Warm-up

D C B A Warm-up • Given Saccheriquadrilateral ABDC in a Hilbert Plane satisfying the acute angle hypothesis. Last time we proved that the perpendicular from C to intersects BD at a point E such that • B * E * D. • In a Hilbert Plane satisfying the acute angle hypothesis, here are some possible conclusions about Saccheriquadrilateral ABDC. Determine whether each is true or false and prove your claim. • The point P at which the midline segment MN of Saccheri quadrilateral ABDC intersects • CE is the midpoint of CE. • PE  MB. • CP PE • CP > AM • There exist infinitely many points X on CE such that • AC > AX. • 6. There exist infinitely many points X on CE such that • AX > AC. N   E  P  M

D C B A Warm-up Given Saccheriquadrilateral ABDC in a Hilbert Plane satisfying the acute angle hypothesis. Last time we proved that the perpendicular from C to intersects BD at a point E such that B * E * D. In a Hilbert Plane satisfying the acute angle hypothesis, here are some possible conclusions about Saccheriquadrilateral ABDC. Determine whether each is true or false and prove your claim. The point P at which the midline segment MN of Saccheri quadrilateral ABDC intersects CE is the midpoint of CE. N   E  P  M

D C B A Warm-up Given Saccheriquadrilateral ABDC in a Hilbert Plane satisfying the acute angle hypothesis. Last time we proved that the perpendicular from C to intersects BD at a point E such that B * E * D. In a Hilbert Plane satisfying the acute angle hypothesis, here are some possible conclusions about Saccheriquadrilateral ABDC. Determine whether each is true or false and prove your claim. The point P at which the midline segment MN of Saccheri quadrilateral ABDC intersects CE is the midpoint of CE. N  False  E Construct PD Since MN is the midline segment of a Saccheri quadrilateral, MN is perpendicular to CD. CNP  DNP by SAS CP  DP But in PDE, PD is opposite right angle PED, making it the longest side of the triangle. Therefore, PD > PE so that CP > PE also.  P  M

D C B A Warm-up Given Saccheriquadrilateral ABDC in a Hilbert Plane satisfying the acute angle hypothesis. Last time we proved that the perpendicular from C to intersects BD at a point E such that B * E * D. In a Hilbert Plane satisfying the acute angle hypothesis, here are some possible conclusions about Saccheriquadrilateral ABDC. Determine whether each is true or false and prove your claim. PE  MB. N   E  P  M

D C B A Warm-up Given Saccheriquadrilateral ABDC in a Hilbert Plane satisfying the acute angle hypothesis. Last time we proved that the perpendicular from C to intersects BD at a point E such that B * E * D. In a Hilbert Plane satisfying the acute angle hypothesis, here are some possible conclusions about Saccheriquadrilateral ABDC. Determine whether each is true or false and prove your claim. PE  MB. False Quad MBEP is a Lambert quadrilateral, and since D is acute, PE > MB by Corollary 3 of the Uniformity Theorem. N   E  P  M

D C B A Warm-up Given Saccheriquadrilateral ABDC in a Hilbert Plane satisfying the acute angle hypothesis. Last time we proved that the perpendicular from C to intersects BD at a point E such that B * E * D. In a Hilbert Plane satisfying the acute angle hypothesis, here are some possible conclusions about Saccheriquadrilateral ABDC. Determine whether each is true or false and prove your claim. 3. CP  PE. N   E  P  M

D C B A Warm-up Given Saccheriquadrilateral ABDC in a Hilbert Plane satisfying the acute angle hypothesis. Last time we proved that the perpendicular from C to intersects BD at a point E such that B * E * D. In a Hilbert Plane satisfying the acute angle hypothesis, here are some possible conclusions about Saccheriquadrilateral ABDC. Determine whether each is true or false and prove your claim. 3. CP  PE. True (CP  PE based on part 1) Construct PD. MN  CD since MN is the midline segment of a Saccheriquadrilateral. 3. Then CPN  DPN (SAS) so that CP  PD . But in right triangle PED, PD > PE by Prop 4.5. Therefore, CP > PE by Prop 3.13(c). N   E  P  M

