Animated Solution for Mathematics - Sets and Relations: Let $R_1$ and $R_2$ be two relation defined as follows: $R_1=\{(a,b)\in R^2:a^2+b^2\in Q\}$ and $R_2=\{(a,b)\in R^2:a^2+b^2\notin Q\}$, where $Q$ is the set of all rational numbers. Then:

Unraveling Transitivity

Welcome, my dear students. Today, we are not just solving a problem; we are embarking on a journey into the heart of logical relations. In the JEE Advanced arena, you will often encounter questions that test your ability to think critically about definitions.

We are given two relations, $R_1$ and $R_2$, defined on the set of real numbers $\mathbb{R}$. Our mission is to determine if they are transitive. Let us peel back the layers of this problem together.

First, let us look at our definitions. We have:

Essentially, $R_1$ is the 'Rational Sum' relation, and $R_2$ is the 'Irrational Sum' relation. To check for transitivity, we must recall the golden rule: a relation $R$ is transitive if, for any three elements $a,b,c$, the existence of the pairs $(a,b)\in R$ and $(b,c)\in R$ implies that the pair $(a,c)$ must also be in $R$.

Think of it as a bridge. If you can walk from $a$ to $b$, and from $b$ to $c$, can you jump directly from $a$ to $c$? If you can find even one scenario where the bridge from $a$ to $c$ collapses, the relation is not transitive.

The Rational Trap

Let us focus on $R_1$. We need to see if $a^2+b^2\in \mathbb{Q}$ and $b^2+c^2\in \mathbb{Q}$ forces $a^2+c^2\in \mathbb{Q}$.

I want you to imagine we are constructing a counter-example. We want the first two conditions to be true, but the final condition to fail. Let us pick $a^2=\sqrt{2}$ and $c^2=\sqrt{2}$.

If we do this, then $a^2+c^2=2\sqrt{2}$, which is clearly irrational! We have already broken the final condition. Now, we just need to find a $b^2$ that makes the first two conditions work. Let $b^2=2-\sqrt{2}$.

Let us check our work:

1. For $(a,b)$: $a^2+b^2=\sqrt{2}+(2-\sqrt{2})=2$. Since $2\in \mathbb{Q}$, $(a,b)\in R_1$. Success!

2. For $(b,c)$: $b^2+c^2=(2-\sqrt{2})+\sqrt{2}=2$. Since $2\in \mathbb{Q}$, $(b,c)\in R_1$. Success!

But look at the final link $(a,c)$: $a^2+c^2=\sqrt{2}+\sqrt{2}=2\sqrt{2}$. This is irrational! Therefore, $(a, c) otin R_1$. The bridge has collapsed. $R_1$ is not transitive.

The Irrational Paradox

Now, let us turn our attention to $R_2$. The condition is that the sum of squares must be irrational. We need to find a case where $(a,b)\in R_2$ and $(b,c)\in R_2$, but $(a, c) otin R_2$.

Let us try a different strategy. We want $a^2+c^2$ to be rational (to break the condition). Let $a^2=1$ and $c^2=1$. Then $a^2+c^2=2$, which is rational.

Now, we need a $b^2$ such that $a^2+b^2$ and $b^2+c^2$ are both irrational. Let $b^2=\sqrt{2}$.

Let us verify:

1. For $(a,b)$: $a^2+b^2=1+\sqrt{2}$. This is irrational, so $(a,b)\in R_2$.

2. For $(b,c)$: $b^2+c^2=\sqrt{2}+1$. This is also irrational, so $(b,c)\in R_2$.

But for the final link $(a,c)$: $a^2+c^2=1+1=2$. Since $2$ is rational, $(a, c) otin R_2$. Again, the bridge is broken! $R_2$ is not transitive.

My friends, what have we learned? We have learned that in the world of mathematics, intuition can be a dangerous guide. We might have assumed that these relations were transitive, but by carefully constructing counter-examples, we proved that neither $R_1$ nor $R_2$ holds the property of transitivity.

Always remember: to disprove a general statement, you do not need a complex proof; you only need one well-chosen example. Keep this spirit of inquiry alive, and you will conquer any problem the JEE throws at you!