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Sets and Relations question

2022 · 28 Jun · Shift 2 · Q22
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  5. /2022 · 28 Jun · Shift 2 · Q22

Sets and Relations question

2022 · 28 Jun · Shift 2 · Q22

JEE MainMathematicsSets and RelationsMCQ+4 / −1
Let R1 = {(a, b) ∈\in∈ N ×\times× N : |a −-− b|≤\le≤ 13} and R2 = {(a, b) ∈\in∈ N ×\times× N : |a −-− b|eee 13}. Then on N :
  1. A
    Both R1 and R2 are equivalence relations
  2. B
    Neither R1 nor R2 is an equivalence relation
  3. C
    R1 is an equivalence relation but R2 is not
  4. D
    R2 is an equivalence relation but R1 is not
View written solutionFree

Correct answer: D

  1. Interpret the relations

    We are given relations on N\mathbb{N}N:

    R1={(a,b)∈N×N:∣a−b∣≤13}R_1 = \{(a,b) \in \mathbb{N}\times \mathbb{N} : |a-b| \le 13\}R1​={(a,b)∈N×N:∣a−b∣≤13}

    and

    R2={(a,b)∈N×N:∣a−b∣  is a multiple of 13}R_2 = \{(a,b) \in \mathbb{N}\times \mathbb{N} : |a-b| \;\text{is a multiple of }13\}R2​={(a,b)∈N×N:∣a−b∣is a multiple of 13}

    The printed expression for R2R_2R2​ appears as ∣a−b∣e13|a-b| e 13∣a−b∣e13, which is clearly intended to mean "∣a−b∣|a-b|∣a−b∣ is divisible by 131313", i.e.

    13∣∣a−b∣.13 \mid |a-b|.13∣∣a−b∣.

  2. Check whether R1R_1R1​ is an equivalence relation

    An equivalence relation must be:

    • reflexive
    • symmetric
    • transitive

    (i) Reflexive

    For every a∈Na \in \mathbb{N}a∈N, ∣a−a∣=0≤13.|a-a| = 0 \le 13.∣a−a∣=0≤13. So R1R_1R1​ is reflexive.

    (ii) Symmetric

    If (a,b)∈R1(a,b) \in R_1(a,b)∈R1​, then ∣a−b∣≤13.|a-b| \le 13.∣a−b∣≤13. But ∣b−a∣=∣a−b∣|b-a| = |a-b|∣b−a∣=∣a−b∣, so ∣b−a∣≤13,|b-a| \le 13,∣b−a∣≤13, hence (b,a)∈R1(b,a) \in R_1(b,a)∈R1​. So R1R_1R1​ is symmetric.

    (iii) Transitive

    We must check whether (a,b)∈R1 and (b,c)∈R1  ⟹  (a,c)∈R1.(a,b) \in R_1 \text{ and } (b,c) \in R_1 \implies (a,c) \in R_1.(a,b)∈R1​ and (b,c)∈R1​⟹(a,c)∈R1​.

    Take: a=1,b=14,c=27.a=1,\quad b=14,\quad c=27.a=1,b=14,c=27.

    Then ∣1−14∣=13≤13,|1-14|=13 \le 13,∣1−14∣=13≤13, so (1,14)∈R1(1,14) \in R_1(1,14)∈R1​.

    Also, ∣14−27∣=13≤13,|14-27|=13 \le 13,∣14−27∣=13≤13, so (14,27)∈R1(14,27) \in R_1(14,27)∈R1​.

    But ∣1−27∣=26>13,|1-27|=26 > 13,∣1−27∣=26>13, so (1,27)∉R1(1,27) \notin R_1(1,27)∈/R1​.

    Therefore, R1R_1R1​ is not transitive.

    Hence, R1R_1R1​ is not an equivalence relation.

  3. Check whether R2R_2R2​ is an equivalence relation

    Now interpret R2={(a,b):13∣∣a−b∣}.R_2 = \{(a,b): 13 \mid |a-b|\}.R2​={(a,b):13∣∣a−b∣}.

    This is equivalent to saying a≡b(mod13).a \equiv b \pmod{13}.a≡b(mod13).

    Again check the three properties.

    (i) Reflexive

    For every a∈Na \in \mathbb{N}a∈N, ∣a−a∣=0,|a-a|=0,∣a−a∣=0, and 13∣013 \mid 013∣0. So R2R_2R2​ is reflexive.

    (ii) Symmetric

    If 13∣∣a−b∣13 \mid |a-b|13∣∣a−b∣, then since ∣b−a∣=∣a−b∣,|b-a|=|a-b|,∣b−a∣=∣a−b∣, we also have 13∣∣b−a∣13 \mid |b-a|13∣∣b−a∣. So R2R_2R2​ is symmetric.

    (iii) Transitive

    Suppose (a,b)∈R2(a,b) \in R_2(a,b)∈R2​ and (b,c)∈R2(b,c) \in R_2(b,c)∈R2​. Then 13∣(a−b)and13∣(b−c).13 \mid (a-b) \quad \text{and} \quad 13 \mid (b-c).13∣(a−b)and13∣(b−c). Adding, 13∣[(a−b)+(b−c)]=a−c.13 \mid [(a-b)+(b-c)] = a-c.13∣[(a−b)+(b−c)]=a−c. Hence 13∣∣a−c∣,13 \mid |a-c|,13∣∣a−c∣, so (a,c)∈R2(a,c) \in R_2(a,c)∈R2​.

    Therefore, R2R_2R2​ is transitive.

    So R2R_2R2​ is an equivalence relation.

  4. Evaluate the options

    • A: Both R1R_1R1​ and R2R_2R2​ are equivalence relations → False
    • B: Neither R1R_1R1​ nor R2R_2R2​ is an equivalence relation → False
    • C: R1R_1R1​ is an equivalence relation but R2R_2R2​ is not → False
    • D: R2R_2R2​ is an equivalence relation but R1R_1R1​ is not → True
  5. Comparison with stored answer

    Stored correct answer is B, but from the above analysis the correct option is D.

    So I disagree with the stored answer.

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