2 Euclidean Geometry - UCI Mathematics

3y ago
26 Views
2 Downloads
453.85 KB
19 Pages
Last View : 2m ago
Last Download : 2m ago
Upload by : Raelyn Goode
Transcription

2Euclidean GeometryWhile Euclid’s Elements provided the first serious attempt at an axiomatization of basic geometry, hisapproach contains several errors and omissions. Over the centuries, mathematicians identified theseand worked towards a correct axiomatic system for Euclidean Geometry. The culmination came withthe publication of David Hilbert’s Grundlagen der Geometrie (Foundations of Geometry) in 1899.2.1Hilbert’s AxiomsWe describe Hilbert’s axioms for plane geometry1 (next page). We will not follow Hilbert strictly, butrather explore how his system can address some of the difficulties/omissions in Euclid.The undefined terms consist of two types of objects (points and lines), and three relations (between, onand congruence.) At various places, definitions and notations are needed. Definition 2.1. AB denotes the line through distinct points A and B. This exists and is uniqueby axioms I-1 and I-2. AB denotes the line segment consisting of distinct points A and B and all the points C on ABsuch that C lies between A and B. This exists and is unique by axioms O-1, O-2 and O-3. The ray AB is the segment AB together with all points C on the line through A and B such thatB lies between A and C. An angle with vertex A consists of a point A together with its sides: two rays AB and AC. Wedenote the angle BAC. A triangle 4 ABC consists of segments AB, BC and CA where A, B, C are non-collinear. Twotriangles are congruent if their sides and angles are congruent in pairs. Lines and m intersect if there exists a point lying on both. Lines are parallel if they do notintersect. Segments and/or rays are parallel if and only if the corresponding lines are parallel. Given a line and points A, B not on , we say that the points lie on the same side of if thesegment AB does not intersect . Otherwise A and B are on opposite sides of .In the usual model of plane geometry, a line extends infinitely in both directions, a segment is bounded,and a ray extends infinitely in one direction.CBA Line ABBBAASegment AB1 Hilbert Ray ABABAngle BACin fact axiomatized 3D geometry: we only give the axioms relevant to planar constructions. These axioms arecomplete in the sense that they are enough to allow one to do everything that is considered Euclidean Geometry, and thatthere is essentially only one model, the ‘usual’ Cartesian Geometry (this is not the same as being complete, though is aboutas good as one can hope for!). The axioms have been shown to be consistent in the absence of the continuity axiom, althoughconsistency cannot be proved once this is included. As stated, the axioms are not quite independent. In particular, axiomO-3 does not require existence (follows from Pasch’s axiom), C-1 does not require uniqueness (follows from the uniquenessin C-4) and C-6 can be weakened. Hilbert’s axioms were revised after their initial publication and we’ve only stated oneversion.1

Hilbert’s Axioms for Plane GeometryUndefined terms1. Points (use capital letters, e.g., A, B, C)2. Lines (use lower case letters, e.g., )3. On (a point A lies on a line )4. Between (a point B lies between points A, C,written A B C)5. Congruence (of segments/angles)Axioms of IncidenceI-1 For any distinct A, B there exists a line onwhich lie A, B.I-2 There is at most one line through distinctA, B (A and B both on the line). Notation: line AB through A and BI-3 On every line there exist at least two distinct points. There exist at least threepoints not all on the same line.Axioms of Order/BetweennessO-1 If A B C, then A, B and C are distinctpoints on the same line and C B A.O-2 For any distinct points A and B, there is atleast one point C such that A B C.O-3 If A, B, C are distinct points on the sameline, exactly one lies between the othertwo.Definitions: segment AB and triangle 4 ABCO-4 (Pasch’s Axiom) Let 4 ABC be a triangleand a line not containing any of A, B, C. If contains a point of the segment AB, thenit also contains a point of either AC or BC.Axioms of Congruence Definition: ray ABC-1 If A, B are distinct points and A0 is a point,then for each ray r from A0 there is aunique point B0 on r such that AB A0 B0 . Moreover AB BA.C-2 If AB EF and CD EF, then AB CD.C-3 If A B C, A0 B0 C’, AB A0 B0 andBC B0 C 0 , then AC A0 C 0 . Definitions: angle ABC, side of line AB C-4 Given BAC and a ray A0 B0 , there is a unique ray A0 C 0 on a given side of A0 B0such that BAC B0 A0 C 0 .C-5 If ABC GH I and DEF GH I, then ABC DEF. Moreover, ABC CBA.C-6 (Side-Angle-Side) Given triangles 4 ABCand 4 A0 B0 C 0 , if AB A0 B0 , AC A0 C 0 ,000and BAC B A C , then the trianglesare congruent.Axiom of Continuity Suppose that the pointson are partitioned into two non-empty subsetsΣ1 , Σ2 such that no point of Σ1 lies between twopoints of Σ2 , and vice versa. Then there exists aunique point O lying on such that P1 O P2if and only if O 6 P1 , O 6 P2 and one of P1 or P2lies in Σ1 and the other in Σ2 .Definition: parallel linesPlayfair’s Axiom Given a line and a point Pnot on , there exists exactly one line through Pparallel to .

