Exercises 1-5 refer to Fano's geometry.

1. The geometry has exactly the same number of points as it has lines.

Ans: True. Solution: There are 7 points and 7 lines.

2.Each pair of lines in the geometry intersects in a point in the geometry.

Ans: True. Solution: Axiom 5 says so.

3. Each two distinct points of the geometry determine a unique line in the geometry..

Ans: True. Solution:This is Axiom 4.

4. Changing one axiom results in Young's geometry.

Ans: True. Solution:You get Young's geometry by changing Axiom 5.

5. The geometry contains at least eight lines.

Ans: False. Solution:Theorem 1.8 declares that there are exactly seven lines in the geometry.

9. Sketch a model for a geometry that satisfies Axioms 1 and 2 of Fano's geometry but not Axiom 3.

10. Sketch a model for a geometry that satisfies Axioms 1 and 3 of Fano's geometry but not Axiom 2.

Solution:

13. Explain why C(7,2) does not give the number of lines in Fano's geometry, but can be used to derive the correct number of lines.

Solution:C(7,2) = 21, counts the number of pairs of points in the geometry. While each pair does determine a line, it is not true that each line only has two points on it. Axiom 2 states that each line has 3 points on it, thus, a single line contributes C(3,2) = 3 pairs of points to the total count. Dividing 21 by 3 will give the correct number of lines, i.e., 7.

14. For Fano's geometry, prove that each point lies on exactly three lines.

Solution:

1. (Without using Thm. 1.8) Pick a line l (exists by Axiom 1). Choose any point P not on l (exists by Axiom 3). Since l has 3 points (Axiom 2), joining P to each of them gives three distinct lines through P (by Axiom 4 there is a line through P and each of these points. If one of these lines contained two points of l, it would have to be l by Thm 1.7, but this contradicts the fact that P is not on l.) If there where another line through P, it would not meet l, contradicting Axiom 5, so there are exactly 3 lines through P. This argument takes care of all points not on l, to deal with a point on l, say Q, choose a line through P which does not contain Q and repeat the argument using Q and this line.
2. (Using Thm. 1.8) Let P be any point. There are 6 other points in the geometry besides P (Thm. 1.8). P is joined to each by a line (Axiom 4). Since each line contains 3 points (Axiom 2), each of these lines contains two points besides P. Thus, there are at least 3 lines through P. Any other line through P would have to contain two points, one on each of two different lines that we have already constructed through P (otherwise Axiom 4 is violated), but this violates Axiom 4. So, there are exactly three lines through P.

Solution:Given a point P, by Axiom 4, any other point of the geometry is joined to P by a line. Thus, all the points are on lines through P.

16. For Fano's geometry, prove that, for any pair of points in the geometry, there exist exactly two lines not containing either point.

Solution:By Exercise 14, there are three lines through each point of the geometry. Given two distinct points, P and Q, there is the line that joins them and two other lines through each of these points. That is a total of 5 lines. Since there are 7 lines in the geometry (Thm 1.8), there are exactly two lines that do not pass through P or Q.

17. For Fano's geometry, prove that , for a set of three lines not all containing the same point, there exists exactly one point in the geometry not on any of the three lines.

Solution:Take two of the lines and let the point where they meet be called P (exists by Thm. 1.7). The third line does not pass through P by assumption, so it must meet the original two lines in points different from P, say Q on one line and R on the other. The three lines, PQ, PR and QR, then form a triangle. Because each of these lines already has one point in common, no pair of them can have another point in common (Thm. 1.7). Since each line has 3 points (Axiom 2), there is another point on each of these lines. Thus we get three more points besides P, Q and R on these lines for a total of 6 points. By Thm 1.8, there are 7 points in the geometry, so there is exactly one point not on any of the three lines.

20. For Young's geometry, prove that two lines parallel to a third line are parallel to each other. Hint: Use Axiom 5 and prove by contradiction.

Solution: Assume that, for distinct lines l, m, n, we have l || m and m || n. If l || n, there is nothing to prove, so assume that l and n meet at a point P. Since P is on l and l || m, P is not on m. We now have a contradiction to Axiom 5 (for Young's geometry) since there are two lines through P that do not intersect m. Thus, we must have l || n.