D C B A Warm-up Given Saccheriquadrilateral ABDC in a Hilbert Plane satisfying the acute angle hypothesis. Last time we proved that the perpendicular from C to intersects BD at a point E such that B * E * D. In a Hilbert Plane satisfying the acute angle hypothesis, here are some possible conclusions about Saccheriquadrilateral ABDC. Determine whether each is true or false and prove your claim. 4. CP > AM. True We proved in part 3 that CP > PE and we proved in part 2 that PE > MB = AM Therefore, CP > AM N   E  P  M

D C B A Warm-up Given Saccheriquadrilateral ABDC in a Hilbert Plane satisfying the acute angle hypothesis. Last time we proved that the perpendicular from C to intersects BD at a point E such that B * E * D. In a Hilbert Plane satisfying the acute angle hypothesis, here are some possible conclusions about Saccheriquadrilateral ABDC. Determine whether each is true or false and prove your claim. 5. There exist infinitely many points X on CE such that AC > AX. N   E  P  M

D C B A Warm-up Given Saccheriquadrilateral ABDC in a Hilbert Plane satisfying the acute angle hypothesis. Last time we proved that the perpendicular from C to intersects BD at a point E such that B * E * D. In a Hilbert Plane satisfying the acute angle hypothesis, here are some possible conclusions about Saccheriquadrilateral ABDC. Determine whether each is true or false and prove your claim. 5. There exist infinitely many points X on CE such that AC > AX. True Since ABEC is a Lambert quadrilateral with acute fourth angle, CE > AB by Corollary 3 of the Uniformity Theorem. Choose the unique point Q on CE with C * Q * E such that EQ  AB, and construct AQ. Then BEQA is a Saccheri quadrilateral. 4. AQE is acute by the Corollary 1 of the Uniformity Theorem. Therefore, its supplement AQC is obtuse. In AQC, AC > AQ by Prop 4.5. For any point X between C and Q,CXA > AQC by the EA Theorem, and so AC > AX.  E  Q

D C B A Warm-up Given Saccheriquadrilateral ABDC in a Hilbert Plane satisfying the acute angle hypothesis. Last time we proved that the perpendicular from C to intersects BD at a point E such that B * E * D. In a Hilbert Plane satisfying the acute angle hypothesis, here are some possible conclusions about Saccheriquadrilateral ABDC. Determine whether each is true or false and prove your claim. 6. There exist infinitely many points X on CE such that AX > AC.  E  X Q 

D C B A Warm-up Given Saccheriquadrilateral ABDC in a Hilbert Plane satisfying the acute angle hypothesis. Last time we proved that the perpendicular from C to intersects BD at a point E such that B * E * D. In a Hilbert Plane satisfying the acute angle hypothesis, here are some possible conclusions about Saccheriquadrilateral ABDC. Determine whether each is true or false and prove your claim. 6. There exist infinitely many points X on CE such that AX > AC. An important observation Since BEQA is a Saccheriquadrilateral, is parallel to . However, is also parallel to . Then any line through A containing a ray between ray and ray must also be parallel to .  E  X Q  So the existence of a Saccheri quadrilateral with acute summit angles implies a negation of Hilbert’s Parallel Axiom!!!!!

Homework – Review Exercises 2, 6, 8, 11, 13, 15, 17, 19, 20 are correct.

Proof of SAA criterion: • (1) Assume AB is not congruent to DE. • Then AB < DE or DE < AB. • If DE < AB, then there is a point G between A and B such that AG  DE. • Then CAG  FDE. • HenceAGC  E. • It follows that AGC  B. • This contradicts a certain theorem. • Therefore, DE is not less than AB. • By a similar argument using point H between D and E, AB is not less than DE. • Hence AB  DE. • Therefore, ABC DEF. F C E B D A G

13. Prove Proposition 4.10 Proposition 4.10 Hilbert’s Euclidean parallel postulate  if kl, m  k, and n  l, then either m = n or mn. 1. Hilbert’s Euclidean parallel postulate if kl, m  k, and n  l, then either m= n or mn. n l Q  k  P m Proposition 4.7 Hilbert’s Euclidean parallel postulate  if a line intersects one of two parallel lines, then it also intersects the other.

