Animated Solution for Mathematics - Sets and Relations: Let $R$ be a relation on $\mathbb{N}\times \mathbb{N}$ defined by $(a,b)R(c,d)$ if and only if $ad(b-c)=bc(a-d)$. Then $R$ is

Welcome, aspiring engineers. Today, we are not just solving a problem; we are peeling back the layers of a mathematical structure. In the world of JEE Advanced, relations are not just sets of ordered pairs; they are the hidden rules that govern how elements interact.

When you see a definition like $(a,b)R(c,d)⟺ad(b-c)=bc(a-d)$, your first instinct might be panic. It looks like a tangled mess of variables. But take a deep breath; every complex equation is just a simple truth wearing a disguise.

Let us look at our given condition: $ad(b-c)=bc(a-d)$. If we expand this, we get $abd-acd=abc-bcd$.

Here is the secret: whenever you see products of variables in a relation, look for a way to normalize them. Dividing by the product of all variables, $abcd$, is a powerful technique. When we perform this division, the equation transforms into:

Watch the magic happen. Terms cancel out, and we are left with $\frac{1}{c}-\frac{1}{b}=\frac{1}{d}-\frac{1}{a}$.

Let us rearrange this to group the $(a,b)$ terms and $(c,d)$ terms. By moving terms around, we get the master condition:

This is the heartbeat of our relation. It is no longer a scary polynomial; it is a clean, elegant statement about reciprocals.

A relation is reflexive if every element is related to itself. In our case, we need $(a,b)R(a,b)$ to hold for all $a,b\in \mathbb{N}$.

Let us plug $(a,b)$ into our master condition. We replace $c$ with $a$ and $d$ with $b$. The condition becomes:

If we bring everything to one side, we get $2(\frac{1}{a}-\frac{1}{b})=0$, which implies $\frac{1}{a}=\frac{1}{b}$, or $a=b$.

But wait! The definition of reflexivity requires this to be true for all pairs $(a,b)$, not just when $a=b$. If we pick $(1,2)$, the condition fails. Thus, the relation is not reflexive.

Symmetry is the most intuitive property. If $(a,b)$ is related to $(c,d)$, is $(c,d)$ related to $(a,b)$?

Let us start with our master condition: $\frac{1}{a}-\frac{1}{b}=\frac{1}{d}-\frac{1}{c}$. If we multiply both sides by $-1$, we get:

Rearranging this gives $\frac{1}{c}-\frac{1}{d}=\frac{1}{b}-\frac{1}{a}$. This is exactly the condition for $(c,d)R(a,b)$. The relation is perfectly symmetric.

Finally, we reach transitivity. We need to check if $(a,b)R(c,d)$ and $(c,d)R(e,f)$ implies $(a,b)R(e,f)$. Let us use the function $f(x,y)=\frac{1}{x}-\frac{1}{y}$. Our condition is $f(a,b)=-f(c,d)$.

If we have $(a,b)R(c,d)$, then $f(a,b)=-f(c,d)$. If we have $(c,d)R(e,f)$, then $f(c,d)=-f(e,f)$.

Substituting the second into the first, we get $f(a,b)=-(-f(e,f))=f(e,f)$. But for transitivity to hold, we need $f(a,b)=-f(e,f)$. We have a sign mismatch!

Let us verify with a counterexample. Let $(a,b)=(2,3)$ and $(c,d)=(3,2)$. Here, $f(2,3)=\frac{1}{6}$ and $f(3,2)=-\frac{1}{6}$.

Now let $(e,f)=(5,30)$. $f(5,30)=\frac{1}{5}-\frac{1}{30}=\frac{1}{6}$.

Since $f(3,2)=-\frac{1}{6}$ and $f(5,30)=\frac{1}{6}$, the relation holds for the pairs. But does $(2,3)R(5,30)$? No, because $f(2,3)=\frac{1}{6}$ and $f(5,30)=\frac{1}{6}$, which are not negatives of each other. Transitivity fails.

We have systematically dismantled the problem. We found it is symmetric, but neither reflexive nor transitive. This journey shows that math is not about memorizing formulas; it is about testing, verifying, and understanding the underlying logic.