Axioms of IncidenceThe axioms of incidence explain how the relation on interacts with points and lines. Hilbert’s presentation differs from Euclid’s in that he states his axioms and definitions when needed. This meansthat results can be built up and models created of geometries that satisfy only the first few axioms,but none later. For example, there is essentially only one model satisfying the incidence axioms (I-1,I-2, I-3) with exactly three points, and two with exactly four points. You should convince yourselfthat this is the case.CDCn { B, C }m { A, C }ACB { A, B}3 points, 3 linesADBA4 points, 6 linesB4 points, 4 linesWe can even prove some very simple theorems in incidence geometry: these only depend on axiomsI-1, I-2 and I-3!Lemma 2.2. If two distinct lines intersect, then they do so in exactly one point.Proof. Suppose A, B are distinct points of intersection. By axiom I-2, there is exactly one line throughA and B. Contradiction.Lemma 2.3. Through any point there exists at least two lines.Try to prove this one yourself.There is not a lot more one can do simply with the incidence axioms, though there are many esotericexamples of incidence geometry, such as the Fano plane. Look them up if you are interested!Axioms of Order: Sides of a Line and Pasch’s AxiomThe axioms of order explain the use of the ternary relation between for points. Their inclusion inHilbert’s axioms is in no small part due to the work of Moritz Pasch, after whom Pasch’s axiom isnamed (c.1882).Once you allow the order axioms in addition to the incidence axioms,geometries with only finitely many points bacome impossible! To seethis note that axioms O-1 and O-2 generate a new point C on the line AB such that A B C. Repeat the exercise to obtain D such thatB C D, etc. You should convince yourself that D 6 A so that wegenuinely obtain a new point. In this fashion we see that any line contains unboundedly many points.3ABCDE

Of the order axioms, Pasch’s axiom (first published in 1882) is the mostimportant. It is necessary to properly define the concept of the side ofa line, which is used in several of Euclid’s arguments.CDefinition 2.4. We say that distinct points A, B lie on the same side of aline if AB does not intersect . The points lie on the opposite side of if the segment intersects . BATheorem 2.5 (Plane Separation). Let A, B, C be points not on a line .1. If A, B lie on the same side of and B, C lie on the same side of , then A, C lie on the same side of .2. If A, B lie on opposite sides of and B, C lie on opposite sides of , then A, C lie on the same side of .3. If A, B lie on opposite sides of and B, C lie on the same side of , then A, C lie on the opposite sides of .Otherwise said, every line separates the plane into exactly two half-planes; the two sides of .BAC CCase 1AA BCase 2BCCase 3Proof. All three parts depend on Pasch’s axiom. We prove the first and leave the others as exercises.Consider the triangle 4 ABC. If intersects AC then, by Pasch’s axiom, it would also intersect one ofthe sides AB or BC. It does neither of these things, whence A, C lie on the same side of .The extreme case is where A, B, C are collinear: we omit this to avoid tediousness.Armed with the above, one can properly define the notion of the inside/interior of a triangle as theintersection of three half-planes. One of Hilbert’s early results is to prove that every triangle has anon-empty interior.Even though the concept of angle is not required until the congruence axioms, it is convenient toconsider the concept of interior point, so that we may properly consider the first of Euclid’s omissions.Definition 2.6. A point I is interior to an angle BAC if:C I lies on the same side of AB as C, and, I lies on the same side of AC as B.Otherwise said, I lies in the intersection of two half-planes.It is now possible to compare angles: for instance BAI BAC.4IAB

