Discovery of Non-Euclidean Geometry

Transcription

1 iscovery of Non-Eucidean Geometry pri 24, Hyperboic geometry János oyai ( ), ar Friedrich Gauss ( ), and Nikoai Ivanovich Lobachevsky ( ) are three founders of non-eucidean geometry. Hyperboic geometry is, by definition, the geometry that assume a the axioms for neutra geometry and repace Hibert s parae postuate by its negation, which is caed the hyperboic axiom. Hyperboic axiom (Negation of Hibert axiom). There exists a ine and a point not on such that at east two distinct ines parae to pass through. Theorem 1.1. In hyperboic geometry, a trianges have ange sum ess than 180, and a convex quadriateras have ange sum ess than 360. In particuar, there is no rectange. roof. Trivia. Theorem 1.2 (Universa Hyperboic Theorem). In hyperboic geometry, for every ine and every point not on there pass through at east two distinct ines parae to. In fact there are infinitey many ines parae to through. t t m S n S n R R Figure 1: Existence of infinite paraes roof. rop segment perpendicuar to with foot on. Erect ine m at perpendicuar to. Then, m are parae. ick a point R on other than, and erect ine t at R perpendicuar to. rop ine n through perpendicuar t, intersecting t at S. If S is on m, then S is the intersection m and t, and subsequenty RS is a rectange, which is impossibe in hyperboic geometry. So point S is not on m. Hence m, n are distinct ines through, both are parae to. See Figure 1. Let R be point on other than, R, and t be ine through R perpendicuar to. There exists ine n through perpendicuar to t, intersecting t at S. If S = S, then RR S S is a rectange, which is impossibe. So S, S are distinct ines. Thus for a points R on other than, the ines S through perpendicuar to are a distinct. 1

2 efinition 1. Two trianges are and said to be simiar if their vertices can be put in on-to-one correspondence so that corresponding anges are congruent, i.e.,,, are congruent to anges,, respectivey. Theorem 1.3 ( criterion for congruence of hyperboic trianges). In hyperboic geometry, if two trianges are simiar then they are congruent. Figure 2: Simiar trianges are congruent in hyperboic geometry roof. Given simiar trianges and. Suppose the statement is not true, i.e., is not congruent to. Then =, =, = ; otherwise = by S. We may assume < and <. Lay off segment on ray r(, ) to have point such that = and, and ay off segment on ray r(, ) to have point such that = and. See Figure 2. Then = by SS. Hence = =, = =. It foows that ines, are parae because of ongruent orresponding nges. So is a convex quadriatera. Since anges, are suppementary, anges, are suppementary, and =, =, then the ange sum of is 360. This is a contradiction. Theorem 1.3 says that in hyperboic geometry it is impossibe to magnify or shrink a triange without distortion. So in hyperboic word photography woud be inherenty surreaistic. nother consequence of Theorem 1.3 is that the ength of a segment may be determined by anges in hyperboic geometry. For exampe, an ange of an equiatera triange determines the ength of a side uniquey. This fact is sometimes referred to that hyperboic geometry has an absoute unit ength. 2 araes that admit a common perpendicuar Given ines, and points,,,... on. rop perpendicuars,,,... from,,,... to with feet,,,... on respectivey. We say that,,,... are equidistant from if a these perpendicuar segments are congruent to one another. See Figure 3. Theorem 2.1 (t most two points equidistant). Given two distinct paraes, in hyperboic geometry. Then any set of points on equidistant from contains at most two points. roof. Suppose it is not true, i.e., there is a set of three points,, on equidistant from. Then quadriateras,, are Saccheri quadriateras (the base 2

