Appendix A
Elementary Real Analysis
Li this appendix, we briefly go through sorne elernentary concepts from analvsis dealing with numbers a'd functions. while most readers will probably be
familiar with this material, it is worth summarizing the basic definitions that
we will be using throughout this text, and thus ensure that everyone is on an
equal footing to begin with. This has the additional advantage in that it also
makes this text virtrra,liy self-contained and all the more useful for self-study.
The reader should feel free to skim this appendix now, and return to certain
sections if and when the need arises.
A.1
S ets
For our purposes, it suffices to assume that the concept of a set is intuitively
clear, as is the notion of the set of integers. In other words, a set is a collection
of objects, each of which is called a point or an element of the set. For
example, the set of integers consists of the nurnbers 0, f 1, +2,. . . and will be
denoted by Z. Furthermore, the set Z+ consisting of the numbers 1,2, .. . will
be called the set of positive integers, while the collection 0, 1,2.... is called
the set of natural numbers (or nonnegative integers). If m and n
f 0 are
integers, then the set of all numbers of the forrn rnf n is called the set of rational
numbers and will be denoted by Q. we shall shortly show that there exist real
numbers not of this form. The most important sets of numbers that we shall be
concenred with are the set IR of real numbers and the set C of complex numbers
(both of these sets will be discussed below).
If S and 7 are sets, then
^9 is said to be a subset of ? if every element of
,9 is also an eleme't of ?, i.e., r € S implies r €T.lf
in addition S f T,the'
s is said to be a proper subset of ?. To denote the fact that s is a subset
of 7, we write S C 7 (or sometimes T I S i' which case ? is said to be a
superset of .9). Note that if S C ? and T C S, then ,S:2.
This fact will be
extremely useful in many proofs. The set containing no elements at all is called
APPENDIXA. ELEMENTARYREAL ANALYSB
the empty set and will be denotedby Z.
Next, considerthe set of all elementswhich are membersof r but not members of ,9. This definesthe set denoted by ? - s and called the complement of
s in 7. (Many authors denotethis set by z\s, but we shall not usethis notation.) In other words,r €T -,s meansthat z € T but:r ( s. rf (as is usually
the case)the set 7 is understoodand s c T, then we write the complementof
5 as S".
Example A.1. Let us prove the useful fact that it A,B € X with Ac C B,
the.njt is true ihat Bc c .4, To slow this, we simply note that r e Bc implies
r e B, which then implies r f A, and hencea € i. This observationis quite
useful in proving many identities involving sets.
Now let st,sz,... be a collectionofsets. (Sucha collectionis calleda
family of sets.) For simplicity we write this collectionas
{sr},,ii € 1. The set
1 is called an index set, and is most frequentlytaken to be the set z+. The
union U;67,siof the collection{s;} is the set of all elementsthat are members
of at least one of the,sr. since the index set is usually understood,we will
simply write this as USr. In other words,we write
USi: {r: r € Si for at leastone e e /}.
This notation will be usedthroughoutthis text, and is to be read as ,,theset of
all r such that a is an element of 51 for at least onei €. 1.,,
Similarly,the intersection nS, of the ,S1is given by
)S i :
{ r: r € 56 for al l i e 1}.
For example,if S,T C X, then S -T :,9o?c whereT. : X -7. Furtherr'ore,
two sets,9r and ,92are said to be disjoint if 51 n Sz: A.
we now use these concepts to prove the extremely useful "De Morgan
Formulas.tt
Theorem A.l. Let {Sa) be a fami,tgof subsetsof somesetT. Then
(i) ust': (n.gi)".
(i,i)nS:: (u,s,)".
Proof. (i) x € Jsic if and only if e is an element of some s;", hence if and only
if lr is not an element of some ,s1, hence if and only if a is not an element of
f1,91,and therefore if and only if ,D€ (n,9,)".
(ii) r e nSr" if and only if r is an element of every ,S;", hence if and only if
r is not an element of any 51, and therefore if and only if r e (Ufi)".
n
while this may seem like a rather technical result, it is in fact directly useful
not only in mathematics, but also in many engineering fields such as digital
electronics where it may be used to simplify various logic circuits.
4.2. MAPPINGS
Finally, if 51 , ,92,. . . , Sn is a collection of sets, we may form the (ordered) set
where each ri € S;. This very useful set is denoted
of all n-tuples (rr,...,rr)
by 51 x . . . x Sn and called the Cartesian product of the Sl.
Example A.2. Ptobably the mpst common example af t:be Cstte{itbptoduct
is the plane R2 : lR,x R. Each point fi € R2 has eaordlnates (r;y) where
o,y € lR. In order to facilitate the generaliaation to R'; we will generally
write = (u,rz) sr ii - (a1,ne). fnis latter nstatiou b lred e@sitely in
more advanced topics such a$ tensor g.nalysis,and 1hffs ls qsuall? ne.confusion
between writing the courponerttsof d as superscripts and thelr beiag intetpreted
as exponents (see Chapter ??)
Exercises
l. Let A,B and C be sets. Prove that
(u) (A - B )n c : (A .c) - (B o c ).
(b) ,4n(B rC): (1n B )r(A o C).
(c) ,4u(B .C): (,4uB ) . (A rC).
(d) (A -B )-c:A -(B uc).
(") A - (B u c) : (A- B) n (A - c).
2. The symmetric
difference
A A B ollwo seLsA and,B is defined by
AAB:(A-B)u(B-A).
Show that
(a),aA B : (AuB) - (A. B) : B L A.
(b) .4n (B
(.4nB) L(Anc).
^C):
(c) ,4uB:(AAB)A(AnB).
(d) _4- B:AA(,4n8).
3. LetR be a nonempty collection of sets with the property thaL A,B €R
are also in R. Show that R must
implies that both AaB
and AAB
contain the empty set, ,4 U B and A - B. (The collection R is called a
ring of sets, and is of fundamentai importance in measure theory and
Lebesgue integration.)
4.2
Mappings
Given two sets ,9 and 7, a mapping or function / from S Lo T is a rule which
assigns a uni,que element g € T to each element r € S. Symbolically, we write
this mapping as / : S + T or f '., * /(r) (this use of the colon should not be
confused with its usage meaning "such that"). The set ,S is called the domain
APPENDIX A. ELE\IENTARY REAL AATAIYS/S
of / and ? is called the range of /. Each point /(r) e T is called the irnage
of z under / ( or t he v a l u e o f f a t r ) , a n d t h e c o l l e c t i o nf ( r ) e T : r
€ 5 ofall
such image points is called the image of /.
In general, whenever a new nrappiug is given, we must check to see that it
is in fact well-defined. This rneans that a given point r € ,9 is mapped into a
uni,quepoint /(z) € T. In otlier words, we must verify that f (r) I /(g) implies
r I A. An equivalent way of saying this is the contrapositive
statement
that z :91 implies f (r) : /(g)
We will use this requirernent several times
throughout the text.
