** 2.2 Incidence Axioms
***Printout*

*Geometry
enlightens the intellect and sets one's mind right.*

—*Ibn Khaldun (1332–1406)*

** **

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 *l*. *S* 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.*

*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 *l*_{m,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*(*x*_{1}*, y*_{1})
and *Q*(*x*_{2}*, y*_{2})
be distinct points in the Cartesian plane. Then *x*_{1} and
*x*_{2}
are either equal or not equal. We first show the existence of a line containing
*P *and *Q*.

Case 1. Assume *x*_{1} =
*x*_{2}. Set *a* =
*x*_{1} =
*x*_{2}. Thus *P* and *Q* are both on line .

Case 2. Assume *x*_{1} and
*x*_{2} are not equal. We need to
find *m* and *b* such that *P* and *Q* are on line *l*_{m,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
*l*_{m,b}. For *Q*, .
Hence *Q* is on line *l*_{m,b}. Therefore, *P* and *Q* are on line *l*_{m,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* = *l*_{a} and *k* =
*k*_{b}. Then *a* = *x*_{1}
= *x*_{2} = *b*. Hence *l* and *k* are the same
line.

Case 2. Assume *l* = *l*_{m,b} and *k* =
*k*_{n,c}. Then
*y*_{1} =
*mx*_{1} +*b*,
*y*_{2} =
*mx*_{2} +*b*,
*y*_{1} =
*nx*_{1} +*c*, and
*y*_{2} =
*nx*_{2} +*c*. Solve
this system of four equations:

*mx*_{1} +*b* =
*nx*_{1} +*c* and
*mx*_{2} +*b* = *nx*_{2}
+*c*

(*m
–* *n*)* x*_{1} = *c* –
*b* and (*m* – *n*)* x*_{2} = *c* –
*b*

(*m
–* *n*)*x*_{1} = (*m – n)x*_{2}

(*m* – *n*)(*x*_{1} –
*x*_{2})
= 0.

Since for nonvertical lines *
x*_{1} and
*x*_{2}
are not equal, we have that *m* = *n*. Thus *c* *– b* = 0, i.e. *c* = *b*. Hence *l* = *l*_{m,b}
= *k*_{n,c }
= *k.*

Case 3. Without loss of generality, assume *l*
= *l*_{a} and *k* = *k*_{n,c}.
Since *P* and *Q* are on *l*, *a* = *x*_{1}
= *x*_{2}. That is, *P* = (*a,
y*_{1}) and *Q* = (*a, y*_{2}). Since *P* and *Q* are on *k*, *y*_{1}* = nx*_{1}* + c = na +
c = nx*_{2}* + c = y*_{2}.
Hence *P* = (*x*_{1}*, y*_{1})
= (*x*_{2}*, y*_{2}) = *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.