Independence of Parallel Postulate

Transcription

1 Independence of arallel ostulate May 14, Consistency of hyperbolic geometry For thousands of years people believe and continue to believe that there is no contradiction in Euclidean geometry, that is, Euclidean geometry is consistent. How about hyperbolic geometry? Is there any contradiction in hyperbolic geometry? Applied geometry (engineering) is about lines that we can draw. ure geometry is about ideal lines which are concepts, not physical objects. hysical experiment can test the validity of physical systems, but cannot verify pure geometry. The only experiment that can perform on ideal lines are thought-experiments. Can we conceive of a non-euclidean geometry? Kant said no at his time (any geometry other than Euclidean geometry is inconceivable). Is hyperbolic geometry consistent? Nowadays this is referred to as a question in metamathematics, that is, a question outside of a mathematical system about the system itself. The question is not about lines or points or other geometric entities; it is about the whole system of hyperbolic geometry. Metamathematical Theorem. If Euclidean geometry is consistent, then so is hyperbolic geometry. To prove Metamathematical Theorem, we have to ask ourselves, What is a line in hyperbolic geometry in fact, what is the hyperbolic plane? The honest answer is that we don t know; it is just a formal system about undefined terms such as points, lines satisfying certain relations such as betweeness, congruence, continuity and hyperbolic parallelism. Then how shall we visualize hyperbolic geometry? In mathematics, as in any field of research, posing the right question is just as important as finding answers. We have seen that the Euclidean parallel postulate is independent of the incidence axioms by exhibiting three-point and five-point models of incidence geometry that are not Euclidean. We want to know whether the parallel postulate is independent of a much larger system of axioms, namely, the neutral geometry. Here we can show that it is, by the same method by exhibiting models for hyperbolic geometry. Beltrami-Klein model (Klein model) We fix once and all a circle in the Euclidean plane with center and radius R. The interior of is the set of points X such that X < R. A chord of is a segment AB in the Euclidean plane joining two endpoints A, B on. The open segment of chord AB, denoted (AB), is called an open chord of. In Klein model points and lines are the interior points and open chords of respectively, and the incidence relation between points and lines are belongness. oin lies in line (AB) means that AB in the Euclidean plane and A B. See Figure 1 1

2 n m l Figure 1: Klein model 3 oincaré model A disk model due to Henri oincaré ( ) also represents a hyperbolic plane whose points are interior points of a Euclidean circle, whose lines are those open circular arcs orthogonal to, and the incidence is the belongness of points and open circular arcs. The interpretation of congruence for segments in the oincaré model is complicated, being based on a way of measuring length that is different from the usual Euclidean way. Congruence of angles has the usual Euclidean meaning and it is the main advantage of the oincaré model over the Klein model. Specifically, if two directed circular arcs intersect at a point A, the number of degrees in the angle they make is by definition the number of degrees in the angle between their tangent rays at A; if one directed circular arc intersects an ordinary ray at A, the number of degrees in the angle they make is by definition the number degrees in the angle between the tangent ray and the ordinary ray at A. Two oincarée lines are parallel if they have no point in common. In oincaré model all axioms of hyperbolic geometry are translated into statements in Euclidean geometry. Hence the oincaré model furnishes a proof that if Euclidean geometry is consistent, so is hyperbolic geometry. The interior upper half-plane also serves as a model for hyperbolic plane, where points are ordinary points in the open upper half-plane and lines are those rays perpendicular to the x-axis and semicircle orthogonal to the x-axis. 4 Model of hyperbolic plane from physics Consider the hyperboloid Σ defined by x + y t = 1, which can be thought as sphere centered at the origin = (0, 0, 0) with radius ı = 1. Let Π denote the plane t = 0 and the disk of Π with center and unit radius. The component Σ := {(x, y, t) R 3 : x + y t = 1, t > 0} can serve as a model for hyperbolic plane, where points are ordinary points of Σ and lines are nonempty intersections of Σ and Euclidean planes through. 5 Inversion in circles roposition 5.1. Given a circle of radius r with center in the Euclidean plane, and a chord T 1 T not to be a diameter of. Let be the intersection of the tangent lines of at

