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REAL MATHEMATICAL ANALYSIS

Lectures by Niranjan Balachandran, IIT Bombay.

Contents

1 Preliminaries: The Real Line 5

1.1 Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2 Natural Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2.1 Axioms for Natural Numbers

N. . . . . . . . . . . . . . . . . . . 6

1.2.2 Addition and Multiplication . . . . . . . . . . . . . . . . . . . . . 7

1.2.3 Order on

N. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2.4 The Well Ordering Principle, and the Euclidean Algorithm . . . . 9

1.2.5 Prime Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.3 Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.3.1 Addition onZ. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.3.2 Multiplication onZ. . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.3.3 Subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.3.4 Order on

Z. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.4 Rational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.4.1 Addition and Multiplication on

Q. . . . . . . . . . . . . . . . . . 18

1.4.2 Order on

Q. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 1.4.3 Qmisses some `numbers' . . . . . . . . . . . . . . . . . . . . . . 21

1.5 Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

1.5.1 Addition and Multiplication on

R. . . . . . . . . . . . . . . . . . 24

1.5.2 Another description for Real Numbers . . . . . . . . . . . . . . . 27

1.5.3 Archimedian Property of

R. . . . . . . . . . . . . . . . . . . . . 28

1

1.6 Cardinality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

1.7 The Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

2 Basic Topology 39

2.1 Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.2 Subsequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

2.3 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

2.4 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

2.5 Induced Topology on Subsets ofR. . . . . . . . . . . . . . . . . . . . . . 52

2.6 Extending from the Reals to arbitrary Metric spaces . . . . . . . . . . . . 54

2.7 Connectedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

2.7.1 A Weird Closed Set inR: . . . . . . . . . . . . . . . . . . . . . . 57

2.7.2 Pathological Examples: . . . . . . . . . . . . . . . . . . . . . . . . 58

2.8 Another Construction ofRfromQ. . . . . . . . . . . . . . . . . . . . . 60

2.9 Returning to the Cantor Set . . . . . . . . . . . . . . . . . . . . . . . . . 64

3 Dierentiation 65

3.1 Dierentiation of Real Valued Functions . . . . . . . . . . . . . . . . . . 65

3.2 The Mean Value Theorems and Consequences . . . . . . . . . . . . . . . 69

3.2.1 A Theorem of Darboux . . . . . . . . . . . . . . . . . . . . . . . . 70

3.2.2 The L'H^opital Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 71

3.3 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

3.3.1 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

3.3.2 Taylor Series and Taylor Approximation . . . . . . . . . . . . . . 80

3.4 Uniform Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

3.4.1 The Metric Space (C[0;1];d) . . . . . . . . . . . . . . . . . . . . . 83

3.4.2 Theorems of Weierstrass . . . . . . . . . . . . . . . . . . . . . . . 85

3.4.3 Baire Category Theorem . . . . . . . . . . . . . . . . . . . . . . . 90

4 Riemann Integration 93

4.1 Integrals according to Riemann and Darboux . . . . . . . . . . . . . . . . 93

4.2 Measure Zero sets and Riemann Integrability . . . . . . . . . . . . . . . . 96

4.3 Consequences of the Riemann-Lebesgue Theorem . . . . . . . . . . . . . 100

4.4 Antiderivatives and some `well known' Integral Calculus techniques . . . 102

4.5 The Integral Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

4.6 Weak version of Stirling's formula forn! . . . . . . . . . . . . . . . . . . 108

4.7 Convergence of sequences of functions and Integrals . . . . . . . . . . . . 110

5 Measures of sets and a peek into Lebesgue Integration 113

5.1 Measure for subsets ofR. . . . . . . . . . . . . . . . . . . . . . . . . . . 113

5.2 Sigma Algebras and the Borel Sigma Field . . . . . . . . . . . . . . . . . 124

5.3 Lebesgue Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

5.4 Lebesgue's Monotone Convergence Theorem . . . . . . . . . . . . . . . . 133

2

Preface

Real Analysis is all about formalizing and making precise, a good deal of the intuition that resulted in the basic results in Calculus. As it turns out, the intuition is spot on, in several instances, but in some cases (and this is really why Real Analysis is important at all), our sense of intuition is so far from reality, that one needs some kind of guarantee, or validation to our heuristic arguments. I thank all my students of this course who very actively and enthusiastically acted as scribes for the lectures for this course. 3

1 Preliminaries: The Real Line

As mentioned in the preface, in order to formalize the results one has studied in a rst Calculus course, one needs to start at the very beginning, in order to ensure that there is no inconsistency in our settings. And we shall begin at the very beginning - with the

Natural numbers.

