7. fill the hole distance, length, and the axioms of continuity

Transcription

1 7. fill the hole distace, legth, ad the axioms of cotiuity

2 78 lesso 7 hilbert s geometry starts with icidece, cogruece, ad order. it is a sythetic geometry i the sese that it is ot cetrally built upo measuremet. owadays, it is more commo to take a metrical approach to geometry, ad to establish your geometry based upo a measuremet. i the metrical approach, you begi by defiig a distace fuctio a fuctio d which assigs to each pairs of poits a real umber ad satisfies the followig requiremets (i) d(p,q), with d(p,q)= if ad oly if P = Q, (ii) d(p,q)=d(q,p), ad (iii) d(p,r) d(p,q)+d(q,r). Oce the distace fuctio has bee chose, the legth of a segmet is defied to be the distace betwee its edpoits. I will follow the covetio of usig the absolute value sig to otate the legth of a segmet, so PQ = d(p,q). The cogruece is defied by sayig that two segmets are cogruet if they have the same legth. Icidece ad order also ca be defied i terms of d: poits P, Q, ad R are all o the same lie, ad Q is betwee P ad R whe the iequality i (iii) is a equality. You see, sythetic geometry takes a back seat to aalytic geometry, ad the sythetic otios of icidece, order, ad cogruece, are defied aalytically. I do ot have a problem with that approach it is the oe that we are goig to take i the developmet of hyperbolic geometry much later o. We have bee developig a sythetic geometry, though, ad so what I would like to do i this lesso is to build distace out of icidece, order, ad cogruece. This is what Hilbert did whe he developed the real umber lie ad its properties iside of the framework of his axiomatic system. Modest expectatios Here we stad with icidece, order, cogruece, the axioms describig them, ad at this poit eve a few theorems. Before we get out of this sectio, I will throw i the last two axioms of eutral geometry, the axioms of cotiuity, too. From all of this, we wat to build a distace fuctio d. Look, we have all dealt with distace before i oe way or aother, ad we wat our distace fuctio to meet coditios (i) (iii) above, so it is fair to have certai expectatios for d. i do t thik it is ureasoable to expect all of the followig.

3 distace legth ad cotiuity 79 () The distace betwee ay two distict poits should be a positive real umber ad the distace from a poit to itself should be zero. that way, d will satisfy coditio (i) above. (2) Cogruet segmets should have the same legth. That takes care of coditio (ii) above, sice AB BA, but it does a whole lot more too. you see, let s pick out some ray r ad label its edpoit O. accordig to the Segmet Costructio Axiom, for ay segmet AB, there is a uique poit P o r so that AB OP. If cogruet segmets are to have the same legth, the that meas AB = d(o,p). Therefore, if we ca just work out the distace from O to the other poits o r, the all other distaces will follow. (3) If A B C, the AB + BC = AC. This is just a part of property (iii) of a distace fuctio. Sice we are goig to develop the distace fuctio o r, we do t have to worry about o-coliear poits just yet (that will come a little later). Relatig back to your work i the last sectio, sice d ever assigs egative values, this meas that AB CD = AB < CD, AB CD = AB > CD. It is up to us to build a distace fuctio that meets all three of these requiremets. The rest of this chapter is devoted to doig just that. The additivity coditio for d. a b a+b

4 8 lesso 7 Divide ad combie: the dyadic poits With those coditios i mid, let s start buildig the distace fuctio d. the picture that i like to keep i my mid as i m doig this is that simple distace measurig device: the good old-fashioed ruler. Not a metric ruler mid you, but a eglish ruler with iches o it. here is oe way that you ca classify the markigs o the ruler. You have the mark. That distace is halved, ad halved, ad halved agai to get the /2,/4, ad /8 marks. Depedig upo the precisio of the ruler, there may be /6 or /32 markigs as well. all the other marks o the ruler are multiples of these. Well, that ruler is the blueprit for how we are goig to build the skeleto of d. First of all, because of coditio (), d(o,o)=. ow take a step alog r to aother poit. ay poit is fie like the ich mark o the ruler, it sets the uit of measuremet. Call this poit P ad defie d(o,p )=. Now, as with the ruler, we wat to repeatedly halve OP. That requires a little theory. def: midpoit a poit M o a segmet AB is the midpoit of AB if AM MB. thm: existece, uiqueess of midpoits Every segmet has a uique midpoit. Proof. Existece. Give the segmet AB, choose a poit P which is ot o AB. accordig to the agle ad segmet costructio Axioms, there is a poit Q o the opposite side of AB from P so that ABP BAQ (that s the agle costructio part) ad so that BP AQ (that s the segmet costructio part). sice P ad Q are o opposite sides of AB, the segmet PQ itersects it. Call that poit of itersectio M. i claim that M is the midpoit of AB. Why? Well, compare MBP ad MAQ. P A M B Q

