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THE REAL NUMBER

Unit 1

The natural numbers, the integers,

the rationals, Dedekind cuts,the reals, completeness

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HOW WILL YOU DESCRIBE ANATURAL NUMBER? THE SET

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A. THE NATURAL NUMBERSObjective:

Construct N, the set of natural numbers: using only few axioms;

un amen a proper es s ou eretained;

allow addition and multiplication; and

allow construction of integers(going to reals).

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1. Peanos Axioms (Dedekind-Peano)

Let N be a set which satisfy the following:1. 1N;2. For each nN, there is a unique

n*N. (n* successor of n)3. For each nN, n* 1.4. For m,nN such that m n, m* n*.

5. IfAN such that 1A andpAimplies thatp*A, thenA=N.

(Principle of Mathematical Induction)

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In symbols . . .

Let N be a set which satisfy the following:

1. 1N; (Some considers 0.)2. nN ! n*N

3. nN n* 1.

4. m,n

N, m

n m*

n*.5. (AN, 1A,pA p*A) A=N

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By 1, N is non-empty.

By 2, each n has a unique successor.

By 3, 1 is the first element.

Consequences

By 4, different ns have differentsuccessors.

(Prove this!) If mN, m 1, then thereexists an nN such that n*=m.

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Note:

The set N satisfying1 to5

numbers.

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Giuseppe Peano(1858-1932)

An Italian mathematicianwho orked on mathematical

.The standard axiomatization of naturalnumbers is named in his honor. He made

key contributions to the modern rigorousand systematic treatment of the method ofmathematical induction.

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2. Definitions. (Arithmetic on N)

ADDITION

n+1 = n*, VnNand n+p* = (n+p)*, Vn,pN

MULTIPLICATION

n1= n, VnNand np* = (np)+n, V n,pN

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Illustrations

n+3 = n+2* n3 = n2*

= (n+2)* = (n2)+n= (n+2)+1 = (n1*)+n

= n+ + = n +n +n= [(n+1)*]+1 = n+n+n= [(n+1)+1]+1

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3. Theorem. (Closure)

N is closed under +.m+n N, Vm,nN

N is closed under .mn N, Vm,nN

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4. Theorems. (Properties of + and on N)

a. Associativity for +

(m+n)+p = m+(n+p), Vm,n,pN

b. Commutativity for +

, ,c. Associativity for

(mn)p = m(np), Vm,n,pNd. Commutativity for

(mn) = (nm), Vm,nN

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4. Theorems. (Properties of + and on N)

e. Distributivity of + over

Vm,n,pN p(m+n) = (pm)+(pn)(m+n)p = (mp)+(np)

. ance at on awsVm,n,pN (p+m =p+n) m=n

(m+p = n+p) m=n(Prove!) (pm =pn) m=n

(mp = np) m=n

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5. Definition. (Less than)

Given m,nN.

mn if pN such that m=n+p.

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Remarks

a. < is neither reflexive norsymmetric.

b. (Prove!) < is transitive.

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6. Law of Trichotomy

V m,nN, exactly one of thefollowing holds:

a.) m = n b.) m < n c.) n < m

(Prove!)

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7. Theorem. (Well-ordering Principle)

A N,A

pA such thatp a , VaA

Every non-empty subset of N

has a least element.

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8. Remark

The PMI and WOP are equivalent.

i.e. PMI implies WOP, and

(Prove!) WOP implies PMI.

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HOMEWORK # 1.

Prove that the Principle ofMathematical Induction follows

Principle.

Deadline: 8 JULY 2009, 4 PMWrite your answers on bond papers.

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WHAT IS AN INTEGER?

An integer is a solution of anequationx+m=n for some m,nN.

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B. THE INTEGERSObjective:

ConstructZ

, the set of integers:

from N in a (more) natural way (than

fundamental properties should beretained;

allow addition and multiplication; and

allow construction of rationals

(going to reals).

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9. Definition. (A relation on NN)

ConsiderN

N={(

m,n

)|m,n

N}

~

(m,n)~(a,b) m+b=n+a

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10. Theorem.

~ is an equivalence relation onN

N.

~ .~ is symmetric.~ is transitive.

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Remark:

An equivalence relation on a setinduces a partition of the set calledas equivalence classes of the set.

In terms of ~ on NN,

[(m,n)]~= {(a,b)|(m,n)~(a,b)}

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11. Definitions. (Integers)

An integeris an equivalence class

induced by ~ on NN.

Z = { [(m,n)]~ }

: set of all equivalence classes

induced by ~ on NN.

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12. Definitions. (+ and on Z)

Z = { [(m,n)]

~

}

ADDITION:

[(m,n)]+[(a,b)] = [(m+a,n+b)]

MULTIPLICATION:[(m,n)][(a,b)] = [(ma+nb,mb+na)]

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Remark:

+ and are WELL-DEFINED

operations on Z.

i.e. The results are independent ofthe choice of elements of theequivalence classes.

