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MODERN GEOMETRY

by: Cassidy Grimes

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MODERN GEOMETRY MATH 532

Cassidy Grimes

GPA 3.51

Staff

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This 2 page Class Notes was uploaded by Cassidy Grimes on Monday October 26, 2015. The Class Notes belongs to MATH 532 at University of South Carolina - Columbia taught by Staff in Fall. Since its upload, it has received 9 views. For similar materials see /class/229539/math-532-university-of-south-carolina-columbia in Mathematics (M) at University of South Carolina - Columbia.

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Date Created: 10/26/15
MATH 532 SPRING 2008 Handout 1 February 5 Axioms for Finite Af ne Geometry Axiom 1 There exists at least 4 points so that when taken any 3 at a time are not co linear Axiom 2 There exists at least one line incident to exactly 71 points Axiom 3 Given two distinct points there is a unique line incident to both of them Axiom 4 Given a line l and a point P not incident to l there is exactly one line incident to P which does not intersect l For the remainder of this seetiori all our results are stated for art a rie geometry of order n Notation We denote by Q R ST the four points guaranteed by Axiom 1 and by Pj 1 S j S n the points on l0 guaranteed by Axiom 2 Lemma 1 There exists a point P0 which lies on exactly 71 1 lines Proof By Axiom 1 we have four points and we know at least two of them cannot belong to lo Denote by P0 to be either one of these two points so P0 is one of QR S T By Axiom 2 there are n distinct points on l0 which we denote by Pj 1 S j S 71 By Axiom 3 there are lines l which are incident to P0 and respectively each of the points Pj These 71 lines must be distinct otherwise the points Pj would not be distinct By Axiom 4 there is one and only one line through P0 which does not intersect l0 call this line in If there are any other lines through P0 then they would have to intersect l0 at a new point Pn11 but l0 has exactly 71 points so there can be no other lines through P0 D Corollary 1 Each line l contains at least one point Proof Consider the point P0 from the previous lemma lf P0 belongs to l then we are done If it does not then by the previous lemma there exists it 1 lines through P0 But then Axiom 4 implies that one and only one of these lines misses l so 71 of them must intersect l Hence l has at least 71 points D Lemma 2 Fix J 1 S J S n and denote by P be any of the points QR S or T If P 74 P then there exists a line in which misses7 both P and PJ Proof Without loss of generality we may suppose that P from the previous lemma is the point Q Consider the line ml which passes through the pair of points R S and the line m2 which passes through the pair of points R T Neither of these lines contains P Q or then we would have a selection of 3 of these 4 special points which would be co linear If both of the lines contained the point PJ then they must in fact be the same line Axiom 3 since they would have two distinct common points R and PJ D Theorem 1 Each point P lies on exactly 71 1 lines Proof Fix the arbitrary point P If P does not lie on l0 then the proof in the rst lemma shows how to construct the unique 71 1 lines through P ie one and only one line through P and the points of P 6 l0 and exactly one line 77parallel to lo through P lf instead P 6 l0 then pick one of the special 4 points P0 Q say which is not on l0 Also let J be the index so that P PJ since P belongs to lo By the previous lemma there is a line m which misses both PJ and P0 Since there are exactly 71 1 lines going through P0 and exactly one of these is parallel to m the rest must intersect m Hence there are exactly 71 points which lie on m otherwise there would be another line through P0 Now m has the property that it misses PJ so the number of lines through PJ with the exception of the one line parallel to m are in one to one correspondence with the points on m as was guaranteed by Axiom 4 Denote these points as Qj 1 S j S 71 Those 71 points are the only points on m Changing our View to that of PJ and recalling that m misses PJ we see that there are 71 lines from PJ and to the points Qj Again by Axiom 4 there is one additional line through PJ which misses7 m ie is parallel to m Hence there are a total of exactly 71 1 lines through PJ which is our point P D Lemma 3 Each line l has exactly 71 7 1 lines which do not intersect l Proof Exercise 14 Homework Solution on Tuesday D Theorem 2 Each line l has exactly 71 points Proof From Axiom 1 ie 4 special points we know that there is some point P0 which is not on l There are n 1 lines through P0 and exactly one of those does not meet Z All the remaining 71 lines do meet l and result in n distinct points of intersection These are the only points on l since the points on l are in one to one correspondence with the lines through P0 which meet l D Theorem 3 There are exactly 712 points and n n 1 lines Proof Fix any line Z By Lemma 3 there are n 7 1 lines which do not intersect l Moreover these lines do not intersect each other why7 By Theorem 2 each line has n points We know each point belongs to one of these lines by Axiom 4 So these n lines which includes l partitions the points into n disjoint sets of points with n points in each of the sets This shows that there are exactly 712 points To count the number of lines we use l0 There are exactly 7171 lines which do not intersect l0 This statement is true by Lemma 3 If we include l0 itself then there are 71 lines in the same direction7 as l0 All other lines must intersect l0 These lines can be separated into disjoint sets indexed by their point of intersection on l0 At each of these 71 points ie each of the P7 there are exactly 71 lines not counting l0 which pass through Pj Hence there are 7 1 1 71 71 n2 71 lines D

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