It may be a bit dicult to formally dene what the natural numbers actually are. But, we shall avoid doing this, by making our approach axiomatic. The more important thing is to ensure that the axioms are no contradictory. As it turns out, very basic axioms about the natural numbers are sucient to set up our understanding of the natural numbers in a wholesome manner.

1.1 Relations

Denition 1SupposeSis a set,then theRelationRonSis dened as a subset of

SS:=f(s1;s2)js1;s22Sg

If(s1;s2)2Rwe denote it ass1s2.

Denition 21. A relation R isRe

exiveifaa ;8a2S

2. A relation R isSymmetricifab)ab ;8a;b2S

3. A relation R isTransitiveifab ;bc)ac ;8a;b;c2S

Denition 3(Equivalence Relation)

A relation is an Equivalence relation i it is Re

exive, Symmetric and Transitive. Proposition 1Supposeis an equivalence relation dened on a setS, theninduces a partition onS. Conversely, for any partition=fSg2of the setS, there is an equivalence relationwhich induces partitionon S. 5 Proof:Given, an equivalence relation, for eacha2S,dene S a=fb2Sjbag

Claim 1fSaga2Sis a partition of S.

Leta6=b, wherea;b2S. ConsiderSa\Sb. SupposeSa\Sb6= . We shall show that S a(Sa\Sb). Similarly,Sb(Sa\Sb) and these will prove thatSa=Sb= (Sa\Sb). SinceSa\Sb6= , there exists somec2(Sa\Sb). Letx2Sa. Then we havexa. However, sincec2Sa, we haveca)ac, by symmetry. Therefore, by transitivity, xc. And sincecb, we also getxb)x2Sb. Now thatx2Saandx2Sb, x2(Sa\Sb). Since this is true for anyx2Sa, we haveSa(Sa\Sb). Similarly, we can show thatSb(Sa\Sb). Thus,Sa=Sb= (Sa\Sb) which is a partition ofSinduced by. Converse: Given =fSg2, a partition ofS, we dene a relationas follows: ab if92 such that a;b2S Now, we prove that it is an equivalence relation onS. Re exivity: (aa)8a2Sas it is in the same partition as itself. Symmetry:ab)ais in the same partition asb)bis in the same partition as a)ba Transitivity:ab;)a;b2Sandbc;)b;c2S. But sinceShas been partitioned by andbbelongs to bothSandS, we must haveS=S. Henceaand cbelong toSand are related under, thus proving the transitivity. These three properties put together makean equivalence relation dened onS.|

1.2 Natural Numbers

1.2.1 Axioms for Natural Numbers

N

1. 02N.

2. For eachn2Nthere is a unique successor forn, denoted byn+ 1.

3. IfSN,satisfying

02N. n2N)n+ 12N. thenS=N. This is referred to asThe Principle of Mathematical Induction.

4. For eachn2Nnf0g, there exists a uniquem2Nsuch thatm+ 1 =n.

6

1.2.2 Addition and Multiplication

We can dene two operations onNcalled `addition, multiplication' that satises the following. We skip the proofs. Theorem 1Givenm;n2N,there exists a binary operation '+' onNsatisfying:

1.0 +n=n,8n2N.

2.m+n=n+m,8m;n2N. (Commutative Property)

3.m+ (n+p) = (m+n) +p,8m;n;p2N. (Associative Property)

Theorem 2Givenm;n;p2N,there exists a binary operation '.' onNsatisfying:

1.1:n=n8n2N.

