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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 iP’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 lan 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 by contradiction. Because a proof by contradiction does not prove a result
directly, it is another type of indirect proof. We provide three examples of proof by contradiction.
The ﬁrst is an example of an application of the pigeonhole principle, a combinatorial technique
that we will cover in depth in Section 5.2. Show that at least four of any 22 days must fall on the same day of the week. Solution: Let p be the proposition “At least four of 22 chosen days fall on the same day of the
week." Suppose that 1p is true. This means that at most three of the 22 days fall on the same day
of the week. Because there are seven days of the week, this implies that at most 2] days could
have been chosen because for each of the days of the week, at most three of the chosen days
could fall on that day. This contradicts the hypothesis that we have 22 days under consideration.
That is, if r is the statement that 22 days are chosen, then we have shown that —p —> (r A r).
Consequently, we know that p is true. We have proved that at least four of 22 chosen days fall 0n the same day of the week. 4 Prove that J2 is irrational by giving a proof by contradiction. Solution: Let p be the proposition “42 is irrational ." To start aproof by contradiction, we suppose
that 1 p is true. Note that —'p is the statement “It is not the case that J2 is irrational,” which
says that J2 is rational. We will show that assuming that  p is true leads to a contradiction. If J2 is rational, there exist integers o and b with J2 = o/b, where o and b have no
common factors [so that the fraction o/b is in lowest terms.) (Here, we are using the fact that every rational number can be written in lowest terms.) Because J2 = 51/23, when both sides of
this equation are squared, it follows that 2=avﬁ. EXAMPLE 1] 1.6 Introduction to Proofs 81 By the deﬁnition of an even integer it follows that n2 is even. We next use the fact that if n2 is
even, a must also be even, which follows by Exercise 16. Furthermore, because a is even, by the deﬁnition of an even integer, a = 2c for some integer .9. Thus,
252 = 4a?
Dividing both sides of this equation by 2 gives 52 = 2C2.
By the deﬁnition of even, this means that b3 is even. Again using the fact that if the square of
an integer is even, then the integer itself must be even, we conclude that b must be even as well. We have now shown that the assumption of rap leads to the equation J2 = 0/17, where
a and b have no common factors, but both a and b are even, that is, 2 divides both a and in.
Note that the statement that J2 = a/b, where a and b have no common factors, means, in
particular, that 2 does not divide both a and :5. Because our assumption of p leads to the
contradiction that 2 divides both a and b and 2 does not divide both a and b, p must be false.
That is, the statement ,0, “J2 is irrational,“ is true. We have proved that J2 is irrational. 4 Proof by contradiction can be used to prove conditional statements. In such proofs, we ﬁrst
assume the negation of the conclusiori. We then use the premises of the theorem and the negation
of the conclusion to arrive at a contradiction. ( The reason that such proofs are valid rests on the
logical equivalence of p —> q and {p A —q} —) F. To see that these statements are equivalent,
simply note that each is false in exactly one case, namely when p is true and q is false.) Note that we can rewrite a proof by contraposition of a conditional statement as a proof
by contradiction. In a proof of p + g by contraposition, we assume that —Iq is true. We then
show that Ip must also be true. To rewrite a proof by contraposition of p —) q as a proof by
contradiction, we suppose that both ,0 and ~19 are true. Then, we use the steps from the proof
of q —+ —p to show that “op is true. This leads to the contradiction p A p, completing the
proof. Example 11 illustrates how a proof by contraposition of a Conditional statement can be
rewritten as a proof by contradiction. Give a proof by contradiction ofthe theorem “lf3n + 2 is odd, then a is odd.” Solution: Let p be “3n + 2 is odd” and q be “it is odd.“ To construct a proof by contradiction,
assume that both p and q are true. That is. assume that 332 + 2 is odd and that n is not odd.
