Sets & RelationshardFree

Relation a/b integer with b in {sqrt 6, sqrt 10, sqrt 15} | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
Let a relation RR be defined on the real numbers such that
R={(a,b):ab∈I; b∈{6,10,15}},R = \left\{\left(a,b\right) : \frac ab \in I;\ b \in \left\{\sqrt6,\sqrt{10},\sqrt{15}\right\}\right\},
 then
Athe domain of RR consists of irrational numbers only
BRR is not a reflexive relation but is a transitive relationcorrect
CRR is an equivalence relation
DR−1R^{-1} is not a reflexive relation but is a transitive relationcorrect
Solution
Step 1: (a,b)∈R\left(a,b\right) \in R needs b∈{6,10,15}b \in \left\{\sqrt6,\sqrt{10},\sqrt{15}\right\} and ab∈Z\dfrac ab \in \mathbb Z, i.e. a=nba = nb for some integer nn. Taking n=0n = 0 gives a=0a = 0, and 06=0∈Z\dfrac0{\sqrt6} = 0 \in \mathbb Z, so
0∈domain of R0 \in \text{domain of } R
 00 is rational, so (1) is false. Step 2: RR is reflexive only if (a,a)∈R\left(a,a\right) \in R for every real aa; but that would need a∈{6,10,15}a \in \left\{\sqrt6,\sqrt{10},\sqrt{15}\right\} for every real aa. So RR is not reflexive, and for the same reason neither is R−1R^{-1}. Step 3: Let (a,b)∈R\left(a,b\right) \in R and (b,c)∈R\left(b,c\right) \in R. Then c∈{6,10,15}c \in \left\{\sqrt6,\sqrt{10},\sqrt{15}\right\}, ab∈Z\dfrac ab \in \mathbb Z and bc∈Z\dfrac bc \in \mathbb Z, so
ac=ab⋅bc∈Z\frac ac = \frac ab\cdot\frac bc \in \mathbb Z
 and the second coordinate cc is in the required set. Hence (a,c)∈R\left(a,c\right) \in R and **RR is transitive.** (In fact bc∈Z\dfrac bc \in \mathbb Z with both b,cb,c in that three-element set gives b=cb = c, since 6/10\sqrt{6/10}, 6/15\sqrt{6/15}, 10/6\sqrt{10/6} and so on are never integers.) So (2) is true, and since RR is not reflexive it is not an equivalence relation, so (3) is false. Step 4:
R−1={(x,y):yx∈Z, x∈{6,10,15}}R^{-1} = \left\{\left(x,y\right) : \frac yx \in \mathbb Z,\ x \in \left\{\sqrt6,\sqrt{10},\sqrt{15}\right\}\right\}
 If (x,y)∈R−1\left(x,y\right) \in R^{-1} and (y,z)∈R−1\left(y,z\right) \in R^{-1} then xx is in the set, yx∈Z\dfrac yx \in \mathbb Z and zy∈Z\dfrac zy \in \mathbb Z, so zx∈Z\dfrac zx \in \mathbb Z and (x,z)∈R−1\left(x,z\right) \in R^{-1} R−1R^{-1} is transitive and not reflexive, so (D) is true. Answer: (2),(4)\left(2\right),\left(4\right)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Sets & Relations · easy
The relation R={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)}R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\} on the set A={1,2,3}A = \{1, 2, 3\} is:
Sets & Relations · easy
The relation RR is defined in A={1,2,3}A = \{1, 2, 3\} by aRbaRb if ∣a2−b2∣≤5|a^2 - b^2| \le 5. Which of the following is false?
Sets & Relations · easy
If AA is the set of even natural numbers less than 88 and BB is the set of prime numbers less than 77, then the number of relations from AA to BB is:
Sets & Relations · easy
The relation RR is defined on the set of natural numbers as {(a,b):a=2b}\{(a, b) : a = 2b\}. Then R−1R^{-1} is given by:

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath