Set Theory Practice Problems

To find the elements of set A, we solve the quadratic equation by factoring the trinomial:

x² − 5x − 6 = (x − 6)(x + 1) = 0

The roots are x = 6 and x = −1. Since the initial condition requires that x ∈ ℕ, we discard the negative value. Therefore:

A = {6}

We list the individual letters that make up the word "vector", omitting duplicates to follow the definition of a set:

The letters are v, e, c, t, o, r. None are repeated within the word, so the set in roster form is:

B = {c, e, o, r, t, v}

The logical connective "∨" (or) indicates that we must include values that satisfy the first equation, the second equation, or both simultaneously within the set of integers ℤ:

Solving the first equation: x² − 9 = 0 ⇒ x² = 9 ⇒ x = 3 or x = −3.

Solving the second equation: 2x − 1 = 9 ⇒ 2x = 10 ⇒ x = 5.

Collecting all integer solutions:

C = {−3, 3, 5}

The compound inequality 2 ≤ x < 7 restricts x to natural numbers greater than or equal to 2 and strictly less than 7. Listing all integers satisfying this condition:

D = {2, 3, 4, 5, 6}

The absolute value inequality |x| < 3 is equivalent to the open interval −3 < x < 3. Looking for solutions within the set of integers ℤ, we take the values between both bounds:

E = {−2, −1, 0, 1, 2}

The elements are the first positive even integers less than or equal to 10. One algebraic way to express this is:

F = {x ∈ ℕ / x is even ∧ 2 ≤ x ≤ 10}

Notice that each term is the square of a consecutive natural number: 1², 2², 3², 4², and 5². We express this rule by setting the range for the base:

G = {x² / x ∈ ℕ ∧ 1 ≤ x ≤ 5}

The elements are all integer multiples of 3, extending indefinitely in both positive and negative directions. Therefore, the property is defined over ℤ:

H = {3k / k ∈ ℤ}

Or alternatively:

H = {x ∈ ℤ / x = 3k ∧ k ∈ ℤ}

The collection contains positive odd integers less than 12. The corresponding set-builder expression is:

I = {x ∈ ℕ / x is odd ∧ x < 12}

Or using the standard formula for an odd integer:

I = {2k − 1 / k ∈ ℕ ∧ 1 ≤ k ≤ 6}

1. True: The letter a is explicitly listed as an element of E.
2. True: The object {a} is treated here as an individual element that belongs directly to E.
3. False: The symbol a represents an element without braces; therefore, it cannot establish a subset relation between sets.
4. True: Since the element a belongs to E, the singleton set {a} is a subset of E.
5. True: Because {a} ∈ E, the set containing that element, which is {{a}}, is a subset of E.
6. True: The empty set ∅ is explicitly listed as an element of E.
7. True: The empty set is, by definition, a subset of every set.
8. True: Since ∅ is an element of E, forming a set with it creates the singleton set {∅}, which is a subset of E.

The union combines all elements present in A or B, listing shared elements only once:

A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8}

The intersection retains only the elements found in both sets at the same time. Comparing the lists, the common numbers are 4, 5, and 6:

A ∩ B = {4, 5, 6}

The difference A − B takes the elements of A and removes any that also belong to B (the values 4, 5, and 6):

A − B = {1, 2, 3}

The difference B − A starts with the elements of B and subtracts those shared with A. This shows that set difference is not commutative:

B − A = {7, 8}

The complement A' gathers all elements in the universal set U that do not belong to A:

A' = U − A = {7, 8, 9, 10}

The symmetric difference is obtained by uniting the elements unique to each set, namely (A − B) ∪ (B − A):

A Δ B = {1, 2, 3, 7, 8}

We first evaluate the operation inside parentheses and then take the intersection with C:

We know that A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8}. Comparing this with C = {2, 4, 6, 8, 10} to identify the common elements:

(A ∪ B) ∩ C = {2, 4, 6, 8}

We first determine the intersection B ∩ C by identifying the elements shared by B and C, which are 4, 6, and 8:

B ∩ C = {4, 6, 8}

Next, we subtract these elements from set A = {1, 2, 3, 4, 5, 6}, removing 4 and 6:

A − (B ∩ C) = {1, 2, 3, 5}

Using the intersection calculated previously, A ∩ B = {4, 5, 6}, the complement of this set collects all elements from U except 4, 5, and 6:

(A ∩ B)' = {1, 2, 3, 7, 8, 9, 10}

We perform each set subtraction separately before uniting the results:

For A − C, we remove from A any elements present in C (2, 4, and 6):

