inquiry-based introduction to proofs

8/12/2019 Inquiry-Based Introduction to Proofs

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An Inquiry-Based

INTRODUCTION TOPROOFS

Jim Hefferonversion 0.99


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NOTATION

N natural numbers {0, 1,2, .. .}Z, Z+ integers {. .. ,2,1,0,1,2,...}, positive integers {1,2,...}

R real numbersQ rational numbers

a| b adividesbamod b the remainder whenais divided byb

a c (mod b) aandchave the same remainder when divided bybgcd(a, b), lcm(a,b) greatest common divisor, least common multiple

a A ais an element of the set A empty set { }

A B Ais a subset ofBA characteristic function of the setAAc complement of the setA

AB,A B union, intersection of the setsAB,A B difference, symmetric difference of the sets

|A

| order of the setA; the number of elements

P(A) power set ofA; the set of all ofAs subsetsx0, x1, . . ., (x0, x1) sequence, ordered pair

lh(x0, x1, . . .) length of the sequenceA0 A1 An1,An Cartesian product of sets, product of a set with itself

f: D C function with domainDand codomain Cid: D D identity map; id(d) = d

fB restriction off to a subset of the domainf1(c), f1(A) inverse image of an element or subset of the codomain

gf function compositionf1 function inverse to f

xy (mod R) (x,y) RwhereRis an equivalence relation

x

equivalence class containingx

P partition of a setA B two sets with the same cardinality

GREEK LETTERS WITH PRONOUNCIATION

character name character name alpha AL-fuh nu NEW beta BAY-tuh , xi KSIGH, gamma GAM-muh o omicron OM-uh-CRON, delta DEL-tuh , pi PIE epsilon EP-suh-lon rho ROW

zeta ZAY-tuh , sigma SIG-muh eta AY-tuh tau TOW (as in cow), theta THAY-tuh , upsilon OOP-suh-LON iota eye-OH-tuh , phi FEE, or FI (as in hi) kappa KAP-uh chi KI (as in hi), lambda LAM-duh , psi SIGH, or PSIGH mu MEW , omega oh-MAY-guh

The capitals shown are the ones that differ from Roman capitals.


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PREFACE

This is a course in mathematical proof. It is for math majors who are US sophomores although

since it requires only high school mathematics it can be used with first year students.APPROACH. This course is inquiry-based (sometimes called Moore method or discovery method).This text is a sequence of exercises, along with definitions and a few remarks. The students workthrough the material together by proving statements or sometimes by providing examples or coun-terexamples. This makes each student grapple directly with the mathematics the instructor onlylightly guides, while the students pledge not to use outside sources talking out misunderstand-ings, sometimes stumbling in the dark, and sometimes having beautiful flashes of insight. Forthese students, with this material, this is the best way to develop mathematical maturity. Besides,it is a lot of fun.

TOPICS. We start with elementary number theory, not logic and sets, for the same reason that thebaseball teams annual practice starts with tossing the ball and not with reading the rulebook. Mathmajors take readily to proving things about divisibility and primes whereas a month of preliminary

material can be less of a lure.But thebackground is good stuff also and students are on board once they see where it is going.In the second and third chapters we do sets, functions, and relations, leveraging the intellectualhabits that we established at the start.

EXE RCIS ES . The exercises are arranged so that, as much as the material allows, nearby exerciseshave about the same difficulty. This difficulty standard rises gradually through the book.

Some exercises have multiple items; these come in two types. If the items are labeled A, B, etc.,then each one is hard enough to be a separate assignment. If the labels are (i), (ii), etc., then theytogether make a single assignment. (In my course I have students put proposed solutions on theboard for the group to discuss. If the items are labelled alphabetically then I ask a different studentto do each one, while for the others I ask a single student to do them all.)

LICENSE. This book is Free; see http://joshua.smcvt.edu/proofs.That site has other material

related to this text, including its LATEX source so that an instructor can tailor the material to theclass.

At the first meeting of the class Moore would define the basic terms and either challenge the

class to discover the relations among them, or, depending on the subject, the level, and the

students, explicitly state a theorem, or two, or three. Class dismissed. Next meeting: "Mr Smith,

please prove Theorem 1. Oh, you cant? Very well, Mr Jones, you? No? Mr Robinson? No? Well,

lets skip Theorem 1 and come back to it later. How about Theorem 2, Mr Smith?" Someone

almost always could do something. If not, class dismissed. It didnt take the class long to

discover that Moore really meant it, and presently the students would be proving theorems and

watching the proofs of others with the eyes of eagles. Paul Halmos

Its a kind of art that may change lives. Peter Schjeldahl

Jim HefferonSaint Michaels CollegeColchester, Vermont USA2013-Spring
http://joshua.smcvt.edu/proofshttp://joshua.smcvt.edu/proofshttp://joshua.smcvt.edu/proofs


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CHAPTER1 NUMBERTHEORY

Webegin with results about the integersZ= { . . . ,2,1,0,1,2,. .. }. Inthischapter number means

integer. Some statements refer to the natural numbers N= {0,1,2,...}.

