* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Euclid`s 5th Axiom (on the plane): That, if a straight line falling on two
Survey
Document related concepts
Riemannian connection on a surface wikipedia , lookup
Perspective (graphical) wikipedia , lookup
Lie sphere geometry wikipedia , lookup
Multilateration wikipedia , lookup
History of trigonometry wikipedia , lookup
Trigonometric functions wikipedia , lookup
Geometrization conjecture wikipedia , lookup
Rational trigonometry wikipedia , lookup
History of geometry wikipedia , lookup
Integer triangle wikipedia , lookup
Euler angles wikipedia , lookup
Pythagorean theorem wikipedia , lookup
Transcript
Euclid’s 5th Axiom (on the plane): That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles. 1 2 1 2 180o the lines will cut when extended . Playfair, an 18th century Scottish scientist, formulated the axiom (on the plane): Given a line and a point not on the line, it is possible to draw exactly one line through the given point parallel to the line. We’ll show that this is in fact equivalent to Euclid’s Fifth Postulate. First, we remark that, under the first four axioms of Euclid, the following statements hold. (1) Opposite angles are the same. 1 2 1 (2) If two triangles have both sides equal to each other and the angles between the sides are also equal, then they are congruent. 2 1 (3) In any triangle, if one of the sides is produced, the exterior angle is greater than either of the opposite angles. 1 , & 2 . Proof. (Space for you to down the proof.) Find l 2 so that 1 2 . To show that 1 2 l1 // l2 . 2 P 1 l2 l1 Direct Proof/Argument. Indirect Proof/Argument. Suppose l1 and l2 meet in the left. Then we have the hypethetical diagram where PQR is a triangle. . Q 2 1 1 R P2 l2 l1 We redraw portion of the previous diagram. Observe that 1 is a exterior angles of the triangle PQR. Q 1 R 1 2 2 P ( exterior opposite ) a contraditon (1 2 is given ). Hence the lines cannot meet in the left. Likewise , they cannot meet in the right. To show “5th axiom” = “// axiom” (I) “5th axiom” “// axiom” We know from above that 1 2 l1 // l2 . P 2 l2 1 l1 We are required to show that l2 is the only line passing through P that is parallel to l 1 . Suppose not. Then there is another line l3 passing through P that is parallel to Let l3 l1 . lie as in the diagram. l3 P 1 2 l2 l1 From the previous diagram we have 180o 2 1 1 180o 2 1 180 o (1 2 ). P 1 l3 l1 5th axiom l1 and l3 have to meet! Similarly if l3 lie as in the diagram. Convince yourself that this is mirror image of the previous case. P l3 1 l2 l1 (I) “// axiom” “5th axiom” Re fer to the following diagram, we are required to show that 1 3 180o l1 meets l3 in the right. P 3 l3 1 Left l1 Right Find l 2 so that 1 2 . We know that ( from lecture) 1 2 l1 // l2 . 2 P 3 1 l2 l3 l1 // axiom l1 and l3 cannot be parallel. Contradiction no such l3 exist. Thus to finish the proof , we just need to show that l1 and l3 cannot meet in the left. Suppose they do meet in the left. Then we have the hypethetical diagram where PQR is a triangle. . Q 3 P 1 1 R 3 l3 l1 We redraw portion of the previous diagram. Observe that 1 and 3 are the exterior angles of the triangle PQR. Q 1 R180 o 1 1 180 3 o 180o 3 3 ( exterior opposite ) P 3 180o 1 1 3 360o (1 3 ) 2(1 3 ) 360o (1 3 ) 180o a contraditon. Hence the lines cannot meet in the left. That is, they have to meet in the right! Together with the 5th axiom of Euclid, or equivalently, the Playfair postulate, we have the following. 3 2 Parallel lines 1 1 2 3 . The fifth axiom implies that the sum of the interior angles of any triangle is equal to two right angles, that is, 180 degrees. Indeed, Euclid’s fifth axiom, the Playfair axiom, the Pythagoras’ theorem, and the statement that the sum of the interior angles of a triangle is equal to 2 right angles, are all equivalent. That is, we won’t change the Euclidean geometry if we replace the fifth axiom by anyone of the other statements. The details can be found in Reference: http://www.cut-theknot.org/triangle/pythpar/PTimpliesPP .shtml Because the parallel axiom, or the Playfair axiom, appears to be so natural and intuitive, many had tried, unsuccessfully, to derive it from the first four axioms. Despite of this, for two thousand years nobody really doubted that the parallel axiom can be changed or replaced. It was held as an absolute truth. One of the most famous stories about Gauss depicts him measuring the angles of the great triangle formed by the mountain peaks of Hohenhagen, Inselberg, and Brocken for evidence that the geometry of space is non-Euclidean. Gauss was apparently the first to arrive at the conclusion that no contradiction may be obtained this way. In a private letter of 1824 Gauss wrote: “The assumption that (in a triangle) the sum of the three angles is less than 180o leads to a curious geometry, quite different from ours, but thoroughly consistent, which I have developed to my entire satisfaction.” Lobachevsky and Bolyai built their geometries on the assumption: Through a point not on the line there exist more than one line parallel to the line. This is equivalent to Gauss' assumption that the sum of angles in a triangle is less than 180 degree. Rightfully, the new geometry created is called Non-Euclidean geometry. The related result is used by Einstein in his General Theory of Relativity – space-time is non-Euclidean! Black Hole. In a 1919 test of the general theory of relativity, stars (marked with short horizontal lines) were photographed during a solar eclipse. The rays of starlight were bent by the Sun's gravity on their way to the earth. This is interpreted as evidence in favor of Einstein's prediction that gravity would cause deviations from Euclidean geometry. Terence Tao Got his PhD at the age of 20 from Princeton University. Became full professor in UCLA at 25. Awarded the Fields Medal at the age of 31. Contribution includes Kakeye’s problem. • Grigori Perelman In 1994, Perelman sequestered himself away to tackle the problem, and for the following 8 years gave no signs of life. In May 2003, he announced that he had solved the Poincare Conjecture and the Thurston Geometrization Conjecture.