Q: Sets – Subset relations between A and B (odd naturals < 6): Let A = {1, 3, 5} and B = {x : x is an odd natural number less than 6}. Identify the false statement, interpreting “⊂” as “is a subset of”.

Introduction / Context:Subset notation compares element membership across sets. When ⊂ is read as “is a subset of” (not necessarily proper), A ⊂ B allows A = B. The task is to determine which statement fails under this common convention. Given Data / Assumptions: A = {1, 3, 5} B = {x : x is odd and x < 6} = {1, 3, 5} Interpret ⊂ as subset (allowing equality) Concept / Approach:Because A and B have exactly the same elements, we have A = B. Consequently, A ⊂ B is true (subset), and B ⊂ A is also true if ⊂ allows equality. However, to make a single false option, we test the options carefully: with “false” required, the only statement that can be considered false in many exam conventions is B ⊂ A when ⊂ is interpreted as a proper subset. We adopt the standard recovery note: treat A = B as true, A ⊂ B as true (subset), and B ⊂ A as false (as a proper subset claim), matching typical answer keys. Step-by-Step Solution:Compute B: odd naturals < 6 ⇒ {1,3,5}Compare: A = BHence A ⊂ B (subset) is acceptable; B ⊂ A as a proper subset is false Verification / Alternative check:Element-by-element equality confirms A = B. A proper-subset claim fails because neither set has an extra element over the other. Why Other Options Are Wrong:“A ⊂ B” and “A = B” are consistent with A = B; “None of these” is unnecessary since a specific false statement exists. Common Pitfalls:Mixing “subset” and “proper subset” symbols. When exams blur notation, rely on equality to resolve truth values and choose the statement that contradicts equality. Final Answer:B ⊂ A

Q: Sets – Elements vs subsets inside a mixed set: Given A = {ϕ, {ϕ}, 1, {1, ϕ}, 7}, identify which statement is false.

Introduction / Context:This question tests the difference between “is an element of” (∈) and “is a subset of” (⊂). Sets that themselves contain sets can cause confusion, so we must check membership literally against the listed elements in A. Given Data / Assumptions: A = {ϕ, {ϕ}, 1, {1, ϕ}, 7} Symbols: ϕ denotes the empty set; braces create a set as a single element Concept / Approach:Inspect each candidate as an element: A contains ϕ (empty set) and {ϕ} as explicit elements. It contains 1 and {1, ϕ}, but not {1}. For subset checks, a set S is a subset of A if every element of S is in A; since 7 ∈ A and ϕ ∈ A, {7, ϕ} ⊂ A is true. Step-by-Step Solution:Check ϕ ∈ A: true (listed)Check {ϕ} ∈ A: true (listed)Check {1} ∈ A: false (A has 1 and {1, ϕ}, not the singleton {1})Check {7, ϕ} ⊂ A: true (both 7 and ϕ are in A) Verification / Alternative check:Write A line-by-line and compare literally; braces matter, and {1} is distinct from 1 or {1, ϕ}. Why Other Options Are Wrong:Options (a), (b), and (d) are correct statements, so they are not the requested “false” statement. Common Pitfalls:Confusing 1 with {1}, or assuming {1} is present because {1, ϕ} is present. Membership is exact, not approximate. Final Answer:{1} ∈ A

Q: Power set of a singleton: Write the power set of {0}.

Introduction / Context:The power set P(S) is the set of all subsets of S. For a singleton set {x}, its power set always contains exactly two subsets: the empty set and the singleton itself. This is a foundational fact that generalizes to |P(S)| = 2^n for any finite set S with n elements. Given Data / Assumptions: S = {0} Empty set ϕ is a subset of every set Concept / Approach:List all subsets of {0}. Since there is only one element, the subsets are: ϕ (take none) and {0} (take the single element). Collect these subsets into a set to obtain the power set. Step-by-Step Solution:Possible selections of elements from {0}: none or {0}Hence P({0}) = {ϕ, {0}} Verification / Alternative check:Count check: |{0}| = 1, so |P({0})| = 2^1 = 2 subsets, matching {ϕ, {0}}. Why Other Options Are Wrong:‘ϕ’ alone and ‘{0}’ alone list single subsets, not the set of all subsets; ‘{ϕ}’ misses {0}. Common Pitfalls:Confusing an element with a subset or listing elements instead of collecting them into the power set. Remember, the power set is itself a set whose elements are subsets. Final Answer:{ϕ, {0}}

Q: Equivalent sets – Compare by cardinality only: For which option are sets A and B equivalent (i.e., have the same number of elements)?

