Q.Show that the relation R in the set $A = \{1, 2, 3\}$ given by $R = \{(1, 2), (2, 1)\}$ is symmetric but neither reflexive nor transitive.

Concept: Relation Properties — checking reflexivity, symmetry, and transitivity for a given relation.

1. Reflexive: For reflexivity, every element must relate to itself. Here, $(1,1), (2,2), (3,3)$ are all missing from $R$. So $R$ is not reflexive.

2. Symmetric: For symmetry, if $(a,b) \in R$ then $(b,a)$ must also be in $R$. We have $(1,2)$ and $(2,1)$ — both are present. No other pairs exist to violate symmetry. So $R$ is symmetric. …

We need to check three properties of a relation on a set: reflexivity, symmetry, and transitivity. Each property has a precise definition, and we test $R$ against each one.

Reflexivity means every element of $A$ must be related to itself. That is, for all $a \in A$, $(a,a) \in R$.

Symmetry means whenever $(a,b) \in R$, then $(b,a)$ must also be in $R$.

Transitivity means whenever $(a,b) \in R$ and $(b,c) \in R$, then $(a,c)$ must be in $R$.

Our set is $A = \{1,2,3\}$ and $R = \{(1,2), (2,1)\}$. Let's test each property.

1. Reflexivity: The pairs we need are $(1,1)$, $(2,2)$, and $(3,3)$. None of these appear in $R$. So $R$ is not reflexive.

2. Symmetry: Look at each pair in $R$.

   • $(1,2)$ is in $R$. Its reverse $(2,1)$ is also in $R$.
   • $(2,1)$ is in $R$. Its reverse $(1,2)$ is also in $R$. Every pair's mirror is present. So $R$ is symmetric.

3. Transitivity: We need to check all possible chains of two pairs. The only pairs in $R$ are $(1,2)$ and $(2,1)$.

   • Take $(1,2)$ and $(2,1)$: the first ends at $2$, the second starts at $2$, so we have $1 \to 2 \to 1$. Transitivity would require $(1,1)$ to be in $R$. But $(1,1)$ is not there. …

Method: Verifying a Small Listed Relation (Symmetric Only)

Use this when a relation is given as a short list of pairs and you must confirm exactly which properties hold.

Steps

Step 1: Reflexive — check every self-pair

List $(1,1),(2,2),(3,3)$ for the base set $\{1,2,3\}$ and see which appear. None do here, so it is not reflexive.

Step 2: Symmetric — mirror each pair

$(1,2)$ has its reverse $(2,1)$ present, and vice versa, so the relation is symmetric.

Step 3: Transitive — check chains, including "there and back" …

Common Mistakes

Mistake 1: Claiming transitivity is "vacuously true"

Why it's wrong: there is a chain $(1,2),(2,1)$, so the condition is triggered and requires $(1,1)$, which is missing. Correct approach: call transitivity vacuous only when no $(a,b),(b,c)$ chains exist at all.

Mistake 2: Thinking a symmetric relation must be reflexive

Why it's wrong: symmetry says nothing about self-pairs, and $(1,1),(2,2),(3,3)$ are all absent here. Correct approach: check reflexivity separately. …