Combinatorics Homework1

Combinatorics Homework1

1. Which of these sentences are propositions? What are the truth values of those that are propositions?

   • Nanjing (南京) is the capital of Jiangsu (江苏).    T
   • Chongqing (重庆) is the capital of Sichuan (四川).    F
   • 2 + 3 = 5.    T
   • 5 + 7 = 10.    F
   • x + 2 = 11.    not a proposition
   • Answer this question.    not a proposition
   • x + y = y + x for every pair of real number x and y.    T
     
     

2. Use truth tables to verify these equivalences.
   (a) \(p ∧ T ≡ p\)        (b) \(p ∨ F ≡ p\)       (c) \(p ∧ F ≡ F\)
   (d) \(p ∨ T ≡ T\)       (e) \(p ∨ p ≡ p\)        (f) \(p ∧ p ≡ p\)
   
   

   Solution:
   We can easily prove the above six propositions from the following two truth tables：

   \[\begin{aligned} &\text { conjunction }\\ &\begin{array}{c|c|c} \hline \hline p & q & p \wedge q \\ \hline \mathrm{T} & \mathrm{T} & \mathrm{T} \\ \mathrm{T} & \mathrm{F} & \mathrm{F} \\ \mathrm{F} & \mathrm{T} & \mathrm{F} \\ \mathrm{F} & \mathrm{F} & \mathrm{F} \\ \hline \end{array} \end{aligned} \qquad \begin{aligned} &\text { disjunction }\\ &\begin{array}{c|c|c} \hline \hline p & q & p \vee q \\ \hline \mathrm{T} & \mathrm{T} & \mathrm{T} \\ \mathrm{T} & \mathrm{F} & \mathrm{T} \\ \mathrm{F} & \mathrm{T} & \mathrm{T} \\ \mathrm{F} & \mathrm{F} & \mathrm{F} \\ \hline \end{array} \end{aligned} \]

   (a) \(p ∧ T ≡ p\) : if \(p\) is false, \(F ∧ T ≡ F\); while \(p\) is true, \(T ∧ T ≡ T\). Therefore \(p ∧ T ≡ p\)
   (b) \(p ∨ F ≡ p\) : if \(p\) is false, \(F ∨ F ≡ F\); while \(p\) is true, \(T ∨ F ≡ T\). Therefore \(p ∨ T ≡ p\)
   (c) \(p ∧ F ≡ F\) : if \(p\) is false, \(F ∧ F ≡ F\); while \(p\) is true, \(T ∧ F ≡ F\). Therefore \(p ∧ F ≡ F\)
   (d) \(p ∨ T ≡ T\) : if \(p\) is false, \(F ∨ T ≡ T\); while \(p\) is true, \(T ∨ T ≡ T\). Therefore \(p ∨ T ≡ T\)
   (e) \(p ∨ p ≡ p\) : if \(p\) is false, \(F ∨ F ≡ F\); while \(p\) is true, \(T ∨ T ≡ T\). Therefore \(p ∨ p ≡ p\)
   (f) \(p ∧ p ≡ p\) : if \(p\) is false, \(F ∧ F ≡ F\); while \(p\) is true, \(T ∧ T ≡ T\). Therefore \(p ∧ p ≡ p\)

3. Let Q(x, y) denote the statement “x is the capital of y.” What are these truth values?

   • Q(“Hangzhou (杭州)”, “Zhejiang (浙江)”)        T
   • Q(“Shenzhen (深圳)”, “Guangdong (广东)”)        F
   • Q(“Qingdao (青岛)”, “Shandong (山东)”)        F
   • Q(“Yinchuan (银川)”, “Ningxia (宁夏)”)        T
     
     

4. Let P(x) be the statement “x can speak Russian” and let Q(x) be the statement “x knows the computer language C++”. Express each of these sentences in terms of P(x), Q(x), quantifiers, and logical connectives. The universe of discourse for quantifiers consists of all students at your school.

