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Sets and Relations question

2023 · 1 Feb · Shift 2 · Q25
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  5. /2023 · 1 Feb · Shift 2 · Q25

Sets and Relations question

2023 · 1 Feb · Shift 2 · Q25

JEE MainMathematicsSets and RelationsMCQ+4 / −1
Let P(S)P(S)P(S) denote the power set of S={1,2,3,….,10}S=\{1,2,3, \ldots ., 10\}S={1,2,3,….,10}. Define the relations R1R_{1}R1​ and R2R_{2}R2​ on P(S)P(S)P(S) as AR1 B\mathrm{AR}_{1} \mathrm{~B}AR1​ B if (A∩Bc)∪(B∩Ac)=∅\left(\mathrm{A} \cap \mathrm{B}^{\mathrm{c}}\right) \cup\left(\mathrm{B} \cap \mathrm{A}^{\mathrm{c}}\right)=\emptyset(A∩Bc)∪(B∩Ac)=∅ and AR2 B\mathrm{AR}_{2} \mathrm{~B}AR2​ B if A∪Bc=B∪Ac,∀A,B∈P(S)\mathrm{A} \cup \mathrm{B}^{\mathrm{c}}=\mathrm{B} \cup \mathrm{A}^{\mathrm{c}}, \forall \mathrm{A}, \mathrm{B} \in \mathrm{P}(\mathrm{S})A∪Bc=B∪Ac,∀A,B∈P(S). Then :
  1. A
    only R2R_{2}R2​ is an equivalence relation
  2. B
    both R1R_{1}R1​ and R2R_{2}R2​ are not equivalence relations
  3. C
    both R1R_{1}R1​ and R2R_{2}R2​ are equivalence relations
  4. D
    only R1R_{1}R1​ is an equivalence relation
View written solutionFree

Correct answer: C

  1. Given set and relations

Let S={1,2,3,…,10},P(S)=power set of S.S=\{1,2,3,\dots,10\}, \qquad P(S)=\text{power set of }S.S={1,2,3,…,10},P(S)=power set of S.

We define relations on P(S)P(S)P(S):

  • A R1 BA\,R_1\,BAR1​B if (A∩Bc)∪(B∩Ac)=∅\left(A\cap B^c\right)\cup\left(B\cap A^c\right)=\emptyset(A∩Bc)∪(B∩Ac)=∅
  • A R2 BA\,R_2\,BAR2​B if A∪Bc=B∪AcA\cup B^c=B\cup A^cA∪Bc=B∪Ac

We must check whether R1R_1R1​ and R2R_2R2​ are equivalence relations.


  1. Analyze R1R_1R1​

We know (A∩Bc)∪(B∩Ac)\left(A\cap B^c\right)\cup\left(B\cap A^c\right)(A∩Bc)∪(B∩Ac) is the symmetric difference of AAA and BBB, i.e. A△B=(A∩Bc)∪(B∩Ac).A\triangle B=(A\cap B^c)\cup(B\cap A^c).A△B=(A∩Bc)∪(B∩Ac).

So the condition for R1R_1R1​ is A△B=∅.A\triangle B=\emptyset.A△B=∅.

But A△B=∅  ⟺  A=B.A\triangle B=\emptyset \iff A=B.A△B=∅⟺A=B.

Hence, A R1 B  ⟺  A=B.A\,R_1\,B \iff A=B.AR1​B⟺A=B.

So R1R_1R1​ is just the equality relation on P(S)P(S)P(S).

Now check equivalence properties:

(i) Reflexive

For every A∈P(S)A\in P(S)A∈P(S), A=A,A=A,A=A, so A R1 AA\,R_1\,AAR1​A.

(ii) Symmetric

If A R1 BA\,R_1\,BAR1​B, then A=BA=BA=B. Hence B=AB=AB=A, so B R1 AB\,R_1\,ABR1​A.

(iii) Transitive

If A R1 BA\,R_1\,BAR1​B and B R1 CB\,R_1\,CBR1​C, then A=BA=BA=B and B=CB=CB=C. Therefore A=CA=CA=C, so A R1 CA\,R_1\,CAR1​C.

Thus, R1R_1R1​ is an equivalence relation.


  1. Analyze R2R_2R2​

Given A∪Bc=B∪Ac.A\cup B^c=B\cup A^c.A∪Bc=B∪Ac. We simplify this relation.

A standard way is to compare membership elementwise.

Take any x∈Sx\in Sx∈S. Let

  • a=1a=1a=1 if x∈Ax\in Ax∈A, else 000
  • b=1b=1b=1 if x∈Bx\in Bx∈B, else 000

Then:

  • x∈A∪Bcx\in A\cup B^cx∈A∪Bc means a∨¬ba\lor \neg ba∨¬b
  • x∈B∪Acx\in B\cup A^cx∈B∪Ac means b∨¬ab\lor \neg ab∨¬a

We need a∨¬b=b∨¬aa\lor \neg b = b\lor \neg aa∨¬b=b∨¬a for every element xxx.

Check all four cases:

aaabbba∨¬ba\lor \neg ba∨¬bb∨¬ab\lor \neg ab∨¬aEqual?
0011Yes
0101No
1010No
1111Yes

Equality holds exactly when a=ba=ba=b.

Therefore, for every x∈Sx\in Sx∈S, x∈A  ⟺  x∈B.x\in A \iff x\in B.x∈A⟺x∈B. Hence, A=B.A=B.A=B.

So, A R2 B  ⟺  A=B.A\,R_2\,B \iff A=B.AR2​B⟺A=B.

Thus R2R_2R2​ is also the equality relation on P(S)P(S)P(S).

Now check equivalence properties:

(i) Reflexive

For every AAA, A∪Ac=S=A∪Ac,A\cup A^c=S=A\cup A^c,A∪Ac=S=A∪Ac, so A R2 AA\,R_2\,AAR2​A.

(ii) Symmetric

If A R2 BA\,R_2\,BAR2​B, then A∪Bc=B∪Ac.A\cup B^c=B\cup A^c.A∪Bc=B∪Ac. Reversing sides gives B∪Ac=A∪Bc,B\cup A^c=A\cup B^c,B∪Ac=A∪Bc, so B R2 AB\,R_2\,ABR2​A.

(iii) Transitive

Since A R2 B  ⟺  A=BA\,R_2\,B \iff A=BAR2​B⟺A=B, if A R2 BA\,R_2\,BAR2​B and B R2 CB\,R_2\,CBR2​C, then A=BA=BA=B and B=CB=CB=C, hence A=CA=CA=C, so A R2 CA\,R_2\,CAR2​C.

Thus, R2R_2R2​ is an equivalence relation.


  1. Evaluate options
  • A: only R2R_2R2​ is an equivalence relation ❌
  • B: both R1R_1R1​ and R2R_2R2​ are not equivalence relations ❌
  • C: both R1R_1R1​ and R2R_2R2​ are equivalence relations ✅
  • D: only R1R_1R1​ is an equivalence relation ❌

  1. Final answer

Both R1R_1R1​ and R2R_2R2​ are equivalence relations.

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