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🔬 Tutorial problems alpha

🔬 Tutorial problems alpha#

Note

This problems are designed to help you practice the concepts covered in the lectures. Not all problems may be covered in the tutorial, those left are for additional practice.

\(\alpha\).1#

Consider the sets \(A = \{2,3,4\}\), \(B = \{2,5,6\}\), \(C = \{5,6,2\}\), and \(D = \{6\}\).

(a) Determine whether each of the following statements are true, false, or ambiguous:

  1. \(4 \in C\),

  2. \(5 \in C\),

  3. \(A \subseteq B\),

  4. \(D \subseteq C\),

  5. \(B = C\),

  6. \(A = B\).

(b) Find each of the following sets:

  1. \(A \cap B\),

  2. \(A \cup B\),

  3. \(A \setminus B\),

  4. \(B \setminus A\),

  5. \(A \triangle B\),

  6. \((A \cup B) \setminus (A \cap B)\),

  7. \(A \cup B \cup C \cup D\),

  8. \(A \cap B \cap C\),

  9. \(A \cap B \cap C \cap D\).

[Sydsæter, Hammond, Strøm, and Carvajal, 2016] Exercises for Section 1.1, Question 1

Consider the sets \(A = \{2,3,4\}\), \(B = \{2,5,6\}\), \(C = \{5,6,2\}\), and \(D = \{6\}\).

Part (a)

  1. Note that \(4 \notin C\), so the claim that \(4 \in C\) is false.

  2. Note that \(5 \in C\), so the claim that \(5 \in C\) is true.

  3. Note that \(3 \in A\) but \(3 \notin B\). Similarly, note that \(4 \in A\) but \(4 \notin B\). Since there is at least one element (in this case more than one element) in \(A\) that does not belong to \(B\), we know that \(A \nsubseteq B\). This means that the claim that \(A \subseteq B\) is false.

  4. Note that \(6 \in D\) and \(6 \in C\). Since \(6\) is the only element in \(D\), we know that every element of \(D\) also belongs to \(C\). Thus \(D \subseteq C\). This means that the claim that \(D \subseteq C\) is true.

  5. Note that (i) \(2 \in B\) and \(2 \in C\), (ii) \(5 \in B\) and \(5 \in C\), and (iii) \(6 \in B\) and \(6 \in C\). Furthermore, there are no other elements that are either in \(B\) alone, \(C\) alone, or both \(B\) and \(C\). As such, every element in \(B\) is also in \(C\), and every element in \(C\) is also in \(B\). This means that both \(B \subseteq C\) and \(C \subseteq B\). The only way that both of these situations can be simultaneously true is for \(B = C\). Thus we can conclude that the claim that \(B = C\) is true.

  6. Recall that we have already shown that \(A \nsubseteq B\). Since the only way for \(A=B\) is for both \(A \subseteq B\) and \(B \subseteq A\), we know that \(A \neq B\). This means that the claim that \(A = B\) is false.

Part (b)

  1. \(A \cap B = \{2,3,4\} \cap \{5,6,2\} = \{2\}\)

  2. \(A \cup B = \{2,3,4\} \cup \{5,6,2\} = \{2,3,4,5,6\}\)

  3. \(A \setminus B = \{2,3,4\} \setminus \{5,6,2\} = \{3,4\}\)

  4. \(B \setminus A = \{5,6,2\} \setminus \{2,3,4\} = \{5,6\}\)

  5. The symmetric difference of the sets \(A\) and \(B\) is \(A \triangle B = (A \setminus B) \cup (B \setminus A) = \{3, 4\} \cup \{5, 6\} = \{3, 4, 5, 6\}\)

  6. \((A \cup B) \setminus (A \cap B) = (\{2,3,4\} \cup \{5,6,2\}) \setminus (\{2,3,4\} \cap \{5,6,2\}) = \{2, 3, 4, 5, 6\} \setminus \{2\} = \{3, 4, 5, 6\}\)

  7. \(A \cup B \cup C \cup D = (A \cup B) \cup (C \cup D) = (\{2, 3, 4\} \cup \{5, 6, 2\}) \cup (\{5, 6, 2\} \cup \{6\}) = \{2,3,4,5,6\}\)

  8. \(A \cap B \cap C = (A \cap B) \cap C = (\{2, 3, 4\} \cap \{5, 6, 2\}) \cap \{5, 6, 2\} = \{2\} \cap \{5, 6, 2\} = \{2\}\)

  9. \(A \cap B \cap C \cap D = (A \cap B \cap C) \cap D = \{2\} \cap \{6\} = \emptyset\)

\(\alpha\).2#

Suppose that \(U\) is the set of all students at a particular university, \(F\) is the set of female students at that university, \(M\) is the set of mathematics students at that university, \(C\) is the set of students in the choir at that university, \(B\) is the set of biology students at that university, and \(T\) is the set of students at that university who play tennis.

(a) Describe the following sets using English prose (that is, in words):

  1. \(F \cap B \cap C\),

  2. \(M \cap F\),

  3. \((M \cap B) \setminus C\), and

  4. \(((M \cap B) \setminus C) \setminus T\).

(b) Write each of the following statements in (mathematical) set terminology.

  1. All biology students are mathematics students.

  2. There are female biology students in the university choir.

  3. No tennis player studies biology.

  4. Those female students who neither play tennis nor belong to the university choir all study biology.

[Sydsæter, Hammond, Strøm, and Carvajal, 2016] Exercises for Section 1.1, Question 2.

Part (a)

  1. The set \(F \cap B \cap C\) is the set of female biology students at that university who are part of the university choir.