DIVISIBILITY

1.1 DEFINITION. For two integers d,nwe say that d divides nif there is an integer ksuch thatd k= n. Here,dis thedivisor,nis thedividend, andkis thequotient. (Alternative wordings are:d is a factor of n, ord goes evenly into n, orn is a multiple of d.) We denote the relationship asd | nifdis a divisor ofnor asdnif it is not.1.2 DEFINITION. A natural number is evenif it is divisible by 2, otherwise it is odd. (Alternativewording is that the numberhas even parityorhas odd parity.)

The notation d | nsignifies a relationship between two numbers. It is different than the fractiond/n, which is a number. We can sensibly ask Does 2 divide 5? but Does 2/5? is not sensible.

1.3 EXE RCIS E. ( INTERACTION OF PARITY AND SIGN) Prove or disprove.A. If a number is even then its negative is even.B. If a number is odd then its negative is odd.C. Ifd | athen d | aandd | a.D. Ifd| athen d| |a|. (Recall that the absolute value|a|of a number is aifa 0 and isaif

a< 0.)1.4 EXE RCIS E. ( INTERACTION OF PARITY AND ADDITION) Prove or disprove.

A. The sum of two evens is even. The difference of two evens is even.

B. The sum of two odds is odd. The difference of two odds is odd.C. Wherea,bZ, the numbera+ bis even if and only ifabis even.D. Generalize the first item to arbitrary divisorsd.

1.5 EXE RCIS E. ( INTERACTION OF PARITY AND M ULTIPLICATION) Prove or disprove.A. The product of two evens is even.B. The quotient of two evens, if it is an integer, is even.C. Generalize the first item to any divisord.

1.6 EXE RCIS E. (DIVISIBILITYPROPERTIES) Letd,m, andnbe integers. Prove each statement.A. (REFLEXIVITY) Every number divides itself.B. Every number divides 0 while the only number that 0 divides is itself.C. (TRANSITIVITY) Ifd | nandn| mthend | m.D. (CANC EL LATIO N) Ford,n Z, if for some nonzero integerawe have thata d| anthend| n.

Conversely, ifd | nthenad | anfor allaZ.E. (COMPARISON) Ford,nZ+, ifnis a multiple ofdthenn d.F. Every number is divisible by 1. The only numbers that divide 1 are 1 and

1.

G. The largest divisor ofais |a|, foraZwitha= 0.H. Every nonzero integer has only finitely many divisors.

1.7 EXE RCIS E. Give the converse of Reflexivity: what conclusion can you make ifa | bandb| a?1.8 EXE RCIS E. Suppose thata,b, cZ.

A. Prove that ifa | bthena| bcfor all integersc.B. Prove that ifa| banda | cthenadivides the sumb+ cand differenceb c.C. (LINEARITY) Generalize the prior item.


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page 2 Introduction to Proofs, Hefferon, version 0.99

INTERLUDE: INDUCTION

Results in the prior section need only proof techniques that are natural for people with a mathe-matical aptitude. However some results to follow require a technique that is less natural, mathe-matical induction. This section is a pause for an introduction to induction.

We will start with exercises about summations (note, though, that induction is not about sum-mation; we use these only because they make helpful initial exercises). For example, many peo-

ple have noticed that the odd natural numbers sum to perfect squares: 1 + 3=4, 1 + 3 + 5=9,1+3+5+7 = 16, etc. We will prove the statement, The sum 1+3+5+ +(2n+1) equals (n+1)2.

That statement has a natural number variable nthat is free, meaning that setting nto be 0,or 1, etc., gives a family of cases:S(0), or S(1), etc. For instance, the statement S(1) asserts that 1+3equals 22. Our induction proofs will all involve statements with one free natural number variable.

These proofs have two steps. For the base stepwe will show that the statement holds for someintial numberiN. Theinductive stepis more subtle; we will show that the following implicationholds.

Ifthe statement holds from the base number up to and includingn= kthenthe statement holds also in the n= k+1 case. ()

The Principle of Mathematical Inductionis that completing both steps proves that the statement istrue for all natural numbers greater than or equal toi: thatS(i) is true, andS(i

+1) is true, etc.

For the example statement about odd numbers and squares, the intuition behind the principleis first that the base step directly verifies the statement for the case of the initial number n= 0. Thenbecause the inductive step verifies the implication () for all k, that implication applied tok= 0gives that the statement is true for the case of the number n= 1. Now, with the statement estab-lished for both 0 and 1, apply () again to conclude that the statement is true for the number n= 2.In this way, induction bootstraps to all natural numbers n 0. Below is an induction argument forthe example statement, with separate paragraphs for the base step and the inductive step.

Proof. We show that 1 +3 + + (2n+1) = (n+ 1)2 by induction. For then= 0 base step note thaton the left the sum has a single term, 1, which equals the value on the right, 12.

For the inductive step assume that the formula is true for n= 0,n= 1, . . . ,n= k, and considerthe n= k+1 case. Thesum is 1+3++(2k+1)+(2(k+1)+1) = 1+3++(2k+1)+(2k+3). By theinductive hypothesis the statement is true in the n

=kcase so we can substitute 1

+3

+ +(2k

+1)+ (2k+3) = (k+1)2 + (2k+3) = (k2 +2k+1)+ (2k+3) = (k+ 2)2. This is the required expressionfor then= k+1 case.