Because it is not odd, we know that it is even. Following the steps in the solution of Example 3
(a proof by contraposition), we can show that if n is even, then 3n + 2 is even. First, because a is
even, there is an integer k such that a = 2k. This implies that 3:: + 2 = 3(2k} + 2 = 61: + 2 =
2(3t'r + 1). Because 3n + 2 is 2:, where t = 3!: + 1, 3n + 2 is even. Note that the statement
“3n + 2 is even" mp, because an integer is even ifand only ifit is not odd. Because both p and
—p are true, we have a contradiction. This completes the proof by contradiction, proving that if 3n + 2 is odd, then a is odd. ‘ Note that we can also prove by contradiction that p —> q is true by assuming that p and
—Iq are true, and showing that q must be also be true. This implies that wq and q are both
true, a contradiction. This observation tells us that we can turn a direct proof into a proof by contradiction. 82  t The Foundations: Logic and Proofs EXAMPLE 12 txlra Examines EXAMPLE 13 I —:\’2 PROOFS OF EQUIVALENCE To prove a theorem that is a biconditional statement, that is,
a statement of the form ,0 +~> q, we show that p —+ q and q —> p are both true. The validity of this approach is based on the tautology
(p H q) <—> [(p * (1) Me > 12)] Prove the theorem “If n is a positive integer, then a is odd if and only if n2 is odd." Solution: This theorem has the form “,0 if and only if q,” where p is “n is 0d ” and q is “:22 is odd.” (As usual, we do not explicitly deal with the universal quantiﬁcation.) To prove this
theorem, we need to show that p —~> q and q —+ p are true.
We have already shown (in Example 1) that p —> 6; is true and (in Example 8) that g —e p is true.
Because we have shown that both p —> g and q —> p are true. we have shown that the theorem is true. * Sometimes a theorem states that several propositions are equivalent. Such a theorem states
that propositions p], pg, [93, . . . , p” are equivalent. This can be written as PI szﬁmﬁpn‘ which states that all n propositions have the same truth values, and consequently. that for all t’
and} with l 5 r' 5 n and l 5 j 5 n, p, and p; are equivalent. One way to prove these mutually
equivalent is to use the tautology [pl6P26"'HPH]H[(P19P2)A(P2_)p3)/\"'A(pn_’Pl}] This shows that if the conditional statements p1 —> pg, ,9; a» p3, . . . . p“ —> [71 can be shown to be true, then the propositions p1, p2, . . . . p,, are all equivalent.
This is much more efﬁcient than proving that p; > p_,— for all i an j with l 5 i 5 n and
Isism When we prove that a group of statements are equivalent, we can establish any chain of
conditional statements we choose as long as it is possible to work through the chain to go from
any one of these statements to any other statement. For example, we can show that p1, p2, and
p; are equivalent by showing that p] —> p3, p3 —> p3, and p: —+ p}. Show that these statements about the integer n are equivalent: pi: n is even.
1):: n — 1 is odd.
p3: n2 is even. Solution: We will show that these three statements are equivalent by showing that the conditional statements p] —> pg, ,0; —> p3, and p3 —> p. are true.
We use a direct proof to show that p] —> p;. Suppose that n is even. Then :1 = 2!: for some integer :17. Consequently, a — l = 2k — l = 20:  l} + 1. This means thatn — l is odd because
it is ofthe form 2m + l, where m is the integer k — 1.
We also use a direct proof to show that p2 —> p3. Now suppose a — l is odd. Then it — 1 = 2!: + 1 for some integer k. Hence, it = 2!: + 2 so that n2 = (2!: + 2)2 = 4;:2 + 8k + 4 :
2(2k2 + 4k + 2). This means that n2 is twice the integer 2!:2 + 4!: + 2, and hence is even. To prove p3 —> p] , we use a proof by contraposition. That is, we prove that if n is not even,
then :32 is not even. This is the same as proving that if n is odd, then a? is odd, which we have
already done in Example 1. This completes the proof. ‘ 1'  Hj’ EXAMPLE 14 Erma 3’
ﬂamllles " llllHS EXAMPLE 15 1.6 Introduction to Proofs 83 COUNTEREXAMPLES In Section 1.3 we stated that to show that a statement of the form
Vx P(x) is false, we need only ﬁnd a counterexample, that is, an example x for which P(x) is
false. When presented with a statement of the form VxPor), which we believe to be false or
which has resisted all proof attempts, we look for a counterexample. We illustrate the use of
counterexamples in Example 14. Show that the statement “Every positive integer is the sum of the squares of two integers" is
false. Solution: To show that this statement is false, we look for a counterexample, which is a par
ticular integer that is not the sum of the squares of two integers. It does not take long to ﬁnd
a counterexample, because 3 cannot be written as the sum of the squares of two integers. To
show this is the case, note that the only perfect squares not exceeding 3 are 02 = 0 and 12 = 1.
