Chapter 1 Elementary Logic

Principles of Mathematics (Math 2450)

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2017-2018 ÈðB@ É®Ë@ Chapter 1 Elementary Logic

The study of logic is the study of the principles and methods used in distinguishing valid arguments from those that are not valid. The aim of this chapter is to help the student to understand the principles and methods used in each step of a proof. The starting point in logic is the term statement (or proposition) which is used in a technical sense. We introduce a minimal amount of mathematical logic which lies behind the concept of proof. In mathematics and computer science, as well as in many places in everyday life, we face the problem of determining whether something is true or not. Often the decision is easy. If we were to say that 2 + 3 = 4, most people would immediately say that the statement was false. If we were to say 2 + 2 = 4, then undoubtedly the response would be “of course” or “everybody knows that”. However, many statements are not so clear. A statement such as “the sum of the ﬁrst n odd integers is equal to n2” in addition to meeting with a good deal of consternation, might be greeted with a response of “Is that really true?” or “why?” This natural response lies behind one of the most important concepts of mathematics, that of proof.

1.1 Statements and their connectives

When we prove theorems in mathematics, we are demonstrating the truth of certain statements. We therefore need to start our discussion of logic with a look at statements, and at how we recognize certain statements as true or false. Deﬁnition. (Statement) QKQ®K, éJ ¯ A statement is a declarative sentence that is either true or false. Remark. A statement is also called a proposition. Example 1. The following sentences are statements

(a) Gaza is a Palestinian city.

(b)2 − 1 equals 3.

(d) IUG is a Palestinian university.

(e) Earth is the closest planet to the sun.

Example 2. The following sentences are NOT statements

(a) How are you?

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(b) Gaza is a beautiful city.

(d) 4+1.

(e) I will come to school next week.

(f) Would you visit us tomorrow.

(g) He lives in Gaza.

(h) x2 = 9.

Remark. Commends, questions, and opinions are not statements.

Compound statements

The statements in Example 1 are simple statements since they do not include any logical connective. A compound statement is statement that has at least one logical connective.

Example 1. The following sentences are compound statements

(a) Gaza is a Palestinian city and Palestine is an arabic country.

(b)2 − 1 equals 3 or 7 is divisible by 2.

(d) 2 divides 6 if and only if 2 × 3 = 6.

(e) π is not a rational number.

Notation. We will denote simple statements by lowercase letters p, q, r, ... and we will denote compound statements by uppercase letters P, Q, R, ....

Fundamental connectives

To form new compound statements out of old ones we use the following ﬁve fundamental connectives:

1. “not” symbolized by ∼.

2. “and” symbolized by ∧.

3. “or” symbolized by ∨.

4. “If..., then ...” symbolized by −→.

5. “...if and only if ...” symbolized by ←→.

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Truth tables( ) YË@ Èð@Yg. A truth table is a mathematical table used in logic which determines the truth values of a compound statement form for all logical possibilities of its components. A truth table has one column for each component and one ﬁnal column for the compound statement. Each row of the truth table contains one possible truth value for each component and the result of the logical operation for those values. We will use ”T” for true and ”F” for false.

Negation

Deﬁnition. (Negation) The connective ∼ is called the negation and it may be placed before any statement p to form a compound statement ∼ p (read: not p or the negation of p). The truth values for ∼ p are deﬁned as follows:

p ∼ p T F F T

Example. Write the negation of each of the following: √ (a) 2 is a rational number.

(b) The sine function is continuous at x = 0.

Solution.

Conjunction

Deﬁnition. (Conjunction) The connective ∧ is called the conjunction and it may be placed between any two statements p and q to form a compound statement p ∧ q (read: p and q or conjunction of p and q). The truth values for p ∧ q are deﬁned as follows:

p q p ∧ q T T T T F F F T F F F F

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Remarks.

(1) The statement p ∧ q is true only when both p and q are true.

(2) In a compound statement with two components p and q there are 2 × 2 = 4 possibilities, called the logical possibilities. In general, if a compound statement has n components, then there are 2n logical possibilities.

Example 1. Indicate which of the following statements is T and which is F:

(a) 1 + 1 = 2 and 3 − 1 = 2.

(b) 5 is an integer and 1 − 3 = 1.

(d) 5 × 2 = 5 and 5 × 3 = 10.

Solution.

Example 2. Construct a truth table for the compound statement p ∧ (∼ q).

Solution.

Remark. The English words but, while, and although are usually translated symbolically with the conjunction connective, because they have the same meaning as and.

Example. Translate the following statement into logical form using connectives: “8 is divisible by 2 but it is not divisible by 3.”

Solution.

Exercise 1.1 (1-24)

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1.2 Three more connectives

In this section we will study the connectives “or”, “If..., then ...”, and “...if and only if ...”.

Disjunction

In English language there is an ambiguity involved in the use of “or”. Inclusive or: The statement “I will get a Master degree or a Ph.D” indicate that the speaker will get both the Master degree and the Ph. D. Exclusive or: But in the statement “I will study mathematics or physics ” means that only one of the two fields will be chosen. In mathematics and logic we can not allow ambiguity. Hence we must decide on the meaning of the word “or”. Definition. (Disjunction) The connective ∨ is called the disjunction and it may be placed between any two statements p and q to form the compound statement p ∨ q (read: p or q or the disjunction of p and q). The truth values for p ∨ q are defined as follows: p q p ∨ q T T T T F T F T T F F F Remark. The statement p ∨ q is true when at least one of p and q is true. Example 1. Indicate which of the following statements is T and which is F: (a) 1 − 1 = 1 or 3 + 3 = 6. (b) 4 = 4 or 4 > 4. √ (c) −1 = 2 or (2)2 = −1. (d) 7 is a prime number or 7 is an odd number. Solution.

