Let A = { 1 , 2 , 3 , 4 , 5 } and R be a relation defined by R = { ( x , y ) : x , y ∈ A , x + y = 5 } . Then, R is

We first write the elements of the set R . i.e. R = { ( 1 , 4 ) , ( 4 , 1 ) , ( 2 , 3 ) , ( 3 , 2 ) } Since, ( 1 , 1 ) ∉ R ⇒ R is not reflexive. Now as, ( 1 , 4 ) ∈ R ⇒ ( 4 , 1 ) ∈ R and ( 2 , 3 ) ∈ R ⇒ ( 3 , 2 ) ∈ R So, R is symmetric Now, ( 1 , 4 ) ∈ R and ( 4 , 1 ) ∈ R Thus, R is not transitive. Hence, R is symmetric but neither reflexive nor transitive.

Mathematics - Sets and Relations Question with Solution | TestHub

MathematicsSets and RelationsQuestions on Symmetric Transitive and Reflexive PropertiesMedium2 minPYQ_2017

MathematicsMediumsingle choice

Let $A = {1, 2, 3, 4, 5}$ and $R$ be a relation defined by $R = {(x, y) : x, y \in A, x + y = 5}$ . Then, $R$ is

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Answer:

Solution:

We first write the elements of the set $R$ .

i.e. $R = {(1, 4), (4, 1), (2, 3), (3, 2)}$

Since, $(1, 1) \notin R \Rightarrow R$ is not reflexive.

Now as, $(1, 4) \in R \Rightarrow (4, 1) \in R$ and $(2, 3) \in R$

$\Rightarrow (3, 2) \in R$

So, $R$ is symmetric

Now, $(1, 4) \in R$ and $(4, 1) \in R$

Thus, $R$ is not transitive.

Hence, $R$ is symmetric but neither reflexive nor transitive.

Stream:BITSATSubject:MathematicsTopic:Sets and RelationsSubtopic:Questions on Symmetric Transitive and Reflexive Properties

⏱ 2mℹ️ Source: PYQ_2017

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