Furthermore, there is no way to get 3 as the sum of two terms each of which is 0 or 1. Conse
quently, we have shown that “Every positive integer is the sum of the squares of two integers” is
false. { Mistakes in Proofs There are many common errors made in constructing mathematical proofs. We will brieﬂy
describe some of those here. Among the most common errors are mistakes in arithmetic and
basic algebra. Even professional mathematicians make such errors, especially when working
with complicated formulae. Whenever you use such computations you should check them
as carefully as possible. (You should also review any troublesome aspects of basic algebra, especially before you study Section 4.1.)
Each step of a mathematical proof needs to be correct and the conclusion needs to follow logically from the steps that precede it. Many mistakes result from the introduction of steps that
do not logically follow from those that precede it. This is illustrated in Examples 15—] 7". What is wrong with this famous supposed “proof” that l = 2'? “Prooﬁ” We use these steps, where a and b are two equal positive integers. Step Reason 1. a = b Given 2. a3 = ab Multiply both sides om) by a 3. 02 — b2 2 ob  bl Subtract b2 from both sides of(2) 4. {a — b)(a + b) = b(a  [9) Factor both sides of(3) 5. o + b = b Divide both sides of(4) by a  b 6. 2b = b Replace a by b in (5) because a = b
and simplify 7’. 2 = 1 Divide both sides of(6} by b Solution: Every step is valid except for one. step 5 where we divided both sides by a — b. The
error is that a — b equals zero; division of both sides of an equation by the same quantity is
valid as long as this quantity is not zero. 4 EXAMPLE 16 What is wrong with this “proof”? “Theorem:” If n2 is positive, then it is positive. 84 I t The Foundations: Logic and Proofs EXAMPLE 17 EXAMPLE 18 M4 “Prooﬁ” Suppose that n2 is positive. Because the conditional statement “If n is positive, then
:22 is positive” is true, we can conclude that a is positive. Solution: Let Ptn) be “it is positive” and (201) be “a: is positive." Then our hypothesis is (2(a).
The statement “If n is positive, then 112 is positive” is the statement Vn(P(n) > Qtnl). From
the hypothesis Q01) and the statement Vn(P(n) —+ Q(n}) we cannot conclude Ptn). because
we are not using a valid rule of inference. Instead, this is an example of the fallacy of afﬁrming
the conclusion. A counterexample is supplied by n = —l for which n2 = 1 is positive. but a is negative. 1
What is wrong with this “proof”? “Theorem:" If n is not positive, then it2 is not positive. (This is the contrapositive of the
“theorem” in Example 16.) "Proof?" Suppose that n is not positive. Because the conditional statement “Ifn is positive, then
it2 is positive” is true, we can conclude that n2 is not positive. Solution: Let Pin} and Q(n) be as in the solution of Example 16. Then our hypothesis is P(n)
and the statement “If n is positive, then it2 is positive” is the statement Vn( P(n) —> Q(n D. From
the hypothesis P(n') and the statement Vn{ Ptn) + Qtn)} we cannot conclude —Q{_ri), because
we are not using a valid rule of inference. Instead, this is an example of the fallacy of denying
the hypothesis. A counterexample is supplied by n : — l. as in Example l6. { Finally, we briefly discuss a particularly nasty type of error. Many incorrect arguments are
based on a fallacy called begging the question. This fallacy occurs when one or more steps of
a proof are based on the truth of the statement being proved. In other words, this fallacy arises
when a statement is proved using itself, or a statement equivalent to it. That is why this fallacy is also called circular reasoning. is the following argument correct? It supposedly shows that n is an even integer whenever n2 is
an even integer. Suppose that n2 is even. Then n2 = 2k for some integer it. Let n = 2! for some integer i.