Example 2. Construct the truth table for the compound statement ∼ [p ∨ (∼ q)]. Solution.

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Deﬁnition. (Equivalent statements) Two statements P and Q, simple or compound, are said to be ( logically) equivalent if P and Q have the same truth values in each of all the logical possibilities. In such case we write P ≡ Q. Example. Show that ∼ [p ∨ (∼ q)] ≡∼ p ∧ q. Solution.

Exercise. Show that ∼ [p ∨ q] ≡ (∼ p) ∧ (∼ q) and ∼ [p ∧ q] ≡ (∼ p) ∨ (∼ q).

Conditional

Deﬁnition. (Conditional) The connective → is called the conditional and it may be placed between any two statements p and q to form the compound statement p → q (read: if p then q). The statement p → q is deﬁned to be equivalent to the statement ∼ [p ∧ (∼ q)].

The truth values of p → q The truth values of p → q are given by the following table p q p → q T T T T F F F T T F F T Remarks. (1) The statement p → q is false only when p is true and q is false.

(2) In a conditional statement p → q, p is called the antecedent or suﬃcient condition and q is called the consequent or necessary condition. Example 1. Determine whether the following statements are T or F: (a) If 2 − 4 = 2, then 2 − 2 = 4.

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(b) If 7 < 9, then 7 < 8.

(d) If 5 < 6, then 5 is even.

Solution.

Example 2. Construct the truth table for the compound statement (p ∨ q) → r.

Solution.

Example 3. Write the following conditional statement as a conjunction: “If 3 + 2 = 5, then 5 − 2 = 3.”

Solution.

Example 4. Is the statement p → q equivalent to the statement ∼ q →∼ p? Explain.

Solution.

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Remark. We use p → q to translate the following statements:

1. If p, then q.

2. p only if q.

3. q if p.

4. p is suﬃcient to q.

5. q is necessary for p.

6. q whenever p.

Biconditional

Deﬁnition. (Biconditional) The connective ↔ is called the biconditional and it may be placed between any two statements p and q to form the compound statement p ↔ q (read: p if and only if q). The statement p ↔ q is deﬁned to be equivalent to the statement (p → q) ∧ (q → p).

The truth values of p ↔ q

The truth values of p → q are given by the following table p q p ↔ q T T T T F F F T F F F T

Remark. The statement p ↔ q is true when both p and q have the same truth values.

Example 1. Determine whether the following statements are T or F:

(a) 1 is odd if and only if 3 is even.

(b) |5| = −5 if and only if 5 > 0. √ (c) 4 = 2 if and only if (2)2 = 4.

(d) 5 > 6 if and only if 5 is even.

Solution.

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Example 2. Construct the truth table for the compound statement (p ∧ q) ↔ p.

Solution.

Remark. We use p ↔ q to translate the following statements:

1. p if and only if q.

2. p is equivalent to q.

3. p is necessary and suﬃcient for q.

Example 3. Translate the given compound statements into a symbolic form using the suggested symbols.

(a) ”A natural number is even if and only if it is divisible by 2.” (E, D)

(b) ”A matrix has an inverse whenever its determinant is not zero.” (I, Z)

Solution.

Exercise 1.2 (1-21)

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1.3 Tautology, Implication, and Equivalence

Tautology ( ) H. ñË@ Ê¢Ó QK Q®K Example. Construct the truth table for p∨ ∼ p.

Solution.

Deﬁnition. (Tautology) A statement is said to be a tautology if it is true for every assignment of truth values to its components.

Example. Show that q → [p ↔ (p ∧ q)] is a tautology.

Solution.

Remark. Tautologies are important both because a statement that has the form of a tautology may be used as a step in a proof, and because tautologies are used to create rules for making deductions in a proof.

Implication

Deﬁnition. (Implication) Let P and Q be two statements, compound or simple. If the conditional statement P → Q is a tautology, then it is called an implication and it is denoted by P ⇒ Q (read: P implies Q).

Example 1. Show that p → p is an implication.

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Solution.

Example 2. Show that (p ∧ q) ⇒ (q ∧ p). Solution.

Exercise: Show that 1. p ⇒ (p ∧ p). 2.( p ∧ q) ⇒ q. Theorem 1.1. Let p and q be any two statements. Then (a) Law of addition: p ⇒ (p ∨ q) (b) Law of simpliﬁcation: (p ∧ q) ⇒ p (c) Disjunctive syllogism (The way that aﬃrms by denying): [(p∨q)∧ ∼ p] ⇒ q.( ) ÉA¯ ù ®¢JÓ AJ ¯ Proof. By truth tables.