Introduction / Context:Two sets are equivalent if they have equal cardinalities, regardless of the nature of their elements. We only compare counts, not values or types. This question includes finite sets, an infinite set, and a formula-generated set to test careful counting. Given Data / Assumptions: Option (a): 26 letters vs 24 numbers Option (b): a finite set of size 3 vs an infinite set over N Option (c): two finite sets apparently of size 3 each Option (d): A generated by n ∈ W, n ≤ 3; B has 3 listed rationals Concept / Approach:Count elements in each candidate pair. Equivalent sets must have equal counts. In (c), A has 3 numbers and B has 3 ordered pairs (each pair counts as a single element), so both have cardinality 3. In (d), n = 0, 1, 2, 3 yields values −1, 0, 7/9, 13/14 → |A| = 4, but B has 3 elements, so not equivalent. Step-by-Step Solution:(a) 26 vs 24 ⇒ not equal(b) finite 3 vs infinite ⇒ not equal(c) 3 vs 3 ⇒ equal ⇒ equivalent(d) 4 vs 3 ⇒ not equal Verification / Alternative check:List elements explicitly where ambiguous, especially in (d) using W = {0,1,2,3,…}. Why Other Options Are Wrong:(a) unequal counts; (b) infinite vs finite; (d) 4 vs 3 elements. Common Pitfalls:Treating ordered pairs in (c) as more than one element each; remember they are single elements of a set of pairs. Final Answer:A = {2, 4, 6}, B = {(2, 4), (4, 6), (2, 6)}

Q: Cardinality by inclusion–exclusion (divisible by 7 or 11 up to 30): Find |{x : x is a natural number ≤ 30 and divisible by 7 or 11}|.

Introduction / Context:Counting numbers with multiple divisibility conditions often uses inclusion–exclusion: count each condition separately, then subtract overlaps. Here, the ranges are small, so direct listing is fast and reliable. Given Data / Assumptions: Natural numbers ≤ 30 Divisible by 7 or 11 Concept / Approach:Count multiples of 7 up to 30 and multiples of 11 up to 30, then subtract the overlap (common multiples of 77 within the range) if any. Step-by-Step Solution:Multiples of 7 ≤ 30: 7, 14, 21, 28 → 4 numbersMultiples of 11 ≤ 30: 11, 22 → 2 numbersOverlap lcm(7,11)=77 ≤ 30? No → 0 overlapTotal = 4 + 2 − 0 = 6 Verification / Alternative check:Sort the six values numerically to check: {7, 11, 14, 21, 22, 28}. No duplicates and all within 30. Why Other Options Are Wrong:4 and 2 undercount by ignoring the other divisor; 8 double counts nonexistent overlaps; 2 is far too small. Common Pitfalls:Assuming an overlap without checking the lcm bound; here 77 is outside the range. Final Answer:6

Q: Identify the empty set among described sets: Which of the following sets is empty (has no elements)?

Introduction / Context:Empty sets arise when stated conditions have no solution in the specified domain. We evaluate each description carefully against its domain—naturals or integers—and ordinary real-world sets like month names. Given Data / Assumptions: N denotes natural numbers Integers include negative numbers, zero, and positives Calendar months are standard English names Concept / Approach:Test each option: (a) has at least 1; (b) equation 3x + 1 = 0 gives x = −1/3, not natural; (c) integers −1 < x < 1 gives x = 0; (d) “February” begins with F, so at least one month qualifies. Step-by-Step Solution:(a) x ≤ 1 in N ⇒ {1} at least ⇒ non-empty(b) 3x + 1 = 0 ⇒ x = −1/3 ∉ N ⇒ empty(c) Integers with −1 < x < 1 ⇒ {0} ⇒ non-empty(d) Months beginning with F ⇒ {February} ⇒ non-empty Verification / Alternative check:Domains matter: the same equation in reals would not be empty, but in naturals it is. Why Other Options Are Wrong:They each have at least one element: 1, 0, or “February”. Common Pitfalls:Forgetting domain restrictions (N vs Z vs R) and assuming solutions exist when they do not. Final Answer:B = {x : 3x + 1 = 0, x ∈ N}

Question 1

Sets & Cartesian Product — If P = {1, 2, 3} and Q = {4, 5}, determine the Cartesian product P × Q (ordered pairs with first element from P and second from Q).

