1. Construction steps: 1.Construct a circle with center A. 2.Choose point P on the circle. 3.Construct radius PA. 4.Choose point B inside the circle. 5.Construct line segment PB. 6.Construct the perpendicular bisector of PB through midpoint M. 7.Label the intersection of PA and the perpendicular bisector, point E. 8.Display the lengths of AE and BE, and their sum. 9.Trace point E 10. Animate point P. 11. Point E will trace an ellipse with foci A and B. An ellipse is the locus (set) of points the sum of whose distances from two fixed points (called foci) is constant. Prove that AE +BE is constant as P moves around circle A. EM ⊥ BP by construction. ∠ EBM and ∠ EPM are congruent right angles. MP ≅ MB by the definition of bisector. EM ≅ EM by the reflexive property. △ PME ≅ △ BME by SAS. AP is the radius of a circle and is constant. AP = AE + EP. BE ≅ EP. BE + AE ≅ AP, and is constant.

2. A Brief Introduction to Non-Euclidean Geometry

3. Euclid’s fifth postulate (the Parallel Postulate) states: Given a line and a point not on the line, exactly one line parallel to the given line may be drawn through the point. What if we disagreed with Euclid’s Parallel Postulate? Would there be any theorems in Euclidean Geometry that we could no longer trust? Another way of asking this question is “What theorems in Euclidean Geometry depend upon the parallel postulate for proof?”

4. For example, consider the statement: Alternate interior angles of parallel lines are congruent. P Q l m

5. If we allow a second line n parallel to line l through P, P Q l m n Then the two angles at point P would be congruent to each other, which they cannot be if n and m are different lines. It turns out that congruent alternate interior angles depend on Euclid’s Parallel Postulate so, if we deny Euclid’s Parallel Postulate, we can no longer trust the alternate interior angles theorem.

6. P Q A B C + +  PCA  ACB  QCB = 180°  A BB Now consider our proof of the theorem: The sum of the measures of the angles of a triangle is 180°. Proof: Construct parallel to. Congruent alternate interior angles So, since the proof that the sum of the measures of the angles of a triangle is 180  rests on congruent alternate interior angles which rests on Euclid’s Parallel Postulate, we can no longer trust the sum of the angles of a triangle theorem if we deny Euclid’s Parallel Postulate. CC

7. Consider the following proof of the theorem: The sum of the measures of the angles of a quadrilateral is 360°. The measures of the angle of  ABD is 180  The measures of the angle of  CBD is 180  Therefore, the sum of the measures of the angles of quadrilateral ABCD is 180  + 180  = 360  Since the proof that the sum of the measures of the angles of a quadrilateral is 360  rests on the sum of the measures of the angles of a triangle is 180  which rests on congruent alternate interior angles, which rests on Euclid’s Parallel Postulate, we no longer accept that the sum of the angles of a quadrilateral is 360  if we deny Euclid’s Parallel Postulate.

8. In order to prove that this quadrilateral is a rectangle, we need to prove that the fourth angle is a right angle, and to do this we need to know the sum of the angles is 360 . Therefore, we cannot even assume rectangles exist if we deny Euclid’s Parallel Postulate. Is this quadrilateral a rectangle?

9. For centuries, mathematicians attempted to prove Euclid’s fifth postulate (the parallel postulate) using the first four. The development of Non-Euclidean Geometry was an outgrowth of these attempts, most often using an indirect proof (proof by contradiction). So what contradictions did mathematicians come up with in assuming the above negation of the Parallel Postulate? Given a line and a point not on the line, exactly one line parallel to the given line may be drawn through the point. Euclid’s Parallel Postulate more than one line One Negation of Euclid’s Parallel Postulate

10. Lets’ examine one attempted proof of the parallel postulate by the Hungarian mathematician Janos Bolyai (1802-1860). Bolyai’s “proof” makes use of the following well known relationship that we discussed in class a few days ago:

11.    A B C   Given a triangle, the circumcenter (center of the circle passing through the vertices of the triangle) is the intersection of the perpendicular bisectors of the sides.  This also assures that a circle can be constructed through any three non-collinear points. 

12. We must prove that n is not parallel to l (i.e. we must prove that n intersects l ). Begin with line PQ and construct lines m and l perpendicular to PQ at points P and Q, respectively. Q P m l Let n be any other line through P. n Then l and m are parallel because they are perpendicular to the same line.

