Complement of a Set and Its Properties

The complement of A is found relative to the universal set U: A′ = U − A. Removing 2 and 4 from U gives A′ = {1, 3, 5}. A set is a subset when every one of its elements belongs to the larger set. Thus {1, 3} is a subset of {1, 3, 5}. The other choices contain 2 or 4, which are members of A and hence cannot belong to A′.

Complements reverse the direction of inclusion. Since \(A\subseteq B\), every element outside \(B\) is certainly outside \(A\) as well. Thus, if \(x\in B^c\), then \(x\notin B\), which implies \(x\notin A\), so \(x\in A^c\). Therefore \(B^c\subseteq A^c\), making option A correct. Option B reverses this result, while option C requires \(A=B\). Option D is not generally true; it would impose an additional condition on the universal set.

Assuming N = {1,2,3,...}, the universal set is U = {1,2,3,4,5,6,7,8,9,10}. The primes in U are A = {2,3,5,7}. The complement A′ contains every member of U that is not prime, giving {1,4,6,8,9,10}. Number 1 is neither prime nor composite, so it correctly belongs to the complement. Option B is A itself, not A′.

The complement of A relative to U is written A′ = U \ A. It contains every element of the universal set U that is not an element of A. Removing 0, 2, and 4 from U leaves 1, 3, and 5. Therefore A′ = {1, 3, 5}, which is exactly B. A itself contains the removed elements, U contains too many elements, and the empty set contains none.

A complement relative to U is normally introduced for a set A that is contained in U. Here A contains d, but d is not an element of U; instead, U contains e, which is not in A. Thus A is not a subset of U, so the stated complement setup is invalid under the usual school-level definition. The other options are false: U is not a subset of A, A is not equal to U, and A is not empty.

To count students in A ∪ B, use the inclusion–exclusion formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Thus, n(A ∪ B) = 18 + 22 − 7 = 33. The universal set has 50 students, so those outside the union number 50 − 33 = 17. The intersection is subtracted because students belonging to both sets were counted twice.

The complement A′ consists of all elements of the universal set U that are not in A. If A′ = ∅, there is no element of U outside A. Since A is already a subset of U, it must contain every element of U. Therefore, A = U. The condition does not imply that either set is empty; it means A covers the whole universal set.

Taking N here as the positive natural numbers, U = {1,2,3,4,5,6,7,8,9,10,11,12}. The multiples of 3 in U are A = {3,6,9,12}. The complement A' contains every element of U that is not a multiple of 3, namely {1,2,4,5,7,8,10,11}. Therefore option B is correct; it has 12 − 4 = 8 elements.

Complements must be taken relative to the universal set U. Thus A' = U − A = {4,5,6}, while B' = U − B = {1,2,6}. The elements common to both complements are found by intersection: {4,5,6} ∩ {1,2,6} = {6}. Therefore, A' ∩ B' = {6}. This also agrees with De Morgan's law: A' ∩ B' = (A ∪ B)'.

First find the intersection of A and B. The only element common to A = {a,b} and B = {b,c} is b, so A ∩ B = {b}. The complement is taken relative to U = {a,b,c,d}; therefore, (A ∩ B)' consists of every element of U except b. Hence, (A ∩ B)' = {a,c,d}, which is option B.

Taking ℕ as the positive natural numbers, U = {1,2,...,20}, so n(U) = 20. The positive divisors of 20 are 1, 2, 4, 5, 10, and 20, giving n(A) = 6. The complement A′ contains the members of U that are not divisors of 20. Hence n(A′) = 20 − 6 = 14, so option C is correct.

Assuming N denotes the positive natural numbers, U = {1,2,3,...,15}, so n(U) = 15. The multiples of 5 that lie in U are A = {5,10,15}, giving n(A) = 3. Since A′ contains the elements of U outside A, n(A′) = n(U) − n(A) = 15 − 3 = 12. Thus option D is correct.

De Morgan’s law states that the complement of an intersection is the union of the complements: \((A\cap B)'=A'\cup B'\). An element is outside \(A\cap B\) whenever it fails to belong to at least one of the two sets. Therefore it belongs to \(A'\cup B'\). Option B is correct; option A incorrectly uses intersection.

