3 free questions from this chapter (the full set has 30). Marking: +4 for correct · −1 for incorrect · 0 for unattempted.
Question 1Level 3 · AdvancedNumerical answer
Let A={1,2,3}. Find the number of relations on A that are symmetric and transitive but NOT reflexive.
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Answer: 10
Key insight: if R is symmetric and transitive, then (a,b)∈R forces (b,a)∈R and hence (a,a)∈R. So R is an equivalence relation on its support D={a:(a,b)∈R for some b}, and it is reflexive on A exactly when D=A. Hence such relations correspond to a choice of D⊆A together with a partition of D. Number of partitions of a k-set: 1,1,2,5 for k=0,1,2,3. Total symmetric and transitive relations =(03)1+(13)1+(23)2+(33)5=1+3+6+5=15. The reflexive ones (D=A) are the 5 equivalence relations. Required number =15−5=10. Trap: the 'proof' that symmetric and transitive implies reflexive fails for an element related to nothing; also do not forget the empty relation. Exam pattern: JEE Main – counting relations on a small set with prescribed propertiesQuestion 2Level 3 · AdvancedSingle correct
How many relations on {1,2,3,4} are reflexive and symmetric, contain the pair (1,2), but are NOT transitive?
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Answer: D
A reflexive symmetric relation on a 4-set is fixed by choosing which of the (24)=6 unordered pairs {a,b} (both orders) it contains. With {1,2} forced, there are 25=32 such relations. Among them, the transitive ones are exactly the equivalence relations in which 1 and 2 lie in the same class. Glue 1,2 into one object; partitions of the three objects {12},3,4 number 5. Required count =32−5=27. Trap: subtracting all 15 equivalence relations on a 4-set gives 17, but those that separate 1 and 2 are not among the 32 in the first place; 49=64−15 ignores the condition on (1,2). Exam pattern: JEE Main – relation types and counting with a forced pairQuestion 3Level 3 · AdvancedSingle correct
Let A={1,2,3,…,20} and R={(a,b)∈A×A:3 divides a2−b2}. The number of elements in R is
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Answer: B
Key insight: a2mod3 is 0 if 3∣a and 1 otherwise, so 3∣a2−b2 iff a,b are both multiples of 3 or both non-multiples. So R is an equivalence relation with only two classes: {3,6,…,18} (6 elements) and the remaining 14 elements. n(R)=62+142=36+196=232. Trap: treating the relation as a≡b(mod3) gives three classes of sizes 6,7,7 and 36+49+49=134; 212 wrongly excludes the 20 diagonal pairs. Exam pattern: JEE Main – size of an equivalence relation defined by divisibility