A − C = {1, 3, 5}

For B − C, we remove from B the elements shared with C (4, 6, and 8):

B − C = {5, 7}

Uniting both resulting sets:

(A − C) ∪ (B − C) = {1, 3, 5, 7}

The intersection A ∩ B corresponds to the entire overlap between circles A and B, including the central region that also overlaps with C. Shading the common area between both circles:

Identifying the elements in the shaded area confirms our result analytically:

A ∩ B = {c, e}

To represent B ∩ C, we identify the overlapping region between circles B and C. This region includes both the area exclusive to B and C as well as the center shared with A:

The elements within the shaded region form the resulting set:

B ∩ C = {e, h}

The operation A ∩ B ∩ C represents the triple intersection. Geometrically, it corresponds to the central region where all three circles overlap simultaneously:

The single element contained in this central region defines the set:

A ∩ B ∩ C = {e}

For the operation (A ∩ B) − C, we begin with the intersection of A and B and remove the portion that lies inside circle C. Only the upper sector of the intersection remains shaded:

Identifying the letter contained in the shaded region:

(A ∩ B) − C = {c}

For the difference A − (B ∪ C), we take circle A and subtract any area covered by B or C. Only the region exclusive to A remains shaded:

The element located in this exclusive region is:

A − (B ∪ C) = {d}

To evaluate (A ∪ B) ∩ C, we consider the entire combined area of A and B, but keep only the portions that also lie inside circle C. This shades both lateral intersections and the central region overlapping C:

Collecting the elements located across the three shaded subregions, we get:

(A ∪ B) ∩ C = {b, e, h}

Set M has cardinality |M| = 3. Therefore, the total number of subsets in P(M) is 2³ = 8.

To ensure no subsets are missed, we list them systematically by size:

• 0-element subset: ∅
• 1-element subsets: {1}, {2}, {3}
• 2-element subsets: {1, 2}, {1, 3}, {2, 3}
• 3-element subset: {1, 2, 3}

Combining all subsets into one collection, we obtain:

P(M) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}

First, we write set A in roster form by listing the prime natural numbers less than 10:

A = {2, 3, 5, 7}

Counting its elements, we find that its cardinality is |A| = 4. Applying the formula for the cardinality of the power set:

|P(A)| = 2^|A| = 2⁴ = 16

Therefore, set A has exactly 16 subsets.

To find the number of subsets of A ∪ B, we first compute the cardinality of the union using the Principle of Inclusion-Exclusion for two sets:

|A ∪ B| = |A| + |B| − |A ∩ B|

Substituting the given values:

|A ∪ B| = 4 + 5 − 3 = 6

Knowing the cardinality of the union, we calculate the number of subsets using base 2 exponentiation:

|P(A ∪ B)| = 2⁶ = 64

Set A ∪ B has 64 subsets in total.

We set up the relationship between the number of subsets and the cardinality n of set K:

2ⁿ = 128

Expressing 128 as a power of 2:

128 = 2⁷

Equating powers with the same base gives n = 7, meaning set K has 7 elements.

To determine the number of proper subsets, we subtract 1 (the set K itself) from the total count of subsets:

Proper subsets = 2ⁿ − 1 = 128 − 1 = 127

Consequently, K has 7 elements and has 127 proper subsets.

Let S be the set of students taking Spanish and F be the set of students taking French within the universal set U of 50 students.

Since 6 students take neither course, we subtract this count from the universal set to find the cardinality of the union—that is, students taking at least one language:

|S ∪ F| = |U| − 6 = 50 − 6 = 44

Applying the Principle of Inclusion-Exclusion for two sets:

|S ∪ F| = |S| + |F| − |S ∩ F|

Substituting the given values:

44 = 32 + 28 − |S ∩ F|

44 = 60 − |S ∩ F|

Solving for the intersection:

|S ∩ F| = 60 − 44 = 16

Therefore, 16 students take both language workshops.

To find the number of students taking only Spanish, we take the total of set S and subtract the students who also take French, represented by the intersection S ∩ F:

|S − F| = |S| − |S ∩ F|

Substituting the values found:

|S − F| = 32 − 16 = 16

Exactly 16 students take only Spanish.

To determine how many students take only one language, we first calculate the group taking French only, following the same procedure:

|F − S| = |F| − |S ∩ F| = 28 − 16 = 12

Having determined both disjoint regions, we add the students taking only Spanish to those taking only French:

Total = |S − F| + |F − S| = 16 + 12 = 28

In total, 28 students take exactly one language.