3 T 1, T, called the pole of T 1 T with respect to. Let be the middle point of T 1 T, and M the middle point of segment. Then (a) and = r. (b) T 1 = T, M = M = MT 1 = MT. T 1 M T Figure : Inversion about circle roof. (a) It is clear that is between and, that is,. Note that T 1 and T 1 are right angles, and the right triangles T 1, T 1 have the common angle T 1 = T 1. We have T 1 T 1. Then the corresponding sides of T 1, T 1 are proportional. Hence : T 1 = T 1 :, where T 1 = r, that is, = r. (b) Clearly, T 1 = T. Since T 1 and T are right triangles, the points, T 1,, T lie on a common circle of diameter. So M = M = MT 1 = MT. Definition 1. Let be a circle of radius r with center in the Euclidean plane. For each point, its inverse with respect to is the unique point on the ray r(, ) such that = r. Let E denote the Euclidean plane and Ẽ = E { }. The map from Ẽ to itself by,, is called the inversion in. roposition 5.. Let, be inverses each other about the inversion in circle with center. Then (a) =. (b) is inside is outside. roof. Trivial by definition. Lemma (Circle-Cut roduct roperty). Let be a point not on a circle. Given three lines passing through ; the first one intersects in a pair of points A 1, A, the second intersects in a another pair of points B 1, B, and the third is tangent to at a point D; see Figure 3. Then A 1 A = B 1 B = D. The constant is called the power of with respect to. roof. Since angles that are inscribed in a circle and subtended the same arc are congruent, then A 1 A B 1 = A1 B B 1 and A A 1 B = A B 1 B. Thus A B 1 B A 1. See Figure 3. So the corresponding sides are proportional, that is, A 1 : B 1 = B : A. 3

4 A A D A 1 A 1 C 1 C B 1 B 1 B B Figure 3: Circle cut Let line intersect at in a pair of points C 1, C with C 1 C. Then D is a right triangle. By the ythagorean theorem we have D = D = ( D )( + D ) = ( C 1 )( + C ) = C 1 C. roposition 5.3 (rthogonality and Cocyleness of Inverse oints). Given a circle of radius r with center, and two points, inverse each other with respect to. Let δ be a circle through, intersecting at T 1, T. Then, δ meet orthogonally if and only if δ passes through. roof. : Let δ intersect orthogonally at T 1, T. Then the tangents to δ at T 1, T pass through. So lies outside δ, and ray r(, ) cuts δ at a point Q. The Circle-Cut roduct roperty implies Q = S 1 = r. Hence Q is the inverse of with respect to, that is, = Q δ. See Figure 4. T 1 T δ Q = Figure 4: oincaré lines through a point. : Let δ pass through. Then the center C of δ lies on the perpendicular bisector of. Note that either or. Clearly, point is outside δ. Then through there exist two tangents to δ at T 1, T. Note that T 1, T are not assumed on. n the one hand, = T 1 = T by the Cut-Circle roduct roperty. n the other hand, = r since, are inverses each other with respect to. Then T 1 = T = r. So T 1 = T 1 and T = T. f course, the tangent ray r(t 1, ) of δ at T 1 is orthogonal to at T 1. This means that δ is orthogonal to. See Figure 4. 4

