Animated Solution for Mathematics - Sets and Relations: Let $R$ be the real line. Consider the following subsets of the plane $R\times R$: $S=\{(x,y):y=x+1\text{ and }0<x<2\}$, $T=\{(x,y):x-y\text{ is an integer}\}$. Which one of the following is true?

A Journey into Relations

Welcome, future engineers! Today, we are going to peel back the layers of a classic problem that tests the very foundation of set theory: the Equivalence Relation.

Imagine you are standing on a vast, infinite plane of real numbers, $\mathbb{R}\times \mathbb{R}$. We have two distinct paths, $S$ and $T$, and our mission is to determine which one behaves like a perfect, balanced system—an equivalence relation.

Before we touch the math, let us define the 'Holy Trinity' of relations. For a relation to be an equivalence relation, it must satisfy three conditions: Reflexivity, Symmetry, and Transitivity.

Think of Reflexivity as the 'Mirror Test': every element must be related to itself.

Symmetry is the 'Reflection Test': if $x$ is related to $y$, then $y$ must be related to $x$.

Finally, Transitivity is the 'Chain Test': if $x$ is related to $y$, and $y$ is related to $z$, then $x$ must be related to $z$. If a relation breaks any of these, it is not an equivalence relation.

The Broken Mirror

Let us look at $S=\{(x,y):y=x+1,0<x<2\}$. This set represents a line segment.

To test for reflexivity, we ask: does every point $(x,x)$ lie on this line? We substitute $y=x$ into the defining equation $y=x+1$.

This gives us $x=x+1$, which simplifies to:

This is a mathematical impossibility! Because $0 eq 1$, the relation $S$ fails the reflexivity test. It is not reflexive, and therefore, it cannot be an equivalence relation. The mirror is broken.

The Integer Grid

Now, consider $T=\{(x,y):x-y\in \mathbb{Z}\}$. This set is much more elegant. Let us test the three pillars.

First, Reflexivity: is $x-x\in \mathbb{Z}$? Yes, $x-x=0$, and $0$ is an integer. The mirror test passes!

Second, Symmetry: if $x-y=k$ (where $k$ is an integer), then $y-x=-k$. Since the negative of an integer is also an integer, $y-x\in \mathbb{Z}$. The reflection test passes!

Third, Transitivity: if $x-y=k_1$ and $y-z=k_2$ (where $k_1,k_2\in \mathbb{Z}$), then:

The sum of two integers is always an integer, so $x-z\in \mathbb{Z}$. The chain test passes!

We have seen that $S$ fails the very first test, while $T$ stands strong, satisfying all three conditions. This is the beauty of mathematics—the structure of $T$ creates a perfect, consistent system, while $S$ is merely a line that misses the mark.

Remember, in your JEE journey, always look for the underlying structure. When you see a condition like $x-y\in \mathbb{Z}$, you are looking at a system that respects the integrity of integers.

Conclusion: $T$ is an equivalence relation, while $S$ is not. Keep practicing, keep visualizing, and you will master these concepts with ease!