Q.Let $L$ be the set of all lines in a plane and $R$ be the relation in $L$ defined as $R = \{(L_1, L_2): L_1 \text{ is perpendicular to } L_2\}$. Show that $R$ is symmetric but neither reflexive nor transitive.

Concept: Relation Properties (reflexive, symmetric, transitive).

Step 1 – Reflexive: A line cannot be perpendicular to itself. So $(L_1, L_1) \notin R$ for any $L_1 \in L$. Hence $R$ is not reflexive.

Step 2 – Symmetric: If $L_1 \perp L_2$, then by definition $L_2 \perp L_1$. So $(L_1, L_2) \in R \implies (L_2, L_1) \in R$. Hence $R$ is symmetric. …

Why This Approach Works

The question asks us to check three properties of a relation: reflexivity, symmetry, and transitivity. Each property has a precise definition, and we test the relation against each one using geometric facts about perpendicular lines.

The key geometric facts you need:

• A line is never perpendicular to itself (perpendicular means meeting at $90^\circ$, which requires two distinct lines).
• Perpendicularity is mutual: if $L_1 \perp L_2$, then $L_2 \perp L_1$.
• If $L_1 \perp L_2$ and $L_2 \perp L_3$, then $L_1$ and $L_3$ are parallel (or coincident), not perpendicular.

Let's test each property one by one.

1. Checking Reflexivity

Definition: A relation $R$ on a set $L$ is reflexive if every element is related to itself. That is, for every line $L_1 \in L$, we must have $(L_1, L_1) \in R$.

For our relation, $(L_1, L_1) \in R$ would mean $L_1$ is perpendicular to itself. But a line cannot be perpendicular to itself — perpendicularity requires two distinct lines intersecting at $90^\circ$. A single line does not make an angle with itself.

So $(L_1, L_1) \notin R$ for any line $L_1$. Therefore, $R$ is not reflexive.

2. Checking Symmetry

Definition: $R$ is symmetric if whenever $(L_1, L_2) \in R$, then $(L_2, L_1) \in R$.

Suppose $(L_1, L_2) \in R$. This means $L_1$ is perpendicular to $L_2$. But perpendicularity is a mutual relationship: if $L_1$ makes a $90^\circ$ angle with $L_2$, then $L_2$ also makes a $90^\circ$ angle with $L_1$. So $L_2$ is perpendicular to $L_1$, which means $(L_2, L_1) \in R$.

Thus, $R$ is symmetric.

3. Checking Transitivity …

Method: Classifying a relation by testing each property

Use this when asked to decide which of reflexive / symmetric / transitive a relation has (it need not have all three).

Steps

Step 1: Test reflexivity

Ask whether $(a,a)\in R$ for every element. To disprove it, one element failing its self-pair is enough (e.g. an object cannot bear the relation to itself).

Step 2: Test symmetry

Assume $(a,b)\in R$ and check whether $(b,a)\in R$ always follows. If the relation is inherently mutual, symmetry holds; otherwise find a counterexample.

Step 3: Test transitivity …

Common Mistakes

Mistake 1: Claiming a line is perpendicular to itself

Why it's wrong: perpendicular means meeting at exactly $90^\circ$, which needs two distinct lines; a line makes $0^\circ$ with itself, so $(L_1,L_1)\notin R$ and $R$ is not reflexive. Correct approach: check the self-pair honestly — reflexivity fails here.

Mistake 2: Believing perpendicularity is transitive …