Animated Solution for Mathematics - Sets and Relations: Let $R$ be a relation on $\mathbb{R}$, given by $R=\{(a,b):3a-3b+\sqrt{7}\text{ is an irrational number}\}$. Then $R$ is

Unraveling Relations

Welcome, my dear student. Today, we are not just solving a problem; we are embarking on a journey into the very foundation of set theory. Relations are the building blocks of functions, and understanding them is like learning the grammar of mathematics.

Let us dissect this problem with the precision of a surgeon and the curiosity of an explorer.

The Reflexive Foundation

We are given a relation $R$ on the set of real numbers $\mathbb{R}$, defined by the condition: $3(a-b)+\sqrt{7}\in \text{Irrational Numbers}$. Our mission is to test the three pillars of relations: Reflexivity, Symmetry, and Transitivity.

Let us start with Reflexivity. For a relation to be reflexive, every element $a$ must be related to itself. That is, the pair $(a,a)$ must belong to $R$.

Let us substitute $b=a$ into our condition:

Is $\sqrt{7}$ an irrational number? Of course it is! Since the condition is satisfied for any arbitrary real number $a$, we can confidently declare: The relation $R$ is reflexive. This is our first victory.

The Symmetry Trap

Now, we move to Symmetry. This is where many students stumble. The rule is simple: if $(a,b)\in R$, then $(b,a)$ must also be in $R$.

But is it always true? Let us test it. If $(a,b)\in R$, then $3(a-b)+\sqrt{7}$ is irrational. Does this force $3(b-a)+\sqrt{7}$ to be irrational?

Let us be clever. We want to find a counter-example where $(a,b)\in R$ but $(b, a) otin R$. Let us choose $a=\frac{\sqrt{7}}{3}$ and $b=0$.

For $(a,b)$:

Since $2\sqrt{7}$ is irrational, $(a,b)\in R$. Now, let us check $(b,a)$:

Zero is a rational number! The condition fails. Therefore, $R$ is not symmetric.

The Transitivity Test

Finally, we tackle Transitivity. If $(a,b)\in R$ and $(b,c)\in R$, must $(a,c)\in R$? Let us test this with a carefully chosen set of numbers. Let $a=\frac{\sqrt{7}}{3}$, $b=1$, and $c=\frac{2\sqrt{7}}{3}$.

First, check $(a,b)$:

This is irrational, so $(a,b)\in R$. Next, check $(b,c)$:

This is also irrational, so $(b,c)\in R$. Now, the moment of truth: check $(a,c)$:

Again, we get $0$, which is rational. Thus, $(a, c) otin R$. The relation is not transitive.

We have systematically dismantled the problem. We found that $R$ is reflexive, but it fails both symmetry and transitivity.

This is the beauty of mathematics—it teaches us to question assumptions and verify every step. Keep practicing, keep questioning, and you will master these concepts in no time!