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13. A more familiar form of Z.

Case 1. m=n

[(m,m)] = {(x,x)|xN}

=0

Case 2. m>n[(m,n)] = {(x+p,x)|xN} ,pN

Case 3. m

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14. Extensions to Z.+ is associative and commutative

on Z. is associative and commutative

on . is distributive over +.

Cancellation laws for + and

(with restrictions) hold on Z.

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14. Extensions to Z.Given m,nZ.

m0 Z+

nZ is negative if n

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15. Theorem.

The equation m+x=n has a

solution in Z for any m,nN.

Consider m[(m+1,1)]and n[(n+1,1)].

Solution: x=[(m,n)] m-n

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Remark:

The equation m+x=n has a

solution in Z for any m,nZ.

Consider m[(a,b)]

and n[(p,q)], a,b,p,qN.

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Exercise.

1. Show that the solution to

m+x=n is unique for anym,nN (or Z).

2. Examine the form (as an equivalenceclass induced by ~) of the additive

identity, additive inverses andmultiplicative identity in Z.

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WHAT IS A RATIONAL

NUMBER?

A raional is a solution of an equationmx=n for some m,nZ, m0.

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C. THE RATIONALSObjective:

Construct Q, the set of rationals: from Z in a (more) natural way (than

us ng equa on ; fundamental properties should be

retained;

allow addition and multiplication; and

allow irrationals and reals.

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16. Definition.(A relation on

Z

Z-{0})

Consider ZZ-{0}={j/k|j,kZ,k0}

-

j/k r/s js = kr

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17. Theorem.

is an equivalence relation

on ZZ-{0}.

In terms of on ZZ-{0},

j/k = [j/k ] = { r/s|j/k r/s}

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18. Definitions. (Rationals)

A rational numberis an equivalence

class induced by on ZZ-{0}.

Q = { [j/k ] }

: set of all equivalence classes

induced by on ZZ-{0}.

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19. Definitions. (+ and onQ

)

Q = { [j/k ]

}

ADDITION:

j/k + r/s = (js+kr) /ks

MULTIPLICATION:j/k r/s =jr/ks

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Remark:

+ and are WELL-DEFINED

operations on Q.

i.e. The results are independent ofthe choice of elements of theequivalence classes.

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20. Extensions toQ

.

+ is associative and commutative

on Q. is associative and commutative

on . is distributive over +.

Cancellation laws for + and

(with restrictions) hold on Q.

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21. Theorem.

The equation kx=j has a solution

in Q for any k,jQ, k 0.

j/k is THE solution to kx=j.

Ifj/k 0, k/j is its multiplicativeinverse

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22. Theorem.

QIS A FIELD.

Read on the ro erties satisfied

by a field.

Note: Q is the smallest fieldcontainingN.

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23. Definitions. (Positive and negativeQ

)

j/k is positive ifj,kZ+ orj,kZ-.is negative ifjZ+,kZ- or

- +, .

0 is neither positive nor negative.

Notations: Q+ : positive rationalsQ- : negative rationals

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24. Definition. (Less than)

Let u,vQ.

u0 if and only if uQ+.

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25. Theorem.

Qis an ordered field.

i. is a field.

ii. The trichotomy holds.iii. u

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Exercise.

1. Prove: If rQ, then there existsnN such that n>r.

2. Prove: If r,sQ such that r

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D. DEDEKIND CUTS AND REALS

Objectives:

Establish existence of H, the set ofirrationals from Q.

onstruct , t e set o rea s: from Q (and H);

fundamental properties should beretained; and

allow addition and multiplication.

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26. Lemmas.

There is no rational number r

such that r2 = 2.

If rQ, then there exists nNsuch that n>r.

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Notes.

On irrationality of rsuch that r2 = 2.

i. In our construction, ris anequivalence classj/k.

.

on some subset of N.

iii. EVERY AMAT/MATH MAJORSHOULD HAVE A (VAGUE) IDEA

OF ITS PROOF!

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27. Definition.

A ray in Q is a non-empty proper

subset of Q, say U, such that:

i. IfxUand yQ, y>x, then yU.ii. Uhas no first element.

A ray is an unbounded open

interval (to +) of Q on the R-line.

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Types of Ray

Let U = Q U.

Uis of Type 1 if Uhas a largest

e emen ro .Uis of Type 2 if Uhas no largest

element in Q.

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28. Remarks:

A ray Uof Type 1 is of the form:

U={xQ|x>r} for some rQ.

In this case,

U={x

Q|x

r

} wherer

is the

largest element

29 fi i i

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29. Definition. (Dedekinds Completeness)

An ordered fieldFis complete if

each ray UinFis a ray of Type 1.

i.e. ray UF,

tFsuch that U={xF|x>t}

30 Th

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30. Theorem.

QIS NOT COMPLETE.