1.9 EXE RCIS E. Prove by induction.

A. 0+1+2+ +n= n(n+1)/2B. 0+1+4+9+ +n2 = n(n+1)(2n+1)/6C. 1+2+4+8+ +2n = 2n+1 1

1.10 EXE RCIS E. Prove by induction.

A. (GEOMETRICSERIES) For allrRwithr= 1 we have 1 + r+ r2 + + rn= (rn+1 1)/(r1).B. (ARITHMETICSERIES) We have b+ (a+b)+ (2a+b)+ + (na+b) = (n(n+1)/2) a+ (n+1) b

for alla,bR.

1.11 EXE RCIS E. Prove by induction thatn< 2n for allnN.1.12 EXE RCIS E. Prove each by induction.

A. For allnN, the numbern2 +nis even.B. For alln 2 the numbern3 nis divisible by 6. Hint:take the base case to be 2.C. IfnN then 4 | 13n1.D. IfnN then (1+ 1/1) (1+ 1/2) (1+ 1/n) = n+1.

1.13 EXE RCIS E. Prove thatn+ 1-term sums of reals commute: a0 + a1 + + an= an+ + a0foralln 1, starting from the assumption that sum of two terms commutes.


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Introduction to Proofs, Hefferon, version 0.99 page 3

1.14 EXE RCIS E. The sequence ofFibonacci numbers0,1,1, 2,3,5,8,13,.. . is defined by the con-dition that each succeeding number in the sequence is the sum of the two numbers before itfn+1= fn+ fn1, subject to the initial conditions f0= 0 and f1= 1. Where does this argument,which purports to show that all Fibonacci numbers are even, go wrong? The base case is clearsince 0 is even. For the inductive step, assume the statement is true for all cases up to and in-cludingn= k. By definition the next case fk+1is the sum of the two prior numbers, which by the

inductive hypothesis are both even. Thus their sum is even.While many induction arguments use the inductive hypothesis only in the n= kcase, some

break from that pattern.

1.15 EXE RCIS E. The game of Nim starts with two piles, each containingnchips. Two players taketurns picking a pile and removing some nonzero number of chips from that pile. The winner isthe player who takes the final chip. Prove by induction on nthat the second player always winsby playing this way: whatever number of chips the first player removes from one pile, the secondplayer removes the same number from the other pile.

1.16 DEFINITION. TheLeast Number Principle(or Well-ordering Principle) is that any nonemptysubset of the natural numbers has a least element.

1.17 EXE RCIS E. Show that the Principle of Induction implies the Least Number Principle. Hint:

show by induction that if a set of natural numbers does not have a least element then that set isempty.

DIVISION

1.18 EXE RCIS E. (DIVISION THEOREM) For any integer divisor b> 0 and dividend a there is aunique pair of integers, the quotient qandremainder r, such thata= bq+ rand 0 r< b. Wedo this argument in steps.

A. Prove thatqand rare unique.B. Show that the statement is true fora= 0. Show also that if the statement is true for a> 0 then

it is true fora< 0.C. Prove the statement for a>0. Hint:you can consider the set of integers {a bq

q Z},note that it has nonnegative elements, apply the Least Number Principle to produce a smallest

nonnegative element, and show that this is the desired remainderr.

Observe thatr= 0 if and only ifb| a. Observe also that the divisor must be positive.1.19 DEFINITION. Wherem> 0, themodulus amod mis the remainder when ais divided bym.Two numbersa,bare congruent modulo m, writtena c (mod m), if they leave the same remain-der when divided bym, that is, if they have the same modulus with respect tom.

1.20 EXE RCIS E. Prove or disprove: (i) amod b= bmod a, (ii) amod b= amod b, (iii)bmodb= 1.1.21 EXE RCIS E. Prove thata b (modm) if and only ifm| (a b), that is, if and only ifaand bdiffer by a multiple ofm(form> 0).1.22 EXE RCIS E. Let a,b,c,d,mbe integers with m> 0 and a b (modm) andc d (mod m).Prove each.

A. a+c b+ d (mod m)B. ac bd (modm)C. an bn (modm) for allnZ+.

1.23 DEFINITION. For any real numberx,itsfloorx is thegreatest integer less than or equal tox.1.24 EXE RCIS E. Prove each.

A. The quotient whenais divided bybis a/b.B. a= b a/b+ amod b.C. b (amod m) = (ba) mod (bm).


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1.25 EXE RCIS E. (PIGEONHOLEPRINCIPLE) Prove each.

A. For a finite list of real numbers, the maximum is at least as big as the average.

B. If you putn-many papers into fewer than n-many pigeonholes then at least one hole gets atleast two papers.

COMMON DIVISORS AND COMMON MULTIPLES1.26 DEFINITION. An integer is a common divisorof two other integers if it divides both of them.The greatest common divisorof two integers is the largest of their common divisors, except that wetake the greatest common divisor of 0 and 0 to be 0. We write gcd( a,b) for the greatest commondivisor ofaandb.