2.m:n=n:m,8m;n2N. (Commutative Property)

3.m:(n:p) = (m:n):p,8m;n;p2N. (Associative Property)

4.m:(n+p) =m:n+m:p,8m;n;p2N. (Distributive Property)

Example 1Prove that 2+2=4.

Proof:

2 + 2 =2 + (1 + 1):::(since2is successor of1);

=(2 + 1) + 1:::(Associative property); =3 + 1:::(3is successor of2); =4:::(4is successor of3): 7

1.2.3 Order on

N

Denition 4We saya > b, if

1.a=b+cfor somec2N,

2.a6=b.

The following theorem follows from the denition of order. The proof is again skipped.

Theorem 3

0>0onNsatises

1.a > b)a+c > b+c8c2Z.

2.a > b and c >0)ac > bc.

3.a > b,b > c)a > c. (Transitivity)

4. Givena;b2N, precisely one ofa > borb > aora=bis satised.

5.0<1<2<3<

Lemma 1Ifc;d2N, andc+d= 0, thenc=d= 0.

Proof:Supposec6= 0. Then,

c=c0+ 1 c+d=(c0+ 1) +d =(c0+d) + 1 =0 So, it follows that 0 is the successor ofc0+d, which is a contradiction to the fact that 0 has no predecessor in

N. Hence proved.|

Remark 1Ifx=x+y, theny= 0.

Note 1Supposea;b2Nand supposea > b, can we haveb > atoo?

If yes, then

a > b)a=b+c and b > a)b=a+d 8 wherec;d2Nandc;d6= 0)(a+b) = (a+b) + (c+d). From earlier remark, we have c=d= 0which shows that only one of the above two is possible, else they are equal. Remark 2We have already seen that0<1<2<3:::. In particular, we have:

For anya2N,a <(a+ 1) +bfor anyb2N;b >0.

1.2.4 The Well Ordering Principle, and the Euclidean Algorithm

We start with an equivalent formulation to the principle of Induction, known as the

Principle of Complete Induction.

1. 02S.

2.f0;1;2;:::;ng S)n+ 12S.

ThenS=N. The proof is a simple consequence of the principle of Mathematical Induc- tion. We skip the proof. Proposition 2The Well Ordering Principle (WOP): Every non empty subset SNcontains a least element, i.e., there existss2Ssuch thats < s0for alls02

S; s06=s.

Proof:For natural numbersa < b, we shall denote by [a;b] the setfa;a+1;:::;b1;bg. Let;(SN. We need to show thatShas a minimal element. SupposeShas no minimal element. LetP(n) be the propositional function:n =2S.

We have two cases:

Case 1: 02S. Since 0 is the least element ofN, it is also the minimal element ofS which is a contradiction. Case 2: 0=2SsoP(0) holds. SupposeP(j) holds for 0jk, i.e., suppose for all j2[0;k] :j =2S. Ifk+ 12Sthenk+ 1 would be the minimal element ofS. Sok+ 1=2Sand so P(k+ 1) also holds. Thus we have proven the following.

1.P(0) holds.

2. For allj2[0;k] :P(j) holds))P(k+ 1) holds.

So by the principle of complete inductionP(n) holds for alln2N. But this meansS is empty which is a contradiction.|

A simple consequence is the following:

9 Theorem 4Euclidean division and the Euclidean algorithm:Given positive integersm;nthere exist unique non-negative integersq;rsuch thatm=qn+r,0r < n. We describe this by saying that the Euclidean algorithm when applied to the ordered pair (m;n)gives a quotient ofqand remainderr.

Proof:DeneS=fmknjk2N0;mkn >0g. Now,SN0andS6= as

m0:n=m2S. By WOP,Shas a least element; call itr=mqn. We claim

1. 0r < n.

2.q;ras determined above, are unique.