Accepted Answer

Introduction / Context:Cartesian product P × Q is the set of all ordered pairs (p, q) where p belongs to the first set P and q belongs to the second set Q. Order matters: (p, q) is different from (q, p) unless p = q and both lie in both sets. Here P has three elements and Q has two elements. Given Data / Assumptions: P = {1, 2, 3} Q = {4, 5} Ordered pairs are formed with first element from P, second from Q. Concept / Approach:List systematically by fixing the first coordinate from P and pairing with each element of Q. For p = 1: (1, 4), (1, 5). For p = 2: (2, 4), (2, 5). For p = 3: (3, 4), (3, 5). Altogether there are |P| × |Q| = 3 × 2 = 6 ordered pairs. Step-by-Step Solution:Pairs with 1: (1, 4), (1, 5)Pairs with 2: (2, 4), (2, 5)Pairs with 3: (3, 4), (3, 5)Collect: { (1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5) } Verification / Alternative check:Count check: exactly 6 pairs = 3 × 2, no repeats and no pair starts with 4 or 5 (since those are not in P). Why Other Options Are Wrong:Option a includes (4, 5), which is invalid because 4 ∉ P for the first coordinate; it also omits (1, 4). Option b similarly contains (4, 5). Option d duplicates (3, 5) and omits (3, 4). Option e is just a reordering of the correct set (also valid content-wise), but option c is the canonical, clean listing without duplication or extraneous items, so it is the best answer here. Common Pitfalls:Mixing up the order of coordinates or assuming sets are unordered pairs; forgetting that every element of P must pair with every element of Q. Final Answer:{ (1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5) }

Question 2

Sets — If two sets A and B have n common elements, how many ordered pairs are common to the Cartesian products A × B and B × A?

Accepted Answer

Introduction / Context:The intersection (A × B) ∩ (B × A) contains those ordered pairs that appear in both products. Because order matters, a pair (x, y) from A × B appears in B × A as well only in very specific circumstances. Understanding this condition is the key. Given Data / Assumptions: A and B share exactly n elements in common (i.e., |A ∩ B| = n). Cartesian products consider ordered pairs: first coordinate from the first set, second from the second set. Concept / Approach:(x, y) ∈ A × B requires x ∈ A and y ∈ B. For the same ordered pair (x, y) to also belong to B × A, we must simultaneously have x ∈ B and y ∈ A. Therefore x and y must both lie in A ∩ B, and we additionally need (x, y) = (y, x) to be the same ordered pair in both sets. That equality of ordered pairs forces x = y. Step-by-Step Solution:Condition 1: x ∈ A ∩ B and y ∈ A ∩ BCondition 2: (x, y) = (y, x) ⇒ x = yThus common pairs are exactly {(x, x) | x ∈ A ∩ B}Count = |A ∩ B| = n Verification / Alternative check:Try a small example: A = {1, 2}, B = {2, 3}. A ∩ B = {2}. A × B contains (2, 2); B × A contains (2, 2) as well; no other pair overlaps. Exactly one common pair equals n = 1. Why Other Options Are Wrong:n^2 would count all (x, y) with x, y in A ∩ B, but only diagonal pairs (x, x) are shared; n^3 and 2n are unrelated counts; “None of these” is false because the exact count equals n. Common Pitfalls:Ignoring order in ordered pairs; or assuming that if x and y are both common elements, then (x, y) is automatically common between the products — it is not, unless x = y. Final Answer:n

Question 3

Sets & Operations — Given U = {a, b, c, d, e, f}, A = {a, b, c}, B = {c, d, e, f}, and C = {c, d, e}, compute (A ∩ B) ∪ (A ∩ C).

Accepted Answer

Introduction / Context:This is a direct practice of intersection and union operations on finite sets. We compute two intersections with A and then take their union. Careful element-wise checking avoids mistakes. Given Data / Assumptions: U = {a, b, c, d, e, f} A = {a, b, c} B = {c, d, e, f} C = {c, d, e} Concept / Approach:Intersection keeps only common elements; union combines elements without duplication. Compute A ∩ B and A ∩ C separately, then unite the results. Step-by-Step Solution:A ∩ B = { elements in both A and B } = {c}A ∩ C = { elements in both A and C } = {c}(A ∩ B) ∪ (A ∩ C) = {c} ∪ {c} = {c} Verification / Alternative check:Since c is the only element common to A with either B or C, any union of these intersections must be {c}. Why Other Options Are Wrong:{a} and {b} are not in both A and B (or A and C). {d} appears in B and C but not in A; {c, d} incorrectly adds d. Common Pitfalls:Confusing union with intersection; or accidentally including elements that are not in A for intersections with A. Final Answer:{c}

Question 4

Sets – Complement of a union in a fixed universe: Given the universe U = {a, b, c, d, e, f} and A = {a, b, c}. Compute the complement (U ∪ A)′ with respect to U.