• There is a student at your school who can speak Russian and who knows C++.

\[\exists x (P(x)\wedge Q(x)) \]

• There is a student at your school who can speak Russian but who doesn’t know C++.

\[\exists x (P(x)\wedge \lnot Q(x)) \]

• Every student at your school either can speak Russian or knows C++.

\[\forall x (P(x) \vee Q(x)) \]

• No student at your school can speak Russian or knows C++.

\[\lnot \forall x ( P(x) \vee Q(x)) \]

1. What rule of inference is used in each of these arguments?

   • Alice is a mathematics major. Therefore, Alice is either a mathematics major or a computer science major.
     Suppose the simple proposition
     p: Alice is a mathematics major;
     q: Alice is a computer science major.
     The above proposition can be translated as \(p \Rightarrow (p \vee q).\) This is the additional law.
     
     
   • Jerry is a mathematics major and a computer science major. Therefore, Jerry is a mathematics major.
     Suppose the simple proposition
     p: Jerry is a mathematics major;
     q: Jerry is a computer science major.
     The above proposition can be translated as \((p \vee q) \Rightarrow p.\) This is the simplification law.
     
     
   • If it is rainy, then the pool will be closed. It is rainy. Therefore, the pool is closed.
     Suppose the simple proposition
     p: It is rainy;
     q: The pool is closed.
     The above proposition can be translated as \((p \rightarrow q) \wedge p \Rightarrow q.\) This is the modus ponens.
     
     
   • If it snows today, the university will close. The university is not closed today. Therefore, it did not snow today.
     Suppose the simple proposition
     p: It snows today;
     q: The university is closed today.
     The above proposition can be translated as \((p \rightarrow q) \wedge \lnot q \Rightarrow \lnot p.\) This is the modus tollens.
     
     
   • If I go swimming, then I will stay in the sun too long. If I stay in the sun too long, then I will sunburn. Therefore, If I go swimming, then I will sunburn.
     Suppose the simple proposition
     p: I go swimming;
     q: I stay in the sun too long;
     r: I sunburn.
     The above proposition can be translated as \((p \rightarrow q) \wedge (q \rightarrow r) \Rightarrow (p \rightarrow r).\) This is the hypothetical syllogism.
   
   

2. Let \(A\), \(B\), and \(C\) be sets. Show that

   • \(A ∪ (B ∪ C) = (A ∪ B) ∪ C.\)
     
     \(\text{Show:} \\ \begin{aligned} x \in A \cup(B \cup C) \Leftrightarrow &x \in A \text { or } x \in(B \cup C) \\ \Leftrightarrow &x \in A \text { or }(x \in B \text { or } x \in C) \\ \Leftrightarrow & x \in A \text { or }x \in B \text { or } x \in C \\ \Leftrightarrow & (x \in A \text { or }x \in B) \text { or } x \in C \\ \Leftrightarrow &x \in (A \cup B) \cup C \end{aligned}\)
     
     

   • \(A ∩ (B ∩ C) = (A ∩ B) ∩ C.\)
     
     \(\text{Show:} \\ \begin{aligned} x \in A \cap(B \cap C) \Leftrightarrow & x \in A \text { and } x \in(B \cap C) \\ \Leftrightarrow &x \in A \text { and }(x \in B \text { and } x \in C) \\ \Leftrightarrow &x \in A \text { and }x \in B \text { and } x \in C \\ \Leftrightarrow &(x \in A \text { and }x \in B )\text { and } x \in C \\ \Leftrightarrow &x \in (A \cap B) \cap C \end{aligned}\)
     
     

   • \(A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).\)
     
     \(\text{Show:} \\ \begin{aligned} x \in A \cup(B \cap C) \Leftrightarrow & x \in A \text { or } x \in(B \cap C) \\ \Leftrightarrow &(x \in A \text { or } x \in B) \text { and }(x \in A \text { or } x \in C) \\ \Leftrightarrow & x \in(A \cup B \text { and } A \cup C) \\ \Leftrightarrow & x \in(A \cup B) \cap(A \cup C) \end{aligned}\)
     
     

3. Determine whether the function \(f : Z × Z → Z\) is onto if

   • \(f (m, n) = m + n.\)       onto
   • \(f (m, n) = m^2 + n^2.\)       false
   • \(f (m, n) = m.\)       onto
   • \(f (m, n) = |n|.\)       false
   • \(f (m, n) = m-n.\)       onto

   Note: A function \(f\) from \(A\) to \(B\) is called onto, or surjective, if and only if for every element \(b ∈ B\) there is an element \(a ∈ A\) with \(f (a) = b\).