Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Sets and Relations question

2023 · 31 Jan · Shift 1 · Q30
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Sets and Relations
  5. /2023 · 31 Jan · Shift 1 · Q30

Sets and Relations question

2023 · 31 Jan · Shift 1 · Q30

JEE MainMathematicsSets and RelationsMCQ+4 / −1
Let R\mathrm{R}R be a relation on N×N\mathrm{N} \times \mathbb{N}N×N defined by (a,b) R (c,d)(a, b) ~\mathrm{R}~(c, d)(a,b) R (c,d) if and only if ad(b−c)=bc(a−d)a d(b-c)=b c(a-d)ad(b−c)=bc(a−d). Then R\mathrm{R}R is
  1. A
    symmetric and transitive but not reflexive
  2. B
    reflexive and symmetric but not transitive
  3. C
    transitive but neither reflexive nor symmetric
  4. D
    symmetric but neither reflexive nor transitive
View written solutionFree

Correct answer: D

  1. Given relation

On N×N\mathbb{N} \times \mathbb{N}N×N, define (a,b) R (c,d)  ⟺  ad(b−c)=bc(a−d).(a,b)\,R\,(c,d) \iff ad(b-c)=bc(a-d).(a,b)R(c,d)⟺ad(b−c)=bc(a−d).

We must check whether RRR is reflexive, symmetric, and transitive.

  1. Simplify the relation

Expand both sides: ad(b−c)=abd−acd,ad(b-c)=abd-acd,ad(b−c)=abd−acd, bc(a−d)=abc−bcd.bc(a-d)=abc-bcd.bc(a−d)=abc−bcd.

So the condition is abd−acd=abc−bcd.abd-acd=abc-bcd.abd−acd=abc−bcd.

Bring all terms to one side: abd−acd−abc+bcd=0.abd-acd-abc+bcd=0.abd−acd−abc+bcd=0.

Factor by grouping: ad(b−c)−bc(a−d)=0,ad(b-c)-bc(a-d)=0,ad(b−c)−bc(a−d)=0, or directly, ab(d−c)+cd(b−a)=0.ab(d-c)+cd(b-a)=0.ab(d−c)+cd(b−a)=0.

A better rearrangement is: abd+bcd=abc+acdabd+bcd=abc+acdabd+bcd=abc+acd d(ab+bc)=c(ab+ad).d(ab+bc)=c(ab+ad).d(ab+bc)=c(ab+ad).

This is not immediately very useful for standard properties, so we test directly.

  1. Reflexivity

For reflexive, we need (a,b)R(a,b)(a,b)R(a,b)(a,b)R(a,b) for every (a,b)∈N×N(a,b)\in \mathbb{N}\times\mathbb{N}(a,b)∈N×N.

Put (c,d)=(a,b)(c,d)=(a,b)(c,d)=(a,b) in the definition: ab(b−a)=ba(a−b).ab(b-a)=ba(a-b).ab(b−a)=ba(a−b). Since ba=abba=abba=ab and a−b=−(b−a)a-b=-(b-a)a−b=−(b−a), ab(b−a)=−ab(b−a).ab(b-a)=-ab(b-a).ab(b−a)=−ab(b−a). Thus 2ab(b−a)=0.2ab(b-a)=0.2ab(b−a)=0. Since a,b∈Na,b\in\mathbb Na,b∈N, we have ab≠0ab\neq 0ab=0, so this requires b−a=0  ⟺  a=b.b-a=0 \iff a=b.b−a=0⟺a=b.

Hence (a,b)R(a,b)(a,b)R(a,b)(a,b)R(a,b) only when a=ba=ba=b, not for all pairs.

So RRR is not reflexive.

  1. Symmetry

Assume (a,b)R(c,d).(a,b)R(c,d).(a,b)R(c,d). Then ad(b−c)=bc(a−d).ad(b-c)=bc(a-d).ad(b−c)=bc(a−d).

