Animated Solution for Mathematics - Sets and Relations: Among the relations $S=\{(a,b):a,b\in \mathbb{R}-\{0\},2+\frac{a}{b}>0\}$ and $T=\{(a,b):a,b\in \mathbb{R},a^2-b^2\in \mathbb{Z}\}$,

Welcome, future engineers. Today, we are not just solving a problem; we are dissecting the very DNA of mathematical relations. In the JEE Advanced arena, relations are not just sets of ordered pairs; they are the rules of engagement for how numbers interact.

We have two contenders today: $S$ and $T$. One is a rigid, well-behaved algebraic structure, and the other is a chaotic, deceptive inequality. Let us begin our journey.

Let us look at the relation $T=\{(a,b):a,b\in \mathbb{R},a^2-b^2\in \mathbb{Z}\}$. This relation is beautiful because it relies on the properties of integers.

To check for symmetry, we assume $(a,b)\in T$, which means $a^2-b^2=k$ for some integer $k$. To check if $(b,a)\in T$, we simply look at $b^2-a^2$.

Algebraically, we find:

Since the negative of an integer is still an integer, the condition holds. Symmetry is confirmed.

Now, consider transitivity. If $(a,b)\in T$ and $(b,c)\in T$, we have $a^2-b^2=k_1$ and $b^2-c^2=k_2$.

When we add these, the $b^2$ terms vanish like magic:

Since the sum of two integers is an integer, $(a,c)\in T$. Thus, $T$ is transitive and acts as a perfect, logical machine.

Now, turn your attention to $S=\{(a,b):a,b\in \mathbb{R}-\{0\},2+\frac{a}{b}>0\}$. This is where students often stumble. The inequality $2+\frac{a}{b}>0$ is equivalent to $\frac{a}{b}>-2$.

This is not a simple equality; it is a region. When we test for symmetry, we ask: if $\frac{a}{b}>-2$, is it guaranteed that $\frac{b}{a}>-2$?

Let us pick a valid pair: $a=-1,b=10$. Then:

So, $(-1,10)\in S$. Now, check the reverse pair $(10,-1)$:

Since $-8<0$, the reverse pair is not in $S$. Symmetry is shattered.

Finally, we examine transitivity for $S$. We need a chain: $(a,b)\in S$ and $(b,c)\in S$, but $(a, c) otin S$.

Let us pick $a=10,b=1$. Then $2+\frac{10}{1}=12>0$, which is valid. Now pick $c=-0.8$.

Check $(b,c)$, which is $(1,-0.8)$:

This is also valid. Now the final test: $(a,c)$, which is $(10,-0.8)$:

This is definitely not greater than zero. The chain is broken. We have proven that $S$ is neither symmetric nor transitive.

You see, the math did not change; our perspective did. We moved from blind calculation to strategic investigation. Keep this mindset, and no JEE problem will ever intimidate you again.