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Equivalence

Deﬁnition. (Equivalence) Let P and Q be two statements, compound or simple. If the biconditional statement P ↔ Q is a tautology, then it is called an equivalence and it is denoted by P ⇔ Q (read: P is equivalent to Q). Remark. P ⇔ Q and P ≡ Q have the same meaning and we will use the two notations interchange- ably. Deﬁnition. (Converse and contrapositive of a conditional statement) Let p and q be statements. (1) The converse of p → q is q → p. (2) The contrapositive of p → q is ∼ q →∼ p. Example. Determine whether p → q and its converse are equivalent or not. Solution.

Theorem 1.2. Let p and q be any two statements. Then (a) Law of double negation: ∼ (∼ p) ≡ p (b) Commutative law: (p ∧ q) ≡ (q ∧ p) and (p ∨ q) ≡ (q ∨ p) (c) Laws of idempotency: (p ∨ p) ≡ p (p ∧ p) ≡ p. (d) Contrapositive law: (p → q) ≡ (∼ q →∼ p) Proof. By truth tables.

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Theorem 1.3. (De Morgan’s laws) Let p and q be any two statements. Then

(a) ∼ (p ∧ q) ≡∼ p∨ ∼ q

(b) ∼ (p ∨ q) ≡∼ p∧ ∼ q

Proof. Done

Deﬁnition. (Denials) A denial of a statement P is any statement equivalent to ∼ P .

Remark. A statement has only one negation, but always has many denials.

Example. Give a useful denial of each statement.

(a)3 ≥ 3.

(b) The function f(x) = sin x is odd and increasing.

(d) If the number 5.1 is an integer, then it is positive.

Solution.

Theorem 1.4. Let p, q, and r be any statements. Then

(a) Associative laws: (p ∧ q) ∧ r ≡ p ∧ (q ∧ r) (p ∨ q) ∨ r ≡ p ∨ (q ∨ r)

(b) Distributive laws: p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r) p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

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Proof. By truth tables.

Theorem 1.5. Let p, q, r, and s be any statements. Then

(a) Constructive dilemmas: [(p → q) ∧ (r → s)] ⇒ [(p ∨ r) → (q ∨ s)] [(p → q) ∧ (r → s)] ⇒ [(p ∧ r) → (q ∧ s)]

(b) Destructive dilemmas: [(p → q) ∧ (r → s)] ⇒ [(∼ q ∨ ∼ s) → (∼ p ∨ ∼ r)] [(p → q) ∧ (r → s)] ⇒ [(∼ q∧ ∼ s) → (∼ p∧ ∼ r)]

Proof. By truth tables.

Theorem 1.6. Let p and q be any two statements. Then

(a) Modus Ponens (The way that aﬃrms by aﬃrming): [(p → q) ∧ p] ⇒ q

(b) Modus Tollens (The way that denies by denying): [(p → q)∧ ∼ q] ⇒∼ p

Example. The following example may help in understanding Modus Ponens and Modus Tollens.

• Statement p → q: If I am in Gaza, then I am in Palestine.

• Modus Ponens: I am in Gaza. Therefore, I am in Palestine.

• Modus Tollens: I am not in Palestine. Therefore, I am not in Gaza.

Exercise 1.3: (1—20)

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1.4 Contradiction

Deﬁnition. (Contradiction) A statement whose truth values are all false for every assignment of truth values to its components is called a contradiction. Example. Show that p ∧ (∼ p) is a contradiction. Solution.

Remark. The negation of a tautology is a contradiction. In other words, if t is a tautology, then ∼ t is a contradiction; conversely, if c is a contradiction, then ∼ c is a tautology. Theorem 1.7. Let t, c, and p be a tautology, a contradiction, and an arbitrary statement, respec- tively. Then (a) p ∧ t ⇔ p, p ∨ t ⇔ t,

(b) p ∨ c ⇔ p, p ∧ c ⇔ c,

Exercise 1.4: (1-5) 1.5 Deductive reasoning ( Q ) ú ÍBYJB@ º®JË@ Deductive reasoning is one of the two basic forms of valid reasoning. While inductive reasoning argues from the particular to the general, deductive reasoning argues from the general to a speciﬁc instance. The basic idea of deductive reasoning is that if something is true for a class of things in general, this truth applies to all legitimate members of that class. The key, then, is to be able to properly identify members of the class. Miscategorizing will result in invalid conclusions. One of the most common and useful forms of deductive reasoning is the syllogism. The syllogism is a speciﬁc form of argument that has three easy steps. • Every X has the characteristic Y .

• This thing is X.

• Therefore, this thing has the characteristic Y .

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Example.

• All numbers ending in 0 or 5 are divisible by 5.

• The number 35 ends with a 5.

• Therefore, 35 is divisible by 5.

Deductive reasoning, also deductive logic or logical deduction or, informally, “top-down” logic, is the process of reasoning from one or more statements (premises) to reach a logically certain conclusion. Thus; deductive reasoning is a method of veriﬁcation that use logic to draw conclusions based on statements accepted as true (premises). It links premises with conclusions. If all premises are true, the terms are clear, and the rules of deductive logic are followed, then the conclusion reached is necessarily true.

Rules of inference The 16 laws summarized in Theorem 1.1 through 1.6 are very useful tools for justifying logical equivalences and implications. We shall call these 16 laws the rules of inference. It should be noted that these rules are selected just for convenient references and are not intended to be independent of each other. For instance, the Contrapositive Law can be established “deductively” by using other laws and relevant deﬁnitions, as the next example shows. Example 1. Prove the Contrapositive Law, (p → q) ≡ [(∼ q) → (∼ p)], using relevant deﬁnitions and other rules of inference.