If ,4 C S, the set f (r) : r € ,4 is called the image of ,4 under / and is denoted
bV f @). If / is a mapping frorn S to ? and AC S, then the restriction of ./
to ,4, denoted bV f lA (or sometimes /a), is the function from ,4 to 7 defined by
f lA : r € A+ f ( r ) e T . I f r ' € T ', t h e n a n y e l e m e n t r € S s u c l i t h a t / ( z ) : r /
is called an inverse image of r/ (this is sometirnes also called a preimage of
r'). Note that in general there rnay be more tlian one inverse irnage for any
particular r' € T. Similarlv, if A' C 7, tlien the inverse irnage of ,4/ is the
subset of S given by r: € S: f (r) e ,4/. We will denote the inverse irnage of ,4'
bvf '(A').
Let / be a mapping froni ,9 to 7. Note that everv element of 7 need not
necessarily be the image of sonie element of ,S. However, if /(S) : 7', then /
is said to be onto or surjective. In other words, / is surjective if given any
r ' €T
t her eex is t s r € S s u c h t h a t / ( r )
In additirin, /issaidto
be
- r '.
one-to-one or injective if r I y iniplies that /(r) I f fu) An alternative
cliaracterizalion is to say that / is injective if f (r): /(9) implies that:r:
g.
If / is both injective and surjective, therr / is said to be bijective. In this
case, given any r'€
T there exists a uni,quer €,9 such that r/ : f (r). If
is
bijective,
then
we
may define the inverse mapping f t , T + S in the
/
following way. For any z'€ 7, we let f -t(*') be that (unique) elernent z € S
such that f (r) : r' .
Example A.3. Consider the function /: JR* IRdefined bV f @): 12. This
mapping is cieariy not surjective since /(a) ) 0 for afly r e IR. Furthermore,
q.
it is also not injective. Indeed, it is clear that2l
-2 but f(2):
f (-Z) Note also that both the domain and range of / are the whole set lR, but that
the image of / is just the subset of all nonnegative real numbers (i.e., the set of
allr €lRwit h0> 0) .
On the other hand, it is easy to see that the mapping g : lR --+ IR defined
bv g@): at-f 6 for any o,b € lR (with a l0) is a bijection. In this case the
inverse mapping is simply given by g-t(r'):
(rt - b)la.
Example A.4. If / is a mapping defined on the collections {,4;} and {,B1} of
sets, then we claim that
f (v A r): u f (A r)
A.2. MAPPINGS
and
f-tPnil - uf' l( Bi'
To prove these relationships we proceed in our usual manrier. Thus r/e have
n' e f Uni if and only if rt - f (a) for solne r e ulr;tre:rce if a.nd,only if I
is in sornef (A), and,therefore if and only if tt e uf (A;)'-This provesthe first
if and only if
statement. As to the secondstatement, we have r e t-ttuB)
t@) e UB;, henceif and only if /(r) is in some Bi,henen if and only if r is in
some/-1(Fr), and thereforeif and only if o € U/-1(Ar)'
several similar relationships that will be referred to again are given in the
exercises.
Now considerthe sets ,S,7 and U along with the mappings/ : S + ? and
g :T
U. We definethe composite mapping (sometimesalso called the
p ro d u c t) g o J :S + U b y
b " f) ( ") : sU@n
go f , and we say that the composition of two
for all r €,9. In general, f
"S I
V,
However, if we also have a mapping h : U
functions is not cornmutative.
then for any r € S we have
(ho (so /))(") : h((eo/)(r)) : nbff@))): (ho s)(f (t))
: ((h g) f)(r)
" "
This means that
ho( so f ) : ( h o g ) " f
and hence the composition of mappings is associative.
As a particular case of the composition of mappings, note that if f : S
is a bijection and /(e) : r' €T where r € ,S, then
T
u " f't)@ ' )-- f (f -' (" ' )): J @): r'
and
U-t " il@): f -' (f (" )): f -r@' ): r.
: 12, then the mapping fu
If we write f
" f-t
for every r' e T. We call 17 the identity
-'"
composition mapping f
f : Is is called
then /o"f-1
thepar t ic ular c as et hat S: 7,
identity mapping.
An extremely important result follows by
has the property lhat l7(rt) : rl
mapping on 7. Similarly, the
the identity mapping on S. In
:1is
: j-t
also called the
" f
noting that (even if S + T)
U-' " s-\@. /)(") : (f-,' " s-l)bj@D) : /-'(s-'(g(/(r))))
: f-' (f (" )):
"
APPENDIX A, ELEMENTARY REAL AATAIYS/S
Since it is also easy to see that (5" f)(f -t og-')(r'):
r', we have shown that
b " f )-' : f -ro s -r.
Exercises
1. Let / be a rnappingof sets. For eachof the following,state any conditions
on / that may be required (e.g.,surjectiveor injective),and then prove
the statement:
(a) .4r c .42 implies f (At) c f (Az).
(b) /(,4)" c l(A) is true if and only if f is surjective.
(c ) /(n ,a ;) c .f (A i ).
(d ) Br C 8 2 i ni pl i esf -' (B t) c f -' (B z).
(e ) ;-1 (n r;) : nf -1(B i ).
(f) f-' (8 " ):
f-' (B )' .
A
2. Given a nonemptyset ,4, we definethe identity mapping i4 : A
by ia(a) : a for every o € A. Let f : A +,4 be any mapping.
(a ) Sh o wth a t /o i ,a : i t" f - f.
(b ) If / i s b i j e cti ve(sothat /-1 exi sts),show thatf of-t: f-| of :i ' A .
(c) Let / be a bijection, and supposethat g is any other mapping with
the propertythat 9o f -- f g : i,a. Showthat S : f-|.
"
A.3
Orderings and Equivalence Relations
Given any two sets S and 7, a subset /? of S x 7 is said to be a relation
and (r,A) e R, then it is common to write
between S and 7. If R c SxT
rRg to show that r and g are "R-related." In particular, consider the relation
syrnbolized by < and defined as having the following properties on a set 5:
( u)
"< <
r ( ryef lex iv ity
);
(b)
for allr,y € S (antisymmetry);
andg i r irnplies a:g
"
(.)
< g and g I z implies r <, z for all r,y,z e S (transitivity).
"
Any relation on a non-empty set S having these three properties is said to be
ordered set. We will
a partial ordering, and S is said to be a partially
sometimes write g t r instead of the equivalent notation r < g. The reason for
including the qualifying term "partial" in this definition is shown in our next
example.
Example A.5. Let S be any set, and let P(^9) be the collection of all subsets
of S (this is sometimes called the power set of S). If A, B and C are subsets
of S, then clearly Ac A so that (a) issatisfied; .4 c B and B c A implies
A : B then satisfies (b); and A c B and B c C implies A c C satisfies (c).
A.3. ORDERINGS AA/D EQUIVALENCE RELATIONS
Therefore C defines a partial ordering on P(S), and the subsets of S are said to
be ordered by inclusion. Note however, that if ,4 c ,9 and B c S but A I B
and B (.4, then there is no relation between A and B, and we say that .4 and
B are not comparable
The terminology used in this example is easily generalized as follows. If S is
any partially ordered set and r, g € S , then we say that r and y are comparable
:r.
ifeit her r ?' gor y - 1
If, in addition to properties (u) ("), a relation R also has the property that
any two elements are comparable, tlien R is sa,id to be a total ordering. In
other words, a total ordering also has the property that
(d) either y -1 g or y 4 r for allrx,y e S.
