MATH 131A Lecture 3: Math 131A Lecture Notation 3
ten Cann lbe up
pbfor 1h Imean
it exists that so aen aforsome new Then we can
get that so until aAnd we know that Athos is in
Sthat means so is not an upper bound forS
then we have researched acontradiction Our assumption
that aoboandthen n.ae bfail Ike
4T Fuat
Proof By contradiction kik'SBo I
Assume exists absuch that na ebfor all
ne NThen Snal ntIN is bounded above
by bSo let so supCS 1By Axiom
Since a0then Sota So t.hu So Al So
So exist no aESsuch that so aCho aSo
By Definition 2Then So no at aLnotDaCS
and so is not supremum
Cmtradiction Eh
week llo ITuesday
Achimedean Property
abso exists ncIN suchthat na b
corollaryit'kidu
Given E0exists ne Nsuchthat th LE
From Archimedeanproperty nabask
proof let aEso bIso thenthere exists na b
Iroof eleibhhesnb
by Archimedean which means that ask where ae
btwhich means that Etn
Theorem 4.7 Denseness of QTYKEksttfk.hn5n
IfabHR and acb thenthereexists arational
number rEQsuch that aercb
TAMEAdt
BIKER Anxieties theAIKA Khl
Proof QAn mnekno
we need to prove that aema eb
aema ebmac man b
Sine we know that acb then means ba0
By Archimedean proporty aobothen na b
let ababIthen we have that nib as 1
Question Thus nb na si Thus it is fairly evident that
there is an integer between na and nb
Proofby professor
since baoso by Archimedean exists nsuchthat
nIbal Iso nbsnat 1
ME2
IrlI7
nanatt nb
Let mmin je Zjsna
Question Thus me In anatt