Solution.

Example 2. Prove the disjunction syllogism, [(p ∨ q) ∧ (∼ p)] ⇒ q, by deductive reasoning.

Solution.

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Example 3. Prove the following Exportation law:

[(p ∧ q) → r] ≡ [p → (q → r)].

Solution.

Example 4. Use deductive reasoning to prove that [(p → r) ∨ (q → s)] ≡ [(p ∧ q) → (r ∨ s)].

Solution.

Remark. If we use truth table to prove the above statement, we need 24 = 16 cases.

Example 5. Prove the tautology (p → q) ⇔ [p → (p ∧ q)] by deductive method.

Solution.

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Example 6. (Exam) Prove [(p → r) ∧ (q → r)] ⇔ [(p ∨ q) → r] by deductive method. Solution.

Example 7. Prove [(p → q) ∨ (p → r)] ∧ p ⇒ (q ∨ r) by deductive method. Solution.

Exercise 1.5 (1-20)

1.6 Quantiﬁcation Rules

Some sentences depend on some variables and become statements when the variables are replaced by a certain values. Deﬁnition. (Open sentence) A sentence containing one or more variables and which becomes a statement only when the variables are replaced by certain values is called an open sentence (or a propositional predicate).

Notation. An open sentence P with variables x1, x2, ··· , xn will be denoted by P (x1, x2, ··· , xn). Example 1. P (x): x + 1 = 0 is an open sentence. P (0) is false but P (−1) is true. Example 2. P (x, y): x + y = 1 is an open sentence. P (0, 1) is true but P (1, 1) is false.

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Definition. (Universe) The set of all objects that can be considered in an open sentence is called the universe or the domain of discourse. Remark. In many cases the universe will be understood from the context. However, there are times when it must be specified explicitly. Definition. (Truth set) The collection of objects from the universe that may be substituted to make an open sentence a true statement is called the truth set of the sentence. Example. Consider the open sentence P (x): x2 < 5. The truth set of this sentence depends on the universe we choose. If we choose the universe to be the set of natural numbers N, then the truth set is {1, 2}. If√ we√ choose the universe to be the set of real numbers R, then the truth set is the open interval (− 5, 5). Definition. (Equivalent open sentences) With a universe specified, two open sentences P (x) and Q(x) are equivalent if they have the same truth set.

Example. Let the universe be the set R. (a) The sentences 4x + 2 = 10 and x = 2 are equivalent open sentences.

(b) The sentences x2 = 4 and x = 2 are not equivalent open sentences.

Universal and existence quantifiers Definition. (Universal and existence quantifiers) Let P (x) be an open sentence. (1) ∀ is called the universal quantifier and the statement (∀x)(P (x)) (read: for all x P (x)) is true when P (x) is true for all x in the universe.

(2) ∃ is called the existential quantiﬁer and the statement (∃x)(P (x)) (read: there exists x such that P (x)) is true when there exists at least one x in the universe such that P (x) is true. Example. Determine whether the following statements are true or false: (a) Let U = {1, 2, 3, 4} be the domain of discourse.

• (∀x)(x + 2 ∈ U) • (∃x)(x + 1 = 4)

(b) Let U = R be the universe. • (∃x)(x ≥ −1) • (∀x)(x ≥ 5) • (∀x)(|x| ≥ 0) • (∃x)(|x| < 0) • (∀x)(x > 0 or x < 0).

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Remark. In general, a sentence of the form “All P (x) are Q(x)” should be symbolized (∀x)[P (x) → Q(x)]. And, in general, a sentence of the form “Some P (x) are Q(x)” should be symbolized (∃x)(P (x) ∧ Q(x)). Example. Translate the following statements into logical language using a quantiﬁer and specify the domain of discourse in each case. (a) All mathematics students study Math 2450. (b) Some arabs are not moslems. (c) The equation x2 + x = 5 has a solution. (d) cos2 x + sin2 x = 1 (e) All integers are rational numbers. (f) Some rational numbers are integers. Solution.

Remarks. 1. “for every”, “for each”, and “for all” have the same meaning in mathematics. 2. “for some”, “there is”, “at least one” and “there exists” have the same meaning in mathematics. 3. In less formal expressions, we often put the quantiﬁer after the sentence. Example. [f(x) = 0 ∀x] ≡ [(∀x)(f(x) = 0)] (∃x)(f(x) = 0) ≡ [f(x) = 0 for some x]

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Equivalent quantiﬁed sentences Deﬁnition.

(1) Two quantiﬁed sentences are equivalent in a given universe if and only if they have the same truth value in that universe.

(2) Two quantiﬁed sentences are equivalent if and only if they are equivalent in every universe.

Example 1. Let the universe be U = N. Show that (∀x)(x > 0) and (∀x)(x ≥ 1) are equivalent. Solution.

2 2 Example 2. Let the universe be U = Z. Show that (∃x)(x = 2) and (∃x)(x = 3) are equivalent. Solution.

Remark. Statements of the form “Every element of the set A has the property P ” and “Some element of the set A has property P ” occur so frequently that abbreviated symbolic forms are desirable.