1.27 EXE RCIS E. Prove.

A. (EXI ST EN CE) Anya,bZ have a greatest common divisor.B. (COMMUTATIVITY) gcd(a,b) = gcd(b, a)C. If either number is nonzero then 0 gcd(a,b) min(|a|, |b|). If both are nonzero then the first

inequality is strict.

D. Ifdis a common divisor ofaand bthen so is |d|. Thus common divisors are limited to theinterval from

gcd(a,b) to gcd(a, b).

E. For anyaZ, both gcd(a, a) = |a| and gcd(a,0) = |a|.F. gcd(a,b) = gcd(|a|, |b|)

1.28 DEFINITION. Two numbers are relatively prime(or coprime, sometimes denoted a b) iftheir greatest common divisor is 1.

1.29 DEFINITION. The least common multipleof two numbers lcm(a,b) is the smallest naturalnumber that is a multiple of each.

1.30 EXE RCIS E. Prove each. (i) (EXI ST EN CE) Any two integers have a least common multiple. Ifthe integers are nonzero then the lcm is positive., (ii) (COMMUTATIVITY) lcm(a,b) = lcm(b, a).1.31 EXE RCIS E. (EUCLIDSALGORITHM) Prove that ifa= bq+ rthen gcd(a,b) = gcd(b,r).

The algorithm associated with this result finds the greatest common divisor for anya,bZ

withb= 0. For instance, to find gcd(803,154) divide the larger by the smaller 803 = 154 5+33 andthen the above result shows that gcd(803,154) = gcd(154,33). Next, since 154 = 33 4+22 the aboveresult gives gcd(154,33) = gcd(33,22). By eye we spot the answer of 11 but continuing as though wehadnt noticed we get 33 = 22 1 + 11, so gcd(33,22) = gcd(22,11). Finally, 22 = 11 2 + 0 gives thatgcd(22,11)= 11, so the algorithm terminates with gcd(803,154) = 11.

We can reverse this calculation. From 33=22 1 + 11 we can express the greatest commondivisor 11 as a combination 11 = 1 331 22. Next, the equation 154 = 33 4+22 lets us express the11 as a combination 11 = 1 331 (1544 33) = 1 154+5 33 of the pair 154,33. Backing up stillfurther, the equation 803 = 154 5+33 gives 11 = 1 154+5 (8035 154) = 5 80326 154, so wecan express the greatest common divisor as a combination of our initial pair 803,154.

1.32 DEFINITION. A number c is a linear combinationof two others aand bif it has the formc

=a

m

+b

nfor somem,n

Z.

1.33 EXE RCIS E. Use Euclids Algorithm to find the greatest common divisor and use that calcula-tion to express the greatest common divisor as a linear combination of the two: (i) 15, 25, (ii) 48,732.

1.34 EXE RCIS E. Prove.

A. Euclids algorithm terminates.

B. The greatest common divisors of two numbers is a linear combination of the two.

C. (BZOUTS LEMMA) The greatest common divisor of two numbers is the smallest positivenumber that is a linear combination of the two.


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1.35 EXE RCIS E. You are given three buckets. The first two are marked 6 liters and 11 liters whilethe last one,which is quite large, is unmarked. Taking water from a nearby pond, use thosebucketsto end with 8 liters in the unmarked one.

1.36 EXE RCIS E. Leta,bZ andmN. Prove each.A. gcd(ma,mb) = mgcd(a, b)B. Ifdis a common divisor ofaand bthen gcd(a/d, b/d)

=gcd(a,b)/d. This implies that di-

viding by the greatest common divisor leaves no common divisors: ifaandbare nonzero thena/g cd(a,b) andb/gcd(a, b) are relatively prime.

C. (EUCLIDSLEMMA) Ifaandbare relatively prime thena| bcimplies thata | c.

PRIMES

1.37 DEFINITION. An integer greater than 1 isprimeif its only divisors are 1 and itself. A numbergreater than 1 that is not prime iscomposite.

1.38 EXE RCIS E. Prove.A. An integern> 1 is composite if and only if it has factorsa,bsuch that 1 < a,b< n.B. Every number greater than 1 has a prime divisor.C. Every composite numbern

>1 has a prime divisorpwithp

n.

D. Show that the prior inequality cannot be made strict.

1.39 EXE RCIS E. (EUCLIDSTHEOREM) There are infinitely many primes.

1.40 EXE RCIS E. Suppose thatpis a prime. Prove each.A. Ifp | abthen eitherp | aorp | b.B. Ifp | a0 a1 an1thenpdivides at least oneai.

1.41 EXE RCIS E. (FUNDAMENTALTHEOREM OFARITHMETIC) Prove.A. Any numbern> 1 can be written as a product of one or more primesn= pe11 p

e22 p

ekk

.

B. That factorization is essentially unique: ifn= pe11 pe22 p

ekk

and the primes are in ascendingorderp1 < p2 < < pkthen any other factorization ofninto a product of primes in ascendingorder will give the same result. Hint: suppose that there are two factorizations into primesn

=p0

ps

1andn

=q0

qt

1in ascending order and use induction ons.