To check symmetry, we need to see whether this implies (c,d)R(a,b),(c,d)R(a,b),(c,d)R(a,b), that is, cb(d−a)=da(c−b).cb(d-a)=da(c-b).cb(d−a)=da(c−b).

Now note: cb(d−a)=bc(−(a−d))=−bc(a−d),cb(d-a)=bc(-(a-d))=-bc(a-d),cb(d−a)=bc(−(a−d))=−bc(a−d), and da(c−b)=ad(−(b−c))=−ad(b−c).da(c-b)=ad(-(b-c))=-ad(b-c).da(c−b)=ad(−(b−c))=−ad(b−c).

So the equation cb(d−a)=da(c−b)cb(d-a)=da(c-b)cb(d−a)=da(c−b) is exactly −bc(a−d)=−ad(b−c),-bc(a-d)=-ad(b-c),−bc(a−d)=−ad(b−c), which is equivalent to bc(a−d)=ad(b−c),bc(a-d)=ad(b-c),bc(a−d)=ad(b−c), true by the given relation.

Therefore RRR is symmetric.

  1. Transitivity

We must check whether (a,b)R(c,d) and (c,d)R(e,f)  ⟹  (a,b)R(e,f).(a,b)R(c,d) \text{ and } (c,d)R(e,f) \implies (a,b)R(e,f).(a,b)R(c,d) and (c,d)R(e,f)⟹(a,b)R(e,f).

To disprove transitivity, one counterexample is enough.

Take: (a,b)=(1,2),(c,d)=(2,1),(e,f)=(2,2).(a,b)=(1,2),\quad (c,d)=(2,1),\quad (e,f)=(2,2).(a,b)=(1,2),(c,d)=(2,1),(e,f)=(2,2).

Check (1,2)R(2,1)(1,2)R(2,1)(1,2)R(2,1)

We need 1⋅1 (2−2)=2⋅2 (1−1).1\cdot 1\,(2-2)=2\cdot 2\,(1-1).1⋅1(2−2)=2⋅2(1−1). That is, 0=0,0=0,0=0, so true.

Check (2,1)R(2,2)(2,1)R(2,2)(2,1)R(2,2)

We need 2⋅2 (1−2)=1⋅2 (2−2).2\cdot 2\,(1-2)=1\cdot 2\,(2-2).2⋅2(1−2)=1⋅2(2−2). That is, 4(−1)=0,4(-1)=0,4(−1)=0, −4=0,-4=0,−4=0, which is false.

So this choice fails. We need a better counterexample.

Let us derive a more usable form.

From ad(b−c)=bc(a−d),ad(b-c)=bc(a-d),ad(b−c)=bc(a−d), expand and rearrange: abd−acd=abc−bcdabd-acd=abc-bcdabd−acd=abc−bcd abd+bcd=abc+acdabd+bcd=abc+acdabd+bcd=abc+acd d(ab+bc)=c(ab+ad).d(ab+bc)=c(ab+ad).d(ab+bc)=c(ab+ad). Thus, for fixed positive integers, dc=ab+adab+bc,\frac{d}{c}=\frac{ab+ad}{ab+bc},cd​=ab+bcab+ad​, not especially convenient.

Instead, search for explicit examples.

Take (a,b)=(1,2)(a,b)=(1,2)(a,b)=(1,2) and (c,d)=(2,4)(c,d)=(2,4)(c,d)=(2,4): 1⋅4(2−2)=2⋅2(1−4)1\cdot 4(2-2)=2\cdot 2(1-4)1⋅4(2−2)=2⋅2(1−4) 0=−12,0=-12,0=−12, false.

Try making b=cb=cb=c; then relation becomes ad(0)=b2(a−d)  ⟹  a=d.ad(0)=b^2(a-d) \implies a=d.ad(0)=b2(a−d)⟹a=d. So if b=cb=cb=c, then we need a=da=da=d.

Try making a=da=da=d; then relation becomes a2(b−c)=bc(0)  ⟹  b=c.a^2(b-c)=bc(0) \implies b=c.a2(b−c)=bc(0)⟹b=c. So if a=da=da=d, then we need b=cb=cb=c.