(a) “Every element of the set A has the property P ” could be restated as “If x ∈ A, then . . .” and symbolized by (∀x ∈ A)(P (x)).

(b) “Some element of the set A has property P ” is abbreviated by

(∃x ∈ A)(P (x)).

Example 1. Consider the deﬁnition of a rational number “The real number x is rational if and only if there are integers p and q, with q 6= 0 such that x = p/q.” It may symbolized: p x is rational if and only if (∃p ∈ )(∃q ∈ )(q 6= 0 ∧ x = ). Z Z q Example 2. The statement “For every rational number there is a larger integer” may be symbolized by (∀x)[(x ∈ Q) → (∃z)(z ∈ Z ∧ z > x)] or (∀x ∈ Q)(∃z ∈ Z)(z > x).

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Rule of quantiﬁer negation

Remark. Let the domain of discourse be U = {a1, a2, ··· , an}. Then

1. The statement (∀x)(P (x)) means P (a1) ∧ P (a2) ∧ · · · ∧ P (an).

2. The statement (∃x)(P (x)) means P (a1) ∨ P (a2) ∨ · · · ∨ P (an). Deﬁnition. (Quantiﬁer negation) (1) ∼ [(∀x)(P (x))] ≡ (∃x)(∼ P (x)) (2) ∼ [(∃x)(P (x))] ≡ (∀x)(∼ P (x)) Example 1. Which of the following is equivalent to the negation of the statement “All functions are continuous” (a) All functions are not continuous. (b) Some functions are continuous. (c) Some functions are not continuous. Solution.

Example 2. Find an equivalent statement to the negation of the statement “Some rational numbers are integers” by using quantiﬁer negation. Solution.

Example 3. State in words the negation of the statement “For all x ∈ Z, if x is divisible by 6, then x is divisible by 3”. Solution.

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Statements with more than one quantiﬁer

Many statements have more than one quantiﬁer. We must deal with each in succession, starting from the left. Remarks. 1. The statement (∀x)(∀y)(P (x, y)) is true when (∀y)(P (x, y)) is true for all x in the domain of discourse.

2. The statement (∀x)(∃y)(P (x, y)) is true when (∃y)(P (x, y)) is true for all x in the domain of discourse.

3. The statement (∃x)(∀y)(P (x, y)) is true when (∀y)(P (x, y)) is true for at least one x in the domain of discourse.

4. The statement (∃x)(∃y)(P (x, y)) is true when (∃y)(P (x, y)) is true for at least one x in the domain of discourse. Example 1. Let U = {1, 2, 3} and let P (x, y): x + y = 6. Then (∀x)(∃y)(P (x, y)) means (∃y)(P (1, y)) and (∃y)(P (2, y)) and (∃y)(P (3, y)) Example 2. Determine which of the following statements is true and which is false using the universe U = R: (a) (∀x)(∃y)(x + y = 0)

(b) (∃x)(∀y)(x + y = 0)

(d) (∀y)(∃x)(x ≤ y)

(e) (∃y)(∀x)(x ≥ y)

(f) (∃x)(∀y)(∀z)(xy = xz) Remarks. Great care must be taken in proofs that contain expressions involving more than one quantiﬁer. Here are some manipulations of quantiﬁers that permit valid deductions. 1.( ∀x)(∀y)P (x, y) ≡ (∀y)(∀x)P (x, y)

2.( ∃x)(∃y)P (x, y) ≡ (∃y)(∃x)P (x, y)

3.( ∃x)(∀y)P (x, y) ⇒ (∀y)(∃x)P (x, y)

4.( ∀x)[P (x) ∧ Q(x)] ≡ [(∀x)P (x) ∧ (∀x)Q(x)]

5.( ∃x)[P (x) ∨ Q(x)] ≡ [(∃x)P (x) ∨ (∃x)Q(x)]

6.[( ∀x)P (x) ∨ (∀x)Q(x)] ⇒ (∀x)[P (x) ∨ Q(x)]

7.( ∃x)[P (x) ∧ Q(x)] ⇒ [(∃x)P (x) ∧ (∃x)Q(x)]

8.( ∀x)[P (x) → Q(x)] ⇒ [(∀x)P (x) → (∀x)Q(x)] Example. Determine which of the following statements is true and which is false using the universe U = R:

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√ (a) (∀x)( x2 = |x|) → (∀x)(x2 = |x|2) (b) (∃x)(x ≥ 0 ∧ |x| < 0) (c) (∀x)(x 6= 0 → x has a multiplicative inverse.) (d) (∀x)(x ≥ 0 ∨ x < 0) (e) (∀x)(x ≥ 0) ∨ (∀x)(x < 0)

Exercise 1.6 (1-5)

Additional Exercises

1. Let U = {a, b}. Prove that (∃x)(∀y)P (x, y) ⇒ (∀y)(∃x)P (x, y). 2. Use quantiﬁers to translate the following statements into symbolic forms. The universe for each is given in parentheses (a) All functions are continuous. (All functions) (b) Some functions are both continuous and diﬀerentiable. (All functions) (c) Every even integer is divisible by 2. (All integers) (d) If x is an even integer, then x + 1 is odd.(All integers) (e) No right triangle is isosceles. (All triangles) (f) Every nonzero real number is positive or negative. (Real numbers) (g) Every integer is greater than −4 or less than 6. (Real numbers) 3. Give an English translation for each statement. The universe is given in parentheses.

