1. Given relation

Let $A$ be the set of all functions $f: \mathbb{Z} \to \mathbb{Z}$.

The relation $R$ on $A$ is defined by

We must check whether $R$ is reflexive, symmetric, and transitive.

2. Check reflexivity

A relation is reflexive if for every $f \in A$, we have $(f,f) \in R$.

Now,

This is equivalent to just

But this is not true for every function $f: \mathbb{Z}\to\mathbb{Z}$.

For example, take

Then

and hence $f(0) \ne f(1)$. So $(f,f) \notin R$.

Therefore, $R$ is not reflexive.

3. Check symmetry

A relation is symmetric if

Suppose $(f,g) \in R$. Then

To have $(g,f) \in R$, we need

But these are exactly the same equalities as above, just written in reverse order.

Hence whenever $(f,g) \in R$, we also get $(g,f) \in R$.

Therefore, $R$ is symmetric.

4. Check transitivity

A relation is transitive if

Assume

Then

and

Combining,

So we get

But for $(f,h) \in R$, we need

These do not follow in general.

So transitivity may fail.

Let us give a counterexample.

Take functions $f,g,h \in A$ such that:

• $f(0)=0, \ f(1)=1$
• $g(0)=1, \ g(1)=0$
• $h(0)=0, \ h(1)=1$

(Values at all other integers can be chosen arbitrarily.)

Then:

• $(f,g) \in R$ because $$f(0)=0=g(1), \qquad f(1)=1=g(0).$$
• $(g,h) \in R$ because $$g(0)=1=h(1), \qquad g(1)=0=h(0).$$

But $(f,h) \notin R$ because that would require

which is false.

Therefore, $R$ is not transitive.

5. Conclusion

Thus $R$ is:

• not reflexive,
• symmetric,
• not transitive.

So the correct option is

6. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

Hence, the derived answer agrees with the stored answer.