Q.Check whether the relation R defined in the set $\{1, 2, 3, 4, 5, 6\}$ as $R = \{(a, b) : b = a + 1\}$ is reflexive, symmetric or transitive.

Showing the 12 most recent of 41 on this concept.

1. CBSE 2020Set 65/1/11 mark
   Q.If for all $a_1, a_2 \in A$, $(a_1, a_2) \in R$ implies $(a_2, a_1) \in R$, then the relation R defined on set A is called a _________ relation.

   ›Reveal solutionSolution

   The property described — whenever $(a_1, a_2) \in R$ then $(a_2, a_1) \in R$ — is the definition of a symmetric relation. The blank should be filled with symmetric.

   Let’s understand why this is the correct classification.

   A relation $R$ on a set $A$ is simply a collection of ordered pairs $(x, y)$ where $x, y \in A$. Different properties of relations describe what patterns these pairs follow. The three most common properties you encounter in exam problems are:

   • Reflexive: Every element is related to itself — $(a, a) \in R$ for all $a \in A$.
   • Symmetric: If $a$ is related to $b$, then $b$ is related back to $a$ — exactly the condition given.
   • Transitive: If $a$ is related to $b$ and $b$ is related to $c$, then $a$ is related to $c$.

   The statement in the question is the textbook definition of symmetry. There is no extra condition — it does not say “for all $a_1, a_2$” means every pair must be present; it only says whenever a pair is present, its reverse must also be present.

   1. Identify the condition: The given statement is:

      For all $a_1, a_2 \in A$, if $(a_1, a_2) \in R$ then $(a_2, a_1) \in R$.

      This is a conditional statement — it does not force any particular pair to exist; it only imposes a requirement on pairs that do exist.

   2. Match to the known property:

      • Reflexive would require $(a, a) \in R$ for every $a$, which is not mentioned.
      • Transitive would involve three elements and a chain condition, not just swapping two elements.
      • Symmetric is exactly: “if $aRb$ then $bRa$”. The phrasing “$(a_1, a_2) \in R$ implies $(a_2, a_1) \in R$” is the formal way to write this.

   3. Check a simple example:

      Let $A = \{1, 2, 3\}$ and $R = \{(1, 2), (2, 1)\}$.

      • For $(1, 2) \in R$, we have $(2, 1) \in R$ — condition holds.
      • For $(2, 1) \in R$, we have $(1, 2) \in R$ — condition holds.
      • No other pairs exist, so the condition is vacuously true for them. This relation is symmetric. If we had $R = \{(1, 2)\}$ only, then $(1, 2) \in R$ but $(2, 1) \notin R$, violating symmetry.

   Watch out

   A common mistake is to confuse symmetry with reflexivity or to think that symmetry requires every element to be related to every other element. Symmetry only governs the direction of existing pairs — it does not create new pairs.

   Tip

   To test symmetry quickly: check if the relation’s matrix (if you think of it as a table) is symmetric about the main diagonal. For a set of ordered pairs, just verify that for every $(x, y)$ you also have $(y, x)$.

   Thus, the blank is filled by the word symmetric.

   ✓Final answer

   The relation is called a symmetric relation.

2. CBSE 2026Set V11 markMCQ
   Q.If a relation $R$ in the set $\{1, 2, 3\}$ be defined by $R = \{(1, 1),(2, 2)\}$ then $R$ is

   (a) symmetric but not transitive

   (b) transitive but not symmetric

   (c) symmetric and transitive

   (d) neither symmetric nor transitive

   ›Reveal solutionSolution

   $R=\{(1,1),(2,2)\}$ is both symmetric and transitive, so the answer is (c).

   Symmetry: whenever $(a,b)\in R$ we need $(b,a)\in R$. Here the only pairs are $(1,1)$ and $(2,2)$, and each is its own reverse, so symmetry holds.

   Transitivity: whenever $(a,b)\in R$ and $(b,c)\in R$ we need $(a,c)\in R$. The only chains available are $(1,1),(1,1)\Rightarrow(1,1)$ and $(2,2),(2,2)\Rightarrow(2,2)$, both already present, so transitivity holds.

