Question

Let x be a nonvoid set.If ρ1 and ρ2 be the transitive relations of x, then-(° denotes the compositions of relations )

• A. ρ1°ρ2 is trasitive relation
• B. ρ1°ρ2 is not transitive relation
• C. ρ1°ρ2 is equivalence relation
• D. ρ1°ρ2 is not any relation on X

Answer 1

The Correct Option is B

Transitivity means that for a relation $R$, if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$ must also hold.

We are asked whether the composition of two transitive relations is necessarily transitive. 

Let’s take a counterexample to demonstrate that it may not be transitive.

Let the set $X = \{1, 2, 3\}$, and define two relations:

• $\rho_1 = \{(1, 2), (2, 3), (1, 3)\}$
• $\rho_2 = \{(2, 3), (3, 1), (2, 1)\}$

Each of these is transitive on its own:

• For $\rho_1$: Since $(1, 2)$ and $(2, 3)$ are in $\rho_1$, and $(1, 3)$ is also in $\rho_1$, it satisfies transitivity.
• For $\rho_2$: Since $(2, 3)$ and $(3, 1)$ are in $\rho_2$, and $(2, 1)$ is also in $\rho_2$, it satisfies transitivity.

Now let’s consider the composition $\rho_1 \circ \rho_2$:

In a composition $\rho_1 \circ \rho_2$, we include $(a, c)$ if there exists a $b$ such that $(a, b) \in \rho_2$ and $(b, c) \in \rho_1$.

Now check for key pairs:

• $(1, 3) \in \rho_1 \circ \rho_2$ and $(3, 1) \in \rho_1 \circ \rho_2$
• But $(1, 1) \notin \rho_1 \circ \rho_2$, so the composition is not transitive.

This violates transitivity: since $(1, 3)$ and $(3, 1)$ are in the relation, then $(1, 1)$ should also be in the relation, but it is not.

Therefore, the composition $\rho_1 \circ \rho_2$ is not necessarily transitive.

Correct option: (B) $\rho_1 \circ \rho_2$ is not a transitive relation

Answer 2

Understanding Transitive Relations and Composition

* Transitive Relation: A relation \(\rho\) on a set \(X\) is transitive if whenever \((a, b) \in \rho\) and \((b, c) \in \rho\), then \((a, c) \in \rho\).

* Composition of Relations: The composition of two relations \(\rho_1\) and \(\rho_2\) on a set \(X\), denoted by \(\rho_1 \circ \rho_2\), is defined as:

\(\rho_1 \circ \rho_2 = \{(a, c) \in X \times X : \exists b \in X \text{ such that } (a, b) \in \rho_2 \text{ and } (b, c) \in \rho_1 \}\)

Counterexample to Show that Composition is Not Necessarily Transitive

To demonstrate that \(\rho_1 \circ \rho_2\) is not necessarily transitive, we will provide a counterexample.

Let \(X = \{1, 2, 3, 4\}\).

Define the relations \(\rho_1\) and \(\rho_2\) on X as follows:

\(\rho_1 = \{(1, 2), (2, 3), (1, 3), (3,4),(3,3),(2,4),(2,2),(4,4)\}\)

\(\rho_2 = \{(4, 1), (1, 2), (4, 2),(1,1),(2,2),(4,4)\}\)

Since \((a,b) , (b,c) \in (\rho)\). implies \((a,c)\in (\rho)\) for some value, they are transitive

\(\rho_1\circ \rho_2\)= {{(4, 2), (1,3), (1, 4),(4,3), (1,2),(4,4),(1,1),(2,2),(4,1)}} If we look to see whether its transitive (4,2) . (2,2)=4,2= But we need to prove it is't always not Transitive we found an Transitive case. To show the composition is not always transitive lets consider another function Let \(X = \{1, 2, 3, 4\}\). Define the relations \(\rho_1\) and \(\rho_2\) on X as follows: \(\rho_1 = \{(1, 2), (2, 3), (1, 3) \}\) \(\rho_2 = \{(3, 4) \}\) With (3,4) if we have (4,X)= then it is not trnasitivve! this can happen if is (2,X) that implies \(\rho_1\) • \(\rho_2\) {{(2, 4)(1,4)}} It has nothing in comon so that It fails to not be transituve. So from These two examples can suggest a \(\rho_1\) • \(\rho_2\) is not always true

Conclusion:

Given two transitive relations \(\rho_1\) and \(\rho_2\) on a set \(X\), their composition \(\rho_1 \circ \rho_2\) is not necessarily a transitive relation.