5 Corollary 5.4 (Inversions in two orthogonal circles map each of them onto itself). A circle δ is orthogonal to a circle δ is mapped onto itself by inversion in. roof. Let be a point on δ and its inverse in. Then,, are collinear. If δ is orthogonal to, then lies on δ by roposition 5.3. Conversely, if the inversion in maps δ onto itself, of course δ passes through,, then δ is orthogonal to by roposition 5.3. Given a circle ; the open disk bounded by is called a oincaré disk. Given two points A, B inside, there exists a unique circle σ through A, B and orthogonal to. The circular arc of σ inside is called a oincaré line or -line (= line in hyperbolic geometry) through points A, B in the oincaré disk bounded by. (Axiom 1-3 of incidence geometry are satisfied.) The circular arc between A and B on the -line through A, B is called a oincaré segment or -segment. Definition 3. Let A, B be two points inside a circle, and, Q the endpoints of the hyperbolic line through A, B. The cross-ratio of A, B is defined as (AB, Q) := A BQ AQ B. The hyperbolic distance between two points A, B is defined as d(a, B) := log(ab, Q). Notice that distance does not depend on the the order of A, B. In fact, if (AB, Q) = x then (BA, Q) = 1/x, so d(b, A) = log(1/x) = log x = log x = d(a, B). Two -segments AB, CD are said to be oincaré-congruent if d(a, B) = d(c, D). With this interpretation, Congruence axiom is immediately verified, for the relation of equal hyperbolic distance is an equivalence relation. Figure 5: Limiting parallel oincaré lines. Let A, B, C be points on a -line inside the oincaré disk bounded by, having end points, Q on. We may assume the order A B C Q on the -line. See Figure 6. We have cross-ratio (AB, Q) = ( A BQ )/( AQ B ) < 1. Likewise, (AC, Q) < 1 5

6 A B C Q Figure 6: oincaré line inside a disk. and (BC, Q) < 1. Their logs are negative and d(a, B) + d(b, C) = log(ab, Q) log(bc, Q) Congruence axiom 3 (addition rule) is satisfied. = log[(ab, Q) (BC, Q)] ( A BQ = log AQ B B ) CQ BQ C ( A = log AQ CQ ) C = d(a, C). Definition 4. Let be a point and k a positive number. The dialation with center and ratio k is the transformation of the Euclidean plane that fixes and maps each point to a unique point on ray r(, ) such that = k. Dilation with center and ratio k maps lines to lines, circles to circles, and preserves angles. In fact, choose rectangular coordinates so that is the origin. The dilation is the map (x, y) (x, y ) = k(x, y). Then x = x /k, y = y /k. For line l : ax + by = c, its image under the dilation is ax /k + by/k = c, which is the line ax + by = kc. For a circle : (x a) + (y b) = r, its image is (x /k a) + (y /k b) = r, which is the circle (x ka) + (y kb) = (kr). roposition 5.5. Let be a circle of radius r and center. Let δ be a circle with center C such that is outside δ. Set k = r /p, where p is the power of with respect to δ. (a) Then the image δ of δ under inversion in is a circle δ, obtained from δ by the dilation with center and ratio k. (b) If δ and its inverse in, then the tangent t to δ at and the tangent t to δ at are symmetric about the perpendicular bisector of segment. roof. (a) Let be a point on δ and its inverse with respect to. Let ray r(, ) meet circle δ at another point Q (with Q = when r(, ) is tangent to δ). Then Q = Q = r p = k. Since, Q r(, ), we have = k Q, that is, the vector is the dilation of the vector Q with ratio k. This means that is on the circle δ obtained from δ by the dilation with center and ratio k. So δ = δ. The center C of δ is the same dilation of the center C of δ, that is, C r(, C) and C = k C. (b) Since is the dilation of Q, the tangent t of δ at is parallel to the tangent t of δ at Q. Let t meet t at R and meet t at S. Then S = QR = S. So S is an isosceles triangle. Hence t, t are symmetric about the perpendicular bisector of. See Figure 7. 6

7 t S t t R Q δ Q δ* δ Figure 7: Image of a circle by inversion in is a circle. Lemma 5. Given a circle with center, and two points, Q such that,, Q are not collinear. Let, Q be inverses of, Q in respectively. Then Q Q. Q Q Figure 8: Image of triangle under inversion in circle is similar. roof. Since r(, ) and Q r(, Q), the triangles Q, Q have the common angle Q = Q. Note that = r = Q Q, that is, This means that Q Q. : Q = Q :. roposition 5.6. Given a circle with center. Let l be a line not passing through. Then the image of l under inversion in is a circle δ through, but having removed, and its diameter through is perpendicular to l. roof. Let be the segment perpendicular to l with foot on l. Let X be an arbitrary point on l. Let, X be the inverse points of, X in respectively. Note that r(, ) and X r(, X). Then X X. So X = X is a right angle. Thus X is on the circle δ with diameter. Corollary 5.7. Given a circle with center. Let δ be a circle passing through. Then the image of δ {} under inversion in is a line l not through, and l is parallel to the tangent of δ at. roof. Trivial. 7

