Q.Let $R$ be the set of all real numbers. Consider the following subsets of the plane $R\times R$:

[!FORMULA] $S=\{(x,y):y=x+1 \text{ and } 0<x<2\} \text{ and } T=\{(x,y): x-y \text{ is an integer}\}$

Then which of the following is true?

Relations — From Intuition to Definition

Think about the people in your class. Some pairs of students are "friends," some are "sitting next to each other," some "have the same birthday month." Each of these is a relation — a way of connecting one person to another. A relation simply tells you, for any two objects, whether they are linked in that particular way.

Now take a step back. Suppose you have two sets: $A = \{1, 2, 3\}$ and $B = \{x, y\}$. A relation from $A$ to $B$ is just a rule that pairs some elements of $A$ with some elements of $B$. For example, "is less than" between numbers and letters doesn't make sense, but we can invent a relation: "the number and the letter appear in the same word." That would pair $(1, o)$? No — we need a concrete rule.

The cleanest way to define a relation is to list the pairs that satisfy it. A relation from $A$ to $B$ is any subset of the Cartesian product $A \times B$.

That's it. If $(a, b) \in R$, we say "$a$ is related to $b$" and often write $a\,R\,b$. If $(a, b) \notin R$, they are not related.

Example to lock it in

Let $A = \{2, 3, 5\}$ and $B = \{4, 6, 10\}$. Define the relation $R$ as "$a$ divides $b$." Then:

• $2$ divides $4$ → $(2,4) \in R$
• $2$ divides $6$ → $(2,6) \in R$
• $3$ divides $6$ → $(3,6) \in R$
• $5$ divides $10$ → $(5,10) \in R$

So $R = \{(2,4), (2,6), (3,6), (5,10)\}$. Every other pair — like $(2,10)$ or $(3,4)$ — is not in $R$.

Special case: relation on a single set …