13. Q P m l Choose a point A between P and Q. A  Let B be the unique point on line PQ such that Q is between A and B and AQ = QB. B  Let C be the unique point on ray AR such that R is between A and C and AR = RC.  C R Construct a perpendicular from A to line n meeting n at a point R.  n

14. Q P m l n A  B   C The center of this circle is the intersection of the perpendicular bisectors of sides AB and AC. Therefore, m is the only line through P that is parallel to l. Since points A, B, and C are the vertices of a triangle, there is a unique circle passing through points A, B, and C. But the perpendicular bisector of side AB is l, and the perpendicular bisector of side AC is n, so l and n must intersect at the center of this circle.  R  Is there a flaw in Bolyai’s “Proof”?

15. 77 Statements Equivalent to Euclid’s Parallel Postulate 77 Statements whose proofs depend on Euclid’s Parallel Postulate.

16. A B C    By using these, Bolyai was implicitly assuming the parallel postulate in attempting to prove the parallel postulate!

17. For centuries, mathematicians attempted to prove Euclid’s fifth postulate (the parallel postulate) using the first four. The development of Non-Euclidean Geometry was an outgrowth of these attempts, most often using an indirect proof (proof by contradiction). What other contradictions did mathematicians come up with in assuming the above negation of the Parallel Postulate? Given a line and a point not on the line, exactly one line parallel to the given line may be drawn through the point. Euclid’s Parallel Postulate more than one line One Negation of Euclid’s Parallel Postulate

18. Theorem: Every right angle is obtuse. Theorem: Rectangles do not exist. Theorem: Parallel lines do not form congruent alternate interior  s.

19. Theorem: Every right angle is obtuse. Theorem: Rectangles do not exist. Theorem: Parallel lines are not always the same distance apart. Theorem: Parallel lines do not form congruent alternate interior  s.

20. Theorem: Every right angle is obtuse. Theorem: Rectangles do not exist. Theorem: Parallel lines are not always the same distance apart. Theorem: In the right triangle shown, a 2 + b 2  c 2. a b c Theorem: Parallel lines form congruent alternate interior angles. Theorem: Parallel lines do not form congruent alternate interior  s.

21. Theorem: Every right angle is obtuse. Theorem: Rectangles do not exist. Theorem: Parallel lines are not always the same distance apart. Theorem: In the right triangle shown, a 2 + b 2  c 2. Theorem: The ratio of the circumference of a circle to its diameter is not constant a b c d and, therefore, not .    Theorem: Parallel lines form congruent alternate interior angles. Theorem: Parallel lines do not form congruent alternate interior  s.

22. Theorem: Every right angle is obtuse. Theorem: Rectangles do not exist. Theorem: Parallel lines are not always the same distance apart. Theorem: In the right triangle shown, a 2 + b 2  c 2. Theorem: The ratio of the circumference of a circle to its diameter is not constant. Theorem: In a triangle, the 3 angle measures never add up to 180  a b c and, therefore, not . d    Theorem: Parallel lines form congruent alternate interior angles. Theorem: Parallel lines do not form congruent alternate interior  s.

23. Theorem: Every right angle is obtuse. Theorem: Rectangles do not exist. Theorem: Parallel lines are not always the same distance apart. Theorem: In the right triangle shown, a 2 + b 2  c 2. Theorem: The ratio of the circumference of a circle to its diameter is not constant. Theorem: In a triangle, the 3 angle measures never add up to 180  a b c and, therefore, not . d    Theorem: All similar triangles are congruent. Theorem: Parallel lines form congruent alternate interior angles. Theorem: Parallel lines do not form congruent alternate interior  s. But are they really contradictions?

24. 77 Statements Equivalent to Euclid’s Parallel Postulate 77 Statements whose proofs depend on Euclid’s Parallel Postulate.

27. Theorem: Every right angle is obtuse. Theorem: Rectangles do not exist. Theorem: Parallel lines are not always the same distance apart. Theorem: In the right triangle shown, a 2 + b 2  c 2. Theorem: The ratio of the circumference of a circle to its diameter is not constant. Theorem: In a triangle, the 3 angle measures never add up to 180  a b c and, therefore, not . d    Theorem: All similar triangles are congruent. Theorem: Parallel lines form congruent alternate interior angles. Theorem: Parallel lines do not form congruent alternate interior  s. So NONE of these is really contradictions!!