3 Figure 3: No more than two equidistant points between two paraes anges are right anges and the sides are congruent). Then the summit anges of the Saccheri quadriateras are congruent, i.e., =, =, =. Thus =. Since, are suppementary, they must be right anges. Hence a,, are rectanges, which is impossibe. Lemma 2. Given a Saccheri quadriatera with base right anges, and equa opposite sides,. Let M, M be the midde points of, respectivey. Then segment MM is perpendicuar to both ines and. roof. raw segments M and M. Note that =, =, and M = M. Then M = M by SS. So M = M. Hence MM = MM by SSS. We then have M M = M M. Subsequenty, M M and M M are right anges. So MM is perpendicuar to the base. Note that MM = MM and M M Figure 4: erpendicuar midde point segment M = M. Then MM = MM by ange addition. Subsequenty, MM and MM are right anges. So MM is perpendicuar to the summit. Theorem 2.2 (ivergent and symmetric paraes). Let, be two ines perpendicuar to a segment MM with M, M. (a) Then MM < XY for a X, Y with XY MM. (b) If M is the midde point of a segment on, then, are equidistant from. (c) If M on and, are segments perpendicuar to with feet,, then <. roof. (a) It is cear that MM < MY for a Y on with Y M. Let X be a point on with X M. Let XX be segment perpendicuar to with foot X on. Then MM XX is a Lambert quadriatera. Thus MM < XX by properties of Lambert quadriateras. Since XX < XY for Y on with Y X. We see that MM < XY. (b) Let, be segments perpendicuar to with,. raw segments M and M. Then MM = MM by SS. So M = M and M M = M M. 3

4 M M Figure 5: ivergent paraes are symmetric Subsequenty, M = M by ange subtraction. Thus M = M by S. Hence = and M = M. (c) Note that M M and M M are Lambert quadriateras. Then M and M are acute anges. So is obtuse, for it is suppementary to M. Hence = M <. Therefore < by the property of. roposition 2.3 (symptotic and monotonic paraes). Given paraes, in hyperboic geometry, no two points of are equidistant from. Let,, be perpendicuar segments to with on and,,. See Figure 6. (a) If <, then <. (b) If <, then <. Figure 6: Monotone distance between asymptotic paraes roof. onsider quadriateras and. (a) Since <, then <. Since the ange sum of is ess than 360, it foows that is acute. So is obtuse. Hence must be acute, since the ange sum of is ess than 360. Of course <, subsequenty, < by the property of. (b) Fix a point on with. For each X on the open ray r(, ) we write X = x and define f(x) = XX, where XX is perpendicuar to with foot X on. We caim that f(x) is a continuous function for x > 0. In fact, fix an x 0 with point X 0 on such that X 0 = x 0. Let X 0 X 0 be segment perpendicuar to with X 0. Note that Then XX XX 0 X 0 X 0 + XX 0, X 0 X 0 X 0 X XX + XX 0. f(x) f(x 0 ) = { XX X 0 X 0 if XX XX 0 X 0 X 0 XX if XX < XX 0 XX 0 = x x 0. eary, f(x) is continuous at x 0. So f(x) is a continuous function for x > 0. Suppose >. Note that =. If <, by intermediate vaue theorem there exists a Y with Y such that Y Y =. Then, Y are equidistant 4

5 from, which is impossibe. If >, by intermediate vaue theorem there exists a point Z with Z such that ZZ =. Then, Z are equidistant from, which is impossibe. We then must have <. 3 Limiting parae rays Given a ine in hyperboic geometry and a point not on. Let m be a ine through parae to with eft ray r(, R). rop perpendicuar segment to with foot on. We consider rays between r(, ) and r(, R), and want to find the critica ray r(, X), caed the eft imiting parae ray to through, that does not meet but any ray between r(, X) and r(, ) meets. Likewise, there is a right imiting parae ray to through on the opposite side of. See Figure 7. Theorem 3.1. Given a ine and a point not on in hyperboic geometry. Let be segment perpendicuar to with foot on. Then there exist two non-opposite rays r(, X), r(, X ) on opposite sides of ine, satisfying the properties: (a) Each of rays r(, X), r(, X ) does not meet. (b) ray r(, Y ) meets if and ony if it is between r(, X) and r(, X ). (c) X = X. roof. Let m be the ine through perpendicuar to. ick a point R on the eft side of m and a point R on the right side of m separated by. raw segments R and R. Then a rays between r(, ) and r(, R) incusive are represented by r(, Y ) with Y R. See Figure 7. R R X X m U T S V V Figure 7: Limiting parae rays (a) Let Σ 1 be the set of points Y r(, R) such that the ray r(, Y ) does not meet, and Σ 2 the compement of Σ 1 in R. It is easy to see that both Σ 1, Σ 2 are convex. So Σ 1, Σ 2 form a edekind cut of R. Then there exists a unique point X R such that Σ 1, Σ 2 are two rays (one of them is an open ray) of R separated by X. We caim that X Σ 1. Suppose X Σ 2, i.e., r(, X) meets at S. ick a point T on such that T S. Then ray r(, T ) is between r(, R) and r(, X). So r(, T ) meets R at U and R U X, i.e., U Σ 2, which is a contradiction. The existence of ray r(, X ) is anaogous. (b) Since R Σ 1 and Σ 2, we see that R X. It is obvious that if a ray r(, Y ) is contained in the open haf-pane opposite to H(m, ) then r(, Y ) does not meet. We then see that a ray r(, Y ) meets if and ony if r(, Y ) is between r(, X) and r(, X ). (c) Suppose that X is not congruent to X, say, X < X. Find point V on such that r(, V ) is between r(, ) and r(, X ), and V = X. Mark a point V on such that V V and V = V. Then V = V by SS. So V = V = X, i.e., r(, X) meets at V, which is a contradiction. 5