(a) (∀x)(x ≥ 1). (U = N) (b) (∃x)(x ≤ 0 ∧ x ≥ 0). (U = R) (c) (∀x)(x is prime ∧ x 6= 2 → x is odd). (U = N) (d) ∼ (∃x)(|x| < 0). (U = R) 4. Determine whether the following statements are true or false where the universe for each statement is the given set U.

2 (a) (∃x)(x + 1 = 0); U = R. x (b) (∀x)( ∈ ); U = . 2 Z Z (c) (∃x)(∀y)(x > y ∨ x < y); U = R. (d) (∃x)(∃y)(x > y ∧ x < y); U = R. (e) (∀x)(∃y)(x + y > 0); U = Z. (f) (∃x)(x > 0 ∧ x < 0); U = Z. (g) (∃x)(x > 0) and (∃x)(x < 0); U = . √ Z (h) (∃x)( x2 < 0); U = . √ R (i) (∀x)( x2 > 0); U = R.

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1.7 Proof of validity

One of the most important task of a logician is the testing of arguments. An argument is the assertion that a statement, called the conclusion, follows from other statements, called the hypotheses or premises. An argument is considered to be valid if the conjunction of the hypotheses implies the conclusion. Example 1. The following is an argument in which the ﬁrst two statements are hypotheses, and the last statement is the conclusion.

• All men are tall.

• Ali is a man.

• Therefore Ali is tall.

The above argument may be symbolized as follows:

• H1. All men are tall.

• H2. Ali is a man. ————————————————

• C ∴ Ali is tall. Example 2. The following is an argument in which the ﬁrst four statements are hypotheses, and the last statement is the conclusion.

• If he studies mathematics, then he will earn a good life.

• If he studies physics, then he will be happy.

• If he will earn a good life or he will be happy, then his university tuition is not wasted.

• His university tuition is wasted.

• Therefore, he studies neither mathematics nor physics.

The above argument may be symbolized as follows: H1. M → E. H2. P → H. H3. (E ∨ H) →∼ W . H4. W . C ∴ ∼ M∧ ∼ P This argument is valid if the the following is true

(M → E) ∧ (P → H) ∧ [(E ∨ H) →∼ W )] ∧ W ⇒∼ M∧ ∼ P

To establish the validity of this argument by means of a truth table would require a table with 32 rows. But we can prove that this argument is valid by deducing the conclusion from the hypotheses in a few steps using the rules of inference.

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Proof.

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Deﬁnition. (Formal proof of validity) A formal proof of validity for a given argument is a sequence of statements each of which is either a hypotheses of the argument or follows from preceding statement by a known valid argument, ending with the conclusion of the argument. Example 1. Construct a formal proof of validity for the following argument

W ∨ (H ∧ L) (W ∨ H) → D/ ∴ W ∨ D. Solution.

Example 2. Construct a formal proof of validity for the following argument

L → M (M ∨ N) → (L → K) ∼ P ∧ L/ ∴ K Solution.

Indirect proof

There are another method of proof called indirect proof or the proof by reduction ad absurdum. An indirect proof of validity is done by including, as an additional hypotheses, the negation of the conclusion, and then deriving a contradiction. As soon as a contradiction is obtained, the proof is complete.

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Example 1. Give an indirect proof of validity for the following argument:

(p ∨ q) → r s → (p ∧ u) q ∨ s/ ∴ r Proof.

Example 2. Give an indirect proof of validity for the following argument:

A ∨ B ∼ B ∨ D/ ∴ A ∨ D Proof.

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Mathematical Proofs

Proofs are essential in mathematical reasoning because they demonstrate that the conclusions are true. Generally speaking, a mathematical explanation for a conclusion has no value if the explanation cannot be backed up by an acceptable proof. A proof is a complete justiﬁcation of the truth of a statement called a theorem. It generally begins with some hypotheses stated in the theorem and proceeds by correct reasoning to the claimed statement. It is nothing more than an argument that presents a line of reasoning explaining why the statement follows from known facts.

Basic Proof Rules

The following four rules provide guidance about what statements are allowed in a proof, and when. (1) At any time state an assumption, an axiom, or a previously proved result. (2) The replacement rule: At any time state a statement equivalent to any statement earlier in the proof. (3) The tautology rule: At any time state a statement whose symbolic translation is a tautology. (4) The modus ponens rule: At any time after p and p → q appear in a proof, state that q is true. Remark. The modus ponens rule is the most fundamental rule of reasoning. It is based on the tautology [(p → q) ∧ p] → q.

Methods of Mathematical Proof

There are several methods of proof. Here we introduce the ﬁve basic methods of proof. • (1) DIRECT PROOF: A direct proof is a logical step-by-step argument from the given conditions to the conclusion. • (2) PROOF BY (EXHAUSTION) CASE ANALYSIS: In this method we prove a statement by dividing it into a ﬁnite number of cases and proving each one separately. • (3) PROOF BY CONTRAPOSITION: A proof by contraposition or contrapositive proof for a conditional statement p −→ q makes use of the tautology (p −→ q) ↔ (∼ q −→∼ p).

