Equal sets and Subsets

Factor the quadratic as \(x^2-4x+3=(x-1)(x-3)\). Thus its natural-number solutions are 1 and 3, so \(A=\{1,3\}\). Since \(B=\{1\}\), every element of B belongs to A, but A has the additional element 3. Therefore B is a proper subset of A: \(B\subset A\) and \(B\neq A\). Option B is false because the sets do not have exactly the same elements, while option D is false because 3 is not in B.

For \(A\subseteq B\), every element of A must belong to B. This fails because 1 belongs to A but not to B. Similarly, \(B\subseteq A\) is false because 4 belongs to B but not to A. The sets are also not equal, although they share the elements 2 and 3. Therefore neither set is a subset of the other, so option D is correct. Common elements alone do not establish a subset relation.

The positive multiples of 6 that are less than 30 are obtained by using 6, 12, 18, and 24. The next multiple is 30, but it is excluded because the condition says less than 30, not less than or equal to 30. Therefore \(A=\{6,12,18,24\}=B\), so option A is correct. Options B and C incorrectly claim a proper-subset relation, while option D is false because 30 is not included.

The positive multiples of 5 below 20 are 5, 10, and 15. The number 20 is not included because the inequality is strict: less than 20. Hence the set described by B is \(\{5,10,15\}\), exactly the same as A. Therefore A and B are equal sets, and option A is correct. Neither proper-subset option applies, and option D is false because 20 is excluded from B.

Because the element 1 must be included, it is already fixed as one member of every required subset. We therefore need to choose the remaining 2 elements from the other 5 elements, namely 2, 3, 4, 5, and 6. The number of choices is C(5, 2) = 5!/(2!3!) = 10. Hence, exactly 10 three-element subsets contain 1.

The condition x ∈ Z means that x must be an integer, and |x| ≤ 2 means that x can range from -2 to 2 inclusive. Therefore, A = {-2, -1, 0, 1, 2}. A set is a subset of A only when every one of its elements belongs to A. Option A contains -2, 0, and 2, all of which are in A, so it is the only valid subset.

Every element of B, namely 2 and 4, belongs to A, and B has fewer elements than A; therefore B is a proper subset of A. Similarly, every element of C, namely 1 and 3, belongs to A, and C is also smaller than A. Hence both B ⊂ A and C ⊂ A are true. B and C are not equal because they contain different elements.

Factor the quadratic equation: x² − 7x + 12 = (x − 3)(x − 4) = 0. Therefore, the natural-number solutions are x = 3 and x = 4, so A = {3, 4}. Since B is also exactly {3, 4}, the two sets have the same elements and hence A = B. The order of elements does not matter in a set.

Use the antisymmetry property of set inclusion. The given relations produce A ⊆ B ⊆ C, while A = C changes this chain into A ⊆ B ⊆ A. Thus both A ⊆ B and B ⊆ A hold, so A = B. Since A = C is already given, all three sets are equal. Option A is correct. Options B and C incorrectly assert proper inclusion, and D contradicts the stated equality A = C.

Solving x² = 16 gives x = 4 or x = −4. Therefore A = {−4, 4}, which is exactly the set B. Sets are equal when they contain the same elements; the order in which those elements are written is irrelevant. Thus A = B, while the proper-subset and empty-intersection statements are false.

The positive even integers less than 10 are 2, 4, 6, and 8. Hence the set-builder description gives B = {2, 4, 6, 8}. Since this is exactly the roster form of A, both sets contain the same elements and A = B. No other positive even integer below 10 exists.

The empty set ∅ is a subset of every set, so ∅ ⊆ {0}. Since B contains the element 0 while A contains no elements, A and B are not equal. Therefore, A is a proper subset of B, written A ⊂ B. Also, 0 is not an element of A because A is empty, so option D is false.

The prime digits less than 10 are 2, 3, 5, and 7. Therefore B = {2, 3, 5, 7}, which has exactly the same elements as A. Hence A = B. The digit 0 is not prime, and 1 is also not prime, so neither can be added to B. Options B and C incorrectly claim a proper-subset relationship, which requires one set to contain an element absent from the other. Option D is false because the intersection is the whole set, not {1}.

Factor the quadratic equation: x² - 5x + 6 = (x - 2)(x - 3). Therefore its solutions are x = 2 and x = 3, so the solution set is A = {2, 3}. Since B is also {2, 3}, the two sets have exactly the same elements and A = B. Option B confuses the constant term with the solution set. Option C is false because a proper subset must be smaller than the other set, while these sets are equal. Option D is false because their intersection is {2, 3}.

Every element of A, namely 1, 2, and 3, is also an element of B. In addition, B contains the element 4, which is not in A. Thus A is contained in B but is not equal to B; this is exactly the definition of a proper subset, written A ⊂ B. Option B contradicts the extra element 4. Options C and D misunderstand the definition: the absence of 4 from A supports the proper-subset relation rather than disproving it.

A positive divisor of 20 is a positive integer that divides 20 without a remainder. The complete list is 1, 2, 4, 5, 10, and 20. Thus A = {1, 2, 4, 5, 10, 20}, which is exactly the displayed set B. Therefore A = B. The list must include both 1 and 20: 1 divides every positive integer, and every positive integer divides itself. Removing either would make the set incomplete, so option D is also incorrect.

In ordinary set notation, an element is recorded only once; repeated listings do not create new elements. Therefore A = {a, b, c} after removing the repeated a and b. This is exactly the set B, so A = B. Option B incorrectly treats a set like an ordered list or multiset. Option C is false because B is not a proper subset of A when the two sets are equal. Option D is false because their intersection is {a, b, c}, not the empty set.

The positive multiples of 3 are 3, 6, 9, 12, 15, and so on. Applying the strict condition x < 15 excludes 15, leaving A = {3, 6, 9, 12}. This is exactly the set B, so A = B. Neither set contains an extra element, and option D is wrong because 15 does not satisfy x < 15. The symbols < and ≤ must be distinguished carefully.

Every element of A, namely 1 and 2, is also present in B, and B contains the additional element 3. Thus A is a proper subset of B. Similarly, every element of B is present in C, while C has the additional element 4, so B is a proper subset of C. Therefore the correct chain is A ⊂ B ⊂ C. The inclusions are proper because the consecutive sets are not equal.

The elements 1, 2, and 3 of A all occur in B, so A is a subset of B. However, B also contains the element {1}, which is a set containing 1; it is not the same object as the number 1. Consequently, B has one additional element and A is a proper subset of B. The sets are not equal, B is not a subset of A, and their intersection is not empty because they share 1, 2, and 3.