{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# 1.6 - 1.6 Introduction to Proofs 75 1.6 Introduction to...

This preview shows pages 1–12. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1.6 Introduction to Proofs 75 1.6 Introduction to Proofs tinlts Introduction In this section we introduce the notion of a proof and describe methods for constructing proofs. A proof is a valid argument that establishes the truth of a mathematical statement. A proof can use the hypotheses of the theorem, if any, axioms assumed to be true, and previously proven theorems. Using these ingredients and rules of inference, the ﬁnal step of the proof establishes the truth of the statement being proved. In our discussion we move from formal proofs of theorems toward more informal proofs. The arguments we introduced in SectiOn 1.5 to show that statements involving propositions and quantiﬁed statements are true were formal proofs, where all steps were supplied, and the rules for each step in the argument were given. However, formal proofs of useful theorems can be extremely long and hard to follow. In practice, the proofs of theorems designed for human consumption are almost always informal proofs, where more than one rule of inference may be used in each step, where steps may be skipped, where the axioms being assumed and the rules of inference used are not explicitly stated. Informal proofs can often explain to humans why theorems are true, while computers are perfectly happy producing formal proofs using automated reasoning systems. The methods of proof discussed in this chapter are important not only because they are used to prove mathematical theorems, but also for their many applications to computer science. These applications include verifying that computer programs are correct, establishing that operating systems are secure, making inferences in artiﬁcial intelligence, showing that system speciﬁca- tions are consistent, and so on. Consequently, understanding the techniques used in proofs is essential both in mathematics and in computer science. Some Terminology Formally, a theorem is a statement that can be shown to be true. In mathematical writing, the term theorem is usually reserved for a statement that is considered at least somewhat important. Less important theorems sometimes are called propositions. (Theorems can also be referred to as facts or results.) A theorem may be the universal quantiﬁcation of a conditional statement with one or more premises and a conclusion. However, it may be some other type of logical statement, as the examples later in this chapter will show. We demonstrate that a theorem is true with a proof. A proof is a valid argument that establishes the truth of a theorem. The statements used in a proof can include axioms (or postulates), which are statements we assume to be true (for example, see Appendix I for axioms for the real numbers), the premises, if any, of the theorem, and previously proven theorems. Axioms may be stated using primitive terms that do not require deﬁnition, but all other terms used in theorems and their proofs must be deﬁned. Rules of inference, together with deﬁnitions of terms, are used to draw conclusions from other assertions, tying together the steps of a proof. In practice, the ﬁnal step of a proof is usually just the conclusion of the theorem. However, for clarity, we will often recap the statement of the theorem as the ﬁnal step of a proof. A less important theorem that is helpful in the proof of other results is called a lemma (plural lemmas or laminate). Complicated proofs are usually easier to understand when they are proved using a series of lemmas, where each lemma is proved individually. A corollary is a theorem that can be established directly from a theorem that has been proved. A conjecture is a statement that is being preposed to be a true statement, usually on the basis of some partial evidence, a heuristic argument, or the intuition of an expert. When a proof of a conjecture is found, the conjecture becomes a theorem. Many times conjectures are shown to be false, so they are not theorems. 76 l -’ The Foundations: Logic and Proofs Extra _ Examples "' ISSESSITIG Ill l—T’G Understanding How Theorems Are Stated Before we introduce methods for proving theorems, we need to understand how many math- ematical theorems are stated. Many theorems assert that a property holds for all elements in a domain, such as the integers or the real numbers. Although the precise statement of such theorems needs to include a universal quantiﬁer, the standard convention in mathematics is to omit it. For example, the statement “If x :— y, where .r and y are positive real numbers, then 12 > yz.” really means “For all positive real numbers x and y. if .r > y, then x3 > yz.” Furthermore, when theorems of this type are proved, the law of universal instantiation is often used without explicit mention. The ﬁrst step of the proof usually involves selecting a general element of the domain. Subsequent steps