28. Theorem: Every right angle is obtuse. Theorem: Rectangles do not exist. Theorem: Parallel lines are not always the same distance apart. Theorem: In the right triangle shown, a 2 + b 2  c 2. Theorem: The ratio of the circumference of a circle to its diameter is not constant. Theorem: In a triangle, the 3 angle measures never add up to 180  a b c and, therefore, not . d    Theorem: All similar triangles are congruent. Theorem: Parallel lines form congruent alternate interior angles. Theorem: Parallel lines do not form congruent alternate interior  s.

29. 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. Letter from the mathematician Farkas Bolyai (1775-1856) to his son, mathematician Janos Bolyai (1802-1860)

30. 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. er

31. Euclidean Geometry Euclid’s five postulates and all the theorems that derive from them.

32. Euclid's 5 postulates 1.A line segment can be drawn joining any two points.  

33. Euclid's 5 postulates 1.A line segment can be drawn joining any two points. 2.Any line segment can be extended indefinitely in a line.  

34. Euclid's 5 postulates 1.A line segment can be drawn joining any two points. 2.Any line segment can be extended indefinitely in a line. 3.Given any line segment, a circle can be drawn having the segment as radius and one endpoint as center.  

35. Euclid's 5 postulates 1.A line segment can be drawn joining any two points. 2.Any line segment can be extended indefinitely in a line. 3.Given any line segment, a circle can be drawn having the segment as radius and one endpoint as center. 4.All right angles are congruent.

36. Euclid's 5 postulates 1.A line segment can be drawn joining any two points. 2.Any line segment can be extended indefinitely in a line. 3.Given any line segment, a circle can be drawn having the segment as radius and one endpoint as center. 4.All right angles are congruent. 5. Given a line and a point not on the line, exactly one line parallel to the given line may be drawn through the point. (Euclid’s Parallel Postulate)  

37. David Hilbert (1862-1943) In the early years of the twentieth century, the German mathematician David Hilbert attempted to fill in the gaps in Euclid’s Geometry.

38. What gaps in Euclid’s Geometry do we mean ? Euclid's 5 postulates 1.A line segment can be drawn joining any two points. 2.Any line segment can be extended indefinitely in a line. 3.Given any line segment, a circle can be drawn having the segment as radius and one endpoint as center. 4.All right angles are congruent. 5. Given a line and a point not on the line, exactly one line parallel to the given line may be drawn through the point. (Euclid’s Parallel Postulate)

39. Construct the angle bisector of  C. Here is how we proved the isosceles triangle theorem The two triangles are congruent by SAS. Therefore,  A is congruent to  B. P 

40. But, which one of Euclid’s postulates says that the diagram can’t be like … P We can still prove the triangles congruent by SAS, but now,  B   CAP (not  CAB).

41. David Hilbert (1862-1943) In the early years of the twentieth century, the German mathematician David Hilbert attempted to fill in the gaps in Euclid’s Geometry. He proposed a series of postulates divided into four categories: Incidence, Betweenness, Congruence, and Continuity axioms. These axioms were designed to replace Euclid’s first four postulates and still enable us to prove all the theorems of Euclidean Geometry., except those based on Euclid’s fifth postulate, (the parallel postulate).

42. The incidence axioms Incidence Axiom 1: For every point P and every point Q not equal to P, there exists a unique line l incident with P and Q. Incidence Axiom 2: For every line l there exist at least two distinct points incident with l. Incidence Axiom 3: There exist three distinct points with the property that no line is incident with all three of them. Any system with lines and points suitably interpreted that satisfies all three incidence axioms is called an “Incidence Geometry.” Equivalent to Euclid’s first two postulates

43. The simplest candidate for model of an incidence geometry: Interpret points as three people: Jay, Kaye, and Emma. Each puts in a cell phone call (line) to the other. Cell phones do not have a “conference call” feature. I-1: For every point P and every point Q not equal to P, there exists a unique line l incident with P and Q. I-2: For every line l there exist at least two distinct points incident with l. I-3: There exist three distinct points with the property that no line is incident with all three of them. J   K  M Is this a model of incidence geometry?

