, 5}?
a) R = {(4, 4), (5, 4), (5, 5)}
b) R = {(4, 4), (5, 5)}
c) R = {(4, 5), (5, 4)}
d) R = {(4, 5), (5, 4), (4, 4)}
View Answer

Answer: d
Explanation: R= {(4, 5), (5, 4), (4, 4)} is symmetric since (4, 5) and (5, 4) are converse of each other thus satisfying the condition for a symmetric relation and it is transitive as (4, 5)∈R and (5, 4)∈R implies that (4, 4) ∈R. It is not reflexive as every element in the set I is not related to itself.

9. (a,a) ∈ R, for every a ∈ A. This condition is for which of the following relations?
a) Reflexive relation
b) Symmetric relation
c) Equivalence relation
d) Transitive relation
View Answer

Answer: a
Explanation: The above is the condition for a reflexive relation. A relation is said to be reflexive if every element in the set is related to itself.
advertisement

10. (a1, a2) ∈R implies that (a2, a1) ∈ R, for all a1, a2∈A. This condition is for which of the following relations?
a) Equivalence relation
b) Reflexive relation
c) Symmetric relation
d) Universal relation
View Answer

Answer: c
Explanation: The above is a condition for a symmetric relation.
For example, a relation R on set A = {1,2,3,4} is given by R={(a,b):a+b=3, a>0, b>0}
1+2 = 3, 1>0 and 2>0 which implies (1,2) ∈ R.
Similarly, 2+1 = 3, 2>0 and 1>0 which implies (2,1)∈R. Therefore both (1, 2) and (2, 1) are converse of each other and is a part of the relation. Hence, they are symmetric.

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

To practice all areas of Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & technical discussions at Telegram SanfoundryClasses.