Remark:had we included 1 among the primes then factorization would not be unique since to anyexisting factorization we could add more 1s.

1.42 EXE RCIS E. Trueorfalse: (i)5719 = 31117, (ii) 47863 = 73557, (iii) 13574183 = 10815251.1.43 EXE RCIS E. Suppose that two numbers have prime factorizations ofa= pe00 p

en1n1 andb=

pf00 p

fn1n1 (to use the same primes, we allow that some of the exponents are zero). Prove each.

A. In the prime factorization of gcd(a,b) the exponent ofpiis min(ei,fi).

B. In the prime factorization of lcm(a,b) the exponent ofpiis max(ei,fi).

1.44 EXE RCIS E. (EXIST EN CE OFIRRATIONALNUMBERS)A. Prove that if a number is a square then in its prime factorization each prime occurs an even

number of times.

B. Prove that

2 is irrational.


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CHAPTER2 SETS

2.1 DEFINITION. Aset is a collection that is definite every object either definitely is contained in

the collection or definitely is not. An object xthat belongs to a set Ais an elementor memberofthe set, writtenx A(to denote thatxis not an element ofAwritex A). Two sets are equal if andonly if they have the same elements.

Read as is an element of rather than in since the latter causes confusion with the subsetrelation defined below. We sometimes say collection as a synonym for set.

We usually specify a set either by listing or by describing its elements. Thus we may write theset of the primes less than ten either as P= {2,3,5, 7} or as P= {pN

pis prime andp< 10} (readthe vertical bar as such that).

2.2 EXE RCIS E. Decide if each is true and justify your decision: (i) {1,3,5} = {5,3,1}, (ii) {2,4,6} ={2,4,6,4}, (iii) {1,3} = {nN

n< 5}, (iv) 0 {1,2,{0} }, (v) 4 {nN n2 < 50}.2.3 DEFINITION. The setBis a subsetof the set Aif every element ofBis an element ofA, that is,ifx

Bimplies thatx

A. This is denotedB

A.

2.4 DEFINITION. The set without any elements is the empty set.

2.5 EXE RCIS E. Decide each, with justification: (i) {1,3,5} {1,3,5,7,9}, (ii) {1,3,5} {1,3,5,7,9},(iii) { 1,3,5}{n N

nis prime}, (iv) {1,2,3,4}, (v) NZ, (vi) {2}{1,{2},3 }, (vii) {2}{1,{2},3}.

2.6 EXE RCIS E. Prove, for all sets A.A. A Aand AB. The empty set is unique: ifAis empty andBis empty then A= B.

2.7 EXE RCIS E. Prove, for any sets A,B, and C.A. (MUTUALINCLUSION) IfA Band B Athen A= B.B. (TRANSITIVITY) IfA Band B CthenA C.

2.8 EXE RCIS E. For each, give example sets or show that no example is possible. (i)A

B,B

C,A C, (ii)A B,B C, A C, (iii)A B B C,A C.

In this book we will always work inside of a universal set, denoted . For instance, in the firstchapter on number theory the universal set is=Z. There, if we say we are considering the set ofthings less than 100 then we are considering the set of integers less than 100.

2.9 DEFINITION. The characteristic functionof a setAis a map A(whose domain is the universalset) such that A(x) = 1 ifx A, and A(x) = 0 ifx A.2.10 EXE RCIS E. (RUSSELLSPAR AD OX) The definition that we gave of sets allows them to containanything, so consider a set that contains all sets. It has itself as an element. Following that idea,consider the set of all sets that have themselves as elements and also the set of all sets that dont.Call the latter setD.

A. Show that assumingDis an element of itself leads to a contradiction.

B. Show that assumingDis not an element of itself also leads to a contradiction.


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OPERATIONS

2.11 DEFINITION. Let Abe a set. ThecomplementofA, denoted Ac, is the set of objects that arenot elements ofA.

Remark:working inside of a universal set makes the complement operation sensible. For instance,in a number theory discussion where =Z, ifAis the set of objects greater than 100 then we can

take the complement ofAand the result is another subset of

, so we are still in number theory.2.12 DEFINITION. LetAand Bbe sets. Their unionisthesetofelementsfromeithersetAB= {x

x Aorx B}. Their intersectionis the set of elements from both sets AB= {x

x Aandx B}.Picture set operations withVenn diagrams.

A B

AB

A B

AB

A

Ac

In each diagram the region inside therectangle depicts the universalset and the region inside each

circle depicts the set. On the left the part in mauve (light purple) shows the union as containingall of the two sets joined, the middle shows the intersection containing only the region common toboth, and on the right the mauve region, the complement, is all but the set A.