It is an indirect proof method in which we ﬁrst give a direct proof of ∼ q −→∼ p and then conclude by replacement that p −→ q. • (4) PROOF BY CONTRADICTION : A proof by contradiction is an indirect proof method. It is based on the tautology p ←→ [∼ p −→ (∼ q ∧ q)]. That is, to prove a statement p is true, we prove that the statement ∼ p −→ (∼ q ∧ q) is true for some statement q. The logic behind such a proof is that if a statement cannot be false, then it must be true. Thus, to prove by contradiction that a statement p is true, we temporarily assume that p is false and then see what would happen. If what happens is an impossibility-that is, a contradiction-then we know that p must be true.

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• (5) PROOF BY MATHEMATICAL INDUCTION: Proof by mathematical induction is a very useful method in proving the validity of a mathematical statement (∀n)P (n) involving integers n greater than or equal to some initial integer n0.

Advances for writing proofs

There is no uniﬁed method to write a mathematical proof. However there are certain techniques that are often useful when writing proofs. Here are some advances that may help in writing correct proofs.

How to start

Begin a proof by rewriting what you are given and what you are asked to prove in a more convenient form. Often this involves converting word to symbols and utilizing the deﬁnitions of the terms used in the statements.

Justify each step

As a general rule, when you write a step in a proof, ask yourself if deducing that step is valid in the sense that it uses one of the four basic proof rules above. If the step follows as a result of the use of a tautology, it is not necessary to cite the tautology in your proof. In fact, with practice you should eventually come to write proofs without purposefully thinking about tautologies. What is necessary is that every step be justiﬁable.

Proving conditional statements p =⇒ q (conditional proof)

The most famous example is the direct proof of statements of the form p =⇒ q which proceeds in a step-by-step fashion from the condition p to the conclusion q. Since p −→ q is false only when p is true and q is false, it suﬃces to show that this situation cannot happen. The direct way to proceed is to assume that p is true and show (deduce) that q is also true. A direct proof of p =⇒ q will have the following form:

Direct proof of p =⇒ q: Assume p. . . Therefore, q.

Proving biconditional statements p ⇐⇒ q

Proofs of biconditional statements p ⇐⇒ q often make use of the tautology

(p ←→ q) ≡ (p −→ q) ∧ (q −→ p).

Proofs of p ⇐⇒ q generally have the following two-part form:

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Two-Part Proof Of p ⇐⇒ q (i) Show p =⇒ q. (ii) Show q =⇒ p. Therefore, p ⇐⇒ q.

In some cases it is possible to prove a biconditional statement p ⇐⇒ q that uses the connective throughout. This amounts to starting with p and then replacing it with a sequence of equivalent statements, the last one being q. With n intermediate statements R1,R2,...,Rn, a biconditional proof of p ⇐⇒ q has the form:

Biconditional Proof Of p ⇐⇒ q p ⇐⇒ R1 ⇐⇒ R2 . . ⇐⇒ Rn ⇐⇒ q.

When to use proof by contradiction

Proof by contradiction is a natural way to proceed when negating the conclusion gives you something concrete to manipulate. A proof by contradiction has the following form:

Proof of p by contradiction: Assume ∼ p. . . Therefore, q. . . Therefore, ∼ q. Hence, q∧ ∼ q a contradiction.

Remark. Two aspects about proofs by contradiction are especially noteworthy. First, this method of proof can be applied to any proposition p, whereas direct proofs and proofs by contraposition can be used only for conditional statements. Second, the proposition q does not appear on the left side of the tautology ∼ p =⇒ (q∧ ∼ q). The strategy of proving p by proving ∼ p =⇒ (q∧ ∼ q) then, has an advantage and a disadvantage. We do not know what proposition to use for q, but any proposition that will do the job is a good one. This means a proof by contradiction may require a spark of insight to determine a useful q.

Proof By Experiment

Although you cannot generally prove statements by experiment, many proofs can be done with the help of experimenting. One typically looks at simple cases to gain insight and this insight results in a proof.

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Consider the statement ”Every odd integer is the sum of two consecutive integers.” Trying a few small cases we have 3 = 1 + 2, 5 = 2 + 3, 7 = 3 + 4. It seems that a general pattern is 2n + 1 = n + (n + 1) and indeed this gives us a proof.

Proving Uniqueness

To prove an object is unique assume that a and b are two objects with the desired property and use this property together with other known information to show that a = b.

Proofs Involving Quantiﬁers

Now we discuss speciﬁcally the proof methods for statements with quantiﬁers.

(1) Proving (∀x)P (x)

To prove a statement of the form (∀x)P (x), we must show that P (x) is true for every object x in the universe. A direct proof is begun by letting x represent an arbitrary object in the universe, and then showing that is true for that object. In the proof we may use only properties of x that are shared by every element of the universe. Then, since x is arbitrary, we can conclude that is (∀x)P (x) true. Thus a direct proof of (∀x)P (x) has the following form:

Direct proof of (∀x)P (x) Let x be an arbitrary object in the universe. (The universe should be named or its objects should be described.) . . Hence P (x) is true. Since x is arbitrary, (∀x)P (x) is true.