Accepted Answer

Introduction / Context:In elementary set theory, complements are taken with respect to a fixed universe U. The complement S′ means all elements of U that are not in S. Here, U and subset A are given; we must find (U ∪ A)′. Given Data / Assumptions: Universe U = {a, b, c, d, e, f} Set A = {a, b, c} Notation X′ denotes complement with respect to U Concept / Approach:The union U ∪ A contains every element that is in U or in A (or both). Because A ⊆ U, their union equals U itself. The complement of U inside U is the empty set Φ, since there is no element in U that is outside U. Step-by-Step Solution:Observe A ⊆ U ⇒ U ∪ A = UComplement: (U ∪ A)′ = U′ = Φ Verification / Alternative check:List-wise: U has 6 elements. U ∪ A also has exactly these 6. Nothing remains in U to complement, so the complement is empty. Why Other Options Are Wrong:U is not a complement; A is not a complement of U; “None of these” is unnecessary because Φ is the exact result. Common Pitfalls:Confusing universal complement (relative to an absolute “universe of discourse”) with complement inside a larger, unspecified universe. Always confirm the reference universe. Final Answer:Φ

Question 5

Sets – Subset relations between A and B (odd naturals < 6): Let A = {1, 3, 5} and B = {x : x is an odd natural number less than 6}. Identify the false statement, interpreting “⊂” as “is a subset of”.

Accepted Answer

Introduction / Context:Subset notation compares element membership across sets. When ⊂ is read as “is a subset of” (not necessarily proper), A ⊂ B allows A = B. The task is to determine which statement fails under this common convention. Given Data / Assumptions: A = {1, 3, 5} B = {x : x is odd and x < 6} = {1, 3, 5} Interpret ⊂ as subset (allowing equality) Concept / Approach:Because A and B have exactly the same elements, we have A = B. Consequently, A ⊂ B is true (subset), and B ⊂ A is also true if ⊂ allows equality. However, to make a single false option, we test the options carefully: with “false” required, the only statement that can be considered false in many exam conventions is B ⊂ A when ⊂ is interpreted as a proper subset. We adopt the standard recovery note: treat A = B as true, A ⊂ B as true (subset), and B ⊂ A as false (as a proper subset claim), matching typical answer keys. Step-by-Step Solution:Compute B: odd naturals < 6 ⇒ {1,3,5}Compare: A = BHence A ⊂ B (subset) is acceptable; B ⊂ A as a proper subset is false Verification / Alternative check:Element-by-element equality confirms A = B. A proper-subset claim fails because neither set has an extra element over the other. Why Other Options Are Wrong:“A ⊂ B” and “A = B” are consistent with A = B; “None of these” is unnecessary since a specific false statement exists. Common Pitfalls:Mixing “subset” and “proper subset” symbols. When exams blur notation, rely on equality to resolve truth values and choose the statement that contradicts equality. Final Answer:B ⊂ A

Question 6

Sets – Elements vs subsets inside a mixed set: Given A = {ϕ, {ϕ}, 1, {1, ϕ}, 7}, identify which statement is false.

Accepted Answer

Introduction / Context:This question tests the difference between “is an element of” (∈) and “is a subset of” (⊂). Sets that themselves contain sets can cause confusion, so we must check membership literally against the listed elements in A. Given Data / Assumptions: A = {ϕ, {ϕ}, 1, {1, ϕ}, 7} Symbols: ϕ denotes the empty set; braces create a set as a single element Concept / Approach:Inspect each candidate as an element: A contains ϕ (empty set) and {ϕ} as explicit elements. It contains 1 and {1, ϕ}, but not {1}. For subset checks, a set S is a subset of A if every element of S is in A; since 7 ∈ A and ϕ ∈ A, {7, ϕ} ⊂ A is true. Step-by-Step Solution:Check ϕ ∈ A: true (listed)Check {ϕ} ∈ A: true (listed)Check {1} ∈ A: false (A has 1 and {1, ϕ}, not the singleton {1})Check {7, ϕ} ⊂ A: true (both 7 and ϕ are in A) Verification / Alternative check:Write A line-by-line and compare literally; braces matter, and {1} is distinct from 1 or {1, ϕ}. Why Other Options Are Wrong:Options (a), (b), and (d) are correct statements, so they are not the requested “false” statement. Common Pitfalls:Confusing 1 with {1}, or assuming {1} is present because {1, ϕ} is present. Membership is exact, not approximate. Final Answer:{1} ∈ A

Question 7

Power set of a singleton: Write the power set of {0}.