2.13 EXE RCIS E. Another tool for illustrating set relationships is this table describing two sets.

x A x B row number0 0 00 1 11 0 21 1 3

Use the bits 0 and 1 instead ofF andTso that each row is the binary representation of its rownumber. That table gives A = {2,3},B= {1,3 }, and = {0,1,2,3}. For each item below, illustratethe statements using these sets. Then pick one item and prove it: (i) the complement of the com-

plement is the original set (Ac)c = A, (ii)A=andA= A, (iii) (IDEMPOTENCE) AA= Aand AA=A, (iv) A B AA B, (v) (COMMUTATIVITY) A B= BAand A B= BA,(vi) (ASSOCIATIVITY) (A B)C=A (BC) and (A B)C=A (BC).2.14 EXE RCIS E. Prove that the following are equivalent: (i)A B, (ii)AB= B, and (iii)AB= A.2.15 DEFINITION. Thedifferenceof two sets is A B= {xA

x B}. Thesymmetric differenceisA B= (AB) (BA).IfAXthenXAis the same asAcwhereXis the universal set =X.2.16 EXE RCIS E. Prove or disprove.

A. AB AB. A B= A BcC. For all pairs of sets,A

B

=B

A.

D. For all pairs of sets,A B= BA.2.17 EXE RCIS E. (DEMORGANSLAWS) Prove.

A. (A B)c = Ac Bc and (A B)c = Ac BcB. (DISTRIBUTIVITY) A (BC) = (AB) (AC) andA (BC) = (A B) (AC)

2.18 DEFINITION. Two sets aredisjointif their intersection is empty.

2.19 EXE RCIS E. A. Prove or disprove: A Band ABare disjoint.B. Find three setsA, B, and C, such thatABCis empty but the sets are not pairwise disjoint,

that is, none ofA B,AC, orBCis empty.


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2.20 DEFINITION. For a finite setA, theorder |A| is the number of elements.2.21 DEFINITION. For a set Athepower setP(A) is the set of all subsets ofA.

2.22 EXE RCIS E. List P({a,b}), P({a,b,c}), P({ a}), andP(). Find the order of each.

2.23 EXE RCIS E. LetA= {, {}}. Prove or disprove.A.

P(A),

P(A)

B. {} P(A), {} P(A)C. { {}} P(A), {{}} P(A)

2.24 EXE RCIS E. Prove or disprove: ifA BthenP(A) P(B).2.25 EXE RCIS E. WhereAis a finite set, prove that |P(A)| = 2|A|.

CARTE SI AN PROD UC T

2.26 DEFINITION. Thesequencex0, x1, . . . , xn1formed from the terms(or elements) x0, x1, . . . ,xn1is an ordered list of objects. Thelengthof a sequence lh(x0, x1, . . . , xn1) is the number ofterms n. Two sequences x0, x1, . . . , xn1 and y0,y1, . . . ,ym1 are equal ifand only if they havethesame length and the same terms, in the same order: n= mand x0 =y0,x1 =y1, . . . ,xn1 =yn1.

2.27 EXE RCIS E. Prove or disprove: (i) {3,4,4,5}= {4,3,5}, (ii)3,4,5 = 4,3,5, (iii)3,4,4,5 =3,4,5.2.28 DEFINITION. For sets A0, A1, . . . , An1 the Cartesian productis the set of all length nse-quencesA0 A1 An1 = {a0, a1, . . . , an1

a0 A0, .. ., andan1 An1}. If the sets are thesame we often writeA Aas An.

A sequence of length two is often called anordered pairand written with parentheses (x0, x1)(similarly we have ordered triples, four-tuples, etc.). Thus, we write Cartesian coordinates of pointsin the plane as members ofR2 = {(x,y)

x,yR}.2.29 EXE RCIS E. Prove that N2 Z2. Generalize.2.30 EXE RCIS E. A. Prove thatA B=iffA=orB=.

B. Show that there are sets so thatA B= BA. Under what circumstances is A B= BA?

C. Is (AB)Cequal toA (BC)?D. Show that this statement is false: A BA Bif and only ifA Aand B B. Patch thestatement to make it true.


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CHAPTER3 FUNCTIONS,REL ATIONS

3.1 DEFINITION. Afunction f(or mapor morphism)from domainset Dto codomainset C, written

f: D C, is a sequence consisting of the two sets along with a graph, a set of pairs (d,c) DCthat iswell-defined: for eachd Dthere is exactly one c Csuch that (d,c) is an element of thegraph. Functions are equal only if they have the same domain, codomain, and graph.

Thus a function associates each element dfrom the domain, called an argumentorinput, withan elementcfrom the codomain, called avalueor output, subject to the condition thatddeter-minesc. We write f(d) = cand say thatcis theimageofdor thatd maps to c.

Abean diagrampictures the domain and codomain as blobs and either shows the function asa whole as a simple arrow or else shows the functions action on individual elements with arrowsthat begin with a bar.

D Cf

f: D C

a f(a)

b f(b)

af(a) andbf(b)

3.2 EXE RCIS E. Decide if each is a function: (i) D={0,1,2}, C={3,4,5}, G= {(0,3),(1,4),(2,5)},(ii)D= {0,1,2}, C= {3,4,5}, G= {(0,3),(1,4),(2,3)}, (iii)D= {0,1,2}, C= {3,4,5}, G= {(0,3),(1,4)},(iv)D= {0,1,2},C= N,G= {(0,3),(1,3),(2,3)}, (v)D=N,C= N,G= {(0,3),(1,4),(2,5)}, (vi)D={0,1,2}, C= {3,4,5}, G= {(0,3),(1,4),(2,4),(0,5)}, (vii)D=N, C=N, G= {(d,c) DC

c= d2}.Remark:we often blur the distinction between a function and its graph. For instance we may

say, a function is an input-output relationship when technically it is the functions graph thatis the pairing. (Weve already done some blurring, in the sentence following the definition.) Thedistinction between function and graph is there only because the graph does not determine thecodomain, so we must specify it separately. (The graph does determine the domain so that infor-mation is redundant.)

