Principles of Mathematics (Math 2450)

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Chapter 1Elementary Logic

The study of logic is the study of the principles and methods used in distinguishing valid argu-ments from those that are not valid.

The aim of this chapter is to help the student to understand the principles and methods usedin each step of a proof. The starting point in logic is the term statement (or proposition) which isused in a technical sense. We introduce a minimal amount of mathematical logic which lie behindthe concept of proof.

1.1 Statements and their connectives

When we prove theorems in mathematics, we are demonstrating the truth of certain statements. Wetherefore need to start our discussion of logic with a look at statements, and at how we recognizecertain statements as true or false.

Definition. (Statement ) 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.

(c) The equation x2 + 1 = 0 has two real solutions.

(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?

(b) Gaza is a beautiful city.

(c) 4+1.

(d) I will come to school next week.

(e) He lives in Gaza.

(f) x2 = 9.

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Compound statements

The statements in Example 1 are simple statements. A combination of two or more simple statementsis a compound statement.

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.

(c) If 5 is an integer, then 5 is a real number.

(d) You will pass this course if and only if you learned how to construct a mathematical proof.

Notation. We will denote simple statements by lowercase letters p, q, r, ... and we will denote com-pound 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 ←→.

Truth tables

An important distinction must be made between a statement and the form of a statement. Astatement form does not have a truth value. Instead, each form has a list of truth values thatdepend on the values assigned to its components. This list is displayed by presenting all possiblecombinations for the truth values of its components in a truth table. We will use ”T” for true and”F” for false.

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

p ∼ pT FF T

Example 1. Write the negation of each of the following:

(a)√

2 is a rational number.

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

(c) 2 divides 6.

Solution:

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

p q p ∧ qT T TT F FF T FF F F

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, calledthe logical possibilities. In general, if a compound statement has n components, then there are2n 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.

(c) 5− 0 = 4 and 5− 1 = 4.

(d) 5× 2 = 5 and 5× 3 = 10.

Solution:

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

Solution:

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

Example 1. Translate the following statement into logical form using connectives: ”8 is divisibleby 2 but it is not divisible by 3.”

Solution:

Exercise 1.1 (1-24)

1.2 Three more connectives

In English language there is an ambiguity involved in the use of ”or”. In mathematics and logic wecan 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 qto form the compound statement p∨ q (read: p or q or the disjunction of p and q). The truth valuesfor p ∨ q are defined as follows:

p q p ∨ qT T TT F TF T TF 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)].

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Solution:

Definition. (Equivalent statements)Two statements P and Q, simple or compound, are said to be ( logically) equivalent if P and Q havethe same truth values in each of all the logical possibilities. In such case we write P ≡ Q.

Example 1. Show that ∼ [p ∨ (∼ q)] ≡∼ p ∧ q.

Solution:

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

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

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The truth values of p→ q

p q p→ qT T TT F FF T TF F T

Remark. The statement p→ q is false only when p is true and q is false.

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

(a) If 2− 4 = 2, then 2− 2 = 4.

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

(c) If 3 > 3, then 4 > 3.

(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 a = 5, then a2 = 25.

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Solution:

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

Solution:

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.

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

The truth values of p↔ q

p q p↔ qT T TT F FF T FF F T

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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:

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 1. Translate the given compound statements into a symbolic form using the suggestedsymbols.

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

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

(c) ”A function is differentiable at a point only if it is continuous at that point.” (D, C).

Solution:

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Exercise 1.2 (1-21)

1.3 Tautology, implication, and equivalence

Example 1. 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 compo-nents.

Example 1. 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 maybe used as a step in a proof, and because tautologies are used to create rules for making deductionsin a proof.

Definition. (Implication )Let P and Q be two statements, compound or simple. If the conditional statement P → Q is atautology, 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.

Solution:

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

Solution:

Exercise: Show that

1. p⇒ (p ∧ p).

2. (p ∧ q)⇒ q.

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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: [(p ∨ q)∧ ∼ p]⇒ q.

Proof.