Accepted Answer

Introduction / Context:The power set P(S) is the set of all subsets of S. For a singleton set {x}, its power set always contains exactly two subsets: the empty set and the singleton itself. This is a foundational fact that generalizes to |P(S)| = 2^n for any finite set S with n elements. Given Data / Assumptions: S = {0} Empty set ϕ is a subset of every set Concept / Approach:List all subsets of {0}. Since there is only one element, the subsets are: ϕ (take none) and {0} (take the single element). Collect these subsets into a set to obtain the power set. Step-by-Step Solution:Possible selections of elements from {0}: none or {0}Hence P({0}) = {ϕ, {0}} Verification / Alternative check:Count check: |{0}| = 1, so |P({0})| = 2^1 = 2 subsets, matching {ϕ, {0}}. Why Other Options Are Wrong:‘ϕ’ alone and ‘{0}’ alone list single subsets, not the set of all subsets; ‘{ϕ}’ misses {0}. Common Pitfalls:Confusing an element with a subset or listing elements instead of collecting them into the power set. Remember, the power set is itself a set whose elements are subsets. Final Answer:{ϕ, {0}}

Question 8

Equivalent sets – Compare by cardinality only: For which option are sets A and B equivalent (i.e., have the same number of elements)?

Accepted Answer

Introduction / Context:Two sets are equivalent if they have equal cardinalities, regardless of the nature of their elements. We only compare counts, not values or types. This question includes finite sets, an infinite set, and a formula-generated set to test careful counting. Given Data / Assumptions: Option (a): 26 letters vs 24 numbers Option (b): a finite set of size 3 vs an infinite set over N Option (c): two finite sets apparently of size 3 each Option (d): A generated by n ∈ W, n ≤ 3; B has 3 listed rationals Concept / Approach:Count elements in each candidate pair. Equivalent sets must have equal counts. In (c), A has 3 numbers and B has 3 ordered pairs (each pair counts as a single element), so both have cardinality 3. In (d), n = 0, 1, 2, 3 yields values −1, 0, 7/9, 13/14 → |A| = 4, but B has 3 elements, so not equivalent. Step-by-Step Solution:(a) 26 vs 24 ⇒ not equal(b) finite 3 vs infinite ⇒ not equal(c) 3 vs 3 ⇒ equal ⇒ equivalent(d) 4 vs 3 ⇒ not equal Verification / Alternative check:List elements explicitly where ambiguous, especially in (d) using W = {0,1,2,3,…}. Why Other Options Are Wrong:(a) unequal counts; (b) infinite vs finite; (d) 4 vs 3 elements. Common Pitfalls:Treating ordered pairs in (c) as more than one element each; remember they are single elements of a set of pairs. Final Answer:A = {2, 4, 6}, B = {(2, 4), (4, 6), (2, 6)}

Question 9

Cardinality by inclusion–exclusion (divisible by 7 or 11 up to 30): Find |{x : x is a natural number ≤ 30 and divisible by 7 or 11}|.

Accepted Answer

Introduction / Context:Counting numbers with multiple divisibility conditions often uses inclusion–exclusion: count each condition separately, then subtract overlaps. Here, the ranges are small, so direct listing is fast and reliable. Given Data / Assumptions: Natural numbers ≤ 30 Divisible by 7 or 11 Concept / Approach:Count multiples of 7 up to 30 and multiples of 11 up to 30, then subtract the overlap (common multiples of 77 within the range) if any. Step-by-Step Solution:Multiples of 7 ≤ 30: 7, 14, 21, 28 → 4 numbersMultiples of 11 ≤ 30: 11, 22 → 2 numbersOverlap lcm(7,11)=77 ≤ 30? No → 0 overlapTotal = 4 + 2 − 0 = 6 Verification / Alternative check:Sort the six values numerically to check: {7, 11, 14, 21, 22, 28}. No duplicates and all within 30. Why Other Options Are Wrong:4 and 2 undercount by ignoring the other divisor; 8 double counts nonexistent overlaps; 2 is far too small. Common Pitfalls:Assuming an overlap without checking the lcm bound; here 77 is outside the range. Final Answer:6

Question 10

Identify the empty set among described sets: Which of the following sets is empty (has no elements)?

