Q.If $A = \{1, 2, 3, 4\}$, define relations on $A$ which have properties of being:

Key idea: A relation is a set of ordered pairs. Reflexive means every element relates to itself; symmetric means if $(a,b)$ is in the relation then $(b,a)$ is also; transitive means if $(a,b)$ and $(b,c)$ are in the relation then $(a,c)$ is also.

(a) Reflexive, transitive, not symmetric.

We need all $(1,1),(2,2),(3,3),(4,4)$. To break symmetry, include $(1,2)$ but not $(2,1)$. For transitivity, if we add $(2,3)$, we must also add $(1,3)$.

One example:

$R = \{(1,1),(2,2),(3,3),(4,4),(1,2),(2,3),(1,3)\}$.

(b) Symmetric, neither reflexive nor transitive.

Symmetric means pairs come in twos. To avoid reflexivity, omit at least one $(a,a)$. To avoid transitivity, include $(1,2)$ and $(2,3)$ but not $(1,3)$.

One example: …

The core idea

A relation on a set is just a set of ordered pairs. The properties — reflexive, symmetric, transitive — are conditions on which pairs must or must not be present. To build a relation with a specific combination of properties, we think about what each property forces us to include, and what it forbids us from including.

Reflexive means every element must be related to itself: $(1,1), (2,2), (3,3), (4,4)$ must all be in the relation.

Symmetric means if $(a,b)$ is in, then $(b,a)$ must also be in.

Transitive means if $(a,b)$ and $(b,c)$ are in, then $(a,c)$ must be in.

The trick is to add pairs carefully so that one property holds while another fails.

(a) Reflexive, transitive but not symmetric

We want a relation that behaves like "less than or equal to" on numbers. That's reflexive ($a \le a$), transitive (if $a \le b$ and $b \le c$ then $a \le c$), but not symmetric (if $a \le b$, we don't necessarily have $b \le a$).

So define $R_1$ as the usual $\le$ order on the numbers $1,2,3,4$:

Check:

• Reflexive: every $(i,i)$ is present. ✓
• Transitive: if $a \le b$ and $b \le c$, then $a \le c$ — all such pairs are included. ✓
• Not symmetric: $(1,2)$ is in, but $(2,1)$ is not. ✓

(b) Symmetric but neither reflexive nor transitive

We want a relation where if $(a,b)$ is in, $(b,a)$ is also in, but some element is not related to itself, and the relation fails transitivity.

A simple choice: "is at distance 1" on the number line. That is, two numbers are related if they differ by exactly 1.

Check:

• Symmetric: every pair appears with its reverse. ✓
• Not reflexive: no $(1,1)$, $(2,2)$, etc. ✓
• Not transitive: we have $(1,2)$ and $(2,3)$, but $(1,3)$ is missing. ✓ …

Method: Building a relation from a property checklist

Use this whenever a question asks you to construct a relation on a finite set with a prescribed mix of reflexive / symmetric / transitive (holding some, deliberately failing others).

Steps

Step 1: Turn each REQUIRED property into forced pairs.

• Reflexive forces every $(a,a)$ to be present.
• Symmetric forces: for each $(a,b)$ you include, also include $(b,a)$.
• Transitive forces: for each chain $(a,b)$ and $(b,c)$, also include $(a,c)$.

Step 2: Turn each FORBIDDEN property into a blocking pair.

• To be not symmetric, include one $(a,b)$ but leave out $(b,a)$.
• To be not reflexive, omit at least one $(a,a)$.
• To be not transitive, plant a chain $(a,b),(b,c)$ but withhold $(a,c)$.

Step 3: Start from the forced pairs, add the smallest set that makes the failures happen, then re-close. …

Common Mistakes

Mistake 1: Reading 'not symmetric' as 'antisymmetric'.

Why it's wrong: antisymmetric means $(a,b)$ and $(b,a)$ together force $a=b$; 'not symmetric' only needs one pair whose reverse is missing. Correct approach: to fail symmetry, just include a single $(a,b)$ and deliberately leave $(b,a)$ out.

Mistake 2: Adding a pair for one property and forgetting the ripple.

Why it's wrong: putting $(2,3)$ into a set that already has $(1,2)$ silently forces $(1,3)$ if you want transitivity — students add pairs without re-closing. Correct approach: after each new pair, re-scan for any reverse or chain the required properties now demand. …