Definition. (Equivalence )Let P and Q be two statements, compound or simple. If the biconditional statement P ↔ Q is atautology, 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 statement.

(1) The converse of p→ q is q → p.

(2) The contrapositive of p→ q is ∼ q →∼ p.

Example 1. Determine whether p→ q and its converse are equivalent or not.

Solution:

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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.

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 1. Give a useful denial of each statement.

(a) 3 ≥ 3.

(b) The number 5.1 is an integer and it is positive.

Solution:

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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)

(c) Transitive law: (p→ q) ∧ (q → r)⇒ (p→ r)

Proof.

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.

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

(a) Modus Ponens: [(p→ q) ∧ p]⇒ q

(b) Modus Tollens: [(p→ q)∧ ∼ q]⇒∼ p

(c) Reduction ad Absurdum: (p→ q) ≡ [(p∧ ∼ q)→ (q∧ ∼ q)]

Proof.

Exercise 1.3: (1—20)

1.4 Contradiction

Definition. (Contradiction )A statement whose truth values are all false for every assignment of truth values to its componentsis called a contradiction.

Example 1. 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,

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(c) c⇒ p, and p⇒ t.

Proof.

Exercise 1.4: (1-5)

1.5 Deductive reasoning

Deductive reasoning is one of the two basic forms of valid reasoning. While inductive reasoningargues from the particular to the general, deductive reasoning argues from the general to a specificinstance. The basic idea is that if something is true for a class of things in general, this truth appliesto all legitimate members of that class. The key, then, is to be able to properly identify members ofthe class. Miscategorizing will result in invalid conclusions.

Example 1.

• 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 is using logic to draw conclusions based on statements accepted as true.The 16 laws summarized in Theorem 1.1 through 1.6 are very useful tools for justifying logicalequivalences and implications. We shall call these 16 laws the rules of inference. It should be notedthat these rules are selected just for convenient references and are not intended to be independent ofeach other.

Example 1. Prove the Contrapositive Law, (p → q) ≡ [(∼ q) → (∼ p)], using relevant definitionsand other rules of inference.

Solution:

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

Solution:

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:

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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:

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)

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1.6 Quantification Rules

Some sentences depend on some variables and become statements when the variables are replacedby a certain values.

Definition. ( Open sentence)A sentence containing one or more variables and which becomes a statement only when the variablesare 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.

Definition. (Universe )The set of all objects that can be considered in an open sentence is called the universe or the domainof discourse.

Remark. In many cases the universe will be understood from the context. However, there are timeswhen 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 truestatement is called the truth set of the sentence.

Example 1. Consider the open sentence P (x) : x2 < 5. The truth set of this sentence depends onthe universe we choose. If we choose the universe to be the set of natural numbers N, then the truthset is {1, 2}. If we choose the universe to be the set of real numbers R, then the truth set is the openinterval (−

√5,√

5).

Definition. (Equivalent open sentences )With a universe specified, two open sentences P (x) and Q(x) are equivalent if they have the sametruth set.

Example 1. 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.

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 truewhen 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 suchthat P (x)) is true when there exists at least one x in the universe such that P (x) is true.

Example 1. Determine whether the following statements are true or false:

(a) Let U = {1, 2, 3, 4} be the domain of discourse.

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• (∀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).

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 besymbolized (∃x)(P (x) ∧Q(x)).

Example 1. Translate the following statements into logical language using a quantifier and specifythe 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:

Remark.

1. ”for every”, ”for each”, and ”for all” have the same meaning in mathematics.

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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 1. [f(x) = 0 ∀x] ≡ [(∀x)(f(x) = 0)](∃x)(f(x) = 0) ≡ [f(x) = 0 for some x]

Definition.

(1) Two quantified sentences are equivalent in a given universe if and only they have the sametruth 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:

Example 2. Let the universe be U = Z. Show that (∃x)(x2 = 2) and (∃x)(x2 = 3) are equivalent.

