Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan 0 Preliminary Notions Throughout this book, we assume that the reader is familiar with the following number systems: • The set of all positive integers N = {1,2,3,···}. • The set of all integers Z = {···,−3,−2,−1,0,1,2,3,···}. • The set of all rational numbers a Q = { : a,b ∈ Z with b 6= 0}. b • The set R of all real numbers. • The set of all complex numbers C = {a+bi : a,b ∈ R} √ where i = −1. We start with this introductory section to present the fundamentals of mathematical logic, mathematical proofs, and set theory. Fundamentals of Mathematical Logic Logic is commonly known as the science of reasoning. We will develop some of the symbolic techniques required later in the book. Definition 0.1 A proposition is any meaningful statement that is either true or false, but not both. 1 We will use lowercase letters, such as p,q,r,···, to represent propositions. We will also use the notation p : 1+1 = 3 to define p to be the proposition 1+1 = 3. Definition 0.2 The truth value of a proposition is true, denoted by T, if it is a true statement and false, denoted by F, if it is a false statement. Statements that are not propositions include questions and commands. Example 0.1 Which of the following are propositions? Give the truth value of the propositions. a. 2+3 = 7. b. Julius Ceasar was president of the United States. c. What time is it? d. Be quiet ! Solution. a. A proposition with truth value (F). b. A proposition wiht truth value (F). c. Not a proposition since no truth value can be assigned to this statement. d. Not a proposition. Definition 0.3 Let p and q be propositions. The conjunction of p and q, denoted p∧q, is the proposition: p and q. This proposition is defined to be true only when both p and q are true and it is false otherwise. The disjunction of p and q, denoted p∨q, is the proposition: p or q. The ’or’ is used in an inclusive way. This proposition is false only when both p and q are false, otherwise it is true. Example 0.2 Let p : 5 < 9 q : 9 < 7. Construct the propositions p∧q and p∨q. 2 Solution. The conjunction of the propositions p and q is the proposition p∧q : 5 < 9 and 9 < 7. The disjunction of the propositions p and q is the proposition p∨q : 5 < 9 or 9 < 7. Definition 0.4 A truth table displays the relationships between the truth values of propositions. Below, we display the truth tables of p∧q and p∨q. p q p ∧ q p q p ∨ q T T T T T T T F F T F T F T F F T T F F F F F F Definition 0.5 The negation of p, denoted ∼ p, is the proposition not p. The truth table of ∼ p is displayed below p ∼ p T F F T Example 0.3 Find the negation of the proposition p : −5 < x ≤ 0. Solution. The negation of p is the proposition ∼ p : x > 0 or x ≤ −5 Definition 0.6 Two propositions are equivalent if they have exactly the same truth values under all circumstances. We write p ≡ q. 3 Example 0.4 a. Show that ∼ (p∨q) ≡∼ p∧ ∼ q. b. Show that ∼ (p∧q) ≡∼ p∨ ∼ q. c. Show that ∼ (∼ p) ≡ p. a. and b. are known as DeMorgan’s laws. Solution. a. p q ∼ p ∼ q p∨q ∼ (p∨q) ∼ p∧ ∼ q T T F F T F F T F F T T F F F T T F T F F F F T T F T T b. p q ∼ p ∼ q p∧q ∼ (p∧q) ∼ p∨ ∼ q T T F F T F F T F F T F T T F T T F F T T F F T T F T T c. p ∼ p ∼ (∼ p) T F T F T F Definition 0.7 Let p and q be propositions. The implication p =⇒ q is the the proposition that is false only when p is true and q is false; otherwise it is true. p is called the hypothesis and q is called the conclusion. The connective =⇒ is called the conditional connective. Example 