13. Prove Proposition 4.10 Proposition 4.10 Hilbert’s Euclidean parallel postulate  if kl, m  k, and n  l, then either m = n or mn. 1. Hilbert’s Euclidean parallel postulate if kl, m  k, and n  l, then either m= n or mn. n l P  Q  k  Q  P m Proposition 4.8 Hilbert’s Euclidean parallel postulate  converse to the alternate interior angles theorem.

13. Prove Proposition 4.10 Proposition 4.10 Hilbert’s Euclidean parallel postulate  if kl, m  k, and n  l, then either m = n or mn. 2. if kl, m  k, and n  l, then either m= n or mn Hilbert’s Euclidean parallel postulate. Given a line l and a point P not on l, we must prove there is at most one parallel to l through P P Construct the perpendicular line t to l through P, and the perpendicular k to line t at P.  Any line through point P other than k must intersect line l by prop 4.7. Therefore, k is the only line through P that is parallel to l. k t Did we use the hypothesis??? l Proposition 4.7 Hilbert’s Euclidean parallel postulate  if a line intersects one of two parallel lines, then it also intersects the other.

Saccheri’s Angle Theorem • For any Hilbert plane, • If there exists a triangle whose angle sum is < 180, then every triangle has an angle sum < 180, and this is equivalent to the fourth angles of Lambert quadrilaterals and the summit angles of Saccheri quadrilaterals being acute. • If there exists a triangle with angle sum = 180, then every triangle has angle sum = 180, and this is equivalent to the plane being semi-Euclidean. • If there exists a triangle whose angle sum is > 180, then every triangle has an angle sum > 180, and this is equivalent to the fourth angles of Lambert quadrilaterals and the summit angles of Saccheri quadrilaterals being obtuse. All Lambert and all Saccheri quadrilaterals are rectangles. Saccehri-Legendre Theorem In an Archimedean Hilbert plane, the angle sum of every triangle is  180

Archimedes’ Axiom: If CD is any segment, A any point, and r any ray with vertex A, then for ever point B ≠ A on r there is a number n such that when CD is laid off n times on r starting at A, a point E is reached such that n · CD  AE and either B = E or B is between A and E. C   D In this illustration, n = 6. A         E B Intuitively, this says we can choose an arbitrary length as a “unit length” and every other segment has a finite length with respect to this unit.

GirolamoSaccheri (1667-1733) was a Jesuit Priest. We saw that the summit angles of his quadrilaterals are congruent to each other, and there are three possible geometries according as those angles are acute, right, or obtuse. Saccheri’s idea was to demonstrate that the acute and obtuse angle cases lead to contradictions, leaving the right angle case – the case where the Saccheri quadrilateral is a rectangle – as the only possibility. By assuming the generally accepted Archimedes’ axiom, Saccheri successfully eliminated the case of the obtuse angle (Saccheri-Legendre theorem). But however hard he tried, Saccheri could not squeeze a contradiction out of the “inimical acute angle hypothesis.” Finally, he exclaimed in frustration: “the hypothesis of the acute angle is absolutely false, because it is repugnant to the nature of the straight line.” D C B A

GirolamoSaccheri (1667-1733) was a Jesuit Priest We saw that the summit angles of his quadrilaterals are congruent to each other, and there are three possible geometries according as those angles are acute, right, or obtuse. Saccheri’s idea was to demonstrate that the acute and obtuse angle cases lead to contradictions, leaving the right angle case – the case where the Saccheri quadrilateral is a rectangle – as the only possibility. By assuming the generally accepted Archimedes’ axiom, Saccheri successfully eliminated the case of the obtuse angle (Saccheri-Legendre theorem). But however hard he tried, Saccheri could not squeeze a contradiction out of the “inimical acute angle hypothesis.” Finally, he exclaimed in frustration: “the hypothesis of the acute angle is absolutely false, because it is repugnant to the nature of the straight line.” It has been claimed by one anonymous writer that in Saccheri’s time, the existence of a valid non-Euclidean geometry was “quite literally, unthinkable – not impossible, not wrong, but unthinkable!”