To prove this claim, note that by denition ofS,r0. Indeed, suppose otherwise. Then m(q+ 1)n=mqnn=rn0, and this impliesm(q+ 1)n2S. Also m(q+ 1)n=rn < rasn >0. Hence,ris not the least element ofS, and this is a contradiction. This proves the rst part. To prove the second, again, suppose otherwise. Letm=q1n+r1andm=q2n+r2 be two such representations and WLOG, letr2> r1. Equating the RHS of the above equations and simplifying, (q1q2)n=r2r1, )r2r1is a multiple ofn. But since,r1;r2< n;r2r1< n, the only possibility is r

2r1= 0. which implies thatr1=r2and it follows thatq1=q2and this proves the

second claim as well. Thus,r;qare unique integers satisfyingr2[0;n) andm=qn+r.| A very important consequence of the Euclidean algorithm is the following. For natu- ral numbersa;bwe say thatndividesmif the corresponding value ofrin the Euclidean algorithm above equals 0. For natural numbersa;bwe say thatdis thegreatest com- mon divisor(and denoted (a;b)) ofa;bifddividesa;band for anyd0that divides both a;bwe also haveddividesd0. A very useful consequence of the Euclidean algorithm is this: Givenm;n;ras in the theorem, (m;n) = (r;n).

1.2.5 Prime Numbers

Denition 5Supposen2N,n >1. We say thatnisprimeifn=ab, )a= 1orb= 1. An equivalent denition is that for every1a < pwe must have (a;p) = 1. Theorem 5(Euclid) The set of Prime Numbers hasno largestelement. Proof:Suppose that there are onlyN2Nprime numbers. Let them bep1;p2;:::pN.

Consider the number

n=p1p2:::pN+ 1 Now, the numbernis none of the prime numbers listed above and so it can be another prime number not in the listed N primes,which then contradicts our assumption. Ifn 10 is not a prime number, thenncan be written asn=abfora;b2Nsuch thata >1 andb >1 (If eitheraorbis 1, thennwill be a prime). Now we note thatp1;p2;:::pN do not divide eitheraorb, which contradicts the Fundamental Theorem of Arithmetic (given below), i.e., there must exist atleast another prime number apart fromp1;p2;:::pN which can dividen. This is due to our incorrect assumption that there exists only N prime numbers. Hence, there is no largest prime number.|

1.3 Integers

informally, an integer is a `number' that can be represented as (ab) wherea;b2N. But this denition is clearly a decient one. We shall see how to make sense of this as follows.

Denition 6Describe a relation 'r' onNNas follows:

(a;b)r(c;d) if and only if a+d=b+c Lemma 2(Cancellation Law forN)Ifx+y=z+y,thenx=z,8x;y;z2N. Proof:We prove it by induction ony. Fory= 0,x+ 0 =z+ 0)x=z.

Suppose it is true fory, i.e, ifx+y=z+y, thenx=z.

Now we need to prove it fory+ 1. Ifx+ (y+ 1) =z+ (y+ 1), then by associativity, we have, (x+y) + 1 = (y+z) + 1.

Since predecessors in

Nnf0gare unique,x+y=y+z. By induction, it follows that x=z. Hence, the lemma holds good.|

Proposition 3

0r0onNNis an Equivalence relation.

Proof:We have (a;b)r(a;b) ifa+b=b+a. Since addition is commutative ,this is satised and

0r0is re

exive. Now consider (a;b)r(c;d). This givesa+d=b+c. Similarly (c;d)r(a;b) this gives c+b=a+d. Since addition on natural numbers is commutative, this also proves the symmetry.

Now considering, (a;b)r(c;d) i.e.,

a+d=b+c and (c;d)r(e;f) i.e., c+f=d+e 11

Adding the above gives

a+d+c+f=b+c+d+e which is (a+f) + (c+d) = (b+e) + (c+d) Using the cancellation law, we geta+f=b+e, thus proving transitivity. Since

0r0is Re

exive, Symmetric and Transitive, it is an Equivalence relation. Since the relationrdenes an Equivalence relation, it partitionsNNand these equivalence classes are calledIntegers, and the set of integers is denoted byZ.

1.3.1 Addition onZ

Consider the integers (a;b) and (c;d). We dene addition in the following manner: (a;b) + (c;d) := (a+c;b+d)

Claim 2(+) is well dened.

