2.2 Incidence Axioms   Printout
Geometry enlightens the intellect and sets one's mind right.
Exit book to another website.Ibn Khaldun (1332–1406)

 

Illustrate three points determine a plane.      Illustrate two lines intersect in a unique point.Why do surveyors, photographers, and artists use tripods? How is the location of a football that has been punted out of bounds determined?  The axioms for Euclidean geometry are such that the mathematics matches our real-world expectations. What axioms or theorems justify the solutions to the two questions?
      The SMSG incidence axioms are Postulates 1 and 5–8; however, since we are only concerned with plane geometry, the only axioms that apply to our study of a neutral geometry are Postulates 1, 5(a), and 6.

 

Postulate 1. (Line Uniqueness) Given any two distinct points there is exactly one line that contains them.

 

Postulate 5(a). (Existence of Points) Every plane contains at least three noncollinear points.

Postulate 6. (Points on a Line Lie in a Plane) If two points lie in a plane, then the line containing these points lies in the same plane.

Definition. A set of points S is collinear if there is a line l such that S is a subset of lS is noncollinear if S is not a collinear set. If {A, B, C} is a collinear set, we say that the points A, B, and C are collinear.

Theorem 2.2. Two distinct lines intersect in at most one point.
Link to lecture on Postulate 5(a).Link to lecture on Postulate 1.
     

Proposition 2.3. The Cartesian plane satisfies SMSG Postulates 1, 5(a), and 6.

Proof. To show Postulate 5(a) is satisfied consider the three points (0, 0), (1, 0), and (0, 1), which are points in the Cartesian plane. The three points are not on the same vertical line, since they do not have the same first coordinate. Suppose they are on a nonvertical line lm,b, then 0 = m(0) + b, 0 = m(1) + b, and 1 = m(0) + b, i.e. b = 1 and b = 0, which is a contradiction. Hence (0, 0), (1, 0), and (0, 1) are three distinct noncollinear points in the Cartesian plane, and SMSG Postulate 5 is satisfied.   
      Next, show SMSG Postulate 1 is satisfied. Let P(
x1, y1) and Q(x2, y2) be distinct points in the Cartesian plane. Then x1 and x2 are either equal or not equal. We first show the existence of a line containing P and Q.
Case 1. Assume
x1 = x2. Set a = x1 = x2. Thus P and Q are both on line .
Case 2. Assume
x1 and x2 are not equal. We need to find m and b such that P and Q are on line lm,b.  (First, do some scratch work to find an m and b that will work. Based on the scratch work, define an m and b, and then show they work.) Set  and  We show that P and Q are on the line . For P,

 

Hence P is on line lm,b. For Q,  . Hence Q is on line lm,b. Therefore, P and Q are on line lm,b. Thus by the two cases, there is at least one line that contains P and Q.
      We need to show there is exactly one line that contains P and Q. Suppose P and Q belong to lines l and k. There are three possible cases (1) both lines are vertical; (2) both lines are nonvertical; or (3) one line is vertical, and the other line is nonvertical.
Case 1. Assume l =
la and k = kb. Then a = x1 = x2 = b. Hence l and k are the same line.
Case 2. Assume l =
lm,b and k = kn,c. Then y1 = mx1 +b, y2 = mx2 +b, y1 = nx1 +c, and y2 = nx2 +c. Solve this system of four equations:

mx1 +b = nx1 +c and mx2 +b = nx2 +c

(m – n) x1 = c – b and (m – n) x2 = c – b

(m – n)x1 = (m – n)x2

(m – n)(x1 – x2) = 0.

Since for nonvertical lines x1 and x2 are not equal, we have that m = n. Thus c – b = 0, i.e. c = b. Hence l = lm,b = kn,c = k.
Case 3. Without loss of generality, assume l =
la and k = kn,c. Since P and Q are on l, a = x1 = x2. That is, P = (a, y1) and Q = (a, y2). Since P and Q are on k, y1 = nx1 + c = na + c = nx2 + c = y2. Hence P = (x1, y1) = (x2, y2) = Q. But this contradicts that P and Q are distinct. Therefore, this case is not possible.
      By the above three cases, there is exactly one line that contains P and Q. Since the two points were arbitrarily chosen, we have that the Cartesian plane satisfies SMSG Postulate 1.
      From how lines are defined for a Cartesian plane and the above proof that Postulate 1 is satisfied, it immediately follows that the Cartesian plane satisfies SMSG Postulate 6.// 

 

Corollary to Proposition 2.3. The Euclidean plane, Taxicab plane, and Max-distance plane satisfy SMSG Postulates 1, 5(a), and 6.

 

Exercise 2.11. Find the axioms from a high school geometry book that correspond to SMSG Postulates 1 and 5.

Exercise 2.12. Prove Theorem 2.2.

Exercise 2.13. Show the Missing Strip Plane satisfies (a) SMSG Postulate 1 and (b) SMSG Postulate 5(a).

 

Exercise 2.14. Show the Poincaré Half-plane satisfies (a) SMSG Postulate 1 and (b) SMSG Postulate 5(a).

Exercise 2.15. Does a Discrete plane satisfy (a) SMSG Postulate 1?  Justify. (b) SMSG Postulate 5(a)? Justify.

Exercise 2.16. Do the Riemann Sphere and Modified Riemann Sphere satisfy (a) SMSG Postulate 1? Justify. (b) SMSG Postulate 5(a)? Justify.

 

2.1.3 Analytic ModelsBack to Analytic Models for Plane Geometry.Next to Distance and Ruler Axioms.2.3 Distance and Ruler Axioms 

Ch. 2 Euclidean/NonEuclidean TOC  Table of Contents

  Timothy Peil  Mathematics Dept.  MSU Moorhead

© Copyright 2005, 2006 - Timothy Peil