Homework has to be turned in this handout. For extra space, use the back pages, or blank pages between. due January 22/23

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1 Math 3181 Name: Dr. Franz Rothe January 15, 2014 All3181\3181_spr14h1.tex Homework has to be turned in this handout. For extra space, use the back pages, or blank pages between. due January 22/23 1 Homework 120 Problem 1.1. As far as two-dimensional geometry is concerned, Hilbert s Proposition 1 reduces to one simple statement: any two different lines either intersect in one point, or are parallel. This statement can be rephrased in many formulations. Encircle the statements below which are such equivalent reformulations; and scratch through the statements which are not. Any two different lines which are not parallel, have a unique point of intersection. There exists at most one parallel to a given line through a given point. On any given line lie two or more points. If two lines have two or more points in common, they are equal. Two lines can but need not have some points in common. Two different lines have at most one point in common. Two different points determine a line going through them uniquely. 1

2 Problem 1.2 (The five-point incidence geometries). Find all (four) non-isomorphic incidence geometries with five points. For each of them Provide a drawing. Tell how many lines exist. Tell how many lines have three or more points, and describe the properties of these lines. Tell which parallel property (elliptic, Euclidean, hyperbolic, or neither) does hold. 2

3 Problem 1.3. How many lines does the hand-shake incidence geometry with an arbitrary number n of points have. How many lines does the straight fan with n points have. Problem 1.4. Which parallel property holds for the hand-shake model with 4 points. Which parallel property holds for the hand-shake model with n 5 points. Which parallel property holds for a straight fan with any numbers n 3 of points.. Problem 1.5. From the axioms of incidence (I.1) (I.2) (I.3), one shows easily that there exist two different lines through every point. Complete the following proof for this statement. Proof. By axiom (I.3b), there exist at least? that do not lie on a line. We call them A, B and C. Let any point P be given. We now distinguish two cases: 1. In the case that point P is? from all three points A, B, C, we draw the three lines?. At least two of them are different since A, B, C do not lie on a line. Thus we are ready. 2. In the case that point P is one of?, we draw the three lines AB, BC and CA. These are three different lines, and two of them go through the given point P. In? P. we have obtained two different lines through the arbitrary point 3

4 Definition 1 (Isomorphism of incidence planes). Two incidence planes are called isomorphic if and only if there exists a bijection between the points of the two planes, and a bijection between the lines of the two planes such that incidence is preserved. Figure 1: Two isomorphic six-point incidence geometries Problem 1.6. Given two incidence geometries, it is not obvious whether they are isomorphic. By corresponding labelling of the points in both geometries, show an isomorphism between the two six-point incidence geometries in the figure on page 4. Color each pair of corresponding sides with the same color, using altogether seven colors. 4

5 Definition 2 (Affine plane). An affine plane is a set of points, and a set of lines, satisfying the axioms: A.1 Every two different points lie on exactly one line. A.2 If point P does not lie on a line l, there exists exactly one line m through the point P that does not intersect l. A.3 There exist three points that do not lie on a line. Axiom (A.2) is just the Euclidean parallel property. Because of the result of the problem below, an affine plane is just an incidence plane with the Euclidean parallel property. Problem 1.7. The reader should convince himself that in an affine plane 1. on every line lie at least two points; 2. in every point intersect at least two lines. Complete the proof for these statements given below and provide drawings for the different cases explained below. Proof. Let A, B, C be the three points not on a line which exist according to axiom (A.3). 1. Given is any line l. At most? of the three lines AB, BC and AC is parallel or equal to the given line l. We may assume that neither line AB nor line AC is parallel or equal to the given line l. Hence they intersect the given line in the points P = AB l and?. If these two points P Q are different, we are ready. We get a more involved argument in the special case that P = Q. In this case, we get A = P = Q. If the line l intersects the line BC, we get a second intersection point R = BC l on the line l and? Still open is only the case that lines BC and l are parallel and the point A lies on the line l. Let m be the parallel to line c = AB through point C. Since m BC l, the line m is not parallel to l. Hence these two lines intersect at point?, which is a second point on the line l. 2. Given is any point P. Assume that P A, B, C. Since these three points do not lie on a line, get get three different? through point P. In the special case that P = A, we get two different lines AB and AC through point P. We argue similarly in the special cases where? 5

6 Problem 1.8. An illustrative affine incidence geometry is the Cartesian plane Z 3 Z 3. In other word, we use analytic geometry and arithmetic modulo 3 for the coordinates. The points are the ordered pairs (a, b) with a, b Z 3. As in analytic geometry, lines are given by linear equations ax + by + c = 0 with a, b, c Z 3 and a, b are not both zero. One defines the slope of a line in the usual way. Lines are parallel if and only if they have the same slope. To find the parallel to a given line through a given point, one uses the point slope equation of a straight line. This procedure shows that the Euclidean parallel property holds. To better understand this model, provide a drawing. How many lines does the model have? Use clearly different colors for lines with different slopes, but give each set of three parallel lines different shades of nearby the same color. 6

7 Lemma 1 (Proclus Lemma).. In any affine plane, a third line intersecting one of two parallel lines intersects the other one, too. a third line parallel to one of two parallel lines, is parallel to the other one, too. Problem 1.9. Explain why Proclus Lemma is an easy consequence of the uniqueness of parallels. Convince yourself that, conversely, Proclus Lemma implies the uniqueness of parallels. 7

8 Problem 1.. In the figure on page 8 are shown two illustrations of the Fano plane: A highly symmetric illustration based on an equilateral triangle, and a more arbitrary chosen one. Denote the seven points with the same names in both drawings, consistently in a way to show the isomorphism. After you have obtained the isomorphism, color the seven lines with seven different colors. Give the corresponding lines in the other illustration the same colors. Figure 2: Fano s seven-point projective plane 8

9 Proposition 1 (Intersections of lines and planes). (ii) Any two different planes have either no point in common, or they have one line and no further points in common. (iii) A plane and a line not lying in this plane have either no or exactly one point in common. Problem Complete the proof of statement (ii) from Proposition 1 above. We use definite names for the objects, explain the steps, and state which axioms are involved. Proof of item (ii). Take any two? planes α β. If they do not intersect, we are ready. Assume that the two planes have? in common. We now use? from Hilbert s axioms of incidence. Hence, the two planes have a? in common. By axioms (I.1) and (I.2), there exists a? a through points A and B. Next we use? from Hilbert s axioms of incidence. Hence every point of line a lies? Reasoning about the same points A and B and line a, axiom (I.6) yields that every point of line a lies in the second plane β, too. In other words, we get the inclusion a α β We need still to check that the two planes α and β do not have any point? line a in common. Take any point C lying in both planes α and β. We assume towards a contradiction that point C does not lie on the line a. Both planes α and β would contain the? Now we use? from Hilbert s axioms of incidence. By this axiom, there exists? through the three points A, B and C. Hence these two planes α and β would be equal, contradicting the assumption α β. Hence the two planes α and β have no point outside the line a in common. Together we get the equality a = α β We have shown that two planes intersecting at one point have a line through this point as their intersection. 9

10 Problem Write a short clear proof for statement (iii) from Proposition 1 above.