Question

DIRECTIONS: For each of the following questions, you may only site propositions 6.1, 7.1, 7.2, and 7.3; the incidence axioms, the betweenness axioms, and any definition from sections 6 & 7 of Hartshorne's textbook. You may also use exercises from the textbook, but only if you (or we as a class) have proved them and their number comes before whatever exercise you are trying to use it to prove. 2. (Exercise 7.1a) Consider an incidence geometry where the betweenness axioms hold. Prove that for any set of 4 distinct points A, B, C, D on a line: If A* B * C and B * C * D, then we get that A * B * D and A* C * D. (Hint: use the incidence & betweenness axioms and line separation.]

Answer 1

Incidence axioms

I1) There exists a line for any two points A and B that contains them both.

I2)There will be  at least two points on a line.

I3)There exist at least three points that do not lie on the same line.

Axioms of betweeness

B1) If a point B lies between points A and C, B is also between C and A, and there exists a line containing the distinct points A, B, C.

B2) If A and C are two points, then there exists at least one point B on the line AC such that C lies between A and B.

B3) Of any three points situated on a line, there is no more than one which lies between the other two.

B4) Pasch's Axiom: Let A, B, C be three points not lying in the same line and let a be a line lying in the plane ABC and not passing through any of the points A, B, C. Then, if the line a passes through a point of the segment AB, it will also pass through either a point of the segment BC or a point of the segment AC.

Proof

A∗B∗C  means B is between A and C.

A,B,C are on a line [from B1]

A,B,D are on a line [from I1]

Then they are on the same line since A,B are included in both

Therefore By B3 only one of the following can hold A∗B∗D or A∗D∗B or B∗A∗D

Take a point F not on the line and draw the line AF by line separation and A∗B∗D we have BD is a line segment on one side or the other of AF as I1 says AF and BD are unique lines by choice of F they intersect only at A this implies that B~D by definition

But B∗A∗D says B does not wiggle D hence it not a possible choice.

Now draw the line DF we can see that from B∗C∗D that line segment BC is on one side or the other of DF as BC and DF are unique lines by I1 that intersect at D and only D

Therefore B~C and from A∗B∗C using the fact that D is the only point of the line AB and on DF we have that AC is connected by a line segment on one side or the other of D as B is between A and C so they must be on the same side of the line DF. But A∗D∗B implies that A cannot wiggle B as they are on opposite side of the line DF. Hence the only possibility by B3 is A∗B∗D