  2. The set \(M \cap F\) is the set of female mathematics students at that university.

  3. \((M \cap B) \setminus C\) is the set of students at the university that study both mathematics and biology but are not in the choir.

  4. \(((M \cap B) \setminus C) \setminus T\) is the set of students at the university that study both mathematics and biology but neither participate in the university choir nor play tennis.

Part (b)

  1. If all biology students were also mathematics students, then we would have \(B \subseteq M\).

  2. If there are some female biology students that participate in the university choir, then \(F \cap B \cap C \neq \emptyset\).

  3. If there are no tennis players that study biology, then \(T \cap B = \emptyset\).

  4. If all female students that neither play tennis nor participate in the university choir study biology, then \(F \setminus (T \cup C) \subseteq B\).

\(\alpha\).3#

A survey revealed that 50 people liked coffee and 40 liked tea. Both of these figures include 35 who liked both coffee and tea. Finally, ten did not like either coffee or tea. How many people in all responded to the survey? Explain your reasoning.

Hint

Drawing a Venn diagram for this question can be very helpful.

[Sydsæter, Hammond, Strøm, and Carvajal, 2016] Exercises for Section 1.1, Question 3.

Let \(U\) be the set of all people that responded to a survey about drinking preferences for caffeinated beverages. Denote the number of elements in the set \(U\) by \(|U|\).

Let \(C\) denote the set of respondents that liked coffee. Denote the number of elements in the set \(C\) by \(|C|\). We are told that \(|C| = 50\).

Let \(T\) denote the set of respondents that liked tea. Denote the number of elements in the set \(T\) by \(|T|\). We are told that \(|T| = 40\).

Clearly the set of respondents that liked both coffee and tea is the set \(C \cap T\). Denote the number of elements in the set \(C \cap T\) by \(|C \cap T|\). We are told that \(|C \cap T| = 35\).

Clearly the set of respondents who did not like either coffee or tea is the set \(U \setminus (C \cup T)\). Denote the number of elements in the set \(U \setminus (C \cup T)\) by \(|U \setminus (C \cup T)|\). We are told that \(|U \setminus (C \cup T)| = 10\).

We want to find \(|U|\).

First note that we can decompose the set \(U\) into four mutually exclusive (and hence disjoint) subsets whose union is equal to \(U\) as follows: \(U = (C \setminus (C \cap T)) \cup (T \setminus (C \cap T)) \cup (C \cap T) \cup (U \setminus (C \cup T))\). Draw a Venn diagram to make sure you understand this decomposition.

Since this decomposition of \(U\) is both mutually exclusive and exhaustive, we must have \(|U| = |C \setminus (C \cap T)| + |T \setminus (C \cap T)| + |C \cap T| + |U \setminus (C \cup T)|\).

Note that \(|C \setminus (C \cap T)| = |C| - |C \cap T| = 50 - 35 = 15\).

Note that \(|T \setminus (C \cap T)| = |T| - |C \cap T| = 40 - 35 = 5\).

Thus we have \(|U| = |C \setminus (C \cap T)| + |T \setminus (C \cap T)| + |C \cap T| + |U \setminus (C \cup T)| = 15 + 5 + 35 + 10 = 65\).

\(\alpha\).4#

Make a complete list of all the different subsets of the set \(\{a,b,c\}\). How many such subsets are there if the empty set and the set itself are included? Repeat this question for the set \(\{a, b, c, d\}\).

[Sydsæter, Hammond, Strøm, and Carvajal, 2016] Exercises for Section 1.1, Question 4

Each element of set \(A\) may enter of not enter a subset. The full list of possibilities is therefore given by the following table, where \(1\) denotes that the element is included in the subset and \(0\) denotes that the element is not included in the subset.

\(a\)

\(b\)

\(c\)

subset

0

0

0

\(\emptyset\)

0

0

1

\(\{c\}\)

0

1

0

\(\{b\}\)

0

1

1

\(\{b,c\}\)

1

0

0

\(\{a\}\)

1

0

1

\(\{a,c\}\)

1

1

0

\(\{a,b\}\)

1

1

1

\(\{a,b,c\}\)

The power set for \(A\) is therefore \(2^A = \{\emptyset,\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},\{a,b,c\}\}\).

Clearly there are three elements in the set \(A\) and eight elements in the set \(2^A\) (that is, eight subsets of the set \(A\)). Note that \(8 = 2^3\).

Note that zeros and ones in the table above can be interpreted as binary numbers, spanning the sequence from \(0\) to \(2^3-1=7\).

The same logic can be applied to the bigger set \(\{a,b,c,d\}\), the powerset of which has \(2^4 = 16\) elements.

\(\alpha\).5#

One thousand people took part in a survey to reveal which newspaper (\(A\), \(B\), or \(C\)) that had read on a (common) specified day. The responses showed that 420 had read \(A\), 316 had read \(B\), and 160 had read \(C\). These figures include 116 who had read both \(A\) and \(B\), 100 who had read both \(A\) and \(C\), and 30 who had read both \(B\) and \(C\). Finally, all of these figures include 16 who had read all three papers.

  1. How many respondents to the survey had read \(A\) but not \(B\)?

  2. How many respondents to the survey had read \(C\), but neither \(A\) nor \(B\)?

  3. How many respondents to the survey had read neither \(A\) nor \(B\) nor \(C\)?

[Sydsæter, Hammond, Strøm, and Carvajal, 2016] Exercises for Section 1.1, Question 8

Solution is similar to question \(\alpha\).3.

Start by find a partition of the set of all respondents in the survey, that is a set of subsets which are jointly exhaustive and pairwise mutually exclusive. Draw a Venn diagram to illustrate the computations.

Correct answers are 1. 304, 2. 46, 3. 334.

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