Let S be a set partiallv ordered ly -( and suppose A c S. It shortld be
clear that ,4 niay be considered to be a pa,rtia,llyordered set by defining a -( b
for a,ll n,,b e A if o I b where o and 6 are considered to be elements of ,9.
(This is similar to the restriction of a mapping.) W-e then say that ,4 has a
partial ordering i induced by the ordering on S. lf A is totally ordered by the
ordering induced by -(, then,4 is frequently called a chain in,9.
Let ,4 be a non-empty subset of a partially ordered set S. An element r € S
is called an upper bound for A if a, I r for all a e ,4. If it so happens that
:r is an elernent of ,4, then r is said to be a largest element of ,4. Sirnilarly,
g € S is called a lower bound for,4 if y 1i a for all a € A,and g is a smallest
element of A if U € A. If ,4 has an upper (lower) bound, then we say that A
is bounded above (below). Note that la,rgestand smallest elements need not
be unique.
Suppose that ,4 is bounded above by a € ,9, and in addition, suppose that
for any other upper bound z of ,4 we have rr -( r. Then we say that a is a
supA.
Ieas t upper bound ( or s upr em u m ) o f , 4 , a n d w e w r i t e a : l u b , 4 :
As expected, if ,4 is bounded below by p e ,5, and if A < 0 for all other lower
bounds g € ,S, then p is calied a greatest lower bound (or infimum), and we
write p : g1b,4 : inf ,4. In other words, if it, exists, the least upper (greatest
lower) bound for ,4 is a smallest (largest) elernent of the set of all upper (lower)
bounds for ,4.
From property (b) above and the definitions of inf and sup we see that, if
they exist, the least upper bound and the great,estlower bound are unique. (For
exanrple, if p and p/ are both greatest lower bounds, then p -2.Bt and B' < P
irnplies that /l : p/.) Hence it is meaningful to talk about lfte least upper bound
and the greatest lower bound.
Let S be a partially ordered set, ancl suppose A c S. An element rr € ,4 is
said to be maximal in A if for any element a €,4 with a I 4,, we have a - rr.
In other words, no elenient of,4 other than a itself is greater than or equal to
a. Similarlv, an elernelt {J e A is said to be minimal in ,4 if for any b € '4
with b < p, we have | : 73. Note that a rna,xirnalelernent may not be a largest
APPENDIX A. ELEMENTARY REAL AAIAIYffS
elernent (since two elements of a partially ordered set need not be comparable),
and there may be many maximal elements in ,4.
we now state Zorn's lemma, one of the most fundamental results in set theory, and hence in all of mathematics. While the reader can hardly be expected
to appreciate the significance of this lemma at the present time, it is in fact
extreniely powerful.
Zorn's Lemma. Let S be a parti,allg ordered set in which euery cha'in has an
upper bound. Then S conta'ins a mari'mal element.
It can be shown (see any book on set theory) that Zorn's lemma is logically
equivalent to the axiom of choice, which states tirat given any non-empty
farnily of non-empty disjoint sets, a set can be formed which contains precisely
one eiement taken from each set in the family. Although this seems like a rather
obvious staternent, it is important to realize that either the axiom of choice or
some other statement equivalent to it must be postulated in the formulation of
the theory of sets, and thus Zorn's lemma is not really provable in the usual
sense. In other words, Zorn's lemnra is frequently taken as an axiom of set
theory. However, it is an indispensable part of sorne of what follows although
we shall have little occasion to refer to it directlv.
Up to this point, we have only talked about one type of relation, the partial
ordering. We now consider another equally important relation. Let ,S be any
set. A relation = on S is said to be an equivalence relation if it has the
foliowing properties for all r, A, z € S:
( a) r t . i
f or all r e 5 ( r e H e x i v i t v ) :
( b) r = y im plie s y = r ( s v n r m e t r y ) :
( c ) r = gandy = z
implies rx zfor allr,y,z€,S
(transitivity).
Note that only (b) differs from the defining relations for a partial orderilg.
of a set S is a family {S,} of non-empty subsets of S such
A partition
tlrat uS1 :,9 and,9, aS; I a tmpliesS, : Si. Suppose r e S and let = be
an equivalence relation on S. The subset of S defined bv ["] : {y : y x r} is
called the equivalence class of r. The most important property of equivalence
relations is contained in the following theorem.
Theorem A.2. The fami,ly of all d,i,st'inctequ'iualenceclasses of a set S forms a
parti.tion of S. (Thi,s i,s called the parti'ti,on ind,uced' by x.) Moreouer, gi'uen any
parti,ti,on of S, there i,s an equiualence relat'ion on S that 'induces thi's parti'tion.
Proof. Let = be an equivalence relation on a set ,S, and let z be any element of
S. Since r x r, tt is obvious that r € [r]. Thus each element of S lies in at least
one non-empty equivalence class. we now show that anv two equivalence classes
are either disjoint or are identical. Let [r1] and [22] be two equivalence classes,
A.4. CARDII\IALITY AND THE REAL NUMBER SYSTEM
and let g be a rnember ofboth classes.In other words, 9: zr and g = 12. Now
choose aly z € [r1] so that z x rr. But this means that z x :x7 x A x 12 so
that any element of [r1] is also an element of [r2], and hence [21] c [r2]. Had
we chosen z e lr2l we would have found that [r2] c [rr]' Therefore lrtl:
l"rl,
and we have shown that if two equivalence classes have any element in common'
then they must in fact be identical. Let iS,) be any partition of S. We define an
equivalence relation on S by letting r x g \f r,g € S1 for any r1A € S. It shorrld
be clear that this does indeed satisfy the three conditions for an equivalence
tr
relation, and that this equivalence relation induces the partition {S;}.
As we will see in Appendix B, this theorem has a direct analogue in the
theory of groups.
Exercises
L. Let Z+ denote the set of positive integers. We write rnln to denote the
fact that m divides n, \.e., n: km for some k € Z+.
(a) Show that I defines a partial ordering onZ+.
(b) Does Z+ corttain either a rnaximal or rninimal elernent relative to this
partial ordering?
(c) Prove that any subset of Z+ containing exactly two elements has a
greatest lower bound and a least upper bound.
(d) For each of the following subsets of Z+ , detetmine whether or not it
is a chain fi Z+ , find a maximal and minimal element, an upper and
lower bound, and a least upper bound:
( i) { 1, 2, 4, 6, 8} .
( ii) { 1, 2, 3, 4, 5} .
( iii) { 3, 6, 9, t 2, 15, 18} .
(iv) {4, 8, 16, 32, 64,L28}.
2. Define a relation = on R by requiring that a = b if lol :
this defines an equivalence relation on 1R..
lbl. Show that
3. For any o,b € IR, let a - b mean nb > 0. Does - define an equivalence
relation? What happens if we use ab ) 0 instead of ab > 0?
A.4
Cardinality
and the Real Number
System
We all have an intuitive sense of what it means to sav that two finite sets
have the same number of elements, but our intuition leads us astray when we
come to consider infinite sets. For example, there are as marry perfect squares
{1,4,9,16,...} among the positive integersas there are positive integers. That
this is true can easily be seen by writing each positive integer paired with its
square:
r,
2,
3,
4,
t2
II
q2
q2
t2
10
APPENDIX A, ELEMENTARY REAL AI{AI,YSS
While it seems that the perfect squares are only sparsely placed throughout the
integers, we have in fact constructed a bijection of ail positive integers with all
of the perfect squares of integers, and we are forced to conclude that in this
sense thev both have the t'same number of elements."