6 The ange X is caed the ange of paraeism at point with respect to, its degree measure is denoted Π( ). We have Π( ) < assification of paraes Theorem 4.1. Given parae ines, in hyperboic geometry. (a) If contains a imiting parae ray to, then, are asymptotic paraes. (b) If does not contain imiting parae ray to, then, are divergent paraes. R X m S T Figure 8: The imiting parae ray is asymptotic and monotonic roof. Fix a point not on and drop a perpendicuar to with foot. Let m be the ine through perpendicuar to. ick a point R on m other than. Let r(, X) be a imiting parae ray to with X R. See Figure 8. (a) Let,,, be points on with and, r(, X). Let,,, be segments perpendicuar to with feet,,,. Note that X is acute, is obtuse, and the ange sum of is ess than 360. Then is acute. Of course <. So < by property of quadriateras with two base right anges. naogousy, is acute and is obtuse. Of course <. Then < by property of quadriateras with two base right anges. We caim for a on open ray r(, X). Suppose >. Let S be a point on such that S = 1 2 ( ). eary, S >. Then S < X by property of quadriatera with two base right anges. Of course S is acute. Since r(, S) is between r(, ) and r(, X), the ray r(, S) meets at T. Note that ST is acute. So S is obtuse, contradicting to that S is acute. We further caim < for two points, on cosed ray r(, X) with. Suppose. There exists a point E on (maybe = ) such that = EE by continuity of distance function. Let M, M be the midde points of E, E respectivey. Then, are divergent paraes. Let F be on such that F M and MF = M. We have F F = >, which is a contradiction. (b) ssume that does not contain any imiting parae ray. If = m, then, are aready divergent paraes. If m, we may assume that a ray r(, Y ) of is between r(, R) and r(, X), where R Y X. It is easy to see that < < by simiar arguments. Since X Y is acute, by ristote s axiom there exists a point on r(, Y ) such that E >, where E is perpendicuar to r(, X) with foot E r(, X). Of course > 6

7 R Y E F X m Figure 9: Non-imiting parae ray is symmetric F > E. So >. Thus, cannot be asymptotic (monotonic) paraes. So, must be divergent (symmetric) paraes. Let, be two distinct points on the same side of a ine such that ines, are parae. Then the figure, consisting of the segment (caed the base) and the rays r(, ) and r(, ) (caed the sides), is caed a biange with vertices and, denoted. See Figure 10. The interior of biange is :=. If, either of rays r(, ), r(, ) is caed an interior ray of. If E G F Figure 10: iange and imiting parae each interior ray r(, ) intersects r(, ), we say that r(, ) is imiting parae to r(, ) and that biange is cosed at, written r(, ) r(, ). Lemma 3. Let be a biange. See Figure 10. (a) If, then r(, ) r(, ) if and ony if r(, ) r(, ). (b) If r(, ) r(, ), so is r(, ) r(, ). roof. (a) ssume r(, ) r(, ). Take a point in the interior of. It is cear that is an interior point of biange. Then r(, ) meets r(, ) at F since is cosed at. Note that is an interior point of F. Then r(, ) is between r(, ) and r(, F ). Thus r(, ) meets F at G with G. y definition r(, ) r(, ). onversey, assume r(, ) r(, ). For each ray r between r(, ) and r(, ), we have r meeting at E between and. ick a point on r such that E. Note that > E = E. There is a ray r(, ) such that = E. Then r(, ) r(e, ). Since r(, ) meets r(, ), we see that r(e, ) must meet r(, ), i.e., r(, ) meet r(, ). Hence is cosed at. (b) Given an interior point Å and consider the ray r(, ). Suppose r(, ) does not meet r(, ). y the coroary of ristote s axiom there exists a point on r(, ) such that <. See Figure 11. Note that r(, ) meets r(, ) at. Then we have triange. Thus > =, which contradicts 7