   (It is not reflexive since $(3,3)\notin R$, but that is not asked.)

   ✓Final answer

   (c) symmetric and transitive

3. CBSE 2026Set A1 markMCQ
   Q.What type of a relation is "less than" in the set of real numbers?

   (a) Only symmetric

   (b) Only transitive

   (c) Only reflexive

   (d) Equivalence

   ›Reveal solutionSolution

   "$<$" on $\mathbb{R}$ is transitive only.

   Consider the relation "$a < b$" on $\mathbb{R}$:

   • Reflexive? $a < a$ is false for every $a$, so NOT reflexive.
   • Symmetric? $a < b$ does not imply $b < a$, so NOT symmetric.
   • Transitive? If $a < b$ and $b < c$ then $a < c$, so YES transitive.

   Hence it is only transitive.

   ✓Final answer

   (b) Only transitive.

4. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $R$ is the relation $\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}$ on $A=\{1,2,3\}$, then which one of the following is true for $R$?

   (a) Reflexive but not symmetric

   (b) Reflexive but not transitive

   (c) Symmetric and transitive

   (d) Neither symmetric nor transitive

   ›Reveal solutionSolution

   $R$ is reflexive (it contains every $(x,x)$) but not symmetric, since $(1,2)\in R$ while $(2,1)\notin R$.

   Given: $R=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}$ on $A=\{1,2,3\}$.

   Reflexivity: $R$ is reflexive if $(x,x)\in R$ for every $x\in A$. Here $(1,1),(2,2),(3,3)$ are all present, so $R$ is reflexive.

   Symmetry: $R$ is symmetric if $(x,y)\in R\Rightarrow(y,x)\in R$. Here $(1,2)\in R$ but $(2,1)\notin R$. So $R$ is not symmetric.

   Transitivity (for completeness): $(1,2)\in R,(2,3)\in R\Rightarrow(1,3)\in R$, which holds. Checking every other pair shows $R$ is in fact transitive too — but among the given options, the one that correctly describes $R$ is "reflexive but not symmetric".

   ✓Final answer

   The correct option is (a) Reflexive but not symmetric.

5. CBSE 2026Set ANNUAL1 markMCQ
   Q.Let $R$ be the relation in the set $N$ given by $R = \{(a, b) : a = b - 2, b > 6\}$, choose the correct answer:

   (a) $(2, 4) \in R$

   (b) $(3, 8) \in R$

   (c) $(6, 8) \in R$

   (d) $(8, 6) \in R$

   ›Reveal solutionSolution

   Check each pair against both conditions $a=b-2$ and $b>6$.

   $R=\{(a,b): a=b-2,\ b>6\}$

   1. $(2,4)$: $a=b-2 \Rightarrow 2=2$ ✓, but $b>6 \Rightarrow 4>6$ ✗. Not in $R$.
   2. $(3,8)$: $a=b-2 \Rightarrow 3=6$ ✗. Not in $R$.
   3. $(6,8)$: $a=b-2 \Rightarrow 6=6$ ✓, $b>6 \Rightarrow 8>6$ ✓. In $R$.
   4. $(8,6)$: $a=b-2 \Rightarrow 8=4$ ✗. Not in $R$.

   ✓Final answer

   Option (c): $(6, 8) \in R$

6. CBSE 2026Set ANNUAL1 markMCQ
   Q.Choose the correct answer : Let $R$ be a relation in the set $\{1,2,3,4\}$ given by $R=\{(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2)\}$. Then

   (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

   Test $R$ against the three defining properties (reflexive, symmetric, transitive) one at a time, directly from its listed ordered pairs.

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

   Reflexive? Need $(a,a)\in R$ for every $a\in\{1,2,3,4\}$: $(1,1)$ ✓, $(2,2)$ ✓, $(3,3)$ ✓, $(4,4)$ ✓. All present — $R$ is reflexive.

   Symmetric? Need: whenever $(a,b)\in R$, also $(b,a)\in R$. Take $(1,2)\in R$: is $(2,1)\in R$? It is not in the list. So $R$ is not symmetric.