8 l X l X X X δ δ Figure 9: Image of a line under inversion in is a circle through center of. roposition 5.8. The directed angle of two circles at their intersection point are preserved by inversion in circle. roof. Let be a circle with center. Let σ, δ be two circles intersecting at a point. Let σ, δ, be the images of σ, δ, under inversion in respectively. Given tangent rays s of σ and t of δ at, and tangent rays s of σ and t of δ at. Then rays s, s are symmetric about the perpendicular bisector of ; so are the rays t, t. Hence the angle between s and t are congruent to the angle between s and t. roposition 5.9. Given a circle with center. Let A, B, C, D be four points distinct from, and A, B, C, D their inverses in respectively. Then the cross-ratio is preserved by inversion in, that is, (AB, CD) = (A B, C D ). roof. Recall (AB, CD) := ( AC BD )/( AD BC ). Consider similar triangles AC C A and AD D A. We have Then Likewise AC : A = A C : C, AD : A = A D : D. Multiplying (1) and (), we obtain AC AD = A C A D D C. (1) BD BC = B D B C C D. () (AB, CD) = AC AD BD BC = A C A D B D B C = (A B, C D ). roposition Given two orthogonal circles, δ with centers, C and radii r, d respectively. Let σ be a circle and l a line, both are orthogonal to. Let σ, l be respectively the images of σ, l under inversion in δ. (a) Then the inversion in δ maps onto, and maps the interior of onto itself. (b) If σ does not pass through C, then σ is a circle orthogonal to. If σ passes through C, then σ is a line passing through the center of. (c) If l does not pass through C, then l is a circle orthogonal to. If l passes through C, then l = C and l = l. (d) Inversion in δ preserves incidence, betweenness, and congruence in the sense of the oincaré disk model inside. 8

9 roof. (a) The former part is implied by the fact that circle orthogonal to an inversion circle is fixed by the inversion, namely, Corollary 5.4. For the latter part, let be a point inside and its inverse in δ. Let r(c, ) intersect at points Q, Q. See Figure 10. Since is Q Q C δ Figure 10: Inversion in δ maps interior of to itself. orthogonal to δ, the point Q must be the inverse of Q in δ by roposition 5.3. Then C C = CQ CQ = d. Note that C,,, Q, Q are collinear and Q Q. We may assume C Q Q without loss of generality. Then CQ < C < CQ. Thus d / CQ > d / C > d / CQ, which is the same as CQ > C > CQ. This means that C Q Q. So is inside. (b),(c) Trivial. (d) The circular arc of σ inside is a -line, and is mapped to the circular arc of σ inside, which is another -line. Given a -line through A, B inside with end points, Q on. The -line through the images A, B of A, B by inversion in δ have end points, Q on. Since (AB, Q) = (A B, Q ), we see that d(a, B) = log(ab, Q) = log(a B, Q ) = d(a, B ). So -segment AB is congruent to -segment A B hyperbolically. Let A B D. Then d(a, D ) = d(a, D) = d(a, B) + d(a, B ) = d(a, B ) + d(b, D ) So B is between A and D. Verification of Congruence Axiom 6 (SAS). Given -triangles ABC and XY Z inside a circle of radius r with center, such that d(a, B) = d(x, Y ), A = X, and d(a, C) = d(x, Z). We need to show that ABC and XY Z are -congruent, that is, the corresponding sides have the same hyperbolic lengths and the corresponding angles have the same measures. Case 1. A = X = (the center of ). Then d(, B) = d(, Y ), d(, C) = d(, Z), and BC = Y Z. Let, Q be end points of -line through points, B and B Q. Then d(, B) = log( BQ / Q B ) = log[(r + B )/(r B ). It follows that B = r(e d(,b) 1)/(e d(,b) + 1). Hence B = Y, C = Z, and BC = Y Z. There exists a unique rigid motion (rotation about, combined with a reflection about a diameter of if necessary) such that the images of B, C under the rigid motion coincide with Y, Z respectively. So -triangles BC and Y Z are -congruent, since rotation about and reflection about diameter of preserve hyperbolic length and angle. 9