The complement is defined by \(A'=U\setminus A\). If \(A'=\varnothing\), then there is no element of \(U\) outside \(A\); hence every element of \(U\) belongs to \(A\), so \(U\subseteq A\). Since the question already gives \(A\subseteq U\), both inclusions imply \(A=U\). Therefore option B is correct.

The universal set U contains the integers 1 through 18, so it has 18 elements. The elements divisible by 4 are A = {4, 8, 12, 16}, giving |A| = 4. The complement A′ contains the elements of U that are not in A. Hence |A′| = |U| − |A| = 18 − 4 = 14, so option C is correct.

First find the intersection. The only element common to A = {p,q,r} and B = {r,s} is r, so A ∩ B = {r}. The complement is taken in U, not in an unrestricted universe. Removing r from U = {p,q,r,s,t} leaves {p,q,s,t}. Hence (A ∩ B)′ = {p,q,s,t}, so option B is correct.

Because A is a subset of B, every element of A is also an element of B. Now take any element x in B'. It is not in B. Since every element of A must lie in B, x cannot be in A either; therefore x belongs to A'. Hence B' ⊆ A'. This is the complement rule that reverses inclusion. Equality is not guaranteed unless A = B.

First find the union of A and B by listing every element appearing in either set: A ∪ B = {1, 2, 3, 5, 6, 7}. The complement is taken relative to U, so remove these union elements from U. The elements left are {4, 8, 9, 10}. Hence option A is correct. Option B is the union itself, while option C is the intersection, not the complement.

The universal set U contains the natural numbers 1 through 30, so n(U) = 30. The positive divisors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30, giving n(A) = 8. The complement A' contains all elements of U that are not in A. Thus n(A') = n(U) − n(A) = 30 − 8 = 22. Hence option C is correct.

Use the complement-counting principle |A′| = |U| − |A|. The integers from −3 through 4 are U = {−3, −2, −1, 0, 1, 2, 3, 4}, so |U| = 8. The condition x² < 4 is equivalent to −2 < x < 2, giving A = {−1, 0, 1} and |A| = 3. Therefore |A′| = 8 − 3 = 5, so option C is correct.

First find the union. A contains a, c, and e, while B contains b, c, and d. Together they contain every element of U, so A ∪ B = {a,b,c,d,e} = U. The complement is defined relative to U. Therefore (A ∪ B)′ = U \ (A ∪ B) = U \ U = ∅. Option B is not correct because c is already included in both A and B, so it cannot remain in the complement.

Use the inclusion–exclusion formula: |A ∪ B| = |A| + |B| − |A ∩ B|. Therefore, |A ∪ B| = 12 + 9 − 5 = 16. The complement contains the elements of U that are not in A ∪ B, so |(A ∪ B)'| = |U| − |A ∪ B| = 20 − 16 = 4. Hence, option B is correct. Subtracting the intersection is essential because common elements would otherwise be counted twice.

First find the intersection: A∩B contains the elements common to both sets, so A∩B={2,4}. The prime symbol denotes complement relative to the stated universal set U. Therefore (A∩B)'=U\(A∩B)={1,2,3,4,5,6,7}\{2,4}={1,3,5,6,7}. Option B is only the intersection, not its complement; the other options omit elements that must remain in the complement.

The union contains every element appearing in A or B: A∪B={2,3,4,6,8,9,10,12}. It has 8 elements. Since the universal set U has 12 elements, its complement has 12−8=4 elements, namely {1,5,7,11}. The same result follows from inclusion–exclusion: n(A∪B)=6+4−2=8 because 6 and 12 are counted in both sets.

A' consists of students who are not in A, so they do not study Mathematics. Similarly, B' consists of students who do not study Physics. Their intersection A'∩B' contains students satisfying both conditions simultaneously: they study neither Mathematics nor Physics. By De Morgan's law, A'∩B'=(A∪B)', which confirms that these are students outside the group studying at least one of the two subjects.