A relation $R$ on set $A = \{1, 2, 3, 4, 5\}$ is defined as $R = \{(x, y) : |x^2 - y^2| < 8\}$. Check whether the relation $R$ is reflexive, symmetric and transitive.

> [!TLDR] > The relation $R$ is reflexive and symmetric but **not** transitive. The condition $|x^2 - y^2| < 8$ is satisfied for all pairs where the numbers are close enough, but transitivity fails because a chain of small differences can add up to a large one. We need to check three properties for $R$ on $A = \{1,2,3,4,5\}$. **Why this approach works.** Instead of blindly testing all 25 ordered pairs, we can think geometrically: $|x^2 - y^2|$ measures how far apart the squares of two numbers are. Since squares grow rapidly, only numbers that are close together will satisfy the inequality. This gives us a quick way to see which pairs belong to $R$ without exhaustive enumeration. Let's work through each property. 1. **Reflexive.** A relation is reflexive if $(a,a) \in R$ for every $a \in A$. Here $|a^2 - a^2| = 0 < 8$ for any $a$, so every element is related to itself. Reflexive holds. 2. **Symmetric.** If $(x,y) \in R$, then $|x^2 - y^2| < 8$. But $|y^2 - x^2| = |x^2 - y^2|$, so the same inequality holds for $(y,x)$. Thus whenever $xRy$, we also have $yRx$. Symmetric holds. 3. **Transitive.** This is where things get interesting. We need: if $xRy$ and $yRz$, does it follow that $xRz$? Let's test with actual numbers. First, let's list all pairs in $R$ to see the pattern. For $x,y \in \{1,2,3,4,5\}$: - $|1^2 - 2^2| = |1-4| = 3 8$, $|1^2 - 5^2| = 24 > 8$ — none of these - $|2^2 - 3^2| = |4-9| = 5 8$ → not in $R$ - $|2^2 - 5^2| = 21 > 8$ - $|3^2 - 4^2| = |9-16| = 7 8$ - $|4^2 - 5^2| = |16-25| = 9 > 8$ → not in $R$ So $R$ contains only the pairs where the numbers differ by exactly 1 (plus all reflexive pairs). That is: $$R = \{(1,1),(2,2),(3,3),(4,4),(5,5),(1,2),(2,1),(2,3),(3,

A relation R on set A = \1, 2, 3, 4, 5\ is defined as R — CBSE Class 12 Mathematics 2024 PYQ

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Concept understanding — Reflexive Transitive Not Symmetric

Reflexive, Transitive, Not Symmetric — A First Look

What kind of relation is reflexive and transitive but not symmetric? "Same grade as" won't do — it's symmetric. We need a relation that only goes one way. Let's build it.

The Intuition: A One-Way Street

The classic example is "divides" on the positive integers:

Reflexive: every number divides itself. $5 \mid 5$ . ✓
Transitive: if $a \mid b$ and $b \mid c$ , then $a \mid c$ . E.g. $2 \mid 4$ and $4 \mid 12$ gives $2 \mid 12$ . ✓
Not symmetric: $2 \mid 4$ is true, but $4 \mid 2$ is false. The relation goes only one way. ✓

The key insight: the relation can go from smaller to larger (or equal), but not back.

The Precise Statement

Let $R$ be a relation on a set $S$ . Then:

\text{Reflexive: } \forall a \in S,\; a\,R\,a

"Reflexive transitive not symmetric" is just a checklist of three properties — not a standard name like "equivalence relation". A relation with these (plus antisymmetry) is a partial order.

Why This Matters

Exams often ask: "Is this relation reflexive? Symmetric? Transitive?" Test each property independently — a relation can be reflexive and transitive but fail symmetry, and that's perfectly fine. For example, on the reals define $x\,R\,y$ if $x \leq y$ : reflexive yes, transitive yes, symmetric no ( $3 \leq 5$ but $5 \not\leq 3$ ).

Watch out

A common mistake: assuming that reflexive + transitive forces symmetry. False — both "divides" and " $\le$ " disprove it. Always test each property separately.

A Quick Table for Clarity

Property	Meaning	Example: "divides" on $\mathbb{N}$
Reflexive	Every element relates to itself	$3 \mid 3$ ✓

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The relation $R$ is reflexive and symmetric but not transitive. The condition $|x^2 - y^2| < 8$ is satisfied for all pairs where the numbers are close enough, but transitivity fails because a chain of small differences can add up to a large one.

We need to check three properties for $R$ on $A = \{1,2,3,4,5\}$ .