10 G β B* C* C σ A B A H α Figure 11: Image of triangle under inversion in circle is similar. Case. General case is reduced to Case 1. Let α be the circle orthogonal to through A, B, and β the circle orthogonal to through A, C. Then α, β meet at A outside, which is the inverse of A in by roposition 5.3. Then, A, A are collinear and A A. Let σ be the circle of radius s with center A, where s = A A A. Then A is inside σ, is outside σ, and, A are inverses each other in σ by definition of inversion in circle. So σ, intersect at two points G, H. Since AA = A A, we have s = ( A A ) A = A A A = A r. Thus A = r + s = H + A H. The ythagorean theorem implies that A H is a Euclidean right triangle with right angle at H. So σ is orthogonal to. See Figure 11. Now the inversion in σ maps inside of onto itself, and in particular maps A to. So the inversion in σ maps the -triangle ABC onto a -triangle B C. Clearly, -triangles ABC and B C are -congruent. Likewise, -triangle XY Z can be mapped onto a -triangle Y Z by an inversion in a circle orthogonal to, and Y Z is -congruent to XY Z. Hence B C and Y Z are -triangles, satisfying d(, B ) = d(, Y ), B C = Y Z, and d(, C ) = d(, Z ). So B C and Y Z are -congruent by Case 1. Therefore BC and Y Z are -congruent. We then have the following theorem. Theorem Two -triangles in the oincaré disk bounded by a circle are -congruent if and only if they can be mapped onto each other by a rotation about the center of, and a succession of inversions in circles orthogonal to and in lines through the center of. Theorem 5.1. Let d be oincaré distance of point to -line l not through inside oincaré disk bounded by circle of radius r. Let Π(d) denote the number of radians in the angle of parallelism of limiting parallel rays through. Then e d = tan[π(d)/]. roof. Without loss of generality we may assume that l is the diameter of and is the segment perpendicular to l with foot on l. Let α be the angle between Euclidean ray r(, Q) and the limiting -line δ parallel to l. Let t be the tangent to δ at. Then t meets l at R. Let S be the common end point of l and δ. Note that both l, δ are tangent to δ. So RS = R S. Set α := Q R = Π(d) and β := RS. Then α + β = π/, that is, β = π/4 α/. Note that = r tan β. Then d = d(, ) = r(r ) log r(r + ) 10 = log r + r = log 1 + tan β 1 tan β,

11 δ t α β l R β S Figure 1: Angle of limiting parallel rays. that is, e d = (1 + tan β)/(1 tan β). Since e ıθ = cos θ + ı sin θ, then e ı(θ 1+θ ) = cos(θ 1 + θ ) + ı sin(θ 1 + θ ), Thus e ı(θ 1+θ ) = e ıθ 1 e ıθ = cos θ 1 cos θ sin θ 1 sin θ + ı(cos θ 1 sin θ + sin θ 1 cos θ ). tan(θ 1 + θ ) = sin(θ 1 + θ ) cos(θ 1 + θ ) = cos θ 1 sin θ + sin θ 1 cos θ cos θ 1 cos θ sin θ 1 sin θ = tan θ 1 + tan θ 1 tan θ 1 tan θ. So tan β = tan(π/4 α/) = (1 tan(α/))/(1 + tan(α/)). Hence e d =, that is, tan(α/) e d = tan(π(d)/). 6 Isomorphism between oincaré model and Klein model Let Σ be a sphere in Cartesian 3-dimensional Euclidean space R 3, given by the equation x 1 + x + x 3 = r. Let be the open disk on the plane x 3 = 0, that is, whose boundary is the circle = {(x 1, x, 0) R 3 : x 1 + x < r }, = {(x 1, x, 0) : x 1 + x = r }. We consider the map F :, induced by the stereographic projection through the lower hemisphere Σ = {(x 1, x, x 3 ) : x 1 + x + x 3 = r, x 3 < 0}. Simplicity let us use complex variable z = x 1 + ıx to denote point (x 1, x ) in. Then F is determined by F (z) z r F (z) =. z r Squaring both sides, we have F (z) / z 1] = 1 F (z) /r, that is, F (z) ( F (z) (1/r + 1/ z ) / z ) = 0. 11