5 distace legth ad cotiuity 8 B i those triagles AMQ BMP MAQ MBP BP AQ (vertical agles) (by costructio) (by costructio) P A M Q so, by a a s, they must be cogruet triagles. that meas that AM MB. it is worth otig that the midpoit of AB has to be betwee A ad B. If it were t, oe of two thigs would have to happe: M A B = MA MB, or A B M = MA MB, ad either way, the segmets MA ad MB could t be cogruet. There are may choices for P, but they each lead to the same midpoit because a segmet ca have oly oe midpoit. Uiqueess. suppose that a segmet AB actually had two midpoits. let s call them M ad M 2, ad just for the sake of coveiece, let s say that they are labeled so that they are ordered as A M M 2 B. P B sice A M M 2, AM AM 2. sice M M 2 B, BM 2 BM. But ow M 2 is a midpoit, so AM 2 BM 2. let s put that together A M Q AM AM 2 BM 2 BM. I the last sectio you proved that is trasitive. This would imply that AM BM which cotradicts the fact that M is a midpoit. Hece a segmet caot have two distict midpoits.

6 82 lesso 7 let s go back to OP. We ow kow that it has a uique midpoit. Let s call that poit P. I order for the distace fuctio d to satisfy coditio (3), OP + P P = OP. But OP ad P P are cogruet, so i order for d to satisfy coditio (2), they have to be the same legth. Therefore 2 OP = ad so OP = /2. Repeat. take OP, ad fid its midpoit. Call it P 2. the OP 2 + P 2 P = OP. agai, OP 2 ad P 2 P are cogruet, so the must be the same legth. therefore 2 OP 2 = /2, ad so OP 2 = /4. By repeatig this process over ad over, you ca idetify the poits P which are distaces of /2 from O. O P3 P 2 P P r With the poits P as buildig blocks, we ca start combiig segmets of legths /2 to get to other poits. I fact, we ca fid a poit whose distace from O is m/2 for ay positive itegers m ad. It is just a matter of chaiig together eough cogruet copies of OP as follows. Begi with the poit P. By the first axiom of cogruece, there is a poit P 2 o the opposite side of P from O so that P P 2 OP. ad there is a poit P 3 o the opposite side of P 2 from P so that P 2 P 3 OP. ad a poit P 4 o the opposite side of P 3 from P 2 so that P 3 P 4 OP. ad so o. this ca be cotiued util m segmets are chaied together

7 distace legth ad cotiuity 83 stretchig from O to a poit which we will label P m. I order for the distace fuctio to satisfy the additivity coditio (3), OP m = OP + P P 2 + P 2 P P m P m. All of these segmets are cogruet, though, so they have to be the same legth (for coditio (2)), so OP m = m OP = m /2 = m/2. Ratioal umbers whose deomiator ca be writte as a power of two are called dyadic ratioals. i that spirit, i will call these poits the dyadic poits of r. O P 2 P 3 P P 3 P 7 P 9 P2 2 2 P 5 P P P2 3 P2 5 2 P 3 2 P 7 fill the hole There are plety of real umbers that are t dyadic ratioals though, ad there are plety of poits o r that are t dyadic poits. how ca we measure the distace from O to them? For starters, we are ot goig to be able to do this without the last two axioms of eutral geometry.