This shows that n is even. Soiution: This argument is incorrect. The statement “let it = 2i for some integer l" occurs in the
proof. No argument has been given to show that n can be written as 21 for some integer i. This
is circular reasoning because this statement is equivalent to the statement being proved. namely,
“a is even.” Of course, the result itself is correct; only the method of proof is wrong. 4 Making mistakes in proofs is part of the learning process. When you make a mistake that
someone else ﬁnds. you should carefully analyze where you went wrong and make sure that
you do not make the same mistake again. Even professional mathematicians make mistakes in
proofs. More than a few incorrect proofs of important results have fooled people for many years
before subtle errors in them were found. Just a Beginning We have now developed a basic arsenal of proof methods. In the next section we will introduce
other important proof methods. We will also introduce several important proof techniques in 1.6 Introduction to Proofs 85 Chapter 4, including mathematical induction, which can be used to prove results that hold for
all positive integers. In Chapter 5 we will introduce the notion of combinatorial proofs. In this section we introduced several methods for proving theorems of the form Vx( P(x) —)
Q (x D, incl uding direct proofs and proofs by contraposition. There are many theorems of this type
whose proofs are easy to construct by directly working through the hypothescs and deﬁnitions
of the terms of the theorem. However, it is often difficult to prove a theorem without resorting
to a clever use of a proof by contraposition or a proof by contradiction, or some other proof
technique. In Section 1.7 we will address proof strategy. We will describe various approaches
that can be used to ﬁnd proofs when straightforward approaches do not work. Constructing
proofs is an art that can be learned only through experience, including writing proofs, having
your proofs critiqued, and reading and analyzing other proofs. Exercises l. 2. ll]. 11. 12. 13.
14.
15. "16. 1?. l8. Use a direct preof to show that the sum of two odd integers
is even. Use a direct proof to show that the sum of two even inte
gers is even. Show that the square of an even number is an even number
using a direct proof. Show that the additive inverse, or negative. of an even
number is an even number using a direct proof. Prove that if m + it and n + p are even integers. where
m. n. and p are integers, then m + p is even. What kind
of proof did you use? Use a direct proof to show that the product of two odd
numbers is odd. Use a direct proof to show that every odd integer is the
difference of two squares. Prove that if n is a perfect square. then it + 2 is not a
perfect square. Use a proof by contradiction to prove that the sum of an
irrational number and a rational number is irrational. Use a direct proof to show that the product of two rational
numbers is rational. Prove or disprove that the product of two irrational num
hers is irrational. Prove or disprove that the product of a nonzero rational
number and an irrational number is irrational. Prove that ifx is irrational, then lg’x is irrational. Prove that ifx is rational and x are 0, then Mr is rational. Use a proof by contraposition to show that ifx + y 3 2.
where .r and y are real numbers. then I a l or y 3 i.
Prove that if m and n are integers and rim is even, then m
is even or u is even. Show that if u is an integer and n3 + 5 is odd, then it is
even using a) a proof by contraposition. b) a proofby contradiction. Prove that if it is an integer and 3H + 2 is even. then it is
even using 20. 21. 22. 23. 24. 25. 26. 27. 28.
29. 3]. 3!. a) a proof by contraposition. b) a proof by contradiction. Prove the proposition P(0}, where P(n] is the proposition
“Ifn is a positive integer greater than 1. then n2 ;» it.”
What kind of proof did you use? Prove the proposition P(l}, where P(n) is the proposi
tion "If n is a positive integer, then :12 z a.” What kind of
proof did you use? Let Pin) be the proposition “Ifa and b are positive real
numbers, then (a + b)” 3 a" + 45”." Prove that P{l) is
true. What kind of proof did you use? Show that if you pick three socks from a drawer contain
ingjust blue socks and black socks, you must get either a
pair of blue socks or a pair of black socks. Show that at least 10 of any 64 days chosen must fall on
the same day of the week. Show that at least 3 of any 25 days chosen must fall in the
same month of the year. Use a proof by contradiction to show that there is no ratio
nal number r for which r3 + r + i = 0. [Hints Assume
that r = nib is a root, where n and b are integers and a {b
is in lowest terms. Obtain an equation involving integers
by multiplying by {23. Then look at whether a and b are
each odd or even] Prove that if n is a p0sitive integer, then It is even if and
only if 7n + 4 is even. Prove that if a is a positive integer, then it is odd if and
only if 531 + 6 is odd. Prove that in2 = n2 ifand only ifm = n orm = —n. Prove or disprove that if m and n are integers such that
mm = 1, then either m =1 and n = l, or else m = —I
and n = — 1. Show that these three statements are equivalent. where
a and b are real numbers: (i) a is less than b. (if) the
average oft: and b is greater than n, and {iii} the average
ofa and b is less than (3. Show that these statements about the integer .r are equiv
alent: (i) 3): + 2 is even, (it) 11' + 5 is odd, {iii} x: is even. 86 32.