12 Σ N z F(z) Σ Figure 13: Stereographic projection Since F (z) 0 when z 0, we further have F (z) = r z /(r + z ), that is, Writing in rectangular coordinates, we have F (z) = r z/(r + z ). F (x 1, x ) = r (x r + x 1 + x 1, x ). Theorem 6.1. The stereographic projection is an isomorphism from the oincaré model to the Klein model. More specifically, given a circle δ orthogonal to, intersecting at, Q. If A is a point on the arc of δ from to Q inside, then the ray r(, A) meets the chord Q of at F (A). roof. Let C be the center of δ with coordinates (c 1, c ). Since, δ meet orthogonally at, Q, the points,, Q, C lie on a circle σ with diameter C, having center ( c 1, c ) and radius 1 c 1 + c. The equation of σ is ( x 1 c 1 ) + ( x c Then, Q are the solutions of the system { x 1 + x = c 1 x 1 + c x x 1 + x = r, i.e., ) c = 1 + c, i.e., x 1 + x = c 1 x 1 + c x. 4 { c1 x 1 + c x = r x 1 + x = r. Clearly, line c 1 x 1 + c x = r passes though, Q; it must be the equation of line Q. Note A F(A) C Q Figure 14: The map between oincaré model and Klein model. that C is a right angle; the ythagorean theorem implies C = C = c 1 + c r. 1

13 The equation (x 1 c 1 ) + (x c ) = C of δ becomes x 1 + x = c 1 x + c x r. Now if A = (a 1, a ) lies on δ and F (A) = (b 1, b ) is the image under F, then and subsequently, Hence b j = a 1 + a + r = c 1 a 1 + c a > 0, r a j r + a 1 + a which means that (b 1, b ) lies on the line Q. = r a j c 1 a 1 + c a, j = 1,. c 1 b 1 + c b = r, roposition 6.. Given a point in side a circle with center and radius r, and its inverse in. Let AB be a chord of with midpoint. Then the circle δ with center and radius A is orthogonal to. Moreover, the inverse of in δ is. roof. Note that A A = r and r = A = d. 7 hilosophical Implications 7.1 What is the geometry of physical space If Euclidean geometry is consistent, so is hyperbolic geometry. Then two geometries are equally consistent. Logically speaking, hyperbolic geometry deserves to be put on an equal footing with Euclidean geometry. Engineering and architecture are evidence that Euclidean geometry is extremely useful for ordinary measurement of distances that are not large. However, the representational accuracy of Euclidean geometry is less certain when deal with larger distances. Because of experimental error, a physical experiment can never prove conclusively that space is Euclidean it can prove only that space is non-euclidean. According to Einstein, space and time are inseparable and the geometry of spacetime is affected by matter, so that light rays indeed curved by the gravitational attraction of masses. Space is no longer conceived of as an empty Newtonian box whose contours are unaffected by the rocks put into it. The problem is much more complicated than Euclidean or non-euclidean neither of the geometries is adequate for the present conception of space. This does not diminish the historical importance of Euclidean and non-euclidean geometries. Einstein said, To this interpretation of [hyperbolic] geometry I attached great importance, for should I not have acquainted with it, I never would have been able to develop the theory of relativity. Henri oincaré said: If geometry were an experimental science, it would not be an exact science. It would be subjected to continual revision.... The geometrical axioms are therefore neither synthetic a priori intuitions nor experimental facts. They are conventions. ur choice among all possible conventions is guided by experimental facts; but it remains free, and is only limited by the necessity of avoiding every contradiction, and thus it is that postulates may remain rigorously true even when experimental laws which have determined their adoption are only approximate. In other words, the axioms of geometry (I do not speak of those of arithmetic) are only definitions in disguise. 13