8 84 lesso 7 These last two axioms, the axioms of cotiuity, are a little more techical tha ay of the previous oes. The first says that you ca get to ay poit o a lie if you take eough steps. The secod, which is ispired by Dedekid s costructio of the real umbers, says that there are o gaps i a lie. the axioms of cotiuity Ct Archimedes Axiom If AB ad CD are ay two segmets, there is some positive iteger such that cogruet copies of CD costructed ed-to-ed from A alog the ray AB will pass beyod B. ct2 Dedekid s Axiom let S < ad S be two oempty subsets of a lie satisfyig: (i) S < S = ; (ii) o poit of S < is betwee two poits of S ; ad (iii) o poit of S is betwee two poits of S <. The there is a uique poit O o such that for ay two other poits P ad P 2 with P S < ad P 2 S the P O P 2 () (2) () Archimedes: Give eough steps, P will be passed. P S S + (2) Dedekid: There is a poit betwee ay two separated partitios of a lie. It is time to get back to the issue of distace o the ray r. so let P be a poit o r. Eve if P is ot itself a dyadic poit, it is surrouded by dyadic poits. I fact, there are so may dyadic poits crowdig P, that the distace from O to P ca be estimated to ay level of precisio usig earby dyadic poits. For istace, suppose we cosider just the dyadic poits whose deomiator ca be writte as 2 : S = {O,P,P 2,P 3,...}. By the Archimedea Axiom, evetually these poits will lie beyod P. If we focus our attetio o the oe right before P, say P m, ad the oe right after, P m +, the O P m P P m +.

9 distace legth ad cotiuity 85 We ca compare the relative sizes of the segmets OP m OP OP m + ad so, if our distace is goig to satisfy coditio (3), OP m < OP < OPm + m < OP < m + Not precise eough for you? Replace S, with S, the set of dyadic poits whose deomiator ca be writte as 2 : S = {O,P,P 2 = P,P 3,P 4 = P 2,...}. Agai, the Archimedea Axiom guaratees that evetually the poits i S will pass beyod P. let P m be the last oe before that happes. The so O P m P P m + OP m < OP < OPm + m /2 < OP < (m + )/2 ad this gives OP to withi a accuracy of /2. cotiuig alog i this way, you ca use S 2, dyadics whose deomiator ca be writte as 2 2, to approximate OP to withi /4, ad you ca use S 3, dyadics whose deomiator ca be writte as 2 3, to approximate OP to withi /8. geerally speakig, the dyadic ratioals i S provide a upper ad lower boud for OP which differ by /2. as goes to ifiity, /2 goes to zero, forcig the upper ad lower bouds to come together at a sigle umber. This umber is goig to have to be OP. ow you do t really eed both the icreasig ad decreasig sequeces of approximatios to defie OP. After all, they both ed up at the same umber. Here is the descriptio of OP usig just the icreasig sequece: for each positive iteger, let P m be the last poit i the list S which is betwee O ad P. I order for the distace fuctio to satisfy coditio (3), we must set OP = lim OP m = lim m /2.

10 / 86 lesso 7 P m 2 2 P m 2+ 2 P m P m P P m + P m + (m 2 + ) 4 m + m / m 2 Capturig a o-dyadic poit betwee two sequeces of dyadic poits. Now do it i reverse Every poit of r ow has a distace associated with it, but is there a poit at every possible distace? Do we kow, for istace, that there is a poit at exactly a distace of /3 from O? The aswer is yes it is just a matter of reversig the distace calculatio process we just described ad usig the Dedekid Axiom. Let s take as our prospective distace some positive real umber x. for each iteger, let m /2 be the largest dyadic ratioal less tha x whose deomiator ca be writte as 2 ad let P m be the correspodig dyadic poit o r. Now we are goig to defie two sets of poits: S < : all the poits of r that lie betwee O ad ay of the P m, together with all the poits of r op. S : all of the remaiig poits of r. so S < cotais a sequece of dyadic ratioals icreasig to x {P m,pm,pm 2 2,Pm 3 3,...},

11 distace legth ad cotiuity 87 ad S cotais a sequece of dyadic ratioals decreasig to x {P m +,P m +,P m 2+ 2,P m 3+ 3,...}. together S < ad S cotai all the poits of the lie through r, but they do ot itermigle: o poit of S < is betwee two of S ad o poit of S is betwee two of S <. Accordig to the Dedekid Axiom, the, there is a uique poit P betwee S < ad S. Now let s take a look at how far P is from O. for all, OP m OP OP m + OP m < OP < OP m + m /2 < OP < (m + )/2 as goes to ifiity, the iterval betwee these two cosecutive dyadics shriks ultimately, the oly poit left is x. so OP = x. Example: dyadics approachig /3 Fidig a dyadic sequece approachig a particular umber ca be tricky busiess. Fidig such a sequece approachig /3 is easy, though, as log as you remember the geometric series formula x = = x if x <. With a little trial ad error, I foud that by pluggig i x = /4, = 4 3. Subtractig oe from both sides gives a ifiite sum of dyadics to /3, ad we ca extract the sequece from that 4 = = 5 6 = = 2 64 =.32825

