GATE CSE 1996 | Question: 2.2

Question

Let $R$ be a non-empty relation on a collection of sets defined by $_{A}R_ B$ if and only if $A \cap B = \phi$. Then, (pick the true statement)

• A. $A$ is reflexive and transitive

• B. $R$ is symmetric and not transitive

• C. $R$ is an equivalence relation

• D. $R$ is not reflexive and not symmetric

Comments

beautiful question!

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A∩B = B∩A = ∅, hence symmetric

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Answer 1

Let $A = \{1,2,3\}$ and $B = \{4,5\}$  and $C = \{1,6,7\}$

now $A\cap B = \emptyset$ and $B\cap C= \emptyset$ but $A\cap C\neq \emptyset$, so $R$ is not transitive.

$A\cap A = A$, so $R$ is not reflexive.

$A\cap B = B\cap A$, so $R$ is symmetric

So, $A$ is false as $R$ is not reflexive or transitive

$B$ is true.

$C$ is false because $R$ is not transitive or reflexive

$D$ is false because $R$ is symmetric

Comments

First, understand the relation. It is given that relation is defined on sets and two sets are only related when their intersection is the empty set.

So let's take an example. A={1,2,3} B={4,5} C={1,6,7}

here A and B are related as  A⋂B = phi

B and C are also related as B⋂C = phi

but A and C are not related as you can see their intersection is not empty

Hence the relation is not transitive.

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Actually what is the relation here ??

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A counterexample for proving $R$ is not transitive is choosing 3 sets $ A,B,C$ such that

$A \cap B = \phi  $ and  $B \cap C = \phi  $ but  $A \cap C \not = \phi  $

Answer 2

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Short trick

Empty set always symmetric relation

The correct option is B R is symmetric and not transitive
(i) Reflexive
A∩A=A≠ϕ
So; (A, A) doesn't belongs to relation R,
∴ Relation R is not reflexive.

(ii) Symmetric
If A∩B=ϕ then B∩A=ϕ is also true.
∴ Relation R is not Symmetric relation.

(iii) Transitive
If A∩B=ϕ and B∩C=ϕ, it need be true that A∩C=ϕ
For example:
A={1,2}, B={3,4}, C={1,5,6}
A∩B=ϕ and B∩C=ϕ but
A∩C={1}≠ϕ
∴ Relation R is not transitive relation.

Answer 3

Answer: A

Let A = {1,2} and B = {3,4}.

Then R = {(1,3),(1,4),(2,3),(2,4)} which is not reflexive and not symmetric.

Comments

There is (1,3) but not (3,1) and there is (1,5) but not (5,1) and likewise for all elements.Therefore it is not symmetric.

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This question is based on relation on sets and not on elements of a set. So, here if A(intersection)B is null then B(intersection)A is also null and hence symmetric.

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A ∩ B = ⏀ , B ∩ A = ⏀ , symmetry holds on aRb when we consider it as empty relation.

but for non-empty relation, I think bRa is also required to have symmetry property.

Answer 4

Option B: $R$ is symmetric and not transitive

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Note here, it’s written that the relation is on collection of sets, i.e. the relation is between sets not between elements of sets. So, (A,B) will be part of R iff A∩B = ∅.

So it cant be reflexive since A∩A != ∅

But it can be symmetric since A∩B = B∩A = ∅ (here A and B are disjoint sets, A={1,2} and B={3,4}

Also, its not sure that it will always be transitive, for eg. A={1,2}, B={3,4} and C={2,5}, here A∩B = ∅, B∩C = ∅ but A∩C != ∅

which makes R, symmetric but not reflexive or transitive.