14 We might think that Euclidean geometry is the most convenient it is for ordinary engineering, but not for the theory of relativity. ne school, which includes Newton, Helmboltz, Russel, and Whitehead, contends that space has an intrinsic metric or standard measurement. The other school, which includes Riemann, oincaré, Clifford, and Einstein, contends that a metric stipulated by convention. The discussion can become very subtle. 7. What is mathematics about? Geometry is not about light rays, but the path of a light ray is one possible physical interpretation of the undefined geometric term line. Bertrand Russell once said that mathematics is the subject in which we do not know we are talking about nor what we say is true. This is because certain primitive terms such as point, line, and plane, are undefined and could just as well be replaced with other terms without affecting the validity of results. Gottlob Frege ( ), who is considered the founder of modern mathematical logic, wrote to Hilbert: I give the name of axioms to propositions which are true, but which are not demonstrated because their knowledge proceeds from a source which is not logical, which we may call space time. The truth of the axiom implies of course that they do not contradict each other. That is to me the criterion of truth and existence. Euclidean and hyperbolic geometries were equally consistent, so they exist and are both true. The discovery that Euclidean geometry was not absolute truth had a liberating effect on mathematicians, who now feel free to invent any set of axioms they wish and deduce conclusions from them. In fact, this freedom may account for the great increase in the scope and generality of modern mathematics. In a 1961 address, Jean Dieudonné remarked on Gauss discovery of non-euclidean geometry: [It] was a turning point of capital significance in the history of mathematics, marking the first step in a new conception of the relation between the real world and the mathematical notions supposed to account for it; with Gauss discovery, the rather naive point of view that mathematical objects were only ideas (in the latonic sense) of sensory objects became untenable, and gradually gave way to a clearer comprehension of the much greater complexity of the question, wherein it seems to us today that mathematics and reality are almost completely independent, and their contacts more mysterious than ever. 7.3 Controversy about foundation of mathematics It would be misleading to say that mathematics is just a formal game played with symbols and having no broader significance. Mathematicians do not arbitrarily make up axioms and deduce their conclusions. Axioms must be lead to interesting and fruitful results. f course, some axioms that appear uninteresting may turn out to have surprising consequences this was the case with the hyperbolic axiom, which was virtually ignored during the lifetimes of Gauss, Bolyai, and Lobachevsky. If, however, axiom systems do not bear interesting results, they become neglected and eventually forgotten. Arguing against the description of mathematics as a formal game, R. Courant and H. Robbins (in their fine book What is Mathematics?) insist that a serious threat to the very life of science is implied in the assertion that mathematics is nothing but a system of conclusions drawn from definitions and postulates that must be consistent but otherwise may be created by the free will of the mathematician. If this description were accurate, mathematics could not attract any intelligent person. It would be a game with definition, rules and syllogisms, without motivation or goal. Hermann Weyl reamrked: The construction of the mathematical mind are at the same time free and necessary. The individual mathematician feels free to define his notions and to 14

15 set up his axioms as he pleased. But the question is, will he get his fellow mathematicians interested in the constructs of his imagination? Leopold Kronecker said: God cerated whole numbers all else is manmade. Axiom of Choice (AC): Given sets S i indexed by i I. There exist elements a i S i for all i I. Continuum Hypothesis (CH): No cardinal number between ℵ 0 (the cardinality of the set of positive integers) and c (the cardinal number of the set of real numbers). Kurt Gödel created a model of the other A-Z axioms in which both AC and CH were true; that demonstrated the impossibility of disproving them. In 1963, models were created in which either AC or CH or both were false. So AC and CH are indepedent of the other Z-F axioms and of each other. There exists an equally valid non-cantorian set theory, just as there is an equally valid non-euclidean geometry. Gödel s Incompleteness Theorem: There will always be valid statements that cannot be demonstrated from systems of axioms that are broad enough to include arithmetic. In other words, Gödel provided a formal demonstration of the inadequacy of formal demonstration. 15