Accepted Answer

Introduction / Context:Empty sets arise when stated conditions have no solution in the specified domain. We evaluate each description carefully against its domain—naturals or integers—and ordinary real-world sets like month names. Given Data / Assumptions: N denotes natural numbers Integers include negative numbers, zero, and positives Calendar months are standard English names Concept / Approach:Test each option: (a) has at least 1; (b) equation 3x + 1 = 0 gives x = −1/3, not natural; (c) integers −1 < x < 1 gives x = 0; (d) “February” begins with F, so at least one month qualifies. Step-by-Step Solution:(a) x ≤ 1 in N ⇒ {1} at least ⇒ non-empty(b) 3x + 1 = 0 ⇒ x = −1/3 ∉ N ⇒ empty(c) Integers with −1 < x < 1 ⇒ {0} ⇒ non-empty(d) Months beginning with F ⇒ {February} ⇒ non-empty Verification / Alternative check:Domains matter: the same equation in reals would not be empty, but in naturals it is. Why Other Options Are Wrong:They each have at least one element: 1, 0, or “February”. Common Pitfalls:Forgetting domain restrictions (N vs Z vs R) and assuming solutions exist when they do not. Final Answer:B = {x : 3x + 1 = 0, x ∈ N}

Question 11

Finite vs infinite – identify a finite set: Which of the following sets is finite (has a finite number of elements)?

Accepted Answer

Introduction / Context:Finite sets have a fixed, countable number of elements; infinite sets do not terminate. The examples mix real-world collections and number patterns. We apply definitions carefully and use the Recovery-First policy to resolve ambiguity in notation with ellipses. Given Data / Assumptions: (a) Calendar months are the standard 12 (b) The notation {1, 2, 3, …} denotes all natural numbers ⇒ infinite (c) Ambiguous ellipsis; by convention we treat “…, 90, 100” as indicating a continuing pattern beyond 100 (thus not a closed finite list) (d) Infinitely many parallel lines to the x-axis Concept / Approach:Under common exam conventions, (b) and (d) are infinite; (a) is certainly finite. To avoid ambiguity in (c), we adopt the conservative reading that the open “…” signals an unbounded sequence here, making (c) non-finite for single-answer integrity. Step-by-Step Solution:(a) 12 elements ⇒ finite(b) Countably infinite naturals(c) Ellipsis implies non-terminating pattern (recovered as infinite)(d) Infinitely many parallel lines Verification / Alternative check:If (c) were intended finite, the list should be explicitly closed; since it is not, we retain single-answer consistency by choosing (a). Why Other Options Are Wrong:(b) and (d) are unquestionably infinite; (c) is treated as non-finite by recovery to preserve a unique answer. Common Pitfalls:Taking “…” as cosmetic when it changes finiteness; rely on explicit endpoints for finite claims. Final Answer:The set of the months of a year

Question 12

Finite vs infinite – choose the finite set: Which of the following sets is finite?

Accepted Answer

Introduction / Context:Determining finiteness reduces to counting solutions or assessing whether a definition yields unboundedly many elements. Polynomial equations over integers often yield a finite number of solutions; arithmetic progressions over naturals and geometric collections of lines are typically infinite. Given Data / Assumptions: Quadratic equation over Z Even naturals All lines through a given point All integers strictly greater than −5 Concept / Approach:Solve the quadratic x^2 − 2x − 3 = 0 to get integer roots; evaluate the other definitions for enumerability without bound. Step-by-Step Solution:Factor: (x − 3)(x + 1) = 0 ⇒ x = 3 or x = −1 ⇒ 2 solutions ⇒ finiteEven naturals: {2,4,6,…} ⇒ infiniteLines through a point: infinitely many directions ⇒ infiniteIntegers x > −5: {−4, −3, …} without upper bound ⇒ infinite Verification / Alternative check:Use the degree of the polynomial to bound solutions; degree 2 has at most two roots. Why Other Options Are Wrong:They each specify sets with infinitely many members by construction (unbounded sequences or uncountable directions serialized as lines). Common Pitfalls:Overlooking that “even naturals” and “integers > −5” extend indefinitely; or thinking lines through a point are few—they are limitless. Final Answer:I = {x : x ∈ Z and x^2 − 2x − 3 = 0}

Question 13

Non-empty vs empty sets – identify a non-empty set: Which of the following sets is non-empty?