8 Figure 11: iange and imiting parae <. So r(, ) must meet r(, ). Hence is cosed at, i.e., r(, ) r(, ). roposition 4.2 (Transitivity of imiting paraeism). If both rays r(, ) and r(, ) are imiting parae to ray r(, ), then r(, ) and r(, ) are imiting parae to each other. E E F Figure 12: Limiting paraes roof. ase 1. Lines and are on opposite sides of ine. See Figure 12. It is cear that r(, ) and r(, ) have no point in common. Let meet at. We may assume. Now for each point interior to, the ray r(, ) meets r(, ) at E since r(, ) r(, ). We may assume E. Then r(e, E ) meets r(, ) at F since r(, ) r(, ), where E E F. Hence r(, ) meets r(, ) at F. Therefore by definition r(, ) and r(, ) are imiting parae to each other. ase 2. Lines and are on the same side of ine. Figure 13: Limiting paraes We first caim that and do not meet. Suppose and meet at point. We may assume that beongs to both rays r(, ), r(, ), and assume,. Take a point such that. Then r(, ) meets r(, ) since r(, ) r(, ), i.e., r(, ) meets r(, ), which is a contradiction. See Figure 13. Let meet at point. We may assume. For each point interior to, the ray r(, ) meets r(, ) at point E. Since r(, ) (= r(, )) meets the triange E, the ray r(, ) meets either E or E. Since r(, ) does not meet r(, ), so r(, ) meet E at F such that F E. For point interior to, the ray r(, ) meets between and, of course r(, ) meets r(, ). Hence r(, ) and r(, ) are imiting parae to each other. 8

9 F E Figure 14: Limiting paraes Two rays r, s are said to be imiting parae, denoted r s, if r s or s r or r s. Then is an equivaence reation on rays in hyperboic geometry. n equivaence cass of rays is caed an idea point or end, viewing it ying on each ray contained in the equivaence cass. Since a point on a ine separates the ine into two opposite rays, and opposite rays are not equivaent, we see that every ine has two ends on it. If, are vertices of two rays r, s with r s. Let R denote the idea point determined by these rays, i.e., R = [r] = [s]. We write r = R and s = R and refer to the cosed biange with side r, s as a singy asymptotic triange R. We sha see that these trianges have some properties in common with ordinary trianges. Lemma 4. In hyperboic geometry if two ines, m are cut by a ine t such that the aternate interior anges are congruent, then, m are divergent paraes. t m Figure 15: symptotic triange roof. Let t meet at and meet m at such that =, where is perpendicuar to m with foot on m and is perpendicuar to with foot on. Then =. So =. Hence, m are divergent parae ines. roposition 4.3. Let R be a singy asymptotic triange with a singe idea point R. Then the exterior anges at, are greater than their respective opposite interior anges, i.e., < ext. R E Figure 16: symptotic triange 9

10 roof. Extend to such that. raw ray r(, ) such that = and extend to E such that E. Then E = =. Thus ines, are divergent paraes. Since r(, ) r(, ), we see that r(, ) r(, ). If r(, ) is between r(, ) and r(, ), then r(, ) meets r(, ), which is a contradiction. So we must have r(, ) between r(, ) and r(, ). This means that <, i.e., >. 10