Proof:Let (a;b) and (a0;b0) belong to one equivalence class and (c;d) and (c0;d0) belong to another. By denition, (a;b) + (c;d) = (a+c;b+d) and (a0;b0) + (c0;d0) = (a0+c0;b0+d0) Also, a+b0=a0+b:::(1) c+d0=c0+d:::(2)

Adding (1) and (2), we get

a+b0+c+d0=a0+b+c0+d

Using Associativity, we can rewrite it as

a+c+b0+d0=a0+c0+b+d Thus, (a+c;b+d) and (a0+c0;b0+d0) belong to the same equivalence class, proving that (+) is well dened.| 12

1.3.2 Multiplication onZ

Denition 7Consider the integers(a;b);(c;d). Then multiplication is dened as (a;b):(c;d) := (ac+bd;ad+bc)

Claim 3(.) is well dened.

Proof:Let (a;b) and (a0;b0) belong to one equivalence class and (c;d) and (c0;d0) belong to another. By denition, (a;b):(c;d) := (ac+bd;ad+bc) and (a0;b0):(c0;d0) := (a0c0+b0d0;a0d0+b0c0) Also, ab=a0b0:::(1) cd=c0d0:::(2)

Multiplying (1) and (2),

(ab):(cd) =ac+bdadbc:::(3) (a0b0):(c0d0) =a0c0+b0d0a0d0b0c0:::(4)

Since LHS of (3) and (4) are equal, we have

ac+bdadbc=a0c0+b0d0a0d0b0c0 (ac+bd) + (a0d0+b0c0) = (ad+bc) + (a0c0+b0d0) Thus, by denition, it follows that (ac+bd;ad+bc) and (a0c0+b0d0;a0d0+b0c0) belong to the same equivalence class, proving that (.) is well dened.|

1.3.3 Subtraction

If (a;b)2Zthen we dene negation onZas follows

(a;b) := (b;a):

Denition 8Subtraction on integers is dened as

(a;b)(c;d) := (a;b) + ((c;d))i:e:; (a;b)(c;d) := (a;b) + (d;c) 13 It can be checked that (-) is also well dened, by using the 'well-dened'ness of (+) and the fact that () can be represented in terms of (+).

Claim 4

NZ. Proof:Dene a setN:=f(a;0)2Zg. Now, consider the following map. f:N7!N i:e:;f:a7!(a;0) We observe thatf(a+b) = (a+b;0) andf(a) = (a;0) ,f(b) = (b;0))f(a+b) = f(a) +f(b)8a;b2N. This map identies Natural Numbers sitting inside the Integers. Proposition 4Addition and Multiplication onZsatisfy the following:

1. They are Commutative, Associative and Addition distributes over Multiplication

2. O:=(0,0) satises (a,b)+O=(a,b)8(a;b)2Z

3. Givenm2Zthere exists a uniquen2Zsuch thatm+n= 0

4. Ifm+x=n+x, thenm=n,8x;m;n2Z

Proof:3:Let us considerm= (a;b)2Z. Letn= (c;d)2Zsuch thatm+n= 0. (a;b) + (c;d) = (0;0) (a+c;b+d) = (0;0) (a+c)(b+d) = 00 (a+c)(b+d) = 0 a+c=b+d ab=dc (ab) =cd (b;a) = (c;d) Hence (c;d) = (b;a) and this shows existence of additive inverse inZ. Now, we prove the uniqueness of additive inverse. Letn1= (c;d)2Zandn2= (e;f)2Z 14 both be additive inverses form. (a;b) + (c;d) = (0;0):::(1) (a;b) + (e;f) = (0;0):::(2) (a+c;b+d) = (0;0):::(using(1)) (a+e;b+f) = (0;0):::(using(2)) (a+c;b+d) = (a+e;b+f):::(from above two) (a+c) + (b+f) = (b+d) + (a+e):::(from definition) a+b+c+f=a+b+d+e:::(using associativity) c+f=e+d:::(cancellation law) cd=ef (c;d) = (e;f) Hence (c;d) and (e;f) represent the same integer and hence, we have proved uniqueness of the inverse.| Proof:4:Let us considerm= (a;b) andn= (c;d)2Z. Letx= (x1;x2)2Zsuch that m+x=n+x. Forx= (0;0), we have (a;b) + (0;0) = (c;d) + (0;0))(a;b) = (c;d).