Theorem 2.7 (Crossbar Theorem). Suppose that I is interior to BAC. Then the ray AI intersects BC.Proof. Extend AB to a point D such that A lies between B and D (axiom O-2). Since C is not on the line BD AB we have a triangle 4 BCD. The line AI intersects one edge of 4 BCD at A and does not cross any vertices. By Pasch’s axiom, itintersects one of the other edges at a point M. The point A is the unique intersection of three distinct lines: AB, AC and I M. However I and M are on the same side2 of AB, thus The segment I M does not intersect AB and so A does not lie on I M. M lies on the ray AI.There are two cases which we show below: all we need to do is to show that the latter is impossible.CCIIMMDDBACorrect arrangementBAThe impossible case If M lies on CD, then I and M must lie on opposite sides of AC. But then I M intersects AC, which itmust do at A. This contradicts the fact that A does not lie on I M.Corollary 2.8. If a line passes through one vertex of a triangle and aninterior point of the triangle, then it must pass through the side oppositethe vertex.CEuclid uses this concept several times, including in his construction of perpendiculars and of angle and segment bisectors(Theorems I. 9 10); we sketch the latter here.Given an angle BAC, construct E such that AB AE; theequilateral triangle (Thm I. 1) 4 DBE on BE is seen to produce the bisector AD after a couple of appeals to Side-Angle-Side.The problem for Euclid is that he gives no argument for why AD should intersect BE, without which his construction goesnowhere!Even with Pasch’s axiom and the crossbar theorem, it requiressome effort to repair Euclid’s proof: in particular, we’d need toshow that D is interior to the angle BAC. . .2 All interior points of the segments CE and BC lie on the same side of AB as C.5 BACEADB

Axioms of CongruenceHilbert uses the concept of congruence for several purposes, in particular:1. To formalize Euclid’s pictorial reasoning (laying one triangle on top of another, etc.).2. To tidy up Euclid’s liberal and confusing use of the word equal.To Hilbert, segments or angles are equal only when they are precisely the same. Certainly equalsegments are congruent, as seen by the reflexivity part of the following result:Lemma 2.9. Congruence of segments (angles) is an equivalence relation.Proof.Reflexivity Let AB be given and apply C-1 to obtain a segment A0 B0 such that AB A0 B0 .We sneakily use this twice and apply C-2:AB A0 B0 and AB A0 B0 AB ABAssume AB CD. By reflexivity, CD CD. C-2 now says that CD AB.Transitivity If AB CD and CD EF, apply symmetry and C-2 to conclude AB EF.SymmetryAxioms C-4 and C-5 say the same thing for angles.The most important congruence axiom is C-6 (Side-Angle-Side/SAS).3 This is Theorem I. 4 of theElements, which relies on Euclid’s unjustified procedure of laying one triangle on top on another. Itwas eventually realized that at least one triangle congruence theorem had to be an axiom. Hilbertassumes SAS and proceeds to prove the remaining congruence theorems. For example:Theorem 2.10 (Angle-Side-Angle/ASA (Euclid I. 26, case I)). Suppose 4 ABC and 4 DEF satisfy ABC DEF,AB DE, BAC EDFThen the triangles are congruent.Proof. Given the assumptions, axiom C-1 gives a unique point G on the ray EF such that EG BC. EDG which, by assumpAxiom C-6 (SAS) says that BAC Ction, is congruent to EDF. Since F and G lie on the same side of DE, axiom C-4 says that theylie on the same ray through D. But then F and G both lie on two distinct lines ( EF and DF): weconclude that F G.By SAS we conclude that 4 ABC 4 DEF.BAGFDE3 Hilbert ultimately states a weaker version of this, merely that one pair of the remaining angles in the triangles arecongruent. The full SAS congruence is then a theorem which Hilbert proves just before proving ASA.6