show that this element has the property in question. Finally, universal generalization implies that the theorem holds for all members of the domain. Methods of Proving Theorems We now turn our attention to proofs of mathematical theorems. Proving theorems can be difﬁcult. We need all the ammunition that is available to help us prove different results. We now introduce a battery of different proof methods. These methods should become part of your repertoire for proving theorems. To prove a theorem ofthe form Vx(P(x) —> Q{x}), our goal is to show that Pk) —> Q(c) is true, where e is an arbitrary element of the domain, and then apply universal generalization. In this proof, we need to show that a conditional statement is true. Because of this, we now focus on methods that show that conditional statements are true. Recall that p —r q is true unless p is true but q is false. Note that when the statement p —+ q is proved, it need only be shown that q is true if p is true. The fo]lowing discussion will give the most common techniques for proving conditional statements. Later we will discuss methods for proving other types of statements. In this sectiOn, and in Section [.7', we will develop an arsenal of many different proof techniques that can be used to prove a wide variety of theorems. When you read proofs, you will often ﬁnd the words “obviously” or “clearly.” These words indicate that steps have been omitted that the author expects the reader to be able to ﬁll in. Unfortunately, this assumption is often not warranted and readers are not at all sure how to ﬁll in the gaps. We will assiduously try to avoid using these words and try not to omit too many steps. However. if we included all steps in proofs. our proofs would often be excruciatingly long. Direct Proofs A direct proof of a conditional statement p —) q is constructed when the ﬁrst step is the assumption that p is true; subsequent steps are constructed using rules of inference, with the ﬁnal step showing that q must also be true. A direct proof shows that a conditional statement p ~+ q is true by showing that if p is true, then q must also be true, so that the combination p true and q false never occurs. In a direct proof, we assume that p is true and use axioms, deﬁnitions, and previously proven theorems, together with rules of inference, to show that q must also be true. You will ﬁnd that direct proofs of many results are quite straightforward, with a fairly obvious sequence of steps leading from the hypothesis to the conclusion. However, direct (.7? DEFINITION 1 EXAMPLE 1 Ema Extittiailes EXAMPLE 2 [.6 Introduction to Proofs 77 proofs sometimes require particular insights and can be quite tricky. The ﬁrst direct proofs we prCSent here are quite straightforward; later in the text you will see some that are less obvious. We will provide examples of several different direchproofs. Before we give the ﬁrst exampie, we need a deﬁnition. The integer n is even if there exists an integer k such that n = 2k, and n is odd if there exists an integer k such that n = 2!: + 1. (Note that an integer is either even or odd, and no integer is both even and odd.) Give a direct proof of the theorem “If a is an odd integer, then :33 is odd.” Solution: Note that this theorem states Vii P{(n) —+ Q00), where PU?) is “n is an odd integer" and (2(a) is “it: is odd." As we have said, we will follow the usual convention in mathematical proofs by showing that P(n) implies Q01), and not explicitly using universal instantiation. To begin a direct proof of this theorem. we assume that the hypothesis of this conditional statement is true, namely, we assume that n is odd. By the deﬁnition of an odd integer, it follows that a = 2!: + l, where k is some integer. We want to show that a: is also odd. We can square both sides of the equation it = 2?: + 1 to obtain a new equation that expresses of. When we do this, we ﬁnd that n2 = (2k + U2 = 4!? + 4k + l = 2(2k2 + 2k) + 1. By the deﬁnition of an odd integer, we can conclude that n2 is an odd integer (it is one more than twice an integer). Consequently, we have proved that if r.- is an odd integer, then :22 is an odd integer. { Give a direct proof that ifm and n are both perfect squares, then not is also a perfect square. {An integer n is a perfect square ifthere is an integer in such that a 2 b2.) Solution: To produce a direct proof of this theorem, we assume that the hypothesis of this conditional statement is true, namely, we assume that m and n are both perfect squares. By the deﬁnition of a perfect square, it follows that there are integers s and r such that m = 32 and a = (2. The goal of the proof is to show that me must also be a perfect square when m and n are; looking ahead we see how we can show this by multiplying the two equations or = 32 and a = r3 together. This shows that mu = szrz, which implies that mm = (3:)2 (using Commutativity and associativity of multiplication}. By the deﬁnition of perfect square, it follows that mn is also a perfect square, because it is the square of st, which is an integer. We have proved that if m and it are both perfect squares, then me is also a perfect square. 