(2) Proving (∃x)P (x)

There are several ways to prove existence theorems-that is, statement of the form (∃x)P (x). In a constructive proof we actually name an object a in the universe such that P (a) is true, which directly veriﬁes that the truth set of is nonempty. Some constructive proofs are quite easy to devise. For example, to prove that ”There is an even prime natural number,” we simply observe that 2 is prime and 2 is even. Another strategy to prove (∃x)P (x) is to show that there must be some object for which P (x) is true, without ever actually producing a particular object. Both Rolle’s Theorem and the Mean Value Theorem from calculus are good examples of this. Existence theorems may also be proved by contradiction.

Disproving

Disproving a statement is proving that a statement is false.

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Although ”proof by example” is not legitimate, you can disprove statements by giving a single example called a counterexample. Consider the statement ”The√ sum of√ two irrational numbers√ is irrational.”√ To disprove this statement we simply observe that 2 and − 2 are irrational but 2 + (− 2) = 0 is rational. Exercise 1.7 (1-9)

1.8 Mathematical Induction

Proof by mathematical induction is a very useful method in proving the validity of a mathematical statement (∀n)P (n) involving integers n greater than or equal to some initial integer n0. Principle of mathematical induction

If P (n) is an open sentence involving the natural number n such that (1) P (1) is true, and

(2) P (k) is true ⇒ P (k + 1) is true for any arbitrary natural number k, then P (n) is true for every natural number n; that is “(∀n)P (n) is true”. In order to apply the principle of mathematical induction to prove a theorem, the theorem must be capable to being broken down into cases, one case for each natural number. Then, we must verify both conditions (1) and (2). The condition (1) is called the base step and its veriﬁcation, that is usually easy, assure us that the theorem is true for at least the case n = 1. The condition (2) is called the inductive step. To verify this condition, we must prove an auxiliary theorem whose hypotheses is “P (k) is true” and whose conclusion is “P (k + 1) is true”. The hypotheses is called the induction hypotheses. Example 1. Use mathematical induction to prove that

n X 1 ∀n ∈ , (3j − 2) = n(3n − 1). N 2 j=1 Proof.

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Example 2. Prove that ∀n ∈ N, 3 + 11 + 19 + ··· + 8n − 5 = 4n2 − n.

Proof.

Example 3. Prove that for all natural numbers n, 3n ≥ 2n + 1.

Solution.

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Remark. Mathematical induction has more general form that can be used to prove the statement (∀n)(P (n)) where the universe may be any set of the form {n0, n0 + 1, ···} for some integer n0. Example 1. Show that n! ≥ 2n, ∀n ≥ 4.

Solution.

Example 2. Prove that 4n − 1 is divisible by 3 for n ≥ 0.

Solution.

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Example 3. (Book) Prove the Generalized De Morgan’s Law ∼ (p1 ∧ p2 ∧ · · · ∧ pn) ⇔∼ p1∨ ∼ p2 ∨ · · · ∼ pn. Solution.

Deﬁnition by induction

The idea of mathematical induction may be used in making definitions involving natural numbers. Example. The definition of powers of unknown number x may be defined by:

1 n+1 n x = x, x = x x, ∀n ∈ N. Thus x2 = xx, x3 = x2x, ··· .

Definition. (Binomial coefficients) Let n be a natural number and r be an integer. The binomial coefficients C(n, r) are defined by

C(0, 0) = 1,C(0, r) = 0 for each r 6= 0 and C(n, r) = C(n − 1, r) + C(n − 1, r − 1).

Example 1. Find C(2, 1)

Solution.

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Example 2. Prove that if r < 0, then for all n ∈ N, C(n, r) = 0. Solution.

Example 3. Prove that for all n ∈ N, if n < r, then C(n, r) = 0. Solution.

Example 4. Prove that for all n ∈ N, C(n, 0) = 1. Solution.

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Theorem 1.8. (The Binomial Theorem) If x and y are two variables and n is a natural number, then

(x + y)n = C(n, 0)xn + C(n, 1)xn−1y + ··· + C(n, r)xn−ryr + ··· + C(n, n)yn.

Proof.

Remark. The method we have used is called the First Principle of Mathematical Induction. There is another form that is called the Second Principle of Mathematical Induction.

Second Principle of mathematical induction

Let n0 be any integer. If P (n) is an open sentence involving the integer n such that 1. P (n0) is true, and 2. P (j) is true for j ≤ k ⇒ P (k + 1) is true for any arbitrary natural number k, then P (n) is true for every integer n ≥ n0. Exercise 1.8 (3-13)

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Additional Exercises

Prove the following statements by mathematical induction: 2 3 n n+1 1. ∀n ∈ N, 2 + 2 + 2 + ··· + 2 = 2 − 2. n X n2(n + 1)2 2. j3 = , n ≥ 1. 4 j=1

n X 1 n 3. = , n ≥ 1. (j + 1)(j + 2) 2n + 4 j=1

n X 1 1 4. ≤ 2 − , n ≥ 1. j2 n j=1

5.4 n > n3, n ≥ 5.

6.( n + 1)! > 2n+2 for n ≥ 5.

7. n! > 3n for n ≥ 7.

8.5 n − 1 is divisible by 4 for n ≥ 0.

9. n3 + 5n + 6 is divisible by 3 for n ≥ 1.

10. Bernoulli’s identity: For every real number x > −1 and every natural number n,

(1 + x)n ≥ 1 + nx.

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