   Transitive? Need: whenever $(a,b)\in R$ and $(b,c)\in R$, also $(a,c)\in R$. Checking every chain:

   • $(1,1),(1,2)\Rightarrow(1,2)$ ✓
   • $(1,1),(1,3)\Rightarrow(1,3)$ ✓
   • $(1,2),(2,2)\Rightarrow(1,2)$ ✓
   • $(1,3),(3,3)\Rightarrow(1,3)$ ✓
   • $(1,3),(3,2)\Rightarrow(1,2)$ ✓
   • $(3,3),(3,2)\Rightarrow(3,2)$ ✓
   • $(3,2),(2,2)\Rightarrow(3,2)$ ✓
   • $(2,2),(2,2)\Rightarrow(2,2)$ ✓, $(4,4),(4,4)\Rightarrow(4,4)$ ✓

   Every chain closes back into $R$ — $R$ is transitive.

   So $R$ is reflexive and transitive, but not symmetric (hence not an equivalence relation either).

   ✓Final answer

   (b) $R$ is reflexive and transitive but not symmetric

7. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $R$ be the relation in the set $N$ given by $R = \{(a,b): a = b - 2, b > 6\}$, then

   (a) $(2, 4) \in R$

   (b) $(3, 8) \in R$

   (c) $(6, 8) \in R$

   (d) $(8, 7) \in R$

   ›Reveal solutionSolution

   Only $(6,8)$ satisfies $a=b-2$ with $b>6$.

   Check each: (a) $(2,4)$: $b=4\not>6$. (b) $(3,8)$: $3 e8-2=6$. (c) $(6,8)$: $6=8-2$ and $8>6$ ✓. (d) $(8,7)$: $8 e7-2=5$.

   ✓Final answer

   (c) $(6,8)\in R$.

8. CBSE 2025Set X11 markMCQ
   Q.A relation $R$ in a set $A$ is called Reflexive relation if

   (a) $(a, a) \in R$ for all $a \in A$

   (b) $(a, a) \in R$ for atleast one $a \in A$

   (c) $(a, b) \in R$ implies $(b, a) \in R$

   (d) $(a, b) \in R$ and $(b, c) \in R$ implies $(a, c) \in R$

   ›Reveal solutionSolution

   Tests the definition of a reflexive relation — correct option is (a).

   A relation $R$ on a set $A$ is reflexive when every element is related to itself. Formally, $(a,a)\in R$ must hold for all $a\in A$ — not just for at least one element. Option (c) states symmetry ($(a,b)\in R\Rightarrow(b,a)\in R$) and option (d) states transitivity, so they describe different properties. Only option (a) is the definition of reflexivity.

   ✓Final answer

   (a) $(a, a) \in R \text{ for all } a \in A$

9. CBSE 2025Set IX1 markMCQ
   Q.A relation $R = \{(a, b) : a = b - 1,\ b \geq 3\}$ is defined on set $N$, then

   (a) $(2, 4) \in R$

   (b) $(4, 5) \in R$

   (c) $(4, 6) \in R$

   (d) $(1, 3) \in R$

   ›Reveal solutionSolution

   Only $(4,5)$ satisfies $a=b-1$ with $b\ge 3$; option (b).

   Concept. A pair $(a,b)$ belongs to $R$ only if it satisfies both conditions: $a=b-1$ and $b\ge 3$.

   • $(2,4)$: $b-1=3\ne 2$. ✗

   • $(4,5)$: $b-1=5-1=4=a$ ✓ and $b=5\ge 3$ ✓. Belongs.

   • $(4,6)$: $b-1=5\ne 4$. ✗

   • $(1,3)$: $b-1=2\ne 1$. ✗

   ✓Final answer

   (b) $(4,5)\in R$.

10. CBSE 2025Set ANNUAL1 markMCQ
    Q.If $A=\{1,2,3,4\}$ and $R=\{(a,b)\mid a+b\ \text{is an odd number},\ a,b\in A\}$ is a relation from $A$ to $A$, then which of the following is true for the relation $R$?