From the textbook (Chapter 5) You may feel that to deny the Euclidean parallel postulate would go against common sense. Albert Einstein once said that “common sense is, as a matter of fact, nothing more than layers of preconceived notions stored in our memories and emotions for the most part before age eighteen.” In this chapter we will examine a few illuminating attempts to prove Euclid’s parallel postulate.. .. It should be emphasized that most of these attempts were made by outstanding mathematicians of their time, not incompetents. And even though each attempt was flawed, the effort was often not wasted: for assuming that all but one step can be justified, when we detect the flawed step, we find another statement which to our surprise is equivalent to the parallel postulate.

Proclus (410-485) – seven centuries after Euclid This [Euclid’s parallel postulate] ought to be struck out of the postulates altogether; for it is a statement involving many difficulties…. The statement that since the two lines converge more and more as they are produced, they will sometime meet is plausible but not necessary…. It is alien to the special character of postulates. Here is Proclus “proof”

Given two parallel lines l and m. Suppose line n intersect line m at point P. We wish to show n intersects l also. Let Q be the foot of the perpendicular from P to l. If n coincides with , then it intersects l at Q and we are finished. Otherwise, one ray of n emanating from P lies between and a ray of m. Take X to be the foot of the perpendicular from Y to m. R X P m  Y n l Z Q

As Y recedes endlessly from P on n (applying Archimedes’ axiom), segment XY will eventually become greater than segment PQ (by Aristotle’s axiom). Aristotle’s Angle Unboundedness Axiom: Given any side of an acute angle and any segment AB, there exists a point Y on the given side of the angle such that if X is the foot of the perpendicular from Y to the other side of the angle, then XY > AB. R X P m  Y n l Z Q

As Y recedes endlessly from P on n (applying Archimedes’ axiom), segment XY will eventually become greater than segment PQ (by Aristotle’s axiom). At that stage, Y must be on the other side of l, so that n must have crossed l. Where is the hidden assumption (the flaw in the proof)? R X P m  Y n l Z Q

As Y recedes endlessly from P on n (applying Archimedes’ axiom), segment XY will eventually become greater than segment PQ (by Aristotle’s axiom). At that stage, Y must be on the other side of l, so that n must have crossed l. Where is the hidden assumption (the flaw in the proof)? R X P m  Y n l Z Q l

Adrien-Marie Legendre (1752-1833) To understand Legendre’s “proof,” we need to again review the Saccehri-Legendre Theorem, part (b) of Saccheri’s angle theorem, the definition of a semi-Euclidean plane, and introduce the definition of the defect of a triangle. Saccehri-Legendre Theorem In an Archimedean Hilbert plane, the angle sum of every triangle is  180 Definition: A Hilbert plane is called semi-Euclidean if all Lambert quadrilaterals and all Saccheri quadrilaterals are rectangles. Saccheri’s Angle Theorem For any Hilbert plane, (b) If there exists a triangle with angle sum = 180, then every triangle has angle sum = 180, and this is equivalent to the plane being semi-Euclidean. Definition: The defect d of a triangle is the difference between 180 and its angle sum. (Based on the Saccehri-Legendre Theorem, d 0 in an Archimedean Hilbert plane.) Legendre will attempt to prove that the angle sum of every triangle is 180 (i.e. d = 0).

Suppose there exists a triangle ABC having defect d ≠ 0. , so one of the angles of ABC must be acute (say A). By the Saccheri-Legendre theorem, d > 0. There exists a ray emanating from B and on the opposite side of from A such that BC  ACB. B P  A C

Suppose there exists a triangle ABC having defect d ≠ 0. , so one of the angles of ABC must be acute (say A). By the Saccheri-Legendre theorem, d > 0. There exists a ray emanating from B and on the opposite side of from A such that BC  ACB (by axiom C-4). Choose point D on so that BD  AC (by axiom C-1). Then ABC  DCB (by SAS) making BC  BCD. Therefore, AC BD and AB DC (by AIA). Therefore, there is a line l through D that intersects in a point B1 ≠ A and in a point C1 ≠ A. This implies D is in the interior of A. B1 B D P   A C C1