0.5 Construct the truth table of the implication p =⇒ q. Solution. The truth table is p q p =⇒ q T T T T F F F T T F F T 4 Example 0.6 Show that p =⇒ q ≡∼ p∨q. Solution. p q ∼ p p =⇒ q ∼ p∨q T T F T T T F F F F F T T T T F F T T T It follows from the previous exercise that the proposition p =⇒ q is always true if the hypothesis p is false, regardless of the truth value of q. We say that p =⇒ q is true by default or vacuously true. In terms of words the proposition p =⇒ q also reads: (a) if p then q. (b) p implies q. (c) p is a sufficient condition for q. (d) q is a necessary condition for p. (e) p only if q. Definition 0.8 Theconverseofp =⇒ q isthepropositionq =⇒ p.Theoppositeorinverse of p =⇒ q is the proposition ∼ p =⇒∼ q. The contrapositive of p =⇒ q is the proposition ∼ q =⇒∼ p. Example 0.7 Show that p =⇒ q ≡∼ q =⇒∼ p. Solution. We use De Morgan’s laws as follows. p =⇒ q ≡ ∼ p∨q ≡ ∼ (p∧ ∼ q) ≡ ∼ (∼ q ∧p) ≡ ∼∼ q∨ ∼ p ≡ q∨ ∼ p ≡ ∼ q =⇒∼ p 5 Example 0.8 Using truth tables show the following: a. p =⇒ q 6≡ q =⇒ p b. p =⇒ q 6≡∼ p =⇒∼ q Solution. a. It suffices to show that ∼ p∨q 6≡∼ q ∨p. p q ∼ p ∼ q ∼ p∨q ∼ q ∨p T T F F T T T F F T F 6= T F T T F T 6= F F F T T T T b. We will show that ∼ p∨q 6≡ p∨ ∼ q. p q ∼ p ∼ q ∼ p∨q p∨ ∼ q T T F F T T T F F T F 6= T F T T F T 6= F F F T T T T Definition 0.9 The biconditional proposition of p and q, denoted by p ⇐⇒ q, is the propositional function that is true when both p and q have the same truth values and false if p and q have opposite truth values. Also reads, ”p if and only if q” or ”p is a necessary and sufficient condition for q.” Example 0.9 Construct the truth table for p ⇐⇒ q. Solution. p q p ⇐⇒ q T T T T F F F T F F F T 6 Example 0.10 Show that the biconditional proposition of p and q is logically equivalent to the conjunction of the conditional propositions p =⇒ q and q =⇒ p. Solution. p q p =⇒ q q =⇒ p p ⇐⇒ q (p =⇒ q)∧(q =⇒ p) T T T T T T T F F T F F F T T F F F F F T T T T Propositions and Quantifiers Another way to generate propositions is by means of quantifiers. Let P(x) be a statement that depends on a variable x and which is defined on some set D. ”If P(x) is true for all values of x ∈ D” then such a statement defines a proposition. For example, suppose that P(n) : 2n is even where n ∈ N. Then the statement ”For all n ∈ N,P(n)” is always true and thus defines a proposition. Such a statement can be written as ∀n ∈ N,{2n is even}. The symbol ∀ is called the universal quantifier. Example 0.11 Write in the form ∀x ∈ D,P(x) the proposition :” every real number is either positive, negative or 0.” Solution. ∀x ∈ R,x > 0,x < 0,or x = 0. The proposition ∀x ∈ D,P(x) is false if P(x) is false for at least one value of x. In this case x is called a counterexample. Example 0.12 Show that the proposition ∀x ∈ R,x > 1 is false. x Solution. A counterexample is x = 1. Clearly, 1 < 2 = 1. 2 2 1 2 7 The notation ∃x ∈ D,P(x) is a proposition that is true if there is at least one value of x ∈ D where P(x) is true; otherwise it is false. The symbol ∃ is called the existential quantifier. Example 0.13 Let P(x) denote the statement ”x > 3.” What is the truth value of the proposition ∃x ∈ R,P(x). Solution. Since 4 ∈ R and 4 > 3 then the given proposition is true. Example 0.14 Write the sets ∩ S and ∪ S using quantifiers. i∈I i i∈I i Solution. Recall that ∩ S = {x|x is an element of S for all i in I}. Using the uni- i∈I i i versal quantifier we obtain ∩ S = {x|∀i ∈ I,x ∈ S }. Similarly, since i∈I i i ∪ S = {x|x is in S for some i ∈ I} then using the existential quantifier we i∈I i i can write ∪ S = {x|∃i ∈ I,x ∈ S }. i∈I i i Example 0.15 a. What is the negation of the proposition ∀x ∈ D,P(x)? b. What is the negation of the proposition ∃x ∈ D,P(x)? Solution. a. ∃x ∈ D,∼ P(x). b. ∀x ∈ D,∼ P(x). Example 0.16 Write the negation of each of the following propositions: a. Every polynomial function is continuous. b. There exists a triangle with the property that the sum of angles is greater than 180◦. Solution. a. There exists a polynomial that is not continuous everywhere. b. For any triangle, the sum of the angles is less than or equal to 180◦. 8 Next we discuss statements that contain multiple quantifiers. A typical example is the definition of a limit. We say that L = lim f(x) if and only if x→a ∀(cid:15) > 0,∃ a positive number δ such that if |x−a| ≤ δ then |f(x)−L| < (cid:15). Example 0.17 a. Let P(x,y) denote the statement ”x+y = y+x.” What is the truth value of the proposition (∀x ∈ R)(∀y ∈ R),P(x,y)? b. Let Q(x,y) denote the statement ”x+y = 0.” What is the truth value of the proposition (∃y ∈ R)(∀x ∈ R),Q(x,y)? Solution. a. The given proposition is always true. b. The proposition is false. For otherwise, one can choose x 6= −y to obtain 0 6= x+y = 0 which is impossible Example 0.18 Find the negation of the following propositions: a. ∀x∃y,P(x,y). b. ∃x∀y,P(x,y). Solution. a. ∃x∀y,∼ P(x,y). b. ∀x∃y,∼ P(x,y) Example 0.19 The symbol ∃ ! stands for the phrase ”there exists a unique”. Which of the following statements are true and which are false. a. ∃ ! x ∈ R,∀y ∈ R,xy = y. b. ∃ ! integer x such that 1 is an integer. x Solution. a. True. Let x = 1. b. False since 1 and −1 are both integers with integer reciprocals As a final application of quantifiers we prove the following result known as DeMorgan’s laws for sets. 9 Theorem 0.1 Let {A } be a collection of sets. Then i i∈I (a) (∪ A )c = ∩ Ac. i∈I i i∈I i (b) (∩ A )c = ∪ Ac. i∈I i i∈I i Proof. (a) Let x ∈ (∪ A )c. Then by the definition of complement we have x ∈ U i∈I i and x ∈/ ∪ A . Thus, x ∈/ A ,∀i ∈ I. Hence, we have x ∈ U and i∈I i i x ∈/ A ,∀i ∈ I and this means that x ∈ Ac,∀i ∈ I. Therefore, x ∈ ∩ Ac. i i i∈I i This proves that (∪ A )c ⊆ ∩ Ac. i∈I i i∈I i Now, let x ∈ ∩ Ac. Then x ∈/ A ,∀i ∈ I. Hence, x ∈/ ∪ A . Since x ∈ U i∈I i i i∈I i and x ∈/ ∪ A then x ∈ (∪ A )c. This shows that ∩ Ac ⊆ (∪ A )c. i∈I i i∈I i i∈I i i∈I i (b) Similar to (a) and is left for the reader. Methods of Mathematical Proofs Mathematical proofs can be classified as either a direct proof or an indirect proof. In a direct proof one starts with the hypothesis of an implication p =⇒ q and then prove that the conclusion is true. Any other method of proof will be referred to as an indirect proof. In this section we study two methods of indirect proofs, namely, the proof by contradiction and the proof by contrapositive. • Proof by contradiction: With this method, a conditional statement ”p =⇒ q” is proved by showing that if p were true and q were not true, then some contradiction (absurdity) would result. We give a couple of problems where we use this method. Example 0.20 Show that if n2 is an even integer then so is n. Solution. Supposethecontrary. Thatissupposethatnisodd. Thenthereisaninteger k such that n = 2k + 1. In this case, n2 = 2(2k2 + 2k) + 1 is odd and this contradicts the assumption that n2 is even. Hence, n must be even Example 0.21 √ Show tthat the number 2 is irrational. 10

MATH 4033 Abstract Algebra I

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