Animated Solution for Mathematics - Sets and Relations: Let $R_1=\{(a,b)\in \mathbb{N}\times \mathbb{N}:|a-b|\le 13\}$ and $R_2=\{(a,b)\in \mathbb{N}\times \mathbb{N}:|a-b|\ne 13\}$. Then on $\mathbb{N}$:

A Journey into Logic

Welcome, future engineers! Today, we are going to peel back the curtain on one of the most elegant yet deceptive topics in Set Theory: Equivalence Relations. Many students look at a problem like this and think, 'Oh, it's just inequalities, I can eyeball this.'

But in the world of JEE Advanced, eyeballing is the quickest way to lose marks. We are going to treat this problem like a detective story, investigating two suspects, $R_1$ and $R_2$, to see if they qualify for the prestigious title of 'Equivalence Relation.'

Before we dive into the math, let's visualize what an Equivalence Relation actually is. Imagine a VIP club. To get in, a relation must pass three strict security checks:

1. Reflexivity: Every element must be friends with itself. Mathematically, $(a,a)\in R$ for all $a\in \mathbb{N}$.

2. Symmetry: If $a$ is friends with $b$, then $b$ must be friends with $a$. Mathematically, $(a,b)\in R⟹(b,a)\in R$.

3. Transitivity: If $a$ is friends with $b$, and $b$ is friends with $c$, then $a$ must be friends with $c$. Mathematically, $(a,b)\in R$ and $(b,c)\in R⟹(a,c)\in R$.

If a relation fails even one of these, it is not an equivalence relation. Let's put our suspects to the test.

The 'Small Step' Trap

Our first suspect is defined as:

Reflexivity: We check $(a,a)$. The distance $|a-a|$ is $0$. Since $0\le 13$, the condition holds. $R_1$ is reflexive.

Symmetry: If $|a-b|\le 13$, then because the absolute value function is symmetric ($|a-b|=|b-a|$), it follows that $|b-a|\le 13$. $R_1$ is symmetric.

Transitivity: This is where the trap lies. We need to see if a chain of 'small steps' forces a 'small jump.' Let's test this with numbers: let $a=1$, $b=13$, and $c=26$.

Check the first link: $(1,13)⟹|1-13|=12$. Since $12\le 13$, this is valid.

Check the second link: $(13,26)⟹|13-26|=13$. Since $13\le 13$, this is also valid.

Now, the final test: $(1,26)⟹|1-26|=25$.

Wait! $25$ is definitely greater than $13$. The direct connection $(1,26)$ is NOT in $R_1$. The chain is broken, and $R_1$ fails transitivity.

The 'Forbidden Line' Trap

Now, let's look at the second suspect:

Reflexivity: We check $(a,a)$. The distance $|a-a|=0$. Since $0 eq 13$, the condition holds. $R_2$ is reflexive.

Symmetry: If $|a - b| eq 13$, then $|b - a| eq 13$. Symmetry holds.

Transitivity: This is the trickiest part. We need to find a counter-example where $(a,b)\in R_2$ and $(b,c)\in R_2$, but $(a, c) otin R_2$.

Let's pick $a=1$, $b=2$, and $c=14$.

Check the first link: $(1,2)⟹|1-2|=1$. Since $1 eq 13$, this is valid.

Check the second link: $(2,14)⟹|2-14|=12$. Since $12 eq 13$, this is also valid.

Now, the final test: $(1,14)⟹|1-14|=13$.

Oh no! The condition for $R_2$ is that the distance must NOT be $13$. But here, the distance is exactly $13$. Therefore, $(1, 14) otin R_2$. The transitivity chain is broken again!

We have rigorously tested both relations. $R_1$ failed transitivity because it allowed 'accumulation' of distance. $R_2$ failed transitivity because it allowed 'bridging' over a forbidden gap.

Neither $R_1$ nor $R_2$ is an equivalence relation. This problem teaches us a vital lesson: never trust your intuition when it comes to relations. Always test the properties, always look for the counter-example, and always verify the logic.