Questions: Sets
Sets26 min readLesson 6 of 7
Q1. State whether the collection of all prime numbers less than 20 is a set, and write it in roster form.
Answer: The collection is well-defined because, for each number less than 20, you can check prime or not. The primes below 20 are 2, 3, 5, 7, 11, 13, 17 and 19, so the set is {2, 3, 5, 7, 11, 13, 17, 19}. Note that 1 is not prime and 2 is the only even prime, so both decisions need care.
Hinglish: Har number ke liye prime hona ya na hona clear check hota hai, isliye ye collection set hai. Primes list karo: 2, 3, 5, 7, 11, 13, 17, 19. 1 prime nahi hota, ye mat bhoolo.
Q2. The set A contains the integers x that satisfy -5 < x <= 2. Write A in roster form.
Answer: The integers strictly greater than -5 start at -4, and the condition x <= 2 stops the list at 2. So A = {-4, -3, -2, -1, 0, 1, 2}. The endpoint -5 is excluded because the left inequality is strict, while 2 is included because the right inequality is weak.
Hinglish: -5 se bada integer -4 se shuru hota hai, aur 2 par weak inequality ki wajah se rukta hai. Answer {-4, -3, -2, -1, 0, 1, 2}. Endpoint check karna zaroori hai.
Q3. Write the set {3, 9, 27, 81} in set-builder form.
Answer: Each element is a power of 3. The exponents are 1, 2, 3 and 4, so the set is {x : x = 3 to the power n, where n in N and 1 <= n <= 4}. Another correct answer is {x : x is a positive power of 3 less than 100}. The rule only has to produce exactly the four listed elements.
Hinglish: Har element 3 ki power hai: 3^1, 3^2, 3^3, 3^4. Isliye set-builder me {x : x = 3^n, n in N aur 1 <= n <= 4} likh sakte hain. Rule se exactly wahi char elements aane chahiye.
Q4. Decide whether the pair {a, e, i, o, u} and {x, y, z} is disjoint.
Answer: The two sets have no common letter. The first contains only vowels and the second contains only consonants, so their intersection is empty. The pair is therefore disjoint. Recall that a set is disjoint from another when the intersection is phi, not when their sizes are equal.
Hinglish: Pehle set me vowels hain, doosre me consonants. Koi common letter nahi hai, isliye intersection phi hai aur pair disjoint hai. Size equal hona zaroori nahi, intersection dekhna hai.
Q5. A = {1, 2, 3, 4, 5} and B = {2, 4}. State whether A is a subset of B, or B is a subset of A, and justify.
Answer: Every element of B, namely 2 and 4, also lies in A, so B is a subset of A. The reverse fails because 1, 3 and 5 lie in A but not in B, so A is not a subset of B. Since B is a subset of A and the sets differ, B is a proper subset of A.
Hinglish: B ke dono elements 2 aur 4 A me hain, isliye B subset of A hai. A ke 1, 3, 5 B me nahi hain, isliye A subset of B nahi. Sets alag hain, to B proper subset of A hoga. Direction dhyan se likho.
Q6. List all subsets of {1, 2, 3} and count them.
Answer: The set has 3 elements, so it has 2 to the power 3 = 8 subsets. Listing by size: the empty set phi, the singletons {1}, {2}, {3}, the pairs {1, 2}, {1, 3}, {2, 3}, and the full set {1, 2, 3}. The full set is a subset of itself by definition, so it belongs in the list.
Hinglish: 3 elements wale set ke 2^3 = 8 subsets hote hain. Size ke hisaab se list karo: phi, teen singletons, teen pairs, aur pura set. Pura set bhi subset hota hai, isse chhodna mat.
Q7. If U = {1, 2, 3, 4, 5, 6, 7, 8} and A = {2, 4, 6, 8}, find A prime and (A prime) prime.
Answer: The complement A prime contains the elements of U that are not in A, so A prime = {1, 3, 5, 7}. The double complement takes the elements of U not in A prime, which returns {2, 4, 6, 8}, that is A. This illustrates the law that complementing twice leaves the set unchanged.
Hinglish: A prime me U ke woh elements hain jo A me nahi: {1, 3, 5, 7}. Dubara complement lene par wapas {2, 4, 6, 8} milta hai, yaani A. Double complement wali law isi ko kehti hai.
Q8. Let A = {1, 2, 3, 4}, B = {3, 4, 5, 6} and C = {4, 5, 6, 7}. Find A intersection (B union C).
Answer: Work inside the bracket first: B union C = {3, 4, 5, 6, 7}. Now intersect with A. The common elements with A = {1, 2, 3, 4} are 3 and 4, so the answer is {3, 4}. Checking with the distributive law, A intersection B = {3, 4} and A intersection C = {4}, and their union is {3, 4}, which matches.
Hinglish: Pehle B union C = {3, 4, 5, 6, 7}. Isko A se intersect karo, to {3, 4} milta hai. Distributive law se check: (A intersection B) union (A intersection C) = {3, 4}. Dono taraf same answer.
Q9. For A = {1, 2, 3, 4, 5} and B = {2, 4, 6}, find A - B, B - A and show they differ.
Answer: A - B keeps the elements of A that are not in B, so A - B = {1, 3, 5}. B - A keeps the elements of B that are not in A, so B - A = {6}. The two results are different sets, which shows that set difference does not commute. It also shows that the elements removed depend on which set is written first.
Hinglish: A - B me A ke woh elements hain jo B me nahi: {1, 3, 5}. B - A me B ke woh elements jo A me nahi: {6}. Dono alag hain, isliye difference commutative nahi hota. Kaunsa set pehle likha hai, wahi decide karta hai.
Q10. Show that if A union B = A intersection B, then A = B.
Answer: Take any element x in A. Then x is in A union B, and since the union equals the intersection, x is in A intersection B, which places x in B. So A is a subset of B. Now take y in B. Then y is in A union B, hence in A intersection B, hence in A. So B is a subset of A. Both inclusions hold, therefore A = B. The argument uses only membership and the two definitions, which is what a set equality proof needs.
Hinglish: Pehle x in A lo. Union me to hai hi, aur union = intersection hai, isliye x intersection me bhi hai, yaani x in B. Isse A subset of B mila. Ab y in B lo aur same tarah B subset of A dikhao. Dono inclusions se A = B. Sirf definitions se proof ho gaya.
Common mistakes on this chapter
Four slips cover most of the lost marks. Writing the set of most talented writers as a set, ignoring the well-defined test. Confusing the symbol for belongs to with the symbol for subset of when a set contains another set as an element. Forgetting that the universal set must be fixed before a complement can be written. Treating the order of letters or numbers in a roster as meaningful, when equal membership is all that matters. Read the answer back once with these four tests in mind.
Quiz
Quiz below: five multiple choice questions on representation, subsets, intervals, operations and De Morgan laws. Attempt them after the notes and the ten questions above, then move to the previous year page.