Animated Solution for Mathematics - Sets and Relations: Let $A=\{1,2,3\}$. The number of relations on A, containing (1, 2) and (2, 3), which are reflexive and transitive but not symmetric, is

My dear student, welcome to the elegant world of discrete mathematics. Today, we are not just solving a problem; we are architects of logic.

We are tasked with constructing relations on the set $A=\{1,2,3\}$ that satisfy a specific set of constraints: they must be reflexive, transitive, and—crucially—not symmetric. Let us peel back the layers of this puzzle together.

Every relation on a set $A$ begins with its 'skeleton.' The condition of reflexivity is our first constraint. It demands that every element in our set must be related to itself.

In the language of set theory, this means the pairs $(1,1)$, $(2,2)$, and $(3,3)$ must be present in our relation $R$. Without these, the relation fails the very first test. Imagine these as self-loops on the nodes of a graph; they are the non-negotiable foundation of our structure.

Next, we are given two specific, mandatory edges: $(1,2)$ and $(2,3)$. These are our 'given' connections. Now, we must invoke the law of transitivity.

Transitivity is the 'flow' of logic: if $a$ relates to $b$, and $b$ relates to $c$, then $a$ must relate to $c$. Look at our current set of pairs: we have $(1,2)$ and $(2,3)$. The logic is inescapable—if $1$ connects to $2$, and $2$ connects to $3$, then $1$ must connect to $3$.

Thus, the pair $(1,3)$ is forced into our relation. We now have our base relation, which we shall call $R_0$:

Now, we must ensure our relation is not symmetric. Symmetry is the 'mirror' property: if $(a,b)\in R$, then $(b,a)$ must also be in $R$.

Our base relation $R_0$ is currently not symmetric because, for example, $(1,2)$ is present, but $(2,1)$ is not. This is perfect! $R_0$ is our first valid relation.

If we add $(2,1)$ to $R_0$, we get $R_1=R_0\cup \{(2,1)\}$. We must check if this breaks transitivity: We have $(2,1)$ and $(1,3)$, which implies $(2,3)$ must be in the relation. It is. We have $(2,1)$ and $(1,2)$, which implies $(2,2)$ must be in the relation. It is.

$R_1$ remains transitive and, crucially, it is still not symmetric because we have $(2,3)$ but not $(3,2)$. Thus, $R_1$ is our second valid relation.

Similarly, if we add $(3,2)$ to $R_0$, we get $R_2=R_0\cup \{(3,2)\}$. Checking transitivity: we have $(1,3)$ and $(3,2)$, which implies $(1,2)$ must be in the relation. It is. $R_2$ is also transitive and not symmetric, making it our third valid relation.

What happens if we try to add more? If we add both $(2,1)$ and $(3,2)$, or if we try to add $(3,1)$, the transitivity requirement forces us to add so many pairs that we eventually construct the Universal Relation $A\times A$.

As we know, the Universal Relation is perfectly symmetric, which violates our final condition. Therefore, we have reached the end of our search.

We have found exactly three relations: $R_0$, $R_1$, and $R_2$. It is a beautiful, finite, and logical conclusion to a fascinating problem. Keep this rigor in your heart, and you will master any challenge the JEE throws your way!