In general, two sets ,9 and 7 are said to have the same cardinality, or to
possessthe sarne number of elements, if there exists a bijection from ,9 to ?. A
set S is finite if it has the same cardinality as either Q or the set {1, 2, . . . , n} for
some positive integer n1 otherwise, S is said to be infinite. However, there are
varying degreesof "infinity." A set S is countable if it has the same cardinality
as a subset of the set Z+ of positive integers. If this is not the case, then we say
that S is uncountable. Any infinite set which is numerically equivalent to (i.e.,
lras the same cardinality as) Z+ is said to be countably infinite. We therefore
say that a set is countable if it is countably infinite or if it is non-ernpty and
finite.
It is somewhat surprising (as was first discovered by Cantor) that the set
Q+ of all positive rational numbers is in fact countable. The elements of Q+
can not be listed in order of increasing size because there is no smallest such
nurnber, and between any two ratiorral numbers there are infinitely many others
(see Theorem A.4 below). To show that Q+ is countable, we shall construct a
Df.lecilonlfotn L ' lo v'
To do this, we first consider all positive rationals whose nurnerator and
denominator add up to 2. In this case we have only 1/1 : 1. Next we list those
positive rationals whose numerator and denominator add up to 3. If we agree
to always list our rationals with numerators in increasing order, then we have
Those rationals whose numerator and denominator add up
ll2 and2ll:2.
t o 4 ar e t hen giv en b y 7 1 3 , 2 1 2 : 1 , 3 1 I : 3 . G o i n g o n t o 5 w e o b t a i n 1 / 4 ,
: l, 412: 2,511: 5.
213,312,
4ll : 4. For6 wehave1/5,214: 112,313
Continuing with this procedure, we list together all of our rationals, omitting
any number already listed. This gives us the sequence
r, rf2,2, rf3,3, lf 4,213,312,4,Lf5,5, ...
which contains each positive rational number exactly once, and provides our
desired bijection.
We have constructed several countably infinite sets of real numbers, and it
is natural to wonder whether there are in fact any uncountably infinite sets. It
was another of Cantors discoveriesthat the set lR of all reai numbers is actually
uncountable. To prove this, let us assume that we have listed (in some Inanner
sirnilar to that used for the set Q+) all the real numbers in decimal form. What
we shall do is construct a decimal .dtdztls ' ' . that is not on our list, thus showing
that the list can not be cornplete.
Consider only the portion of the nurnbers on our list to the right of the
decima,l point, and iook at the first number on the list. If the first digit after
the decimal point of the first number is a 1, we let d1 : 2; otherwise we let
& - l. No niatter how we choosethe rernaining d's, our nurnber will be different
from the first on our list. Now look at the second digit after the decimal point
4.4. CARDINALITY
AND THE REAL NUMBER SYSTEM
11
of the second number on our list. Again, if this second digit is a 1, we let
dz : 2; otherwise we let d2 : 1. We have now constructed a number that
differs from the first two numbers on our list. Continuing in this manner, we
construct a decimal .dtdzds... that differs from every other number on our list,
contradicting the assumption that all real numbers can be listed, and proving
that IR is actually uncountabie.
Since it follows from what we showed above that the set Q of all rational
numbers on the real line is countable, and since we just proved that the set IR
is uncountable, it follows that a set of irrational numbers must exist and be
uncountably infinite.
From now on we will assume that the reader understands what is meant by
the real number system, and we proceed to investigate some of its rnost useful
properties. A cornplete axiomatic treatment that justifies what we already know
would take us too far afield, and the interested reader is referred to, e.g., l?].
Let S be any ordered set, and let
c ,9 be non-empty and bounded above.
'4
We say that S has the least upper bound property if sup,4 exists in ,S. In
the special case of S : IR, we have the following extreniely important axiorn.
Archimedean
Axiom,
Euerg non-ernpty set of real numbers whi,ch has an
upper (Iower) bound has a least upper bound (greatest lower bound).
The usefulness of this axiom is demonstrated in the next rather obvious
though irnportant result, sometimes called the Archimedean
property of the
real number line.
Theorem A.3.
that na > b.
Let a, b € IR and supposea ) 0. Then there eri,stsn € Z+ such
Proof . Let,S be the set of all real rmmbers of the form na where n is a positive
integer. If the theorern were false, then b would be an upper bound for ,5. But
by the Archimedean axiom, S has a lea-stupper bound ri: sup,9. Since a > 0,
we have o - a <o and o - a can not be an upper bound of ,S (bv definition of
a). Therefore, there exists an z) € Z+ suchthat ma e ,5 and o- a < mo. But
therr a < (rn* 1)a e S which contradicts the fact that cr: supS.
n
One of the most useful facts about the reai line is that the set Q of all
rational numbers is dense in lR. By this we mean that given any two distinct
real nurnbers, we can always find a rational number between them. This means
that any real number may be approximated to an arbitrary degree of accuracy
by a rational number. It is worth proving this using Theorem A.3.
Theorem 4.4. Suppose r,A e R and assume tltat r I g. Then there ertsts a
rat'ional number p € Q such that r < p < g.
L2
APPEI,IDIXA. ELEMENTARYREAL ANALYSIS
Proof. Since r < A we have gr- r ) 0. In Theorem A.3, choose a : A - r and
b : 1 so there exists n €Z+ such that n(A - r) ) 1 or, alternatively,
Ilnrlng.
Applying Theorem A.3 again, we let a : 1 and both b : nt: and b : -nr to
find integers rrtrr,Ttr2€Z+ sttch that m1 ) nr andfi1,2) -nr.
Rewriting the
second of these as -m2 < nrl we combine the two inequalities to obtain
-mz 1nr
< m7
so that nr lies between two integers. But if nr lies between two integers, it
nrust lie between two consecutive integers m - 1 and m for some m € Z where
-1nz 1 m 4 mt. Thus m - 1 4 nr < rn implies that m I | + nr ard nr 1 m.
We therefore obtain
n r <m 3 7 l n r
or , equiv aier r t ly( s inc en +O ) ,
r <m f n
1nu
<y .
!
Corollary.
Suppose r, g € lR and assume that r < y. Then there eri,st 'integers
m € Z and k ) 0 such that r I ml2k < y.
Prool. Simply note that the proof of Theorem A.4 could be carried through
if we choosean integer /c ) 0 so that 2k(g - z) > 1, and replacen by 2k
I h ro u e h o u l .
n
In addition to the real number system IR we have been discussing, it is
convenient to introduce the extended real number system as follows. To
the real number system IR, we adjoin the symbols *oo and -oo which are
defined to have the property that -oo < r < lx: for all r € lR. This is of
great notationai convenience. We stress however, that neither *oo or -oo are
considered to be elements of IR.
Suppose ,4 is a non-empty set of real numbers. We have already defined
sup,4 in the case where A has an upper bound. If A is non-empty and has no
-cn.
upper bound, then we say that sup,4: *oo, and if A: A, then sup A:
Similarly, if A+ a and has no lower bound, then inf ,4 - -N, andif. A:
a,
t heninf A: * oo.
