Animated Solution for Mathematics - Sets and Relations: 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 Journey into Function Mapping

Welcome, future engineer. Today, we are not merely solving a problem; we are dissecting the DNA of a mathematical relation. When you first look at a problem involving a set $A$ of all functions $f:\mathbb{Z}\to \mathbb{Z}$, it is natural to feel a sense of intimidation.

The set of all functions is infinite, complex, and abstract. But here is the secret of the JEE Advanced: the most complex-looking problems often hide the most elegant, simple truths. Let us peel back the layers of this relation $R$ together.

Decoding the Cross-Connection

The relation $R$ is defined by the condition:

Think of this as a 'cross-connection'. The function $f$ is not looking at $g$ directly; it is looking at $g$ through a mirror.

The value of $f$ at zero must match the value of $g$ at one, and the value of $f$ at one must match the value of $g$ at zero. Imagine two people standing across from each other, holding up signs. Person $f$ holds a sign at position $0$ that must match the sign Person $g$ holds at position $1$.

The Reflexivity Trap

Now, let us test for reflexivity. A relation is reflexive if every element is related to itself. In our case, we need to check if $(f,f)\in R$ for every function $f$ in our set $A$.

If we substitute $g$ with $f$ in our definition, the condition becomes:

Ask yourself: does this hold for every function? Consider the identity function, $f(x)=x$. Here, $f(0)=0$ and $f(1)=1$.

Since $0 eq 1$, the identity function is not related to itself. Because we found even one counter-example, the relation fails the test of reflexivity. It is not reflexive.

The Symmetry Proof

Next, we investigate symmetry. A relation is symmetric if $(f,g)\in R$ implies $(g,f)\in R$. Let us assume $(f,g)\in R$. This gives us two powerful equations:

Now, we want to see if $(g,f)\in R$. According to our definition, this requires:

Look closely at the equations we already have. If we simply read $f(1)=g(0)$ backwards, we get $g(0)=f(1)$. If we read $f(0)=g(1)$ backwards, we get $g(1)=f(0)$.

The conditions match perfectly! The relation is symmetric. It is like a mirror; if $f$ sees $g$, then $g$ must see $f$.

The Transitivity Investigation

Finally, we arrive at the most challenging part: transitivity. Transitivity requires that if $(f,g)\in R$ and $(g,h)\in R$, then $(f,h)\in R$.

Let us introduce a third function, $h$. We have our first set of connections:

We have our second set of connections:

We need to find the relationship between $f$ and $h$. Let us trace the path for $f(0)$. We know $f(0)=g(1)$. But from our second set of equations, we know $g(1)=h(0)$. By the transitive property of equality, we conclude $f(0)=h(0)$.

Now, let us trace the path for $f(1)$. We know $f(1)=g(0)$. From our second set of equations, we know $g(0)=h(1)$. Therefore, $f(1)=h(1)$.

So, what have we found? We found that $f(0)=h(0)$ and $f(1)=h(1)$. But wait! For $(f,h)\in R$, the definition requires $f(0)=h(1)$ and $f(1)=h(0)$.

Our result ($f(0)=h(0)$ and $f(1)=h(1)$) is a straight, parallel connection. It is not the cross-connection required by the relation. Therefore, the relation is not transitive.

We have systematically dismantled the problem. We proved it is not reflexive using the identity function. We proved it is symmetric by rearranging the equations. And we proved it is not transitive by showing that two cross-connections result in a straight link.

The relation is symmetric, but neither reflexive nor transitive. You have navigated the logic, visualized the connections, and arrived at the truth. This is the essence of JEE Advanced mathematics—not just calculation, but clear, logical visualization.