Question

Let $ A $ be the set of all functions $ f: \mathbb{Z} \to \mathbb{Z} $ and $ R $ be a relation on $ A $ such that $$ R = \{ (f, g) : f(0) = g(1) \text{ and } f(1) = g(0) \} $$ Then $ R $ is:

• A. Symmetric and transitive but not reflective
• B. Symmetric but neither reflective nor transitive
• C. Reflexive but neither symmetric nor transitive
• D. Transitive but neither reflexive nor symmetric

Hint:

Always verify reflexivity, symmetry, and transitivity with definitions and counterexamples.

Answer 1

The Correct Option is B

This problem requires an examination of the properties (reflexive, symmetric, transitive) of a relation \(R\) defined on the set \(A\), which comprises all functions \(f: \mathbb{Z} \to \mathbb{Z}\). The relation \(R\) is formally defined as \(R = \{ (f, g) : f(0) = g(1) \text{ and } f(1) = g(0) \}\).

Concept Used:

For a relation \(R\) on a set \(A\):

1. Reflexive: \( \forall a \in A, (a, a) \in R \).
2. Symmetric: \( \forall a, b \in A, \text{ if } (a, b) \in R, \text{ then } (b, a) \in R \).
3. Transitive: \( \forall a, b, c \in A, \text{ if } (a, b) \in R \text{ and } (b, c) \in R, \text{ then } (a, c) \in R \).

The following steps will evaluate the relation \(R\) against these properties.

Step-by-Step Solution:

Step 1: Check for Reflexivity

Reflexivity mandates that for every function \(f \in A\), the condition \((f, f) \in R\) must hold. This translates to:

\[ f(0) = f(1) \quad \text{and} \quad f(1) = f(0) \]

The requirement simplifies to \(f(0) = f(1)\) for all functions \(f: \mathbb{Z} \to \mathbb{Z}\). Since \(A\) contains all such functions, we can find a function where this is not true. For instance, consider \(f(x) = x\). For this function, \(f(0) = 0\) and \(f(1) = 1\). As \(f(0) eq f(1)\), the condition \((f, f) \in R\) is not satisfied for this \(f\). Therefore, the relation is not reflexive.

Step 2: Check for Symmetry

Symmetry requires that if \((f, g) \in R\), then \((g, f) \in R\) must also hold. Assuming \((f, g) \in R\) implies:

\[ f(0) = g(1) \quad \text{and} \quad f(1) = g(0) \]

Now, let's examine the condition for \((g, f) \in R\):

\[ g(0) = f(1) \quad \text{and} \quad g(1) = f(0) \]

The conditions defining \((f, g) \in R\) are identical to those defining \((g, f) \in R\). Thus, if the first set of equalities holds, the second set is automatically satisfied. Consequently, the relation is symmetric.

Step 3: Check for Transitivity

Transitivity requires that if \((f, g) \in R\) and \((g, h) \in R\), then \((f, h) \in R\) must hold. Assume \((f, g) \in R\) and \((g, h) \in R\). From \((f, g) \in R\):

\[ f(0) = g(1) \quad \text{and} \quad f(1) = g(0) \quad \cdots (1) \]

From \((g, h) \in R\):

\[ g(0) = h(1) \quad \text{and} \quad g(1) = h(0) \quad \cdots (2) \]

To check if \((f, h) \in R\), we need to verify if \(f(0) = h(1)\) and \(f(1) = h(0)\). Using equations (1) and (2):

From \(f(1) = g(0)\) and \(g(0) = h(1)\), we deduce \(f(1) = h(1)\).

From \(f(0) = g(1)\) and \(g(1) = h(0)\), we deduce \(f(0) = h(0)\).

The condition for transitivity is \(f(0) = h(1)\) and \(f(1) = h(0)\). Let's construct a counterexample. Define three functions:

• Let \(f\) be such that \(f(0) = 1\) and \(f(1) = 2\).
• For \((f, g) \in R\), we require \(g(1) = f(0) = 1\) and \(g(0) = f(1) = 2\).
• For \((g, h) \in R\), we require \(h(1) = g(0) = 2\) and \(h(0) = g(1) = 1\).

This gives us \(f(0)=1, f(1)=2\), and \(h(0)=1, h(1)=2\). Now, we check the condition for \((f, h) \in R\): Is \(f(0) = h(1)\)? We have \(f(0) = 1\) and \(h(1) = 2\). Since \(1 eq 2\), this condition is not met.

As we found a scenario where \((f, g) \in R\) and \((g, h) \in R\) but \((f, h) otin R\), the relation is not transitive.

Final Computation & Result:

Based on the analysis:

• The relation is not reflexive because the condition \(f(0)=f(1)\) is not universally true for all functions \(f\).
• The relation is symmetric because the conditions for \((f,g) \in R\) are identical to those for \((g,f) \in R\).
• The relation is not transitive, as demonstrated by the counterexample.

Therefore, the relation \( R \) is Symmetric but neither reflexive nor transitive.