Suppose o, b € lR with a ( b. Then the closed interval fa, bl from a to b is
the subset of lR defined bv
fa,bl:
Similarly, the open interval
{r e R : a < r < b ) .
(o, b) is defined to be the subset
(a,b): {r: o,1 r < b} .
13
A.4. CARDINALIIIY AND THE REAL NUMBER SYSTEM
we may also define the open-closed and closed-open intervals in the obvious
way. The infinity symbols *oo thus allow us to talk about intervals of the
form (-oo, b],[o,*oo) and (-oo, *oo).
Another property of the sup that will be neededlater on is contained in the
following theorem. By way of notation, we define lR+ to be the set of all real
numbers) 0, and R+ : IR+U {0} to be the set of all real numbers} 0.
Theorem A.5. Let A and,B be non-empty bounded,setsof real numbers, and
define the sets
A+ B :{ x + g :n€A and,A eB I
and
AB :{ n a : n€A andgeB }.
Then
(i,) For all A, B c lR ne haue stp(A + B) : sup,4 + sup B.
(tt) For oll A,B € lR+ ue haaewp(AB) < (tup, )(supB).
P ro o f. (i ) L e t a :
supA, p:
rl g
s uP B , and suppose
€,4+ B . Then
r+ A< a* y< o* 0
so that o+0 is an upper bound for,4+8. Now note that givene > 0, thereexists
r €,4 such that o - €12 < r (or elseo would not be the least upper bound).
Similarly,there existsgr€ B such thaLB - el2 < A. Then o't 13- e < r + 3/so
that o * 0 must be the least upper bound for A+ B.
(ii) If z € A r- lR+
must have r { sup,A, and if y € B r_lR+ we have
g ( supB. Henceny <-e(sup,4)(supB) for all xg € AB, andthereforeA + a
and B I o implies
< (sup,4)(sup
B).
sup(,4-B)
The reader should verify that strict equality holds if ,4 c lR+ and B c IR+. tr
The last topic in our treatment of real numbers that we wish to discussis
the absolutevalue. Note that if r € IRand 12 : a, then we alsohave('r)' : o.
number r such I'hat t2 : a,
We d,efineJa, for o ) 0, to be the uniqueposi'ti'ue
and we call r the square root of a.
Supposer,g ) }and let 12 : a atd 92 : b. Then, : rt, a : Jband we
ob which impliesthat
have (Jat/b)' : @a)' - ,'a':
'/A: JaJb.
For any a € IR, we define its absolute value lal by lal :
followsthat l-ol : lol, and hence
vt:{_i;l:: :
,/o". lt th"n
APPENDIX A. ELEMENTARY REAL ANA-LYSIS
I4
This clearly implies that
a<lal .
In addition, if a,b >0 and o ( b, then we have (t/")' : a 1 b : (r4)2 so that
la < t/b.
The absolute value has two other useful properties. First, we note that
t-=
: l"'b' : vFrta': lollbl
laul-- I1ab12
Second,we seethat
l a+ bl 2:(a+ b)2
: o2+ b2 + 2ab
< lo l' + lb l2
+ 2 la b l
: 1al2
+ z lollbl
+ lbl'?
: (lo l+lb l)' ?
and therefore
lo+bl <lal+lbl.
Using these results, many other useful relationships may be obtained. For
example,
l"l : l"+b-b l < la + b l+ | -b l : lo + b l* lb ls ot h a t
lo l-la l 5 lo + b l.
Others are to be found in the exercises.
ExampleA.6, tetusshowthat if e > 0,therrfrl < e if andoaly il-e 1r 1e'
Indeed, we seethat il r > O, then lol = n I €t a*d if t < 0, theu l*l * -* < e
which impliee -:e 1* 10 (we again use the fact that a 4 6 irrulig :b < '")'
combining these reeults shows that l*l < e implies -E < & 4. e. we leave it to
the readeito teverse the argument and cornplete the proof.
A particular caseof this iesult that will be of use later or comesfrom letting
:
o-b. We then seethat lo - bl < e if and only if -e < a-b < e' Rearranging,
,
this may be written in the form b- e <a <b+ e. The readershould draw a
picture of this relationship.
Exercises
1. Prove that if A and B are countablesets,then A x B is countable.
2. (a) A real number r is said to be algebraic (over the rationals) if it
satisfiessome polynomial equation of positive degreewith rational
coefficients:
rn I an-trn-t + ' " + a1r ! ag: Q.
Given the fact (which we will prove in Chapter C) that each polynomial equation has only finitely many roots, show that the set of all
aleebraicnumbers is countable.
4,5. INDUCTION
15
(b) We say that a real number r is transcendental if it is not algebraic
(the most common transcendentalnumbersare n and e). Using the
fact that the reals are uncountable,show that the set of all transcendental numbers is also uncountable.
3. If a.b > 0. show that t/ab < (a +b)12.
4. For any o, b € IR,show that:
(u) llol-lallsla+bl .
(u) Ilol-lbll<lo-bl
tr
A.5
(a) If ,4 C IR is nonempty and bounded below, show sup(-, ) : - inf,4.
(b) If ,4 c IR is nonempty and bounded above, show inf(-,4) : -sup,4.
Induction
Another important concept in the theory of sets is called "well-ordering." In
particular, we say that a totally ordered set ,S is well-ordered
if euery nonempty subset A of S has a smallest element. For example, consider the set S
of all rational numbers in the interval [0, 1]. It is clear that 0 is the smallest
element of ,S, but the subset of ,9 consisting of all rational numbers ) 0 has no
smallest element (this is a consequenceof Theorem A.4).
For our purposes, it is an (apparently obvious) axiom that every non-empty
set of natural numbers has a smallest element. In other words, the natural
numbers are well-ordered. The usefulness of this axiom is that it allows us to
prove an important property called induction.
Theorem A.6. Assume that for all n € Z+ we are gi,uen an assertion A(n),
and assume it can be shown that:
(i,) A(I) is true;
(1,1,)
If A(n) is true, then A(n * l) i,s true.
Then A(n) is true for all n e Z+ .
Proof. lf we let S be that subset of Z+ for which ,4(n) is not true, then we
must show that,9 : A. According to our well-ordering axiom, if S I A then
,S contains a least element which we denote by N. By assumption (1), we must
have I{ I 1 and hence -|y' > 1. Since N is a least element, N - 1 ( S so that
A(N - 1) must be true. But then (ii) implies that ,4(N) must be true which
contradicts the definition of N.
n
APPENDIX A. ELEMENTARY REAL ANAI,YffS
16
Example A.7. Let n > 0 be an integer. We define n-factorial, written n!, to
be the number
nl : n(n - l X n - 2)... (2X 1)
with 0! defined to be 1. The binomial
/n\
\*l
coefficient
([) is defined by
n!
:
ki6-El
where n and k are nonnegativeintegers. We leaveit to the reader (seeExercise
A.6.1) to showthat
(" \:(
" \
\k/
\" - kl
and
(
" \
/ " \ _ / n + r\
k )
\ r - r l* \ r / : \
What we wish to prove is the binomial theorem:
n
( r + d": t
/_\
{ ' ,' l*nr "- *.