Why this approach works. Instead of blindly testing all 25 ordered pairs, we can think geometrically: $|x^2 - y^2|$ measures how far apart the squares of two numbers are. Since squares grow rapidly, only numbers that are close together will satisfy the inequality. This gives us a quick way to see which pairs belong to $R$ without exhaustive enumeration.

Let's work through each property.

Reflexive. A relation is reflexive if $(a,a) \in R$ for every $a \in A$ . Here $|a^2 - a^2| = 0 < 8$ for any $a$ , so every element is related to itself. Reflexive holds.
Symmetric. If $(x,y) \in R$ , then $|x^2 - y^2| < 8$ . But $|y^2 - x^2| = |x^2 - y^2|$ , so the same inequality holds for $(y,x)$ . Thus whenever $xRy$ , we also have $yRx$ . Symmetric holds.
Transitive. This is where things get interesting. We need: if $xRy$ and $yRz$ , does it follow that $xRz$ ? Let's test with actual numbers.

First, let's list all pairs in $R$ to see the pattern. For $x,y \in \{1,2,3,4,5\}$ :
- $|1^2 - 2^2| = |1-4| = 3 < 8$ → $(1,2)$ and $(2,1)$ in $R$
- $|1^2 - 3^2| = |1-9| = 8$ — not less than 8, so $(1,3)$ not in $R$
- $|1^2 - 4^2| = 15 > 8$ , $|1^2 - 5^2| = 24 > 8$ — none of these
- $|2^2 - 3^2| = |4-9| = 5 < 8$ → $(2,3)$ and $(3,2)$ in $R$
- $|2^2 - 4^2| = |4-16| = 12 > 8$ → not in $R$
- $|2^2 - 5^2| = 21 > 8$
- $|3^2 - 4^2| = |9-16| = 7 < 8$ → $(3,4)$ and $(4,3)$ in $R$
- $|3^2 - 5^2| = |9-25| = 16 > 8$
- $|4^2 - 5^2| = |16-25| = 9 > 8$ → not in $R$
So $R$ contains only the pairs where the numbers differ by exactly 1 (plus all reflexive pairs). That is:

R = \{(1,1),(2,2),(3,3),(4,4),(5,5),(1,2),(2,1),(2,3),(3,2),(3,4),(4,3)\}

Now check transitivity. Take $x=1$ , $y=2$ , $z=3$ : …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set 65/2/11 markMCQ
$R$
›Reveal solutionSolution

On $A=\{1,2,3\}$ , $R$ is not reflexive (missing $(3,3)$ ), not symmetric (has $(1,2)$ but not $(2,1)$ ), and is transitive. So $R$ is transitive only — option (D).

Check each property of $R=\{(1,1),(2,2),(1,2)\}$ on $A=\{1,2,3\}$ .

Reflexive? Requires $(a,a)\in R$ for every $a\in A$ , i.e. $(1,1),(2,2),(3,3)$ . Since $3\in A$ but $(3,3)\notin R$ , $R$ is not reflexive.

Symmetric? Requires $(b,a)\in R$ whenever $(a,b)\in R$ . Here $(1,2)\in R$ but $(2,1)\notin R$ , so $R$ is not symmetric.

Transitive? Requires $(a,c)\in R$ whenever $(a,b),(b,c)\in R$ . The only linking pairs are: …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Let R = {(4, 4), (6, 6), (7, 7), (4, 6), (6, 4), (4, 7), (6, 7)} be a relation defined on A = {4, 6, 7}, then this relation R is ................. .$
(a) Reflexive, not symmetric and not transitive
(b) Reflexive, symmetric and transitive
(c) Reflexive and transitive but not symmetric
(d) Neither Reflexive, nor symmetric and nor transitive
›Reveal solutionSolution

Check reflexivity, symmetry and transitivity directly against the listed ordered pairs.

Reflexive: $(4,4),(6,6),(7,7)$ are all present in $R$ , so $R$ is reflexive.

Symmetric: $(4,6)\in R$ and $(6,4)\in R$ is fine, but $(4,7)\in R$ while $(7,4)\notin R$ . So $R$ is NOT symmetric.

…

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CBSE 2025Set ANNUAL1 mark
$R$
›Reveal solutionSolution

Use the transitivity of the order relation $\le$ on real numbers.