Suppose there exists a triangle ABC having defect d ≠ 0. , so one of the angles of ABC must be acute (say A). By the Saccheri-Legendre theorem, d > 0. There exists a ray emanating from B and on the opposite side of from A such that BC  ACB (by axiom C-4). Choose point D on so that BD  AC (by axiom C-1). Then ABC  DCB (by SAS) making BC  BCD. Therefore, AC BD and AB DC (by AIA). Therefore, there is a line lthrough D that intersects in a point B1 ≠ A and in a point C1 ≠ A. This implies D is in the interior of A. Let’s show that A * B * B1. Suppose B1 is on segment AB. Then A and B1 would be on the same side of . Since , points A and C1 are on the same side of . Then B1 and C1 are on the same side of by axiom B-4. But D lies in the interior of A, so B1 * D * C1. B1 Similarly, we can show A * C * C1 . Contradiction - therefore, A * B * B1 . B Since ABC  DCB, they must have the same defect d. D P   A C C1 Claim: The defect of AB1C1 > 2d. (Let’s assume it for now and come back and prove it once we finish.)

The defect of AB1C1  2d. Lets’ repeat the WHOLE process using AB1C1 as the initial triangle. Then the defect of AB2C2  4d. Repeating this process n times, we obtain a triangle with defect greater than or equal to 2nd. But as n increases, 2ndincreases without limit, yet the defect of a triangle cannot be more than 180. CONTRADICTION – so the defect of ABC must be 0, making its angle sum 180 B2 B1 D2  B D P  A C C1 C2

Proof that the defect of AB1C1  2d. (Remember that since ABC  DCB so they have the same defect, d.) Prove that 180 – (a + e + k) > 2d. We have also proven that the defect of a triangle is the sum of the defects of the triangles into which it has been divided. Let the defect of BB1D = d1 and the defect of CC1D = d2 e a + b + c = 180 – d e + f + g = 180 – d1 h + k + m = 180 – d2 x + y + z = 180 – d g add these f x b z h a y c m k a + b + c + e + f + g + h + k + m + x + y + z = 720 – (2d + d1 + d2) B1 + 180 + a + e + k = 720 – (2d + d1 + d2) 180 + 180 B D P  Therefore, a + e + k = 180 – (2d+ d1 + d2), which can be rewritten as 180 – (a + e + k) = (2d + d1 + d2) > 2d. Therefore, the defect of AB1C1 = (2d + d1 + d2) > 2d.  A C C1

Suppose there exists a triangle ABC having defect d ≠ 0. , so one of the angles of ABC must be acute (say A). By the Saccheri-Legendre theorem, d > 0. There exists a ray emanating from B and on the opposite side of from A such that BC  ACB (by axiom C-4). B P  A C

Suppose there exists a triangle ABC having defect d ≠ 0. , so one of the angles of ABC must be acute (say A). By the Saccheri-Legendre theorem, d > 0. There exists a ray emanating from B and on the opposite side of from A such that BC  ACB (by axiom C-4). Choose point D on so that BD  AC (by axiom C-1). Then ABC  DCB (by SAS) making BC  BCD. Therefore, AC BD and AB DC (by AIA). This implies D is in the interior of A. Therefore, there is a line l through D that intersects in a point B1 ≠ A and in a point C1 ≠ A. B1 B D P   A C C1

Suppose there exists a triangle ABC having defect d ≠ 0. , so one of the angles of ABC must be acute (say A). By the Saccheri-Legendre theorem, d > 0. There exists a ray emanating from B and on the opposite side of from A such that BC  ACB (by axiom C-4). Choose point D on so that BD  AC (by axiom C-1). Then ABC  DCB (by SAS) making BC  BCD. Therefore, AC BD and AB DC (by AIA). This implies D is in the interior of A. Therefore, there is a line l through D that intersects in a point B1 ≠ A and in a point C1 ≠ A. Let’s show that A * B * B1. Suppose B1 is on segment AB. Then A and B1 would be on the same side of . Since , points A and C1 are on the same side of . Then B1 and C1 are on the same side of by axiom B-4. But D lies in the interior of A, so B1 * D * C1. B1 Similarly, we can show A * C * C1 . Contradiction - therefore, A * B * B1 . B Since ABC  DCB, they must have the same defect d. D P   A C C1 The defect of AB1C1 > 2d.