Geometry Without CirclesFor reasons we’ll deal with shortly, Hilbert barely mentions circles, and yet constructions using circlesare at the heart of Euclid’s presentation. Hilbert therefore has alternative constructions of some ofEuclid’s basic results. We give sketches of a few.Definition 2.11. A triangle is isosceles if two of its edges are congruent. The third edge is known asthe base.Theorem 2.12 (Euclid I. 5). Isosceles triangles have congruent base angles.Euclid’s argument is famously unpleasant, relying on a complicated construction. Hilbert does thingsfar more speedily.Proof. Suppose WLOG that AB AC so that 4 ABC is isosceles. Consider a ‘new’ triangle 4 A0 B0 C 0 4 ACB where the base points areswapped: i.e.A0 A,B0 C,A A0C0 BWe have: BAC CAB (axiom C-5) BAC B0 A0 C 0 . AB AC AB A0 B0 and AC A0 C 0 .SAS says that ABC A0 B0 C 0 ACB.B C0C B0The proof is very sneaky: just relabel the original triangle and use SAS!The converse to this statement is also true (see the homework). Dropping a Perpendicular Given a point C not on the line AB, useaxioms (C-4 and C-1) to transfer the angle PAB to the other side ofthe line at A, thus creating a new point D.Since C and D lie on opposite sides of the line, it must intersect the lineat some point M.Otherwise, SAS shows that the triangles 4 MAC and 4 MAD are congruent whence the supplementary angles AMC and AMD are congruent also.Since these sum to a straight edge, they must be right angles.The only reason this construction might fail is if, by chance, M A.But then the same argument applied to 4 BAC and 4 BAD shows thatwe have a right angle at A.7CABMD

The SSS congruence This is done similarly to Euclid.Suppose 4 ABC and 4 DEF have sides congruent in pairs,then the triangles are congruent.Proof. Transfer the angle EDF to A on the other side of ABfrom C to obtain the point G (axioms C-4 and C-1).SAS shows that 4 BAG 4 EDF and so BG EF BC.Joining CG we now have two pairs of isosceles triangles withvertices at A and B and congruent base angles at C and G.Summing angles at C and G and applying SAS shows that4 ABC 4 ABG 4 DEF as required.ECFDBAA little work is necessary to properly define angle sum andto deal with the case where the sum is actually a subtraction,but we omit these details.GThe Exterior Angle Theorem Euclid’s approach requires a bisector, which he obtains from circles.Hilbert does things a little differently.Proof. Given 4 ABC, extend AB to D such that AB BD. If δ γ we can apply SAS to the triangles 4 ACB 4 DBC which forces e β. Clearly D (on AB) lies on the same side of AC as B, but yet β and δ, and thus γ and e, sum to astraight edge. Contradiction. We conclude that δ 6 γ. Taking the vertical angle at B, we see that anexterior angle δ to a triangle cannot be congruent to either opposite interior angle α or β.CCη eγ eαAβδBαDAStep 1: δ γβEδBDStep 2: δ 6 γIn the second picture, assume δ γ. Transfer the angle to C as shown where η δ: by the crossbartheorem, we obtain an intersection point E. But now δ is an exterior angle of 4 EBC and congruentto an interior angle η of the same triangle, which contradicts the above. We conclude the full exteriorangle theorem.The Exterior Angle Theorem also proves that the sum of any two angles in a triangle is strictly lessthan a straight edge: α β δ β.Moreover, suppose that the above picture for the SSS congruence comprised two isosceles triangles: thefact that the base angles must now be less than a right angle proves that CG is in fact the perpendicularbisector of AB: in this manner, one may bisect both angles and segments in Hilbert’s geometry, anderect perpendiculars, without relying on circles. In particular, this fixes Euclid’s Theorems I. 9 10.8

The SAA congruence If AB DE, ABC DEF, and BCA EFD, then the triangles are congruent:GC4 ABC 4 DEF Proof. Choose the unique G on BC such that BG EF.SAS says that 4 ABG 4 DEF.If G 6 C, then the triangle 4 AGC has an exterior angle and anopposite interior angle both congruent to ABC, which contradicts the exterior angle theorem.BAA similar picture could be drawn when G lies on BC, though the argument is unchanged.Is Euclid now fixed? At this point Hilbert has proved all of Euclid’s results in Book I of the Elementsprior to the application of the parallel postulate. Including Playfair’s Postulate allows the remainderof the results in Book I to be completed, including Pythagoras’ Theorem.There is one remaining major problem with Euclid: his use of circles.Comparing Lengths: Circles and ContinuityNeither Hilbert nor Euclid require an absolute notion of length, although additional axioms can beadded to accommodate this. However they both require a method of comparing lengths.Definition 2.13. Let segments AB and CD be given. Using axiom C-1, construct the unique point E on AB such thatAE CD. We write CD AB if E lies between A and B.Any two segments are then comparable: given AB andCD, precisely one of the following is true:AB CD,CD AB,BEACDAB CDA similar approach can be done for angles using a discussion of interior points: indeed this wasalready necessary in order to prove some of the previous results.We now proceed to define circles.Definition 2.14. Let O and R be distinct points. The circle C with center Oand radius OR is the collection of points A such that OA OR.A point P lies inside the circle C if P O or OP OR.A point Q lies outside if OR OQ.Since segments are comparable, every point lies either inside, outside or on agiven circle.9RAPOQ