Solution:

Remark. Statements of the form ”Every element of the set A has the property P” and ”Some elementof 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 onlyif there are integers p and q, with q 6= 0 such that x = p/q.” It may symbolized:

x is rational if and only if (∃p ∈ Z)(∃q ∈ Z)(q 6= 0 ∧ x =p

q).

Example 2. The statement ”For every rational number there is a larger integer” may be symbolizedby

(∀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 functionsare 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 numbersare integers” by using quantifier negation.

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, startingfrom 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 ofdiscourse.

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

3. The statement (∃x)(∀y)(P (x, y)) is true when (∀y)(P (x, y)) is true for at least one x in thedomain 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 thedomain 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 universeU = R:

(a) (∀x)(∃y)(x + y = 0)

(b) (∃x)(∀y)(x + y = 0)

(c) (∃x)(∃y)(x2 + y2 = −1)

(d) (∀y)(∃x)(x ≤ y)

(e) (∃y)(∀x)(x ≥ y)

(f) (∀y)(∃x)(∀z)(xy = xz)

Remarks. Great care must be taken in proofs that contain expressions involving more than onequantifier. 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) ∨ (∀x)Q(x)]⇒ (∀x)[P (x) ∨Q(x)]

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

Example 1. Determine which of the following statements is true and which is false using the universeU = R:

(a) (∀x)(√x2 = |x|)→ (∀x)(x2 = |x|2)

(b) (∃x)(x ≥ 0 ∧ |x| < 0)

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(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 eachis 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 state-ment is the given set U .

(a) (∃x)(x2 + 1 = 0); U = R.

(b) (∀x)(x

2∈ Z); U = 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 assertionthat a statement, called the conclusion, follows from other statements, called the hypotheses orpremises.

Example 1. The following is an argument in which the first two statements are hypotheses, andthe 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, andthe 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

An argument is considered to be valid if the conjunction of the hypotheses implies the conclusion.Thus the above 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 32rows. But we can prove that this argument is valid by deducing the conclusion from the hypothesesin a few steps using the rules of inference.

proof:

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Another way to prove the argument:

1. M → E (hypotheses)

2. P → H (hypotheses)

3. (E ∨H)→∼ W (hypotheses)

4. W/ ∴ ∼M∧ ∼ P

5.

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10.

11.

Definition. (Formal proof of validity )A formal proof of validity for a given argument is a sequence of statements each of which is either ahypotheses of the argument or follows from preceding statement by a known valid argument, endingwith 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

(C → O) ∧ (B → L)C ∨B(M → R) ∧ (S → A)C ∧O(L→∼ H) ∧ (A→ H)M ∨ S / ∴ R ∨ S

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Proof:

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 theconclusion, and then deriving a contradiction. As soon as a contradiction is obtained, the proof iscomplete.

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

(p ∨ q)→ rs→ (p ∧ u)q ∨ s/ ∴ r

Proof:

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

A ∨B∼ B ∨D/ ∴ A ∨D

Proof:

Mathematical Proofs

A proof is a complete justification of the truth of a statement called a theorem. It generally beginswith some hypotheses stated in the theorem and proceeds by correct reasoning to the claimed state-ment. It is nothing more than an argument that presents a line of reasoning explaining why thestatement 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 sentence equivalent to any statement earlier in theproof.

(3) The tautology rule: At any time state a sentence 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 thetautology

[(p→ q) ∧ p]⇒ q.

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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 givenconditions to the conclusion.

• (2) PROOF BY (EXHAUSTION) CASE ANALYSIS: In this method we prove astatement by dividing it into a finite number of cases and proving each one separately.

• (3) PROOF BY CONTRAPOSITION: A proof by contraposition or contrapositive prooffor a conditional sentence 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 thenconclude by replacement that p =⇒ q.

• (4) PROOF BY CONTRADICTION : A proof by contradiction is an indirect proofmethod. It is based on the tautology p⇐⇒ [∼ p −→ (∼ q ∧ q)]. That is, to prove a statementp 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 andthen see what would happen. If what happens is an impossibility-that is, a contradiction-thenwe know that p must be true.