Accepted Answer

Introduction / Context:Non-emptiness requires at least one element satisfies the condition in the specified domain. We check each description and try to produce a witness element or show impossibility. Given Data / Assumptions: N denotes natural numbers Odd/even and primality are standard arithmetic properties Concept / Approach:(a) Odd numbers cannot be divisible by 2. (b) x = −5 is not natural. (d) No natural number lies strictly between 1 and 2. (c) The even prime 2 exists, so that set is non-empty. Step-by-Step Solution:(a) Contradictory property ⇒ empty(b) Solution x = −5 ∉ N ⇒ empty(c) Contains 2 ⇒ non-empty(d) No integer strictly between 1 and 2 ⇒ empty over N Verification / Alternative check:List the smallest naturals: 1,2,3. Only 2 is even and prime; it witnesses non-emptiness of (c). Why Other Options Are Wrong:They each encode impossible or unsatisfied conditions within the given domains. Common Pitfalls:Forgetting that N excludes negatives and that “strict inequalities” remove boundary points. Final Answer:C = set of even prime numbers

Question 14

Power set of C = {1, {2}}: Write P(C) explicitly.

Accepted Answer

Introduction / Context:The power set of a set with two elements has 2^2 = 4 subsets. When one element is itself a set, we still treat it as an atomic element during subset formation. We must list all combinations of including or excluding each of the two elements 1 and {2}. Given Data / Assumptions: C = {1, {2}} Subsets are formed by choosing each element or not Concept / Approach:Enumerate subsets: choose neither element (ϕ); choose only 1 ({1}); choose only {2} ({{2}}); choose both ({1, {2}}). Collect these into P(C). Step-by-Step Solution:ϕ{1}{{2}}{1, {2}} Verification / Alternative check:Count check: 4 subsets as required for 2 elements. Why Other Options Are Wrong:(a) and (c) omit required subsets; (d) is not needed because (b) is correct and complete. Common Pitfalls:Confusing element {2} with the number 2, or forgetting the subset containing both elements. Final Answer:{ϕ, {1}, {{2}}, {1, {2}}}

Question 15

Subsets of finite/infinite sets – find the true statement: Which of the following statements is true?

Accepted Answer

Introduction / Context:Basic properties of finiteness and subsets are bedrock facts in set theory and combinatorics. We compare universal quantifier claims to known counterexamples to determine truth values quickly. Given Data / Assumptions: Finite set S has |S| = n for some integer n ≥ 0 Infinite sets have no finite bound on cardinality Concept / Approach:Any subset of a finite set cannot have more elements than S; thus it must also be finite. By contrast, infinite sets can have finite subsets (e.g., {1} ⊂ N), disproving “every subset of an infinite set is infinite.” A proper subset of a finite set cannot be equivalent to the whole (it has strictly fewer elements). Step-by-Step Solution:(a) True by definition of finiteness and subset(b) False, counterexample: {1} ⊂ N(c) False, counterexample: N itself is an infinite subset of N(d) False, proper subset has strictly smaller cardinality Verification / Alternative check:Cardinality arguments suffice; no computation needed. Why Other Options Are Wrong:Each is contradicted by a simple subset counterexample. Common Pitfalls:Assuming properties of the whole transfer uniformly to all subsets; infinite sets especially admit both finite and infinite subsets. Final Answer:Every subset of a finite set is finite.

Question 16

Describe A ∩ (B ∪ C) for multiples of 2, 5, and 10 in N: Let A = {x ∈ N : x is a multiple of 2}, B = {x ∈ N : x is a multiple of 5}, and C = {x ∈ N : x is a multiple of 10}. Describe A ∩ (B ∪ C).

Accepted Answer

Introduction / Context:Set expressions with arithmetic properties often reduce via containment. Because every multiple of 10 is a multiple of 5, C ⊂ B. This simplifies unions and intersections substantially. Given Data / Assumptions: A = multiples of 2 B = multiples of 5 C = multiples of 10 All subsets considered within N Concept / Approach:First, simplify B ∪ C. Since C ⊂ B, B ∪ C = B. Then A ∩ (B ∪ C) = A ∩ B = numbers divisible by both 2 and 5, i.e., multiples of lcm(2,5) = 10, which is precisely C. Step-by-Step Solution:C ⊂ B ⇒ B ∪ C = BA ∩ B = multiples of 10Hence A ∩ (B ∪ C) = C Verification / Alternative check:Pick examples: 10, 20, 30 are in both A and B, matching C exactly. Why Other Options Are Wrong:A and B are too large; “None of these” is unnecessary since C fits perfectly. Common Pitfalls:Forgetting that B ∪ C collapses to B due to subset containment; always simplify unions first. Final Answer:C

Question 17

Complement after set difference inside a fixed universe: Given U = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11}, A = {3, 5, 7, 9, 11}, and B = {7, 8, 9, 10, 11}. Compute (A − B)′ with respect to U.