    (i) Reflexive

    (ii) Symmetric

    (iii) Transitive

    (iv) Equivalent

    ›Reveal solutionSolution

    Check each property directly: $R$ turns out to be symmetric only.

    $A=\{1,2,3,4\}$, $R=\{(a,b): a+b \text{ is odd}\}$.

    Reflexive? $(a,a)$ needs $a+a=2a$ to be odd — but $2a$ is always even. So $R$ is not reflexive.

    Symmetric? If $a+b$ is odd, then $b+a=a+b$ is the same sum, also odd. So $(a,b)\in R \Rightarrow (b,a)\in R$. $R$ is symmetric.

    Transitive? Take $(1,2)\in R$ (since $1+2=3$ odd) and $(2,3)\in R$ (since $2+3=5$ odd). For transitivity we'd need $(1,3)\in R$, but $1+3=4$ is even. So $R$ is not transitive.

    Hence only the symmetric property holds.

    ✓Final answer

    (ii) Symmetric.

11. CBSE 2025Set ANNUAL1 markMCQ
    Q.Let $A = \{a, b, c\}$ and $R = \{(a, a), (a, b), (b, a)\}$, then $R$ is

    (a) reflexive and symmetric but not transitive

    (b) reflexive and transitive but not symmetric

    (c) symmetric and transitive but not reflexive

    (d) an equivalence relation

    ›Reveal solutionSolution

    R is symmetric (the only cross-pair (a,b)/(b,a) both appear) and not transitive ((b,a) & (a,b) would force (b,b), which is missing) — matching option (a) once we note a reflexivity caveat below.

    Step 1 — Reflexive? Full reflexivity on A = {a, b, c} needs (a,a), (b,b), (c,c) all in R. Only (a,a) is present; (b,b) and (c,c) are not. So strictly, R is not fully reflexive on {a,b,c}.

    Step 2 — Symmetric? The only pair with a 'partner' is (a,b), and its reverse (b,a) is also in R. There is no pair in R whose reverse is missing. So R is symmetric.

    Step 3 — Transitive? Take (b,a) ∈ R and (a,b) ∈ R: transitivity would require (b,b) ∈ R. It is not. So R is not transitive.

    Step 4 — Match to options. (d) is out (not an equivalence relation, since not transitive). (b) and (c) both claim R is transitive, which we disproved. That leaves (a) as the only remaining choice, and its two correct claims (symmetric, not transitive) both check out.

    ✓Final answer

    (a) reflexive and symmetric but not transitive. (R is definitely symmetric and definitely not transitive; note that strict reflexivity on {a,b,c} would also require (b,b) and (c,c), which are not in R — a known imprecision in how this frequently-repeated question is worded, but (a) is the only option consistent with R's actual symmetric/non-transitive behaviour.)

12. CBSE 2025Set ANNUAL1 markMCQ
    Q.Relation R = {(x, y) : x < y² where x, y ∈ R} is:

    (a) Reflexive but not symmetric.

    (b) Symmetric and transitive but not Reflexive.

    (c) Reflexive and Symmetric.

    (d) Neither reflexive nor symmetric nor transitive.

    ›Reveal solutionSolution

    Test each property of $R=\{(x,y): x<y^2\}$ with concrete counterexamples — all three fail.

    Reflexive? Need $x<x^2$ for every $x\in\mathbb R$. Take $x=\dfrac12$: is $\dfrac12 < \dfrac14$? No. So $R$ is not reflexive.

    Symmetric? Need $x<y^2 \Rightarrow y<x^2$. Take $x=0,\,y=1$: $0<1^2$ is true, so $(0,1)\in R$. But is $1<0^2=0$? No. So $(1,0)\notin R$, and $R$ is not symmetric.

    Transitive? Need $x<y^2,\ y<z^2 \Rightarrow x<z^2$. Take $x=4,\,y=-3,\,z=1$: $4<(-3)^2=9$ ✓, and $-3<1^2=1$ ✓, but $4<1^2=1$? No. So $R$ is not transitive.

    Since all three properties fail, $R$ is neither reflexive, symmetric, nor transitive.

    ✓Final answer

    Neither reflexive nor symmetric nor transitive — option (d).