Do not think that a function must have a formula. The final item in the prior exercise has aformula but for other items Gjust has arbitrary pairings.

3.3 EXE RCIS E. LetDand Cbe finite sets. How many maps are there fromDto C?

A function may have multiple arguments; one example is x,y f x2 2y2. We write this mapf: R2 R as f(x,y) rather than f(x,y). The number of arguments is the functionsarity.3.4 DEFINITION. For f: D C, therangeis the set {y C

there is ax Dsuch thatf(x) =y}, de-noted Ran(f) (or f(D)).

3.5 EXE RCIS E. For each item in Exericse3.2, if it is a function then find its range.

3.6 DEFINITION. Let f: D C. The restrictionof f to B D is the function fB: B Cwhoseaction is given byfB(b)

=f(b) for all b

B. A synonym is that fis an extensionoffB. The image

of the setBunder f, denoted Im(f) (or f(B)), is the range of the functionfB.3.7 EXE RCIS E. Find the image of each set under the real function f(x) = tan(x): (i) [/4 ../2) ={xR

/4 x


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COMPOSITION

3.11 DEFINITION. The compositionof two functions f: D Cand g:C Bis gf: D Bgivenbygf(d) = g(f(d)).

D C Bf g

g f

Note that, although when reading the expression g ffrom left to right the gcomes first, thefunction that is applied first is f. Note also that the codomain of fis the domain ofg(we some-times allow a composition when the domain ofgis not the entire codomain, but instead is a su-perset of the range off).

Readgfaloud as gcircle f or gcomposed with f (or gfollowing f).3.12 EXE RCIS E. LetD= {0,1,2},C= { a,b,c,d}, andB= {,,}. Suppose that f: D Cis givenby 0 a, 1 c, 2 dand thatg:C Bis given bya ,b , c , andd .

(i) Computegfon all arguments or show that the composition is not defined., (ii) Computefgon all arguments or show it is not defined., (iii) Find the range of f, g, and any defined composi-tions.

3.13 EXE RCIS E. Let f: R Rbe f(x)=x2 and let g: R Rbe g(x)=3x+ 1. Find the domain,codomain, and a formula for each ofgf and f g.3.14 EXE RCIS E. Prove each.

A. (ASSOCIATIVITY) h (gf) = (h g)fB. Function composition need not be commutative.

INVERSE

The definition of a function specifies that for every argument in the domain there is exactly one

associated value. The definition puts no such condition on the elements of the codomain.3.15 DEFINITION. A function is one-to-one(or aninjection) if for each value there is at most oneassociated argument, that is, iff(d0) =f(d1) implies that d0 = d1for elements d0,d1ofthe domain.A function is onto(ora surjection) if for each value there is at leastone associated argument, that is,if for each elementcof the codomain there exists an elementdof the domain such that f(d) = c.A function that is both one-to-one and onto, so that for every value there is exactly one associatedargument, is acorrespondence(orbijection, orpermutation).

3.16 EXE RCIS E. Letf: R R be f(x) = 3x+1 andg: R R beg(x) = x2 +1.A. Show that fis one-to-one.B. Show thatfis onto.

C. Show thatgis not one-to-one, and also that it is not onto.

3.17 EXE RCIS E. Proving these two items shows that, where DandCare finite sets, if there is acorrespondencef: D Cthen the two sets have the same number of elements.A. Iffis onto then |C| |D|.B. Iffis one-to-one then |C| |D|.

3.18 EXE RCIS E. Prove.A. A composition of one-to-one functions is one-to-one.B. A composition of onto functions is onto.

C. A composition of correspondences is a correspondence.

3.19 EXE RCIS E. A. Prove that ifgfis onto thengis onto.


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B. Prove that ifgfis one-to-one then fis one-to-one.C. Do the other two cases hold?

3.20 DEFINITION. Anidentity functionid: D Dhas id(d) = dfor alld D.3.21 DEFINITION. A function inverseto f: D C is f1 :C Dsuch that f1 f is the identitymap onDand f f1 is the identity map on C.

3.22 EXE RCIS E. A. Let f: Z Z be f(a) = a+3 and letg: Z Z beg(a) = a 3. Show thatgisinverse to f.

B. Leth: Z Z be the function that returnsn+1 ifnis even, and returnsn1 ifnis odd. Finda function inverse toh.

C. Lets: R+ R+be s(x) = x2. Find the map that is inverse to s.D. Show thatt: R R given byt(x) = x2 has no inverse.

3.23 EXE RCIS E. Find a domain and codomain, and a pair of maps f: D Candg:C D, suchthatgfis the identity but f gis not.

Ifgf is the identity function then we say thatgis a left inverseoff, or what is the same thing,that f is aright inverseofg. Ifgis both a left and right inverse off then, as above, we say just thatit is an inverse off(or for emphasis, atwo-sided inverse).

