Animated Solution for Mathematics - Sets and Relations: Which of the following is *not* correct for relation $R$ on the set of real numbers?

Imagine you are standing on the infinite expanse of the real number line, $\mathbb{R}$. Every point is a potential partner in a dance of logic. In mathematics, a relation $R$ is simply a rule that decides which pairs of these points are 'connected'.

Today, we are going to act as detectives, investigating four different rules to see which one fails to hold up under scrutiny. Our mission is to identify the incorrect statement among the options provided.

Before we dive into the specific options, let's sharpen our tools. A relation on a set is defined by three fundamental properties: Reflexivity, Symmetry, and Transitivity.

Reflexivity asks: "Is every element related to itself?" Mathematically, for all $x\in \mathbb{R}$, is $(x,x)\in R$?

Symmetry asks: "If $x$ is related to $y$, is $y$ also related to $x$?" That is, does $(x,y)\in R⟹(y,x)\in R$?

Transitivity is the most demanding: "If $x$ is related to $y$, and $y$ is related to $z$, does it force $x$ to be related to $z$?" If any of these fail, the relation lacks that specific property.

Testing the Rules

Let's start with the relation defined as $(x,y)\in R⟺|x-y|\le 1$.

Is it reflexive? If we replace $y$ with $x$, we get $|x-x|=0$, and since $0\le 1$, it is indeed reflexive.

Is it symmetric? The absolute value function is beautifully indifferent to order; $|x-y|=|y-x|$. So, if $|x-y|\le 1$, then $|y-x|\le 1$. This statement is correct.

Now, consider the relation $|x|-|y|\le 1$. Reflexivity is easy: $|x|-|x|=0\le 1$.

But symmetry? Let's test $x=0$ and $y=5$. $|0|-|5|=-5$, which is $\le 1$. So $(0,5)\in R$.

But swap them: $|5|-|0|=5$, which is definitely not $\le 1$. So $(5, 0) otin R$. This relation is not symmetric.

Why the Relation Fails Transitivity

We arrive at the claim that the relation $(x,y)\in R⟺0<|x-y|\le 1$ is both symmetric and transitive. We already know it is symmetric because $|x-y|=|y-x|$.

But transitivity? This is where the trap lies. Let's use our number line. Let $x=1$, $y=2$, and $z=3$.

The distance between $1$ and $2$ is $1$, which satisfies $0<1\le 1$. So $(1,2)\in R$.

The distance between $2$ and $3$ is $1$, which also satisfies $0<1\le 1$. So $(2,3)\in R$.

For the relation to be transitive, we must have $(1,3)\in R$. But the distance between $1$ and $3$ is $2$.

Since $2>1$, the condition $0<|x-z|\le 1$ is violated. The chain is broken! The relation is not transitive.

By methodically testing each property, we have unmasked the culprit. The statement claiming that the relation $(x,y)\in R⟺0<|x-y|\le 1$ is transitive is the incorrect statement.

Remember, in the world of JEE Advanced, you don't need to be afraid of these abstract definitions. Treat them as physical constraints on a number line, and you will always find the truth.