• (5) PROOF BY MATHEMATICAL INDUCTION: Proof by mathematical induction isa very useful method in proving the validity of a mathematical statement (∀n)P (n) involvingintegers 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 techniquesthat are often useful when writing proofs. Here are some advances that may help in writing correctproofs.

How to start

Begin a proof by rewriting what you are given and what you are asked to prove in a more convenientform. Often this involves converting word to symbols and utilizing the definitions of the terms usedin 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 thesense that it uses one of the four basic proof rules above. If the step follows as a result of the use ofa tautology, it is not necessary to cite the tautology in your proof. In fact, with practice you shouldeventually come to write proofs without purposefully thinking about tautologies. What is necessaryis that every step be justifiable.

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Proving conditional statements p =⇒ q

The most famous example is the direct proof of statement of the form p −→ q which proceeds in astep-by-step fashion from the condition p to the conclusion q. Since p −→ q is false only when p istrue and q is false, it suffices to show that this situation cannot happen. The direct way to proceedis 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:

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 sentence p ←→ q that uses the connectivethroughout. This amounts to starting with p and then replacing it with a sequence of equivalentstatements, the last one being q. With n intermediate statements R1, R2, . . . , Rn, a biconditionalproof of p←→ q has the form:

Biconditional Proof Of p⇐⇒ qp⇐⇒ 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 somethingconcrete to manipulate.

A proof by contradiction has the following form:

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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 methodof proof can be applied to any proposition p, whereas direct proofs and proofs by contrapositioncan be used only for conditional sentences. Second, the proposition q does not appear on the leftside 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 anyproposition that will do the job is a good one. This means a proof by contradiction may require aspark of insight to determine a useful q.

Proof By Experiment

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

Consider the statement ”Every odd integer is the sum of two consecutive integers.” Trying afew small cases we have 3 = 1 + 2, 5 = 2 + 3, 7 = 3 + 4. It seems that a general pattern is2n + 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 usethis 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 theuniverse. A direct proof is begun by letting x represent an arbitrary object in the universe, and thenshowing that is true for that object. In the proof we may use only properties of x that are sharedby 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:

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Direct proof of (∀x)P (x)Let x be an arbitrary object in the universe.

(The universe should be named or its objects 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 aconstructive proof we actually name an object a in the universe such that P (a) is true, which directlyverifies that the truth set of is nonempty. Some constructive proofs are quite easy to devise. Forexample, to prove that ”There is an even prime natural number,” we simply observe that 2 is primeand 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 MeanValue 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.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 state-

ment 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 mathematicalstatement (∀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 that1. 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. ”(∀n)P (n) is true”

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Example 1. Use mathematical induction to prove that

∀n ∈ N,n∑

j=1

(3j − 2) =1

2n(3n− 1).

proof:

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.

proof:

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Definition by induction

The idea of mathematical induction may be used in making definitions involving natural numbers.Example 1. The definition of powers of unknown number x may be defined by;;

x1 = x, xn+1 = xnx, ∀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

andC(n, r) = C(n− 1, r) + C(n− 1, r − 1).

Example 1. Find C(2, 1).

Solution:

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

proof

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

Solution:

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

Solution:

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.

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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 2n ≥ 2n + 1, ∀n ≥ 3.

Solution:

Example 2. Show that n! ≥ 2n, ∀n ≥ 4.4n − 1 is divisible by 3 for n ≥ 0.

Solution:

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Remark. The method we have used is called First Principle of Mathematical Induction. There isanother 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)

Additional Exercises

Prove the following statements by mathematical induction:1. ∀n ∈ N, 2 + 22 + 23 + · · ·+ 2n = 2n+1 − 2.

2.n∑

j=1

j3 =n2(n + 1)2

4, n ≥ 1.

3.n∑

j=1

1

(j + 1)(j + 2)=

n

2n + 4, n ≥ 1.

4.n∑

j=1

1

j2≤ 2− 1

n, n ≥ 1.

5. 4n > n3, n ≥ 5.

6. (n + 1)! > 2n+2 for n ≥ 5.

7. n! > 3n for n ≥ 7.

8. 5n − 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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