7-\K/
We proceedby induction as follows.
For n: l, we have
/1\
^
(o)"''
/1\
+ (t,)"'u'- : @+ a)l
so the assertionis true for n = 1. We now assumethe theorem holds for n, and
proceedto show that it also holds for n.* 1. We have
f n
/ \
'l
@+ u)"+': (n+ u)@+ s)": (t + s)
- l): (l\**r"-r1
L?:o\k/
J
n
/- \
n
( *)
/^\
t ( i' l*x+rrn-n+ t {',ol*xo"-r'+t
E6\tc/
ffi\K/
By relabelling the summation index, we seethat for any function / with domain
e q u a lto { 0 ,1 ,...,n } w e have
n
n+L
rk=i0t r l : /( 0+) /( r )+ "' + !( n ) :fh=7 r tt - tl.
We use this fact in the firet sum in (*), and separa.teout the ft - 0 term in the
secondto obtain
nl T t\n/\
(* + y)n*' : f, ( r? .,\*or"- o*t + an+7
ni
?
\k - 1/
k_l
+1an-k+l.
f?)
\ k / "-
L7
4.6. COMPLEX NUMBERS
We now separate out the k = n * I term from the first sum in this expreesion
and group terms to find
(, * v)"+t :
ant,+ un+,.
. (J)'1,.r,i;n-h+t
E [(,.: ,)
: vn*7
+ yn*,.
E f It),00,*,-o
-I ("l,t)*o'"*'*r
a,swas to be shown.
,4..6 Complex Numbers
At this time we wish to formally define the complex number system C, although most readers should already be familiar with its basic properties. The
motivation for the introduction of such numbers comes from the desire to soive
equations such as 12 + | :0 which leads to the square root of a negative number. We may proceed by manipulating square roots of negative numbers as if
they were square roots of positive numbers. However, a consequenceof this is
-I, while on the other hand
that on the one hand, (/lT)'-
(=)'
: ,l-vf--v: vGTF
: t/+t: t.
In order to avoid paradoxical manipulations of this type, the symbol i was
introduced by Euler (in 1779) with the defining property that i2: -1. Then,
: i1fi. Using this notation, a complex number z € C
if a > 0, we have J4
r e lR is called the real part of z
is anumber of the form z: rli.gwhere
(written Rez), and g e IR is called the imaginary part of z (written Imz).
Two complex numbers r -t i,g and u I i,u are said to be equal if r : u and
g : o. Algebraic operations in C are defined as follows:
addition:
s ubt r ac t ion:
multiplicationt
division:
(r + iy) + (u f i,u) : (r + z) + i.(g + a).
-o).
( r + ig) - ( u +i , u ) :
( r - u ) +i ( g
(r + ig)(u -l iu) :
(ru - gu) -l i.(ru * yu).
(r+i,a) _@+ia)@-iu)
(u-t i,u) (u * i.a)(u- iu)
_ (ru * yu)* ilAu - ur)
u 2 +u 2
It should be clear that the results for multiplication and division may be obtained by formally multiplying out the terms and using the fact that i2 : -1.
18
APPENDIX A, ELEMENTARY REAL ANAIYSIS
The complex conjugate z* (or sometimesZ) of a complex number z :
x * ig is definedto be the complexnumber z* : fr - ig. Note that if z,w € C
we have
\z+ w )
\* : z* -tw
, *
\zw ):zu
z-l z* :2P tez
z - z* :2i ,Imz
The absolute value (or modulus) lzl of a complexnumber z : r I iA is
defined to be the real number
ltl - J"4
at : Qz*)r/2.
By analogy to the similar result for real numbers,if z,w € C then (using the
f ac t l, hat z : x + iy i m p l i e s R e z : ,
+ A' : lrl)
< G
lz + wl2-- ("-r w)(z-tw).
:22
:
<
:
:
tZU
+Z
U +WW
+ lwl2
lzl2+ 2Re(zw*)
lrl' + 2 lz w"+l lwl2
+ 2 lz llwl+ lwl2
1 z l2
(lzl+ knl)2
and hence taking the square root of both sides yields
lz+ wl < lz l+ lu l.
Let the sum 21 + . .' + z, be denoted by DLr
known as Schwartz's inequality.
Theorem A.7 (Schwartzts Inequality).
plea numbers. Then
z.i. The following theorem is
Let a1, . . . ,an andb1,. . . ,bn be com-
r')
l7"u,l=
LEr.,r')(Dr,,
B:Lilbil"
Proof. Write (suppressing
the limits on the sum) ,4 :Diloil2,
and C : D1a1b;. If B : 0, then bi : 0 for allT : 1,...,n and thereis nothing
19
A,6. COMPLEXNUMBERS
to prove,so we aasumethat B 10. We then have
o =T l B a i -cb 6 l 2
,^
: s1
l@"i
- cbr)@ai- c*bt)
7
: B"Llaal2- BC*D"na;- BCDaibt+lcl2f
laul'
-- 82A - B lcl' - B lc lt + lc f a
: B (A B- lcl,).
But B > 0 so that AB -Pf
) 0 and hencelCl2 < AB.
!
It is worth our going through some additional elementary properties of complex numbers that will be needed on occasion throughout this text. Purely for
the sake of logical consistency, let us first prove some basic trigonometric relationships. our starting point will be the so-called "law of cosines" which
states that c2 : a2 +b2 -2abcosd (see the figure below).
A special case of this occurs when d : rl2, in which ca,sewe obtain the
theorem a2 + b2 : c2. (While most readers should
famous Pythagorean
already be familiar with these results, we prove them in Section ??.)
Now consider a trianele inscribed in a uni't circle as shown below:
APPENDIX A, ELEMENTARY REAL ANALYflS
20
The point P has coordinates(rp,gp) : (coso,sino), and Q has coordinates
tri@e,Ae): (cosp,sinp). Applying the Pythagoreantheorem to the right
un!t" *ittr hypotenusedefined by the points P and Q (and noting rfi + a'ze:
r2p* A2p: 1), we seethat the squareof the distancebetweenthe points P and
Q is given by
(PQ)" : (*q -
-r (ue - a")'
"p)'
: @b + a2d+ @2p-ra2p)
- 2(rprq-r apye)
: 2 - 2(cosacosp* si nasi nP ).
on the other hand, we can apply the law of cosinesto obtain the distance PQ,
p)' Equatingtheseexpressions
in which casewe find that (PQ)' :2-2cos(ayields the o*tt t"tttlos(o
_ 0) : coso cos,6* sin a sinp.
Replacing 0 bv -0 we obtain
cos(o* 0) : coto" osP - si nasi nB .
If we let a : r l2,then we have cos(trf2 - B) : sinB, and if we now replace
B by r 12 - 13,wefind that cosB : sin(tr12- B). Finally, we can use theselast
results to obtain formulas for sin(a + p). In particular, we replacep by a + 13
to obtain
sin(o * 0) : cos(tr12 - (a + l3))
:cos(trl 2-o-B )
: cos(nl2- o)cosp -t sin(tr12- o)sinB
: si nocos0* cosasi nB
Again, replacing B by -P yields
si n(o- 0) : si nocosP- cosasi nB .
