Euclid s 5 th Axiom (on the plane):

Transcription

1 Euclid s 5 th Axiom (on plane): That, if a straight line falling on two straight lines makes interior angles on same side less than two right angles, two straight lines, if produced indefinitely, meet on that side on which are angles less than two right angles.

2 o lines will cut when extended.

3 Playfair, an 8 th century Scottish scientist, formulated axiom (on plane): Given a line and a point not on line, it is possible to draw exactly one line through given point parallel to line. We ll show that this is in fact equivalent to Euclid s Fifth Postulate.

4 First, we remark that, under first four axioms of Euclid, following statements hold. () Opposite angles are same. 2

5 (2) If two triangles have both sides equal to each or and angles between sides are also equal, n y are congruent.

6 2 (3) In any triangle, if one of sides is produced, exterior angle is greater than eir of opposite angles., &. 2

7 Proof. (Space for you to down proof.)

8 Find l 2 so that 2. To show that 2 l // l 2. P 2 l 2 l

9 Direct Proof/Argument. Indirect Proof/Argument.

10 Suppose Then we where PQR is l and have l 2 meet in hypetical a triangle. left. diagram Q. 2 2 P l 2 R l

11 .. PQR triangle of angles exterior a is that Observe diagram previous of portion redraw We 2.,. ). ( ) ( 2 2 right in meet cannot y Likewise left in meet cannot lines Hence given is contraditon a opposite exterior Q P R

12 To show 5th axiom = // axiom (I) 5th axiom // axiom We know from above l // l 2 2. that P 2 l 2 l We are required to show that l 2 is only line passing through P that is parallel to. l

13 Suppose not. Then re is anor line passing through P that is l 3 l parallel to. Let l 3 lie as in diagram. l 3 P 2 l 2 l

14 l P l 3 ). ( o o o have we diagram previous From! 5 3 meet to have l and l axiom th

15 Similarly if l 3 lie as in diagram. Convince yourself that this is mirror image of previous case. P l 2 l 3 l

16 (I) // axiom 5th axiom Re fer to following diagram, we are required to show that 3 80 o l meets l3 in right. P l 3 3 l Left Right

17 Find l We know that ( from lecture) 2 2 so that l // l P 2 3 l 3 l 2 // axiom l Contradiction no such and l 3 cannot l 3 be exist. l parallel.

18 l P 3 3 l..., 3 triangle a is PQR where diagram hypetical have we Then left in meet do y Suppose left in meet cannot l and l that show to need just we proof finish to Thus Q R 3.

19 .. 3 PQR triangle of angles exterior are and that Observe diagram previous of portion redraw We o 80 o!,.. 80 ) ( 360 ) 2( ) ( ) ( right in meet to have y is That left in meet cannot lines Hence contraditon a opposite exterior o o o o o Q P R

20 Toger with 5 th axiom of Euclid, or equivalently, Playfair postulate, we have following. 2 3 Parallel lines 2 3.

21 The fifth axiom implies that sum of interior angles of any triangle is equal to two right angles, that is, 80 degrees.

22 Indeed, Euclid s fifth axiom, Playfair axiom, Pythagoras orem, and statement that sum of interior angles of a triangle is equal to 2 right angles, are all equivalent. That is, we won t change Euclidean geometry if we replace fifth axiom by anyone of or statements. The details can be found in Reference:

23 Because parallel axiom, or Playfair axiom, appears to be so natural and intuitive, many had tried, unsuccessfully, to derive it from first four axioms. Despite of this, for two thousand years nobody really doubted that parallel axiom can be changed or replaced. It was held as an absolute truth.

24 One of most famous stories about Gauss depicts him measuring angles of great triangle formed by mountain peaks of Hohenhagen, Inselberg, and Brocken for evidence that geometry of space is non-euclidean.

25 Gauss was apparently first to arrive at conclusion that no contradiction may be obtained this way. In a private letter of 824 Gauss wrote: The assumption that (in a triangle) sum of three angles is less than 80 o leads to a curious geometry, quite different from ours, but thoroughly consistent, which I have developed to my entire satisfaction.

26 Lobachevsky and Bolyai built ir geometries on assumption: Through a point not on line re exist more than one line parallel to line. This is equivalent to Gauss' assumption that sum of angles in a triangle is less than 80 degree. Rightfully, new geometry created is called Non-Euclidean geometry.

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28 The related result is used by Einstein in his General Theory of Relativity space-time is non-euclidean! Black Hole.

29 In a 99 test of general ory of relativity, stars (marked with short horizontal lines) were photographed during a solar eclipse. The rays of starlight were bent by Sun's gravity on ir way to earth. This is interpreted as evidence in favor of Einstein's prediction that gravity would cause deviations from Euclidean geometry.

30 Terence Tao Got his PhD at age of 20 from Princeton University. Became full professor in UCLA at 25. Awarded Fields Medal at age of 3. Contribution includes Kakeye s problem.

31 Grigori Perelman In 994, Perelman sequestered himself away to tackle problem, and for following 8 years gave no signs of life. In May 2003, he announced that he had solved Poincare Conjecture and Thurston Geometrization Conjecture.