44. I-1: For every point P and every point Q not equal to P, there exists a unique line l incident with P and Q. I-2: For every line l there exist at least two distinct points incident with l. I-3: There exist three distinct points with the property that no line is incident with all three of them. A B C D In this example, the 4 vertices of the tetrahedron are the points and the 6 edges are the lines. Is this a model of incidence geometry?

45. In this example, the 8 vertices of the cube are the points and the 14 visible edges are the lines. Is this a model of incidence geometry? I-1: For every point P and every point Q not equal to P, there exists a unique line l incident with P and Q. I-2: For every line l there exist at least two distinct points incident with l. I-3: There exist three distinct points with the property that no line is incident with all three of them.

46. Interpret points to be the letters in the set {B, A, L, D}. Interpret lines as all possible subsets of size three. Is this a model of an incidence geometry? I-1: For every point P and every point Q not equal to P, there exists a unique line l incident with P and Q. I-2: For every line l there exist at least two distinct points incident with l. I-3: There exist three distinct points with the property that no line is incident with all three of them. Points B, A, L, D Lines {B,A,L}, {B,A,D}, {B,L,D}, {A,L,D}

47. In this example, the 5 vertices of the figure are the points and the 10 connectors are the lines. Is this a model of an incidence geometry? I-1: For every point P and every point Q not equal to P, there exists a unique line l incident with P and Q. I-2: For every line l there exist at least two distinct points incident with l. I-3: There exist three distinct points with the property that no line is incident with all three of them. Why didn’t I draw the diagram like this?

48. David Hilbert (1862-1943) Hilbert’s incidence, betweenness, congruence, and continuity axioms can replace Euclid’s first four postulates. Hilbert’s axioms prevent...

49. But what about Euclid’s fifth postulate, the parallel postulate? Euclid’s Parallel Postulate Given a line and a point not on the line, exactly one line parallel to the given line may be drawn through the point.

50. In this example, the 4 vertices of the tetrahedron are the points and the 6 edges are the lines. Euclid’s Parallel Postulate Given a line and a point not on the line, exactly one line parallel to the given line may be drawn through the point. Does Euclid’s parallel postulate hold in this model? A B C D

51. In this example, the 5 vertices of the figure are the points and the 10 connectors are the lines. Euclid’s Parallel Postulate Given a line and a point not on the line, exactly one line parallel to the given line may be drawn through the point. Does Euclid’s parallel postulate hold in this model?

52. J   K  M Euclid’s Parallel Postulate Given a line and a point not on the line, exactly one line parallel to the given line may be drawn through the point. Does Euclid’s parallel postulate hold in this model?

53. Proclus (410-485) – seven centuries after Euclid (and 14 centuries before Bolyai) This [Euclid’s parallel postulate] ought to be struck out of the postulates altogether; for it is a statement involving many difficulties…. It is alien to the special character of postulates. Here is Proclus “proof” of the Parallel Postulate (given a modern twist)

54. P m n l Given two parallel lines l and m. Suppose line n intersects line m at point P, and suppose n is also parallel to l. Q Let Q be the foot of the perpendicular from P to l.  X Take X to be the foot of the perpendicular from Y to m. Since m and n intersect at point P, a point Y can be chosen on n such that ray PY lies between ray PQ and a ray PR on line m.  Y  R 

55. The further Y is chosen from point P, the longer segment XY will become. As Y recedes along n endlessly from P, the length of segment XY will eventually become greater than segment PQ. P m n l  X  Y R  Q  X  Y 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)? Contradiction, since n was assumed parallel to l.

57. The further Y is chosen from point P, the longer segment XY will become. As Y recedes along n endlessly from P, the length of segment XY will eventually become greater than segment PQ. P m n l  X  Y R  Q  X  Y 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)? l Contradiction, since n was assumed parallel to l.

58. No contradiction to the other four Euclidean postulates (or to any theorem based on them) has ever been reached. nor can it be. In attempting an indirect proof in this way, mathematicians for centuries replaced Euclid’s Parallel Postulate (they hoped temporarily) with: Given a line and point not on the line, more than one parallel can be drawn to the line through the point. is called Hyperbolic Geometry A system in which Hilbert’s incidence, betweenness, congruence, and continuity axioms hold (they are equivalent to Euclid’s first four postulates), ………………………………………….. nor can it be. AND the new postulate And you can prove a lot of strange theorems in Hyperbolic Geometry