(The reader may already know that theseresults are simple to derive using the
Euler formula exp(*id) : cosd f isind which follows from the Taylor series
expansionsof expr, sina and cosr, along with the definitioni'2 : -I.)
It is often of great use to think of a complex number z : r I ig as a point
in the rgr-plane.If we define
, : lr l: \ F 4
and
tan0: a/r
then a complex number may also be written in the form
z : r * i a : r(cos0* i si nd) : rexp(i d)
(seethe figure below).
21
A.6, COMPLEX NUMBERS
Given two complex numbers
zt : rt(cos0t * esin d1)
and
z z :rz(cos9zti si nd2)
we can use the trigonometric addition formulas derived aboveto show that
z122: rp2lcos(01-l 0z)+ i sin(d1+ 0z)1.
In fact, by induction it should be clear that this can be generalizedto (see
ExerciseA.6.5)
z \2 2 " ' z n : r7 r2 " ' r' [c o s (d 1-f 0z * " ' + 0)*
In the particular casewhere z1
i si n(d1-f 0z* " ' + 0" )1.
z', we find that
z ' :rn ( cosn?+ i si nn9).
This is often calledDe Moivrets theorem.
One of the main uses of this theorem is as follows. Let tu be a complex
number, and let z : lrn (where rz is a positive integer). We say that tr.ris
annth root of z, and we write this as w: z7/'. Next, we observefrom De
M o i v re ' sth e o re mth a t w ri ti n g z: r(cose* zsi nd) and tu: s(cos/* i si n@ )
yields (assumingthat z I 0)
r(cos0 * z sind) : sn(cosnd t i'sinnS).
But cosd: cos(d+2ktr) for k : 0, +1, +2,. . ., and thereforer : En and n$ :
0 *2kr. (This followsfrom the fact that lf z1 : rr * iAr: rr(cosdr* isindl)
and z2 : rz I iAz : rz(cos?z'l i,sin?z), then z1 : zz implies rt : tz and
Ar : A2 so that 17 : rzt and hencefu : 02.) Then s is the real positive nth
Sincethis expressionfor / is the sameif any
root of r, and $:0ln*2ktrln.
two integersk differ by a multiple of n, we seethat there are preciselyn distinct
solutionsof z: wn (when z * 0), and theseare given by
. :,yt/"lcos(?1_2kr)lnf
w h e re k :0 . 1 .....n -1 .
isin(d + 2kr)lnl
APPENDIX A. ELEMENTARY REAL AJVA,LYSIS
Exercises
1. (a) Show
(" \-( :
\
"
\"- n)
\ti
(b) Show
/n+1\
/"\
/
"
\ r ):\r)*\*-'l
2 . Pro v eb y i n d u cti onthe formul aI+ 2+
\
" ' l n:n(n+ I)12.
3. Prove by induction the formula
1* a*
:=
t' + -..1:t' -1
l -r
wherez is any real number f 1.
4. Prove by induction that for any complex numbers z7t . . . t zn we have:
(a)
l " t" r " ' z" l :
l zl l l rzl ' ' ' l r" l .
(b)
ln
I
n
l\- r,"l|- < \lz,l
"' .
/-/'
l Z-.
I i.:7
|
i:l
(c)
ttul n\1/z n
f t,'t.
+
t "' t < {\Eilt ' , , t /' ) <-,:,
\/nt =,
lHint: For the first half of this inequalityyou will needto showthat
2lzl lz2l< lrrlt + lrrl'.l
5. Prove by induction that for any complex numbers 27,. . . ,2, we have
ZIZ2'
' ' Zn
:
r Lr 2 " 'r , [ c o s ( d 1 * 0 z '1 "
' a 0") * isin(d1 * 0z-t ' " + 0")]
where zi : ri exq(iqi).
A.7
Additional Properties of the Integers
The material of this section will be generalized in Sections C.1 and C.2 to the
theory of polynomials. However, it will also be directly useful to us in our
discussion of finite fields in Section C.6. Most of this section should be familiar
to the reader from very elementary courses.
4.7. ADDTTTONALPROPtrRTIESOF THE IATTECERS
Our first topic is the division of an arbitrary integer a € Z by a positive
in t eger b e Z+ . For ex am ple, we c a n d i v i d e 1 1 b y 4 t o o b t a i n 1 1 : 2 . 4 +3 .
As another example, 7 divided by 2 yields -7 - -4 .2 + I. Note that each
of these examples may be written in the form a : eb * r where q € Z and
0 ( r < b. The number q is called the quotient, and the nurnber r is called the
remainder in the division of o by b. In the particular case that r : 0, we say
that b divides a and we write this as b a. lf r f 0, then b does not divide o,
and this is written as b ! a. If an integer p €'L+ is not divisible by any positive
integer other than 1 and p itself, then p is said to be prime.
It is probably worth pointing out the elementary fact that if a I b and a c,
therr a | (m,b+ nc) for any m,n eZ. This is becarisea I b implies $: q1a, alnd
a I c inrplies c : eza. Thrs mb I nc - (rrqt + nq2)a so that n, | (mb + nc,).
Theorem A.8 (Division Algorithm).
If a e Z and b e Z+, then there enist
un'iquei,ntegersq andr such that q:qb*r
where 0 < r ( b.
P r oof . Def ine S:
0: n € Z }. l n o t h e r w o r c l s , S c o n s i s t s o f a l l
{ , - nb}
norinegative integers of the form a -bn. It is easy to see that S + g. Indeed,
if a > 0 we simply choose n : 0 so that o € S, and if a < 0 we choose rl : o so
tl r at a- ba: a( I
b) eS ( s inc e a <0 a n d 1 - b <0 ) .
SinceSisanonernpty
subset of the natural numbers, we may apply the well-orderirig property of the
natural nurnbers to conclude that S contains a least elernent r ) 0. If we let q
be the valrre of n corresponding to this r, then we have o,- qb : v sv o : qb * r
where 0 ( r. We must show that r ( b. To see this, suppose that r ) b. Then
o - ( q + l) b : a - s b - b :
r - b) 0
so that a - (s + l)b e S. But b > 0 so that
whichcontra.,..,,;:::J,::
,:
.^,i',:
:",:,";
s Hence
r<b
To prove uniqueness, we suppose that we
-: may write a - qtb * 11 and 6 :
qzblrz where 0 ( 11 { b and 0 <r'2 < b. Equating these two formulas yields
qtb * r r : q2b-l 12 or (q1 - q2)b : r'2 - r r t and tlierefore b | (r, - 11). Using the
fac t t hat 0 ( 11 ( b and 0 112 1b , w e s e e t h a t 1 2 - r t 1 b - r y { 6 . S i m i l a r l y
we hav e 11- 12 1b
12 t hor - b l r z - r t .
This rneansthat -b 112-11 1b.
Therefore 12 - rr is a rnultiple of b that lies strictly between -b and b, and
tlr us wem us t hav e 12- r r : 0.
Thel (q-qz)b:0withbf
0, and hence
: rz and qr: q2 which completes the proof
0
also.
This
shows
that
11
et Qzof uniqueness.
!
