Math 102A Hw 3 P a (2 points)

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1 Math 102 Hw 3 P a (2 points) If any pair of these lines are equal, the conclusion is immediate, so assume that we have three distinct lines such that l m and m n. Suppose, on the contrary, that l meets n at point P. P does not lie on m, because l m. Hence we have two distinct parallels n and l to m through P, which contradicts the uclidean parallel property of the affine plane. b (1 point) l is always parallel to itself (reflexive property) so if l = n, then the statement yields no new information. c (1 point) This question is not quite right, as elliptic geometry will be a counterexample. There should be some additional assumptions here. d (2 points) Let the set of points be {,,,, } and the set of lines be all two letter subsets. The Incidence xioms are readily verifed. y looking at figure 1 we can see that there are parallel lines. urthemore lines, and, but lines and share a common point, thus they are not parallel. So we do not have transitivity of parallelism in this model. igure 1 1

2 13 (2 points) part 1 y I3, every model for incidence geometry needs to have at least 3 noncollinear points, say {,, }. These pairwise determine 3 lines, and (see figure 2, first picture). Since each of these lines contains only 2 points, we need to add at least 3 more points into our model, say {,, } each lying on one of these lines (see figure 2, second picture). There is now a lack of lines in the new model. y I1 for each pair of points there needs to be a unique line passing through them. onsider first the pairs {, }, {, } and {, }. or each of these 3 pairs, the corresponding lines, and cannot contain any other of the already existing points (see figure 2, third picture). or example if contained the point then the lines and would intersect in 2 (or more) points contradicting proposition 2.1. G G igure 2 Therefore in order to have all lines contain at least 3 points, we need an additional point, say G. To keep the number of points at a minimum, we arrange G to lie on each of, and (see figure 2, third picture). inally, a check reveals that there are left over pairs of points with no lines passing through them, namely the pairs {, }, {, } and {, }. To remedy this, we add in another line into our model, one which contains all three of these points (see figure 2, fourth picture). It is easy to verify that all the axioms of incidence geometry hold. There are 7 points and 7 lines in this model. Observe that this is the projective plane associated to the 4 point affine plane, this is called the ano plane. 2

3 part 2 To construct the minimal geometry where the parallel postulate holds and where every line has at least three points, let s again start with the bare minimum of any model: three points {,, } and their associated lines, and (figure 3, first picture). We now continue the construction by focusing first on the parallel axiom. or the line and the point there has to be a line parallel to and containing. We add a point to our model and let be that line. Likewise, be the line parallel to and passing through. In addition, let be the line through the points {, } (figure 3, second picture). G H I igure 3 Neither of the lines so far contains 3 points. So we add the points {,, G, H, I} to ensure each line has 3 points (figure 3, third picture). We arrive at an interpretation which satisfies the parallel axiom and where each line has exactly 3 points. Unfortunately it fails the first incidence axiom. or example the points G and H don t have any lines passing through them. To remedy this we add four more lines into the picture (figure 3, fourth picture). case by case check shows that this is indeed a model of incidence geometry where the parallel axiom holdsand where every line has precisely 3 points. 3

4 14 a (1 point) To show this statement is not a theorem of incident geometry, we need only provide a model of Incidence geometry where such statement does not hold, i.e. a counterexample. Let the set of points be {,, } and the set of lines be all two letter subsets. The Incidence xioms are readily verifed. ut statement S fails, since lines and are two distinct lines, yet there is no point that does not lie on either or. b (2 point) Let l and m be any two distinct lines in a projective plane. Suppose that all points lie on either l or m. y the elliptical parallel property lines l and m meet at one point, call it Q. Since l m then there exist points Q in l that does not lie in m and point Q in m that does not lie in l. y I1 there exist a line through and, call this line n. Then n l since does not lie in l, likewise n m since does not lie in m. y the strengthened I2 every line has at least three distinct points lying on it. So n has a third point, call it. If lies on l there would be two distinct lines, l and n, through the points and, contrary to I1. Likewise if lies on m there would be two distinct lines passing through the points and. So point does not lie on either l or m, contrary to assumption that all points lie on l or m. Thus for any two distinct lines in a projective plane, there exist a point that does not lie on either of them. c (1 points) Let l and m be any two distinct lines in a finite projective plane. Let the points i lie on l and the points j lie on m. y statement S there exist a point P that does not lie on l or m. So for any i, I1 produces a unique line through the points i and P, call it n i. y the elliptic parallel property all lines meet, thus n i meets m at some point m j. Thus for each point i in l the line n i has found a point m j in m. So the number of points in l is less than or equal of those of m. y reversing the rolles of l and m, we find that the number of points in m is less than or equal of those of l. Since we are in a finite projective plane then the number of points in l and m are the same. Since l and m are arbitrary, we have that all lines have the same number of points lying on them. 4

5 d (1 points) Let l and m be any two distinct lines in a finite affine plane, and let be its projective completion. Then by adding an extra point at infinitiy to both l and m, the augmented lines l and m are still distinct and now lie in, a finite projective plane. y part c) we know that l and m have the same number of points lying on them. So by removing the one point at inifinity to both l and m, we get back l and m and now lie back in, and so again have the same number of points. So all lines in a finite affine plane have the same number of points lying on them. P.145 (2 points) 1 Incorrect 2 orrect 3 orrect 4 Incorrect 5 Incorrect 5