4 Proof by Contraposition Direct proofs lead from the hypothesis of a theorem to the conclusion. They begin with the premises, continue with a sequence of deductions, and end with the conclusion. However, we will see that attempts at direct proofs often reach dead ends. We need other methods of proving theo- rems of the form Vx( P{x} + Q(x )J. Proofs of theorems of this type that are not direct proofs, that is, that do not Start with the hypothesis and end with the conclusion, are called indirect proofs. An extremely useful type of indirect proof is known as proof by contraposition. Proofs by contraposition make use of the fact that the conditional statement p —> .5; is equivalent to its contrapositive, ﬁg —) mp. This means that the conditional statement p -—r or can be proved by showing that its contrapositive, -|q —> —-p, is true. In a proof by contraposition of ,o -+ q. we take mg as a hypothesis, and using axioms, deﬁnitions, and previously proven theorems, together with rules of inference, we Show that -p must follow. We will illustrate proof by contraposition 78 I r‘ The Foundations: Logic and Proofs EXAMPLE 3 Extra [Hanniles ' EXAMPLE 4 EXAMPLE 5 i-P’R with two examples. These examples show that proof by contraposition can succeed when we cannot easily ﬁnd a direct proof. Prove that ifrr is an integer and 3n + 2 is odd, then rr is odd. Solution: We ﬁrst attempt a direct proof. To construct a direct proof, we ﬁrst assume that 3n + 2 is an odd integer. This means that 3n + 2 = 2k + l for some integer it. Can we use this fact to show that n is odd? We see that 3r; + l = 2k, but there does not seem to be any direct way to conclude that n is odd. Because our attempt at a direct proof failed, we next try a proof by contraposition. The ﬁrst step in a proof by contraposition is to assume that the conclusion of the conditional statement “If 3n. + 2 is odd, then r1 is odd” is false; namely, assume that r: is even. Then, by the deﬁnition of an even integer, r1 = 2k for some integer k. Substituting 2!: for n, we find that 3r: + 2 = 3(2k) + 2 = 6!: + 2 = 2(3k + 1). This tells us that 3n + 2 is even (because it is a multiple of 2), and therefore not odd. This is the negation of the hypothesis of the theorem. Because the negation of the conclusion of the conditional statement implies that the hypothesis is false, the original conditional statement is true. Our proof by contraposition succeeded; we have proved the theorem “If 3n + 2 is odd. then r? is odd." { Prove that if n = ab, where o and b are positive integers, then a 5 J5 or b 5 J5. Solution: Because there is no obvious way of showing that a _<_ Jr? or b 5 Jr? directly frOm the equation rr = ab, where o and b are positive integers, we attempt a proof by contraposition. The ﬁrst step in a proof by contraposition is to assume that the conclusion of the conditional statement “If rt = ab, where a and b are positive integers, then a 5 Jr? or b 5 Jr?" is false. That is, we assume that the statement (a 5 Jr?) \r (b 5 J3) is false. Using the meaning of disjunction together with De Morgan’s law, we see that this implies that both a 5 J5 and h 5 Jr? are false. This implies that o. > J5 and b > Jr? We can multiply these inequalities together (using the fact that if 0 < r and 0 < v, then so < w} to obtain ob > J? = n. This shows that ab 79 r2, which contradicts the statement n = ab. Because the negation of the conclusion of the conditional statement implies that the hy- pothesis is false. the original conditional statement is true. Our proof by contraposition suc- ceeded; we have proved that if rt 2 ob, where a and b are positive integers, then a 5 Jr? or ngE. 4 VACUOL'S AND TRNIAL PROOFS We can quickly prove that a conditional statement p —> q is true when we know that p is false, because ,0 —> q must be true when p is false. Consequently, if we can show that p is false, then we have a proof. called a vacuous proof, of the conditional statement p —r q. Vacuous proofs are often used to establish special cases of theorems that state that a conditional statement is true for all positive integers [i.e.. a theorem of the kind VrrP(rr'}, where P0?) is a propositional function]. Proof techniques for theorems of this kind will be discussed in Section 4. I. Show that the proposition P(O) is true, where He) is “lfrr :> I, then r22 > rr“ and the domain consists of all integers. Solution: Note that P{0) is “IfO } 1, then 02 > 0." We can show P{0) using a vacuous proof, because the hypothesis 0 > 1 is false. This tells us that P[0) is automatically true. { I- 3’9 1 .6 Introduction to Proofs 79 Remark: The fact that the conclusion of this conditional statement, 02 > 0, is false is irrelevant to the truth value of the conditional statement, because a conditional statement with a false hypothesis is guaranteed to be true. We can also quickly prove a conditional statement