Suppose we are given two integers a,b € Z where we assulne that o and b
are rrot both zero. We say that an integer d e Z+ is the greatest common
divisor of o and b it d I a and d I b, and if c is any other integer with the
p r oper t y t hat c la and c lb, t henc l d . We d e n o t e t h e g r e a t e s t c o m l n o n d i v i s o r
APPENDIXA. ELEMENTARYREAL ANALYSIS
24
of o and b by gcd{a, b}. Our next theorem showsthat the gcd alwaysexists and
is unique. Fhrthermore, the method of proof showsus how to actually compute
the gcd.
Theorem A.9 (Euclidean Algorithm). /Jra,b € z are not both zero, then
there esi,stsa unique positi,aeinteger d e Z+ such that
(i ,)d ,l a a n d d l b .
(i t) tf c e % i s s u chthat cl a and'cl b, thencl d.
Proof. First assumeb > 0. Applying the division algorithm, there exist unique
integersQ1and **r'::
"t
with 0 < rt 1b.
1br rt
If 11 :0, then b I o and we may take d: b to satisfyboth parts of the theorem.
If 11 10, then we apply the division algorithm again to b and 11, obtaining
b: qzrr I rz
w i th 0 < 12 111'
continuing this procedure,we obtain a sequenceof nonzeroremainders11, 12,
...,1 6 w h e re
c r:q ftl 11
b : qzrt -f rz
1 1 :q sr2+ rs
with0(11
(b
w i t h 0 <1 2
111
w i t h 0 <1 3
<- 1 2
(*)
:
r k_ 2 :
r k- r
:
qkrh-7 -f Th
w i t h 0 <r p 1 r p - 1
qk+7rk
That this process must terminate with a zero remainder as shown is due to the
fact that each remainder is a nonnegative integer with 11 ) 12 ) "' . We have
denoted the last nonzero remainder by r;r.
We now claim that d: r*. Since 16-1 : ek+trkt we have r;, I 16-1. Then,
using r7r-2 : qkrk-t * 16 along with rr I 16 and rx I rx-t, we have rn I rn-z'
Continuing in this manner, we seethat rt l rx-r, r* |,x-2,. . ., r k I rr,16 | b and
16 | a. This shows that rk is a common divisor of a and b. To show that 16 is in
fact the greatest common divisor, we first note that if c I o and c I b' then c I 11
because rt : a - q1b. But now we see in the same way that c I 12, a,r'd working
our way through the above set of equations we eventually arrive at c I 16. Thus
rp is a gcd as claimed.
If b < 0, we repeat the above process with o and -b rather than a and b'
Since b and -b have the same divisors, it follows that a gcd of {o, -b} will be a
gcd of {4, b}. (Note we have not yet shown the uniqueness of the gcd.) If b : 0,
then we can simply let d': lol to satisfy both statements in the theorem.
A.7. ADDITIOI,{AL PROPERTIES OF THE INTEGERS
25
As to uniquenessof the gcd, supposewe have integersdr and dz that satisfy
both statementsof the theorem. Then applying statement(ii) to both d1 and
d2, we must have dt I dz and dz I dr. But both dr and d2 are positive, and
tr
hencedy : i'2.
Corollary. If d : gcd{a,b} wherea and b are not both zero, then d,= am * bn
for somen'Lrne Z.
Proof. Referring to equations (*) in the proof of Theorem A.9, we note that the
equation for rp - 2 may be solvedfor r;, to obtain rk: rk-2 - r6-1q7r.Next,
the equationrk-s: qk-rrk-2-lrx-t may be solvedfor rt-1, and this is then
substitutedinto the previousequationto obtain rk: rk-2(l+qn-flil-rk-sqk.
Working our way up equations(*), we next eliminaterk-2 to obtain r7,in terms
of rp-s and r1-4. Continuing in this manner,we eventuallyobtain 16 in terms
n
ofb a n d a .
If a,b € Z and gcd{a,b} : t, then we say that o and b are relatively prime
(or sometimescoprime). The last result on integersthat we wish to prove is
t h e re s u l tth a t i f p i s p ri m ea n d p l ab (w herea,b e Z), thenei therpl aor pl b.
Theorem A.LO. (i) Supposea,b,c € Z wherea lbc and,a and b are relati,uelg
pri me . T h e n a l c .
(1 ,1If.)p i ,sp ri ,rn e
anda
, 1 ,...,a n €Z w i th pl at...an, thenpl aafor some
i: l r. . .,n .
Proof. (i) By the corollary to TheoremA.9 we havegcd{a,b} : 1 : am I bn
for some m,n € Z. Multiplying this equationby c we obtain c : amc * bnc.
But a I bc by hypothesisso clearly a I bnc. Sinceit is also obvious that a I amc,
we seethat a I c.
(ii) We proceedby inductionon rz,the casen : 1 being trivial. We therefore
a s s u m eth a t n > 1 a n d p I a t..-an. l f p I ar...en-rt then p I o1 for some
'i : 1,. . . ,n_I by our inductionhypothesis.On the otherhand,if p l ay " 'an-1
1 sincep is prime. We then apply part (i) with
then gcd{p,ar,...,(t,-r}:
!
a:p ,b a 7 .' .o 2 -1 a n d c --a n t o concl udel hatpl a" .
Exercises
1. Find the gcd of the followingsetsof integers:
(a ) { 6 ,1 a } .
(b ) { -7 5 ,1 0 5 } .
(c ) { 1 4 ,2 1 ,3 5 } .
APPEI,{DIX A. ELEMENTARY REAL ANALYflS
26
2 . Find the gcd of each set and write it in terms of the given integers:
(a ) { 1 0 0 1 ,3 3 }.
(b ) { -e o ,1 3 8 6} .
(c ) { -2 8 6 0 , -2 310}.
3. Supposepis prime andp! o wherea e Z. Provethat a andp are relatively
prime.
A
Prove that if gcd{o,b} : 1 and c mod a, then gcd{b,c} : 1.
5 . lf a,b e Z+, then m € Z+ is calledthe least common multiple (abbrev i a te dl c m ) i f a l matdbl m, andi f c € Z i s suchthat o I c and b I c, then
a: p1" t ..' pk" k and b: pTtt" ' pkto w herep1,...,P k are
rn I c . Su p p o se
distinct primes and each 81and tt are ) 0.
(a ) P ro v eth a t a I b i f and onl y i f si 1 ti for al l i :1,...,k.
(b) For eachi : I,. . . ,k let ui : min{s;' li} and ui : max{si,t';}. Prove
that gcd{o,b) : ptu' '''pkun and lcm{g,b} : pto' ' ' 'pk'n
Prove the F\rndamental Theorem of Arithmetic: Every integer ) 1
can be written as a unique (except for order) product of primes. Here is
an outline of the proof:
(a) Let
S:
{ a e Z :o >
l and a cannot bew ri tten as aproduct of pri mes.}
(In particular, note that S contains no primes.) Show that S : a
by assumingthe contrary and using the well-orderedproperty of the
natural numbers.
(b) To prove uniqueness,assumethat n ) 1 is an integer that has two
as n: Pt'''Ps: Qt'" g, whereall the pi and
differentexpansions
qj are prime. Show that p1 | qi for somej - 1,...,1 and hence
that p1 : q7. Thus p1 and qj aaube canceledfrom each side of the
equation. Continue this argument to cancel one p1 with one {7, and
then finally concludingthat s: L