Now test simple valid related pairs.

Take (1,2)(1,2)(1,2) and (2,1)(2,1)(2,1): 1⋅1(2−2)=2⋅2(1−1)=0,1\cdot1(2-2)=2\cdot2(1-1)=0,1⋅1(2−2)=2⋅2(1−1)=0, so yes.

Now choose a third pair related to (2,1)(2,1)(2,1). Since relation is symmetric, maybe (1,2)(1,2)(1,2) again, but then transitivity would hold trivially. Need another one.

Check (2,1)R(3,2)(2,1)R(3,2)(2,1)R(3,2): 2⋅2(1−3)=1⋅3(2−2)2\cdot2(1-3)=1\cdot3(2-2)2⋅2(1−3)=1⋅3(2−2) 4(−2)=0,4(-2)=0,4(−2)=0, false.

Check (2,1)R(1,3)(2,1)R(1,3)(2,1)R(1,3): 2⋅3(1−1)=1⋅1(2−3)2\cdot3(1-1)=1\cdot1(2-3)2⋅3(1−1)=1⋅1(2−3) 0=−1,0=-1,0=−1, false.

Check (2,1)R(4,2)(2,1)R(4,2)(2,1)R(4,2): 2⋅2(1−4)=1⋅4(2−2)2\cdot2(1-4)=1\cdot4(2-2)2⋅2(1−4)=1⋅4(2−2) −12=0,-12=0,−12=0, false.

Let us solve for pairs related to (1,2)(1,2)(1,2).

Condition: 1⋅d(2−c)=2c(1−d).1\cdot d(2-c)=2c(1-d).1⋅d(2−c)=2c(1−d). So d(2−c)=2c−2cdd(2-c)=2c-2cdd(2−c)=2c−2cd 2d−cd=2c−2cd2d-cd=2c-2cd2d−cd=2c−2cd 2d+cd=2c2d+cd=2c2d+cd=2c d(c+2)=2cd(c+2)=2cd(c+2)=2c d=2cc+2.d=\frac{2c}{c+2}.d=c+22c​.

For c,d∈Nc,d\in\mathbb Nc,d∈N, this gives:

  • c=1⇒d=23c=1 \Rightarrow d=\frac23c=1⇒d=32​ not natural
  • c=2⇒d=1c=2 \Rightarrow d=1c=2⇒d=1
  • c=4⇒d=86c=4 \Rightarrow d=\frac{8}{6}c=4⇒d=68​ not natural and for larger ccc, not integer generally.

So the only natural-number pair related to (1,2)(1,2)(1,2) is (2,1)(2,1)(2,1).

Similarly, pairs related to (2,1)(2,1)(2,1) satisfy 2d(1−c)=c(2−d).2d(1-c)=c(2-d).2d(1−c)=c(2−d). Expand: 2d−2cd=2c−cd2d-2cd=2c-cd2d−2cd=2c−cd 2d=2c+cd=c(d+2)2d=2c+cd=c(d+2)2d=2c+cd=c(d+2) c=2dd+2.c=\frac{2d}{d+2}.c=d+22d​. For natural numbers, only d=2d=2d=2 gives c=1c=1c=1. So the only pair related to (2,1)(2,1)(2,1) is (1,2)(1,2)(1,2).

Thus these do not help produce a counterexample.

Let us find a whole chain by trying equal pairs (k,k)(k,k)(k,k). For (a,b)=(k,k)(a,b)=(k,k)(a,b)=(k,k) and (c,d)(c,d)(c,d): kd(k−c)=kc(k−d).kd(k-c)=kc(k-d).kd(k−c)=kc(k−d). If k≠0k\neq 0k=0, divide by kkk: d(k−c)=c(k−d)d(k-c)=c(k-d)d(k−c)=c(k−d) dk−dc=ck−cddk-dc=ck-cddk−dc=ck−cd dk=ckdk=ckdk=ck d=c.d=c.d=c. So (k,k)(k,k)(k,k) is related to every (m,m)(m,m)(m,m).