Existence of a ray of Type 2 in Q.Let V= {xQ|x>0,x2>2 }

a. Show: Vis a ray in Q.b. Show: Vhas no largest element in Q.

R k

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Remarks:

1. A ray in Q of Type 1 is arational ray.

2. A ray in Q of Type 2 is anirrational ray.

31 D fi iti (R l b )

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31. Definition. (Real number)

A real numberis a ray in Q

: t e set o rea num ers

32 D fi iti ( d R)

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32. Definition. (+ and on R)

Let U,VR.

ADDITION

U+V = { rQ |r >u+v for someuU, vV}

MULTIPLICATION

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MULTIPLICATION

If 0U, 0V,UV = { rQ |r >uv for some

uU, vV}If 0U, 0V,

UV = { rQ |r >uy for someuU, yV,y0 }

If 0U, 0V,UV = { rQ |r >xy for some

x

U, y

V}

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Remarks:

1. U+Vand UVare rays in Q.

. + an on are cons stentwith + and on Q.

33 Definitions

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33. Definitions.

A real number Uispositive if

Ucontains some positiverational number.

se, s nega ve.Less than: Let U,VR.

U< V provided VU

34 Theorem

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34. Theorem.

R is an ordered field.

i. R is a field.

ii. The trichotomy holds.iii. U

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35. Lemma.

Let rQ, VR.

rVif and only if r >V.

36 Theorem (Dedekinds Theorem)

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36. Theorem. (Dedekind s Theorem)

R IS COMPLETE.

i.e.Every ray in R is a ray of Type 1.

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Richard Dedekind(1831-1916)

A German mathematicianwho worked in abstract,

theory and the foundations of the realnumbers. He is known for his works

in redefining the irrationals,construction of reals from rationalsand algebraic number fields.

Exercise (EASY!)

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Exercise. (EASY!)

PROVE:

Two real numbers a and b are

equal if and only if for everyreal number >0, |a b|< .

E. EQUIVALENT FORMS OF

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Q

COMPLETENESS OF R

Objectives: Make the com leteness of R

usable!

Derive some other properties of

R from completeness.

People involved

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People involved

Augustin Louis Cauchy (17891857)

Bernhard Bolzano (17811848)Niels Henrik Abel 18021829

Peter Lejeune Dirichlet (1805-1859)KarlWeierstrass (18151897)

Bernhard Riemann (18261866)

37. Definitions. (Bounds)

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37. Definitions. (Bounds)

Let SR, u,lR.

u is an upper boundof S

u , S

l is a lower boundof S l x, xS

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Uis a least upper boundof S Uis an upper bound of S, and

U u, u upper bound of S.

L is agreatest lower boundof S

L is a lower bound of S, andL l, l lower bound of S.

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Notations:

M l.u.b. of S M= sup S

L g.l.b. of S L = inf S

38. Theorem. (LUB Property)

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( p y)

Every non-empty subset of Rwhich has an upper bound has

(GLB) Every non-empty subset

of R which has a lower boundhas a greatest lowerbound.

Remarks:

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The L.U.B. property isconsidered as theAxiom of

Completeness in someconstruction.

Remarks:

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Dedekinds Theorem implies

L.U.B. Property.

, . . .Dedekinds Theorem. (Prove!)

Dedekinds Theorem and L.U.B.Property are equivalent.

39. Lemma.

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IfM=sup S and yy.

i.e. everyx

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p y

If a,bR with a>0, b>0, thenthere is an nN such that na>b.

Alternative

IfxR, then there is an nNsuch that n>x.

41. Corollary.

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Given >0, there exists an nNsuch that 1/n < .

Proof:By Archimedean Property witha= and b=1.

Remark:

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R is a complete

field.

Exercise. Compute, without proofs, the

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GLB and LUB of thefollowing sets:

(a) { n N | n2 < 10 }

(b) { n/(m+n) |m,n N }(c) { n/(2n+1) |n N }

(d) { n/m |m,n N, m+n 10 }

42. Theorem.

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Every open interval (a,b) of Rcontains a rational number.

Also, every open interval (a,b) of

R contains an irrationalnumber. (Prove!)

43. Definition. (Dense)

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A D

R is dense in R providedDI for every intervalI.

Q is dense in R.

H is dense in R.

Remarks.

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Every open interval (a,b) of R

contains infinitely manyrational numbers.

Also, every open interval (a,b) of

R contains infinitely manyirrational numbers.

Remarks:

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Dedekinds Theorem, L.U.B.Pro ert and Archimedean

Property are equivalent.

Exercise.

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1. LetA be bounded below, and

define B={ bR |b is a lowerbound forA }.

= .

2. Assume thatA and B arenonempty, bounded above,

and satisfy BA.Show supB supA.

3. Use the Archimedean Property

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of R to rigorously prove thatinf{ 1/n|n N} = 0.

4. Consider the o en interval

(0,1), and let S={(x, y)|0

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END

UNIT 1