33. 34. 35. 36. 1.7 Proof Methods and Strategy I .’ The Foundations: Logic and Proofs Show that these statements about the real number x are
equivalent: (1')): is rational, (it) x,r’2 is rational, and (iii)
3r — l is rational. Show that these statements about the real number x are
equivalent: (i) x is irrational, (it) 3:: + 2 is irrational, (iii)
.1: i2 is irrational. Is this reasoning for ﬁnding the solutions of the equation s/Zr2 w l = .r correct? U} «2::2 — 1 = x is given; (2) 37. 38. I «W: to be equivalent by showing that pl 9 p4, p; 9 p3, and
PI 9 P31 Show that the propositions p].p2. pg. 334, and p5 can be
shown to be equivalent by proving that the conditional
statements ,0. —> p4, p3 —> pl, p4 —> p3, p2 —> p5, and
p3 —> p3 are true. Find a counterexample to the statement that every positive
integer can be written as the sum of the squares of three 213 — I = .r2, obtained by squaring both sides of ( l); (3} integers .r2 e l = 0, obtained by subtracting 3:2 from both sides of 39. Prove that at least one of the real numbers a. , a2, . . . , an
(2}; (4)0: — l}(.r + l) = 0, obtained by factoringthe left is greater than or equal to the average of these numbers.
hand side ofr2 — l; (5) x : l or x = — l, which follows What kind ofproof did you use? because ab = 0 implies that a = 0 or b = 0. Are these steps for ﬁnding the solutions of in: + =
3 — x correct? {i} «xx + 3 = 3 — .r is given; (2} x +
3 = 13 — ﬁx + 9, obtained by squaring both sides of
(I); {3) 0 = x2 ~ 7x + 6. obtained by subtracting x +
3 from both sides of (2); (4) 0 = (x — no: — 6),
obtained by factoring the right~hand side of (3}:
(5) x = l or x = 6, which follows from {4) because
ab = 0 implies that n = 0 or b : 0. Show that the propositions pl. 132, p3, and p4 can be shown 40. 41. 42. Use Exercise 39 to show that if the ﬁrst [0 positive inte—
gers are placed around a circie, in any order, there exist
three integers in consecutive locations around the circle
that have a sum greater than or equal to 17. Prove that if n is an integer, these four statements are
equivalent: (i) n is even. (it) n + l is odd, (iii) 3n + l is
odd, (iv) 3:! is even.
Prove that these four statements about the integer n are
equivaient: {r} n? is 0ch (ii) 1 — n is even. {at} n3 is odd,
(iv) :12 + l is even. Introduction HSSGSSI‘IIBIII In Section 1.6 we introduced a variety of different methods of proof and illustrated how each
method is used. In this section we continue this effort. We will introduce several other important
proof methods, including proofs where we consider different cases separately and proofs where
we prove the exiStence of objects with desired properties. In Section 1.6 we only brieﬂy discussed the strategy behind constructing proofs. This
strategy includes selecting a proof method and then successfully constructing an argument step
by step, based on this method. In this section, after we have developed a wider arsenal of proof
methods, we will study some additional aspects of the art and science of proofs. We will provide
advice on how to ﬁnd a proof of a theorem. We will describe some tricks of the trade, including
how proofs can be found by working backward and by adapting existing preofs. When mathematicians work, they formulate conjectures and attempt to prove or disprove
them. We will brieﬂy describe this process here by proving results about tiling checkerboards
with dominoes and other types of pieces. Looking at tilings of this kind, we will be able to
quickly formulate conjectures and prove theorems without ﬁrst developing a theory. We will conclude the section by discussing the role of open questions. In particular, we
will discuss some interesting problems either that have been solved after remaining open for
hundreds of years or that still remain open. Exhaustive Proof and Proof by Cases Sometimes we cannot prove a theorem using a single argument that holds for all possible cases.
We now introduce a method that can be used to prove a theorem, by considering different cases ...
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This note was uploaded on 12/16/2009 for the course MATH 1014 taught by Professor Ganong during the Spring '09 term at York University.
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