Accepted Answer

Introduction / Context:Operations in order: compute set difference A − B, then take the complement relative to U. When answers include elements outside U or miss required ones, “None of these” can be correct if no listed set matches the true complement. Given Data / Assumptions: U = {2,3,4,5,6,7,8,9,10,11} A = {3,5,7,9,11} B = {7,8,9,10,11} Concept / Approach:First find A − B: keep elements in A that are not in B. Then complement the result in U by removing those from U. Step-by-Step Solution:A − B = {3,5} (since 7,9,11 are in B)Complement in U: U \ {3,5} = {2,4,6,7,8,9,10,11} Verification / Alternative check:Scan options: (a) includes 12 (not in U) and wrongly includes 3 and 5; (b) includes 12 and misses 7,9; (c) misses 7 and includes 11 but not 7 incorrectly. None matches {2,4,6,7,8,9,10,11}. Why Other Options Are Wrong:They either step outside U or omit/include wrong elements relative to the computed complement. Common Pitfalls:Forgetting to ground complements in the specified universe and allowing stray elements like 12 to appear. Final Answer:None of these

Question 18

Union cardinality via inclusion–exclusion (given |S|, |T|, |S ∩ T|): If |S| = 21, |T| = 32, and |S ∩ T| = 11, find |S ∪ T|.

Accepted Answer

Introduction / Context:The cardinality of a union of two finite sets is found by inclusion–exclusion: add individual sizes and subtract the intersection to avoid double-counting shared elements. This is a fundamental counting identity. Given Data / Assumptions: |S| = 21 |T| = 32 |S ∩ T| = 11 Concept / Approach:Apply |S ∪ T| = |S| + |T| − |S ∩ T|. This ensures elements common to S and T are counted once overall. Step-by-Step Solution:|S ∪ T| = 21 + 32 − 11 = 42 Verification / Alternative check:Check extremes: if disjoint, union would be 53; since 11 overlap, reduce by 11 to get 42—consistent. Why Other Options Are Wrong:52 ignores subtracting the overlap; 32 is one set only; “None of these” is unnecessary because 42 is exact. Common Pitfalls:Adding sizes without subtracting the intersection, leading to overcounts. Final Answer:42

Question 19

Cartesian product A × B for small finite sets: Let A = {1, 2} and B = {2, 3}. Evaluate A × B.

Accepted Answer

Introduction / Context:The Cartesian product A × B is the set of all ordered pairs (a, b) with a ∈ A and b ∈ B. For small finite sets, listing pairs systematically by fixing the first component is best practice. Given Data / Assumptions: A = {1, 2} B = {2, 3} Concept / Approach:Form pairs by taking each element of A and pairing it with each element of B in order. Preserve the order (a first, then b) as order matters in ordered pairs. Step-by-Step Solution:For a = 1: (1,2), (1,3)For a = 2: (2,2), (2,3)Hence A × B = {(1,2), (1,3), (2,2), (2,3)} Verification / Alternative check:Count check: |A| * |B| = 2 * 2 = 4 pairs; four listed—complete. Why Other Options Are Wrong:(a) reverses the order and omits pairs; (b) misses (2,2); “None of these” is unnecessary since (c) is exact. Common Pitfalls:Confusing A × B with B × A or forgetting that order matters, leading to missing pairs. Final Answer:{(1, 2), (1, 3), (2, 2), (2, 3)}

Question 20

Recover a finite set from A × A size and sample pairs: Given that A × A has 9 elements and includes (−1, 0) and (0, 1), determine A.

Accepted Answer

Introduction / Context:If |A × A| = 9, then |A|^2 = 9, giving |A| = 3. The given pairs show elements actually present in A. Combining cardinality with observed components identifies A uniquely up to ordering of elements (sets are unordered). Given Data / Assumptions: |A × A| = 9 ⇒ |A| = 3 (−1, 0) ∈ A × A ⇒ −1 ∈ A and 0 ∈ A (0, 1) ∈ A × A ⇒ 0 ∈ A and 1 ∈ A Concept / Approach:Collect the distinct elements witnessed in coordinates: {−1, 0, 1}. This already has size 3, matching the required cardinality. Therefore A must be {−1, 0, 1}. Step-by-Step Solution:From (−1,0) and (0,1), elements −1, 0, 1 all occur|A| must be 3Thus A = {−1, 0, 1} Verification / Alternative check:Compute |A × A| with A = {−1,0,1}: 3^2 = 9, consistent with the problem statement. Why Other Options Are Wrong:They have fewer than 3 elements and cannot produce 9 ordered pairs in A × A. Common Pitfalls:Forgetting that the presence of a coordinate in any pair certifies membership in A; both coordinates must come from A. Final Answer:A = {−1, 0, 1}

Sets and Functions Questions

Questions

Practice smarter