Now take (a,b)=(1,1),(c,d)=(2,2).(a,b)=(1,1),\quad (c,d)=(2,2).(a,b)=(1,1),(c,d)=(2,2). Then (1,1)R(2,2)(1,1)R(2,2)(1,1)R(2,2) is true. Also take (e,f)=(3,3).(e,f)=(3,3).(e,f)=(3,3). Then (2,2)R(3,3)(2,2)R(3,3)(2,2)R(3,3) is true, and (1,1)R(3,3)(1,1)R(3,3)(1,1)R(3,3) is also true. So still no contradiction.

Let us simplify algebraically in another way.

Starting from ad(b−c)=bc(a−d),ad(b-c)=bc(a-d),ad(b−c)=bc(a−d), expand: abd−acd=abc−bcdabd-acd=abc-bcdabd−acd=abc−bcd ab(d−c)=cd(a−b).ab(d-c)=cd(a-b).ab(d−c)=cd(a−b). So equivalently, (a,b)R(c,d)  ⟺  ab(d−c)=cd(a−b).(a,b)R(c,d) \iff ab(d-c)=cd(a-b).(a,b)R(c,d)⟺ab(d−c)=cd(a−b).

Now divide by abcdabcdabcd (valid since all are natural numbers): d−ccd=a−bab.\frac{d-c}{cd}=\frac{a-b}{ab}.cdd−c​=aba−b​. Hence 1c−1d=1b−1a.\frac1c-\frac1d=\frac1b-\frac1a.c1​−d1​=b1​−a1​. Rearrange: 1a+1c=1b+1d.\frac1a+\frac1c=\frac1b+\frac1d.a1​+c1​=b1​+d1​.

This is the key form.

So define ϕ(x,y)=1x+1y.\phi(x,y)=\frac1x+\frac1y.ϕ(x,y)=x1​+y1​. Then (a,b)R(c,d)  ⟺  ϕ(a,c)=ϕ(b,d),(a,b)R(c,d) \iff \phi(a,c)=\phi(b,d),(a,b)R(c,d)⟺ϕ(a,c)=ϕ(b,d), but more precisely from the rearrangement, (a,b)R(c,d)  ⟺  1a+1c=1b+1d.(a,b)R(c,d) \iff \frac1a+\frac1c=\frac1b+\frac1d.(a,b)R(c,d)⟺a1​+c1​=b1​+d1​. Equivalently, 1a−1b=1d−1c.\frac1a-\frac1b=\frac1d-\frac1c.a1​−b1​=d1​−c1​. That is, (a,b)R(c,d)  ⟺  (1a−1b)=−(1c−1d).(a,b)R(c,d) \iff \left(\frac1a-\frac1b\right)= -\left(\frac1c-\frac1d\right).(a,b)R(c,d)⟺(a1​−b1​)=−(c1​−d1​).

Let F(a,b)=1a−1b.F(a,b)=\frac1a-\frac1b.F(a,b)=a1​−b1​. Then (a,b)R(c,d)  ⟺  F(a,b)=−F(c,d).(a,b)R(c,d) \iff F(a,b)=-F(c,d).(a,b)R(c,d)⟺F(a,b)=−F(c,d).

Now properties become easy.

Reflexive

Need F(a,b)=−F(a,b)F(a,b)=-F(a,b)F(a,b)=−F(a,b), so F(a,b)=0F(a,b)=0F(a,b)=0, i.e. a=ba=ba=b. Not true for all pairs. Hence not reflexive.

Symmetric

If F(a,b)=−F(c,d)F(a,b)=-F(c,d)F(a,b)=−F(c,d), then F(c,d)=−F(a,b)F(c,d)=-F(a,b)F(c,d)=−F(a,b). Hence symmetric.