The relation is $R=\{(a,b):a\le b\}$ on $\mathbb{R}$ . To prove transitivity, assume

$(a,b)\in R\quad\text{and}\quad(b,c)\in R.$

By definition of $R$ this means

$a\le b\quad\text{and}\quad b\le c.$

The usual order on real numbers is transitive, so …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Let R be the relation in the set {1, 2, 3, 4} given by R = {(1,2), (2,2), (1,1), (4,4), (1,3), (3,3), (3,2)}. Choose the correct answer in the given options.$
(a) R is reflexive and symmetric but not transitive.
(b) R is reflexive and transitive but not symmetric.
(c) R is symmetric and transitive but not reflexive.
(d) R is an equivalence relation.
›Reveal solutionSolution

Check reflexivity (every (a,a) present), symmetry (every pair's reverse present) and transitivity (chain closure) directly against the listed pairs.

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

Reflexive: We need (1,1), (2,2), (3,3), (4,4) ∈ R. All four are present, so R is reflexive.

Symmetric: (1,2) ∈ R but (2,1) ∉ R, so R is NOT symmetric.

…

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CBSE 2020Set 65/2/11 markMCQ
$\{1, 2, 3\}$
›Reveal solutionSolution
The relation $R = \{(1, 2), (2, 1), (1, 1)\}$ on the set $\{1, 2, 3\}$ is symmetric, but it is neither reflexive nor transitive. Therefore, option (C) is correct.

To determine the properties of the given relation $R$ on the set $A = \{1, 2, 3\}$ , we need to check if it satisfies the definitions of reflexivity, symmetry, and transitivity. Understanding these definitions precisely is key to avoiding common errors.

A relation $R$ on a set $A$ is:

Reflexive if for every element $a \in A$ , the ordered pair $(a, a)$ is in $R$ . This means every element must be related to itself.

Symmetric if for every pair $(a, b) \in R$ , the pair $(b, a)$ is also in $R$ . This means if $a$ is related to $b$ , then $b$ must also be related to $a$ .

Transitive if for every $a, b, c \in A$ , whenever $(a, b) \in R$ and $(b, c) \in R$ , it must follow that $(a, c) \in R$ . This means if $a$ is related to $b$ and $b$ is related to $c$ , then $a$ must also be related to $c$ .

Let's examine $R = \{(1, 2), (2, 1), (1, 1)\}$ on the set $A = \{1, 2, 3\}$ step by step.

Check for Reflexivity:

For $R$ to be reflexive, every element in $A$ must be related to itself. That is, $(1, 1)$ , $(2, 2)$ , and $(3, 3)$ must all be present in $R$ .

We see that $(1, 1) \in R$ .

However, $(2, 2) \notin R$ .

Also, $(3, 3) \notin R$ . Since $(2, 2)$ and $(3, 3)$ are not in $R$ , the relation $R$ is not reflexive.

Check for Symmetry:

For $R$ to be symmetric, for every pair $(a, b) \in R$ , the reverse pair $(b, a)$ must also be in $R$ .

Let's check each pair in $R$ :

For $(1, 2) \in R$ , we need to check if $(2, 1) \in R$ . Yes, $(2, 1) \in R$ .

For $(2, 1) \in R$ , we need to check if $(1, 2) \in R$ . Yes, $(1, 2) \in R$ .

For $(1, 1) \in R$ , we need to check if $(1, 1) \in R$ . Yes, $(1, 1) \in R$ . Since for every ordered pair $(a, b)$ in $R$ , the pair $(b, a)$ is also in $R$ , the relation $R$ is symmetric.

Check for Transitivity:

For $R$ to be transitive, if $(a, b) \in R$ and $(b, c) \in R$ , then $(a, c)$ must also be in $R$ . We need to check all such combinations.

Consider $(1, 2) \in R$ and $(2, 1) \in R$ . Here, $a=1, b=2, c=1$ . According to the definition of transitivity, $(a, c) = (1, 1)$ must be in $R$ . We see that $(1, 1) \in R$ . This case holds.

Consider $(2, 1) \in R$ and $(1, 2) \in R$ . Here, $a=2, b=1, c=2$ . …
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Q.A relation RRR on set A={1,2,3,4,5}A = \{1, 2, 3, 4, 5\}A={1,2,3,4,5} is defined as R={(x,y):∣x2−y2∣<8}R = \{(x, y) : |x^2 - y^2| < 8\}R={(x,y):∣x2−y2∣<8}. Check whether the relation RRR is reflexive, symmetric and transitive.

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Q.A relation $R$ on set $A = \{1, 2, 3, 4, 5\}$ is defined as $R = \{(x, y) : |x^2 - y^2| < 8\}$ . Check whether the relation $R$ is reflexive, symmetric and transitive.

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