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

Concept: Reflexive, Symmetric, Transitive — checking each property individually.

Step 1 — Reflexive:

For reflexivity, every element must be related to itself. Here $(1,1), (2,2), (3,3)$ are all present. So $R$ is reflexive.

Step 2 — Not symmetric:

Symmetry requires that if $(a,b) \in R$, then $(b,a) \in R$.

We have $(1,2) \in R$ but $(2,1) \notin R$. Hence $R$ is not symmetric.

Step 3 — Not transitive: …

We need to check three properties: reflexivity, symmetry, and transitivity. Each has a precise definition, and we test $R$ against them one by one.

1. Reflexivity — A relation $R$ on a set $A$ is reflexive if every element of $A$ is related to itself. That means for each $a \in A$, the pair $(a,a)$ must be in $R$.

   Here $A = \{1,2,3\}$. We check:

   • $(1,1) \in R$
   • $(2,2) \in R$
   • $(3,3) \in R$ All three are present. So $R$ is reflexive.

2. Symmetry — $R$ is symmetric if whenever $(a,b) \in R$, then $(b,a) \in R$ as well.

   Look at the pairs in $R$ that are not of the form $(a,a)$:

   • $(1,2) \in R$, but $(2,1)$ is not in $R$.
   • $(2,3) \in R$, but $(3,2)$ is not in $R$. Since we found a counterexample, $R$ is not symmetric.
   Watch out

   A common mistake is to think that because $(1,1)$ is symmetric with itself, the whole relation is symmetric. Symmetry must hold for every pair — one missing reverse pair breaks it.

3. Transitivity — $R$ is transitive if whenever $(a,b) \in R$ and $(b,c) \in R$, then $(a,c) \in R$.

   Check all possible chains of two pairs: …

Method: Testing properties of a relation given as an explicit list of pairs

Use this when the relation is handed to you as a finite set of ordered pairs and you must check reflexive / symmetric / transitive.

Steps

Step 1: Reflexive — scan for every diagonal pair

List the elements of the base set and confirm each $(a,a)$ appears in $R$. If any is missing, $R$ is not reflexive.

Step 2: Symmetric — check the reverse of each non-diagonal pair

For every $(a,b)\in R$ with $a\ne b$, look for $(b,a)$. A single missing reverse pair breaks symmetry.

Step 3: Transitive — check every two-step chain …

Common Mistakes

Mistake 1: Calling the relation symmetric because the diagonal pairs are "symmetric"

Why it's wrong: $(1,1),(2,2),(3,3)$ being present says nothing about symmetry; the test is whether the reverse of every off-diagonal pair is present, and $(1,2)\in R$ while $(2,1)\notin R$. Correct approach: check each non-diagonal pair's reverse — one missing reverse breaks symmetry.

Mistake 2: Missing the broken transitivity chain $1\to2\to3$ …