Transitive

If (a,b)R(c,d)(a,b)R(c,d)(a,b)R(c,d) and (c,d)R(e,f)(c,d)R(e,f)(c,d)R(e,f), then F(a,b)=−F(c,d),F(a,b)=-F(c,d),F(a,b)=−F(c,d), F(c,d)=−F(e,f).F(c,d)=-F(e,f).F(c,d)=−F(e,f). So F(a,b)=F(e,f).F(a,b)=F(e,f).F(a,b)=F(e,f). For (a,b)R(e,f)(a,b)R(e,f)(a,b)R(e,f) we would need F(a,b)=−F(e,f),F(a,b)=-F(e,f),F(a,b)=−F(e,f), which need not hold unless F(a,b)=0F(a,b)=0F(a,b)=0. So in general, not transitive.

A concrete counterexample:

  • (1,1)R(2,2)(1,1)R(2,2)(1,1)R(2,2) because both have F=0F=0F=0.
  • (2,2)R(1,1)(2,2)R(1,1)(2,2)R(1,1) because both have F=0F=0F=0.
  • But this still gives transitivity true.

Need nonzero FFF values with sign alternation.

Take F(1,2)=1−12=12.F(1,2)=1-\frac12=\frac12.F(1,2)=1−21​=21​. Then we need a pair with value −12-\frac12−21​, e.g. F(2,1)=12−1=−12.F(2,1)=\frac12-1=-\frac12.F(2,1)=21​−1=−21​. So (1,2)R(2,1).(1,2)R(2,1).(1,2)R(2,1). Now take a third pair with F=12F=\frac12F=21​ again, say (1,2)(1,2)(1,2). Then (2,1)R(1,2).(2,1)R(1,2).(2,1)R(1,2). But for transitivity we would need (1,2)R(1,2),(1,2)R(1,2),(1,2)R(1,2), which is false since relation is not reflexive there.

Indeed,

  • (1,2)R(2,1)(1,2)R(2,1)(1,2)R(2,1) is true.
  • (2,1)R(1,2)(2,1)R(1,2)(2,1)R(1,2) is true.
  • (1,2)R(1,2)(1,2)R(1,2)(1,2)R(1,2) is false.

Hence RRR is not transitive.

  1. Conclusion

RRR is:

  • not reflexive,
  • symmetric,
  • not transitive.

Therefore the correct option is D: symmetric but neither reflexive nor transitive.\boxed{\text{D: symmetric but neither reflexive nor transitive}}.D: symmetric but neither reflexive nor transitive​.

  1. Comparison with stored answer

Stored correct answer: D

This matches our derived answer.

PreviousNext

More from Sets and Relations

  • Among the relations S={(a,b):a,b∈R−{0},2+ba​>0} and T={(a,b):a,b∈R,a2−b2∈Z}…2023 · MCQ
  • The sum of all the elements of the set {α∈{1,2,.....,100}:HCF(α,24)=1} is ​.2022 · Numerical
  • Let A={1,2,3,4,5,6,7}. Define B={T⊆A: either 1otinT or 2∈T} and C={T⊆A:T the sum of all the elements of T is a prime number }. Then the number of elements in the set B∪C is ​…2022 · Numerical
  • Let A={1,2,3,4,5,6,7} and B={3,6,7,9}. Then the number of elements in the set {C⊆A:C∩Beqϕ} is ​.2022 · Numerical
  • Let A = {n ∈ N : H.C.F. (n, 45) = 1} and Let B = {2k : k ∈{1, 2, ......., 100}}. Then the sum of all the elements of A ∩ B is ​.2022 · Numerical
  • Let A=i=1∑10​j=1∑10​min{i,j} and B=i=1∑10​j=1∑10​max{i,j}. Then A + B is equal to ​.2022 · Numerical
  • Let R1​ and R2​ be two relations defined on R by aR1​b⇔ab≥0 and aR2​b⇔a≥b Then,2022 · MCQ
  • For α∈N, consider a relation R on N given by R={(x,y):3x+αy is a multiple of 7 }. The relation R is an equivalence relation if and only if :2022 · MCQ