The defect of AB1C1  2d. Lets’ repeat the WHOLE process using AB1C1 as the initial triangle. Then the defect of AB2C2  4d. Repeating this process n times, we obtain a triangle with defect greater than or equal to 2nd. But as n increases, 2ndincreases without limit, yet the defect of a triangle cannot be more than 180. CONTRADICTION – so the defect of ABC must be 0, making its angle sum 180 B2 B1 D2  B D P  A C C1 C2

Letter from the Hungarian mathematician FarkasBolyai (1775-1856) to his son, mathematician Janos Bolyai (1802-1860) who was embarking on his own exploration of the parallel postulate You must not attempt this approach to parallels. I know this way to its very end. I have traversed this bottomless night, which extinguished all light and joy of my life. I entreat you, leave the science of parallels alone….I thought I would sacrifice myself for the sake of truth. I was ready to become a martyr who would remove the flaw from geometry and return it purified to mankind. I accomplished monstrous, enormous labors; my creations are far better than those of others and yet I have not achieved complete satisfaction….I turned back when I saw that no man can reach the bottom of the night. I turned back unconsoled, pitying myself and all mankind. I admit that I expect little from the deviation of your lines. It seems to me that I have been in these regions; that I have traveled past all reefs of this infernal Dead Sea and have always come back with broken mast and torn sail. The ruin of my disposition and my fall date back to this time. I thoughtlessly risked my life and happiness.

Janos Bolyai’s response to his father It is now my definite plan to publish a work on parallels as soon as I can complete and arrange the material and an opportunity presents itself; at the moment I still do not clearly see my way through, but the path which I have followed gives positive evidence that the goal will be reached, if it is at all possible; I have not quite reached it, but I have discovered such wonderful things that I was amazed, and it would be an everlasting piece of bad fortune if they were lost. When you, my dear father, see them, you will understand; at present I can say nothing except this: that out of nothing I have created a strange new universe. All that I have sent you previously is like a house of cards in comparison with a tower. I am no less convinced that these discoveries will bring me honor than I would be if they were completed.

JnosBolyai From the textbook Jnos Bolyai did publish his discoveries , as a 26 page appendix to a mathematical treatise by his father (the Tentamen, 1831). Farkas sent a copy to his friend, the German mathematician Carl Friedrich Gauss, undisputedly the foremost mathematician of his time. FarkasBolyai had become friends with Gauss 35 years earlier when they were both students in Gttingen . . . Jnos was 13 years old when he mastered the differential and integral calculus. His father wrote to Gauss begging him to take the young prodigy into his household as an apprentice mathematician. Gauss never replied to his request perhaps because he was having enough trouble with his own son Eugene, who had run away from home). Fifteen years later, when Farkas mailed the Tentamen to Gauss, he certainly must have felt that his son had vindicated his belief in him, and Jnos must have expected Gauss to publish his achievement. FarkasBolyai Letter from Gauss to FarkasBolyai: If I begin with the statement that I dare not praise such a work, you will of course be startled for a moment: but I cannot do otherwise; to praise it would amount to praising myself; for the entire content of the work, the path which your son has taken, the results to which he is led, coincide almost exactly with my own meditations which have occupied my mind for from thirty to thirty-five years. . . . My intention was, in regard to my own work, of which very little up to the present has been published, not to allow it to become known during my lifetime. Most people have not the insight to understand our conclusions . . . .

Another actor in the historical drama came along to steal the limelight from both J. Bolyai and Gauss: the Russian mathematician NilolaiIvanovichLobachevsky (1792-1856). He was the first to actually publish an account of non-Euclidean geometry, in 1829. Lobachevsky initially called his geometry “imaginary,” then later “pangeometry.” His work attracted little attention on the continent when it appeared because it was written in Russian. . . Nevertheless, Lobachevsky courageously continued to publish further articles in Russian and then a treatise in 1840 in German, which he sent to Gauss. In an 1846 letter to Schumacher, Gauss reiterated his own priority in developing non-Euclidean geometry but conceded that “Lobachevsky carried out the task in a masterly fashion and in a truly geometric spirit.” At Gauss’ secret recommendation, Lobachevsky was elected to the Gttingen Scientific Society. (Why didn’t Gauss recommend Jnos?) It was not until after Gauss’ death in 1855, when his correspondence was published, that the mathematical world began to take non-Euclidean ideas seriously.

Homework Read Chapter 5 (it’s fascinating) Review exercises 1 – 18 (they will test how well you read the chapter) Exercises 5, 7

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