p —-> q if we know that the conclusion q is true. By showing that q is true, it follows that p —> 9 must also be true. A proof of p —> g that uses the fact that q is true is called a trivial proof. Trivial proofs are often important when special cases of theorems are proved (see the discussion of proof by cases in Section 1.7) and in mathematical induction, which is a proof technique discussed in Section 4.1. EXAMPLE 6 Let P(n) be “lfo and b are positive integers with o 3 b, then a" 3 17"," where the domain consists of all integers. Show that P(0) is true. Solution: The proposition P(0) is“If a 3 b, then a0 3 b0.” Because a0 = b0 = 1,the conclusion of the conditional statement “If a 3 b, then a” 3 13"" is true. Hence, this conditional statement, which is P({}), is true. This is an example of a trivial proof. Note that the hypothesis, which is the statement “a 3 b,” was not needed in this proof. ‘ A LITTLE PROOF STRATEGY We have described two important approaches for proving theorems of the form Vx(P(x) —> Q(x)): direct proof and proof by contraposition. We have also given examples that show how each is used. However, when you are presented with a theorem of the form Vx(P(x) —> Q(x)), which method should you use to attempt to prove it? We will provide a few rules of thumb here; in Section 1.7 we will discuss proof strategy at greater length. When you want to prove a statement of the form Vx{P(x) —> Q(x)), ﬁrst evaluate whether a direct proof looks promising. Begin by expanding the deﬁnitions in the hypotheses. Start to reason using these hypotheses, together with axioms and available theorems. If a direct proof does not seem to go anywhere, try the same thing with a proof by contraposition. Recall that in a proof by contraposition you assume that the conclusion of the conditional statement is false and use a direct proof to show this implies that the hypothesis must be false. We illustrate this strategy in Examples 7 and 8. Before we present our next example, we need a deﬁnition. DEFINITION 2 The real number r is rational if there exist integers p and q with q sé 0 such that r = p/q. A real number that is not rational is called irrational. EXAMPLE 7 Prove that the sum of two rational numbers is rational. (Note that if we include the implicit quantiﬁers here, the theorem we want to prove is “For every real number r and every real number 5, if r and s are rational numbers, then r + s is rational.) Extra Solution: We ﬁrst attempt a direct proof. To begin, suppose that r and s are rational numbers. From Hamnles the deﬁnition of a rational number, it follows that there are integers p and q, with q ¢ 0, such that r = p/q. and integers t and u, with u 3e 0, such that s = t /u. Can we use this information to show that r + s is rational? The obvious next step is to add :- = p/q and s = t/u, to obtain t‘ r+.=£+_:n+qﬂ q n gt: Because q 7E 0 and u 72 0, it follows that qu # 0. Consequently, we have expressed r + s as the ratio of two integers, pit + or and qu, where qt: 72 0. This means that r + s is rational. We have proved that the sum of two rational numbers is rational; our attempt to ﬁnd a direct proof succeeded. 4 80 l l The Foundations: Logic and Proofs EXAMPLE 8 EXAMPLE 9 Extra "Q EnamnIes EXAMPLE 10 l-an Prove that if n is an integer and n3 is 0ch then it is odd. Solution: We ﬁrst attempt a direct proof. Suppose that n is an integer and n2 is odd. Then, there exists an integer k such that n3 = 2:? + 1. Can we use this information to show that n is odd? There seems to be no obvious approach to show that n is odd because solving for it produces the equation it 2 iv’ 2!? + l, which is not terribly useful. Because this attempt to use a direct proof did not bear fruit. we next attempt a proof by contrapositi on. We take as our hypothesis the statement that n is not odd. Because every integer is odd or even, this means that n is even. This implies that there exists an integer k such that n = 2k. To prove the theorem, we need to show that this hypothesis implies the conclusion that n2 is not odd, that is, that n2 is even. Can we use the equation it = 2% to achieve this? By squaring both sides ofthis equation, we obtain n2 = 4!? = 2(2k2), which implies that n3 is also even because :12 = 2!, where r = 2&2. We have proved that if n is an integer and n2 is odd, then :1 is odd. Our attempt to ﬁnd a proof by contraposition succeeded. { Proofs by Contradiction Suppose we want to prove that a statement p is true. Furthermore. suppose that we can ﬁnd a contradiction q such that -'Ip —> g is true. Because q is false. but -p —r q is true, we can conclude that - p is false, which means that p is true. How can we ﬁnd a contradiction c; that might help us prove that p is true in this way? Because the statement r A —-r is a contradiction whenever r is a proposition, we can prove that p is true if we can show that —1p —> (r A -r} is true for some proposition r. Proofs of this type are called proofs b...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern