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

The problem asks us to determine the properties (reflexive, symmetric, transitive) of the relation \(R\) defined on the set \(A\) of all functions \(f: \mathbb{Z} \to \mathbb{Z}\). The relation is given by \(R = \{ (f, g) : f(0) = g(1) \text{ and } f(1) = g(0) \}\).

Concept Used:

Let \(R\) be a relation on a set \(A\).

1. Reflexive: For all \(a \in A\), \((a, a) \in R\).
2. Symmetric: For all \(a, b \in A\), if \((a, b) \in R\), then \((b, a) \in R\).
3. Transitive: For all \(a, b, c \in A\), if \((a, b) \in R\) and \((b, c) \in R\), then \((a, c) \in R\).

We will test the given relation \(R\) against these three properties.

Step-by-Step Solution:

Step 1: Check for Reflexivity

For \(R\) to be reflexive, the condition \((f, f) \in R\) must hold for every function \(f \in A\). The condition for \((f, f) \in R\) is:

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

This simplifies to the condition that \(f(0) = f(1)\) must be true for all functions \(f: \mathbb{Z} \to \mathbb{Z}\). 
However, \(A\) is the set of all functions from \(\mathbb{Z}\) to \(\mathbb{Z}\). 
We can define a function for which this condition is not true. For example, consider the function \(f(x) = x\). 
For this function, \(f(0) = 0\) and \(f(1) = 1\). Here, \(f(0) \neq f(1)\). 
Since we have found a function \(f \in A\) for which \((f, f) \notin R\), the relation is not reflexive.

Step 2: Check for Symmetry

For \(R\) to be symmetric, if \((f, g) \in R\), then \((g, f) \in R\) must also be true. Assume \((f, g) \in R\). This means:

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

Now, let's check the condition for \((g, f) \in R\). This requires:

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

The conditions for \((f, g) \in R\) are identical to the conditions for \((g, f) \in R\). 
If the first set of equalities holds, the second set automatically holds. Therefore, the relation is symmetric.

Step 3: Check for Transitivity

For \(R\) to be transitive, if \((f, g) \in R\) and \((g, h) \in R\), then \((f, h) \in R\) must be true. 

Assume \((f, g) \in R\) and \((g, h) \in R\). From \((f, g) \in R\), we have:

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

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

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

We need to check if \((f, h) \in R\), which means we need to check if \(f(0) = h(1)\) and \(f(1) = h(0)\). 
Using equations (1) and (2), we can establish relationships: From \(f(1) = g(0)\) and \(g(0) = h(1)\), we get \(f(1) = h(1)\). From \(f(0) = g(1)\) and \(g(1) = h(0)\), we get \(f(0) = h(0)\). So we have \(f(0) = h(0)\) and \(f(1) = h(1)\). 
The condition for transitivity is \(f(0) = h(1)\) and \(f(1) = h(0)\). 
Let's construct a counterexample to show that this is not always true. 
Let's define three functions based on their values at 0 and 1:

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

So we have \(f(0)=1, f(1)=2\), and \(h(0)=1, h(1)=2\). 
Now, let's check the condition for \((f, h) \in R\): Is \(f(0) = h(1)\)? We have \(f(0) = 1\) and \(h(1) = 2\). 
Since \(1 \neq 2\), the condition is not met. 
Since we found a case where \((f, g) \in R\) and \((g, h) \in R\) but \((f, h) \notin R\), the relation is not transitive.

Final Computation & Result:

Based on the analysis:

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

Thus, the relation \( R \) is Symmetric but neither reflective nor transitive.

Answer 2

- Reflexive For \( f \in A \), \( fRf \Rightarrow f(0) = f(1) \). But not true for all \( f \). 
So not reflexive. 
- Symmetric If \( fRg \Rightarrow f(0) = g(1), f(1) = g(0) \), then clearly \( gRf \). 
So symmetric. 
- Transitive 
Suppose \( fRg \) and \( gRh \Rightarrow f(0) = g(1), g(0) = f(1), g(0) = h(1), g(1) = h(0) \). 
This implies \( f(0) = h(0) \) and \( f(1) = h(1) \), which does not satisfy \( fRh \Rightarrow \) Not transitive.