The Axiom of Continuity Why does Hilbert barely mention circles? Part of his goal was to see whyand when each axiom is required; he wants to build as much geometry as possible without invokingcomplicated axioms. A major weakness of Euclid is that many of his proofs rely on the intersectionsof lines and circles. To use circles in this manner requires the axiom of continuity, undeniably the mostdifficult of Hilbert’s axioms.The following statements can be proved from the Axiom of Continuity.Theorem 2.15.1. (Elementary Continuity Principle) If C is a circle such that P is inside and Q is outside, then the segment PQ intersects C in exactly one point.2. (Circular Continuity Principle) If C and D are circles such that D contains at least one point insideand one point outside C , then the circles intersect in exactly two points; these points lie on oppositesides of the line joining the circle centers.The idea of the first is to partition the segment PQ into two pieces:Σ1 consists of the points lying inside or on C ,Σ2 consists of the points lying outside C .Σ1POΣ2QOne shows that Σ1 and Σ2 satisfy the assumptions of the Axiom, whichguarantees the existence of the unique point O . Finally O is shown tolie on the circle itself.A full discussion of this (even a sketch of the second argument) lies b

2 Euclidean Geometry While Euclid’s Elements provided the first serious attempt at an axiomatization of basic geometry, his approach contains several errors and omissions. Over the centuries, mathematicians identified these and worked towards a correct axiomatic system for Euclidean Geometry. The culmination came with

Related Documents:

course. Instead, we will develop hyperbolic geometry in a way that emphasises the similar-ities and (more interestingly!) the many differences with Euclidean geometry (that is, the 'real-world' geometry that we are all familiar with). §1.2 Euclidean geometry Euclidean geometry is the study of geometry in the Euclidean plane R2, or more .

Feb 05, 2010 · Euclidean Parallel Postulate. A geometry based on the Common Notions, the first four Postulates and the Euclidean Parallel Postulate will thus be called Euclidean (plane) geometry. In the next chapter Hyperbolic (plane) geometry will be developed substituting Alternative B for the Euclidean Parallel Postulate (see text following Axiom 1.2.2).

Euclidean Spaces: First, we will look at what is meant by the di erent Euclidean Spaces. { Euclidean 1-space 1: The set of all real numbers, i.e., the real line. For example, 1, 1 2, -2.45 are all elements of 1. { Euclidean 2-space 2: The collection of ordered pairs of real numbers, (x 1;x 2), is denoted 2. Euclidean 2-space is also called .

Yi Wang Chapter 4. Euclidean Geometry 64 2. Similar triangles only exist in Euclidean geometry. in non-Euclidean geometry, similar triangles do not exist, unless they are congruent. Lemma 4.25 Consider ABC with D and E on sides AB and AC. if DEkBC then AD AB

Lecture Notes in Modern Geometry RUI WANG The content of this note mainly follows John Stillwell’s book geometry of surfaces. 1 The euclidean plane 1.1 Approaches to euclidean geometry Our ancestors invented the geometry over euclidean plan

TRS Support Desk UCI A&FS - Payroll Division eec@uci.edu timesheet.uci.edu 7 The TRS main page will be displayed with access options. 4. Select the Employee option. To access TRS directly, employees should first navigate to the main page for TRS (https://timesheet.uci.edu) and select the appropriate User Type.

Gain access to private university events, work with a personal concierge to utilize university resources, and cheer on the UCI Anteaters in action by attending prime sporting events. Once you've been granted a seat on the board, you'll receive a welcome kit filled with UCI items - Program Brochure, UCI Magnet, and UCI Sticker to name a few!

Answer questions developed by the test maker . Language Arts – Reading Directions Time 35 minutes 20 Questions This is a test of some of the skills involved in understanding what you read. The passages in this test come from a variety of works, both literary and informational. Each passage is followed by a number of questions. The passages begin with an introduction presenting .