3.24 EXE RCIS E. LetD= {0,1,2 } and C= {,,}. Also let f, g: D Cbe given byf(0) =, f(1) =, f(2) = , andg(0) = ,g(1) = , g(2) = . Then: (i) verify that f is a correspondence, (ii) con-struct an inverse for f, (iii) verify thatgis not a correspondence, (iv) show that ghas no inverse.

3.25 EXE RCIS E. Prove each.A. A function has an inverse if and only if that function is a correspondence.

B. If a function has an inverse then that inverse is unique.C. The inverse of a correspondence is a correspondence.

D. Iff andgeach is invertible then so is g f, and (gf)1 =f1 g1.

RELATIONS

A function, or more precisely the functions graph, is a set ofntuples subject to the condition

that the input determines the output. The generalization that we get by dropping the condition isuseful.

3.26 DEFINITION. Arelationon setsA0, . . . ,An1is a subset R A0 An1. If all ofthe sets arethe sameA0 = A1 = =Athen we say it is a relation on A.

The number of sets nis the arityof the relation and we say it is n-ary. If the arity is 2 then it is abinary relation. In this case, where (a0, a1) Rwe say thata0isR-relatedtoa1, sometimes writtena0Ra1.

3.27 EXE RCIS E. List five elements of each relation: (i) {(x,y) N2xand yhave the same parity},

(ii) less-than


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3.43 DEFINITION. A binary relation Ris antisymmetricif (x,y) Rand (y, x) R implies that x=y.A relation is apartial orderingif it is reflexive, antisymmetric, and transitive.

3.44 EXE RCIS E. Prove each.A. The usual less than or equal to relation on the real numbers is a partial order.B. The relation divides onN is a partial order.C. For any set Athe relation

on P(A) is a partial order.

3.45 EXE RCIS E. Can a relation be both symmetric and antisymmetric?


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CHAPTER4 INFINITY

Recall Exercise3.17,that if two sets correspond then they have the same number of elements.

4.1 DEFINITION. Two sets have the same cardinality(or are equinumerous) if there is a correspon-dence from one to the other. We write A B.4.2 DEFINITION. A set isfiniteif it hasnelements for somenN, that is, if it has the same cardi-nality as {iN

i< n} = {0,1,..., n1}. Otherwise the set isinfinite. A set isdenumerableif it hasthe same cardinality as N. A set iscountableif it is either finite or denumerable.

4.3 EXE RCIS E. Prove that the relation is an equivalence.4.4 EXE RCIS E. Prove.

A. The set of integers is countable.B. The set NN is countable.

4.5 EXE RCIS E. Prove that the following are equivalent for a set A: (i)Ais countable (ii) Ais emptyor there is an onto function from N toA(iii) there is a one-to-one function from Ato N.

4.6 EXE RCIS E. Prove that the set of rational numbers is countable.

4.7 EXE RCIS E. Prove that these infinite sets are not countable.A. P(N)B. R


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CHAPTERA PEANO AXIOMS

Particularly in the first chapter a person struggles with when to consider a statement sufficiently

justified and soon comes to wonder what the axioms are like. Here we give the most often usedaxiom system for the natural numbers, to convey a sense of that.This system was introduced by Dedekind in 1888 and tuned by Peano in 1889. In addition to

the usual logical and set symbols such as = and , with the traditional properties, our language willuse at least two symbols, 0 andS, whose properties are limited by the conditions below.

A.1 AXI OM . (EXI ST EN CE OF A NATU RA L NUMB ER) The constant 0 is a natural number.

A.2 AXI OM . (ARITHMETICAL PROPERTIES) ThesuccessorfunctionShas these properties.A. (CLOSURE) For allaN, its successorS(a) is also a natural number.B. (ONE-TO-ON E) For alla,bN, ifS(a) = S(b) thena= b.C. (ALMOST ONTO) For allaN, ifa= 0 then there is abNwithS(b) = a. In contrast, nocN

has 0 as a successor.

These properties give infinitely many natural numbers: 0, S(0), S(S(0)), etc. Of course, the

notation 0, 1, 2, etc., is less clunky.

A.3 AXI OM . (INDUCTION) Suppose thatKis a set satisfying both (i) 0 Kand (ii) for all nN, ifn KthenS(n) K. ThenK=N.(In this book we use an induction variant that changes condition (ii) to: for all kN, ifn K for0n k then S(k)K. Condition (ii) above is often calledweak inductionwhile the versionwe use isstrong induction. There are technical differences but for our purposes the two variantsare interchangeable. We prefer the strong variant because while it is more awkward to state, it issometimes easier to apply.)

From those axioms we can for instance define addition by recursion using successor

add(a,n) =

a ifn= 0S(add(a,m)) if n

=S(m)

and then define multiplication by recursion using addition.

mul(a,n) =

0 ifn= 0add(mul(a,m), a) ifn= S(m)

inquiry-based introduction to proofs

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