12 88 lesso 7 Segmet additio, redux for ay two poits P ad Q, there is a uique segmet OR o the ray r which is cogruet to PQ. Defie d(p, Q) = OR. With this setup, our distace fuctio will satisfy coditios () ad (2). That leaves coditio (3) a lot of effort wet ito tryig to build d so that coditio would be satisfied, but it is probably a good idea to make sure that it actually worked. let s close out this lesso with two theorems that do that. thm: a formula for distace alog a Ray If P ad Q are poits o r, with OP = x ad OQ = y, ad if P is betwee O ad Q, the PQ = y x. Proof. If both P ad Q are dyadic poits, the this is fairly easy. First you are goig to express their dyadic distaces with a commo deomiator: OP = m/2 OQ = m /2. the OP is built from m segmets of legth /2 ad OQ is built from m segmets of legth /2. to get PQ, you simply have to take the m segmets from the m segmets so PQ is made up of m m segmets of legth /2. that is PQ =(m m) 2 = y x. O m' copies m copies P Q 2 / Measurig the distace betwee two dyadic poits.

13 distace legth ad cotiuity 89 P m P m + P Q O P m+ P m Measurig the distace betwee two o-dyadic poits. If oe or both of P ad Q are ot dyadic, the there is a bit more work to do. i this case, P ad Q are approximated by a sequece of dyadics P m ad P m where m lim 2 = x & lim m 2 = y. ow we ca trap PQ betwee dyadic legths: P m + P m PQ P m P m + P m + P m < PQ < P m P m + m m 2 < PQ < m + m 2 as approaches ifiity, PQ is stuck betwee two values both of which are approachig y x. Now while this result oly gives a formula for legths of segmets o the ray r, it is easy to exted it to a formula for legths of segmets o the lie cotaiig r. I fact, this is oe of the exercises for this lesso. The last result of this lesso is a reiterpretatio of the Segmet Additio Axiom i terms of distace, ad it cofirms that the distace we have costructed does satisfy coditio (3). thm: segmet additio, the measured VeRsio If P Q R, the PQ + QR = PR. Proof. The first step is to trasfer the problem over to r so that we ca start measurig stuff. So locate Q ad R o r so that:

14 9 lesso 7 Q R P O Q R OR OQ start measurig stuff. So locate Q ad R o r so that: O Q R, PQ OQ, QR Q R. Accordig to the Segmet Additio Axiom, this meas that PR OR. ow we ca use the last theorem, QR = Q R = OR OQ = PR PQ. Just solve that for PR ad you ve got it.

15 distace legth ad cotiuity 9 exercises. Our method of measurig distace alog a ray r ca be exteded to the rest of the lie. I our costructio each poit o r correspods to a positive real umber (the distace from O to that poit). suppose that P is a poit o r op. there is a poit Q o r so that OP OQ. If x is the positive real umber associated with Q, the we wat to assig the egative umber x to P. ow suppose that P ad P 2 are ay two poits o the lie ad x ad y are the associated real umbers. show that d(p,p 2 )= x y. 2. Write /7, /6, ad /5 as a ifiite sum of dyadic ratioals. 3. Sice writig this, it has come to my attetio (via Greeberg s book []) that Archimedes Axiom is actually a cosequece of Dedekid s Axiom. You ca prove this yourself as follows. If Archimedes were ot true, the there would be some poit o a ray that could ot be reached by via ed-to-ed copies of a segmet. I that case, the ray ca be divided ito two sets: oe cosistig of the poits that ca be reached, the other of the poits that caot. By icludig the opposite ray i with the set of poits that ca be reached, you get a partitio of a lie ito two sets. Prove that these sets form a Dedekid cut of the lie. The by Dedekid s Axiom there is a poit betwee them. Now cosider what would happe if you took oe step forward or backward from this poit. Refereces [] Marvi J. Greeberg. Euclidea ad No-Euclidea Geometries: Developmet ad History. W.H. Freema ad Compay, New York, 4th editio, 28.

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