Now, we induct onx1.

Assume thatm+x=n+x)m=nis true forx= (x1;x2). We now prove this to be true forx= (x1+ 1;x2). (a+x1;b+x2) = (c+x1;d+x2))(a;b) = (c;d):::(given) (a+x1)(b+x2) = (c+x1)(d+x2))(a;b) = (c;d) (ab) + (x1x2) = (cd) + (x1x2))(a;b) = (c;d) (ab) + (x1x2) + 1 = (cd) + (x1x2) + 1)(a;b) = (c;d) (ab) + ((x1+ 1)x2) = (cd) + ((x1+ 1)x2))(a;b) = (c;d) Hence, we have shown that the statement holds forx= (x1+1;x2), thus completing the induction.|

1.3.4 Order on

Z

Letm;n2ZWe saym > nif

1.m=n+p, for somep2Z

2.m6=n.

15 The following theorem is in the same spirit as the corresponding one for the natural numbers.

Theorem 6

0>0onZsatises

1.a > b,ab >0.

2.a > b)a+c > b+c8c2Z.

3.a > b and c >0)ac > bc.

4.a > b) b >a.

5.a > b;b > c)a > c.

6. Givena;b2Zprecisely one ofa > borb > aora=bis satised.

We return to the notion of divisibility that was introduced for the natural numbers. We extend the same to the integers in the following manner. We say thatajb(for integers a;bif there exists an integercsuch thatb=ac. We also dene the greatest common divisor in exactly the same manner as in the case of the natural numbers. Theorem 7Ifa;b2Zand suppose(a;b) = 1. Then there exist integersm;nsuch that

1 =am+bn.

Proof:We will sketch the proof. Consider the setS:=fam+bnjm;nZg. The following can be proved in a straightforward manner.

1. 02S.

2.x;y2S)a+b2S. Similarly,x2S) x2S.

3.x2S;2Z)x2S.

In particular considerS0:=S\N. This is clearly non-empty by the above observations, and the fact thata;b2S. Now, letdbe the least element ofS0which exists by the WOP. Consider the remainderrby applying the Euclidean algorithm to (a;d) (i.e. pick the appropriate non-negative integer). It is easy to check thatr2Sas well since both a;d2S. By the minimality ofd, it follows thatr= 0, i.e.dja. By the same reason,djb.

Hencedj(a;b) = 1, so we must haved= 1.|

16

1.4 Rational Numbers

Again, informally, by the rational numbers, we denote `numbers of the form pq , where p;q2Zandq6= 0'. But as we have seen before with the Integers, the path to formalizing this involves setting up the right kind of equivalence relation on pairs of integers.

Denition 9We dene a RelationonZ(Zrf0g)as

a==bc==d ifad=bc, wherea;b;c;d2Zand b;d6= 0. Remark 3The integers satisfy the property that there areno zero divisors, i.e. ifa;b2Zandab= 0,then eithera= 0orb= 0or botha;b= 0. Indeed, ifa;b >0 then we have seen from the properties of the natural numbers thatab >0. If saya <0, then(a)b >0and this also implies thatab <0, so in particular, it is not zero. The same argument works in the other cases as well. Proposition 5dened onZ(Zrf0g)is an Equivalence Relation. Proof:We prove the three properties of equivalence relations. 1.Re exivity

We have

a==ba==b )ab=baand this is true.

2.Symmetry

We have , if

a==bc==d )ad=bc, then c==da==b )bc=ad, which is true, sincead=bc.

3.Transitivity

If a==bc==d )ad=bc, and c==de==f 17 )fc=ed, we get (ad)(fc) = (bc)(ed)quotesdbs_dbs14.pdfusesText_20

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