Animated Solution for Mathematics - Sets and Relations: Let $R=\{(1,3),(4,2),(2,4),(2,3),(3,1)\}$ be a relation on the set $A=\{1,2,3,4\}$. The relation R is

Welcome, future engineers! Today, we are diving into the elegant world of set theory and relations. A relation $R$ on a set $A$ is essentially a set of rules that dictates how elements of $A$ interact with one another.

Think of it as a map of connections in a network. We are given the set $A=\{1,2,3,4\}$ and the relation $R=\{(1,3),(4,2),(2,4),(2,3),(3,1)\}$.

To truly understand this, let us visualize these pairs as directed paths between nodes in a graph. Imagine four cities labeled $1,2,3,$ and $4$. The pairs in $R$ are the one-way roads connecting these cities. Our goal is to analyze the structural properties of this network.

Reflexivity

First, let us examine reflexivity. A relation is reflexive if every element is related to itself. Mathematically, this means for every $a\in A$, the pair $(a,a)$ must be in $R$.

Visually, this would look like a loop at every single city, allowing you to travel from a city back to itself. Let us check our list: do we have $(1,1)$? No. Do we have $(2,2)$? No.

In fact, none of the required self-loops exist. Because even one missing pair is enough to disqualify the property, we can confidently state that $R$ is not reflexive.

Symmetry

Next, we test for symmetry. A relation is symmetric if, for every pair $(a,b)\in R$, the reverse pair $(b,a)$ is also in $R$. It is the mathematical equivalent of a two-way street.

If you can drive from city $a$ to city $b$, you must be able to drive back from $b$ to $a$. Let us inspect our pairs. We have $(1,3)$, and we see $(3,1)$ exists. That is a two-way street!

We have $(4,2)$, and we see $(2,4)$ exists. So far, so good. But wait—look at the pair $(2,3)$. Is there a return path $(3,2)$?

Scanning our set $R$, we find no such pair. This missing return path is our counterexample. Because this single street is one-way, the entire relation fails the symmetry test. Thus, $R$ is not symmetric.

Transitivity

Now, let us tackle transitivity. This property is about efficiency. A relation is transitive if, whenever we have a path from $a$ to $b$ and a path from $b$ to $c$, there must be a direct shortcut from $a$ to $c$.

In our network, if you can travel $1\to 3$ and $3\to 1$, there must be a direct road $1\to 1$. As we established earlier, $(1,1)$ is missing. Therefore, the chain $1\to 3\to 1$ does not have the required shortcut. The relation is not transitive.

Is it a Function?

Finally, we ask: is this relation a function? A function is a very strict, deterministic machine. If you input a value, you must get exactly one output.

Think of a vending machine: if you press the button for $2$, you should always get the same item. If pressing $2$ sometimes gives you a soda and sometimes gives you a water, the machine is broken.

Looking at our relation, the input $2$ maps to $3$ (the pair $(2,3)$) and also maps to $4$ (the pair $(2,4)$). Because the input $2$ has two different outputs, this relation is not a function.

We have systematically dismantled the properties of $R$. It is not reflexive, not transitive, and not a function.

Most importantly, it failed the symmetry test due to the missing $(3,2)$ pair. By breaking down these definitions, we see that the correct description is that the relation is not symmetric. Keep practicing this methodical approach, and you will master these concepts in no time!