## I. Betweenness Exercise 3

**A) Part 1**

1) We are given that A*B*C (Hyp.)

2) We know that for line segment AB to be a part of line
segment AC, any point x on segment

AB must also exist on segment AC; the statements A*x*B and A*B*C must hold true

(Proposition 3.3).

3) Given A*B*C, the line segments AB, AC, and BC can be
re presented as the following sets

(suggested by Prof. Smith):

a) Line segment AB is the set of points AB = {x on line AB: A*x*B or x=A or x=B}

(Def. line segment).

b) The line segment AC is the set of points AC = {x on line AC: A*x*C or x=A or
x=C}

(Def. line segment).

c) The line segment BC is the set of points BC = {x on line BC: B*x*C or x=B or
x=C}

(Def. line segment).

4) Three cases exist for the placement of x on AB and its
according relationship to AC ( Step 2):

a) Case 1. If x=A, then x lies on segment AC (Step 3-b).

b) Case 2. If x=B, then x lies on segment AC because we are given that A*B*C
(Steps

1, 3-b).

c) Case 3. If A*x*B, then x lies on segment AC because we are given that A*B*C
and

have specified that x lies on segment AB (Steps 1, 2, 3-b).

5) Therefore, because any point x on line segment AB is
included in the set of points composing

line segment AC, AB AC (Conc.)

Q.E.D.

**A) Part 2**

Accordingly, because Betweenness Axiom 1 guarantees
symmetry and distinction given

A*B*C, the set of points on line segment CB is included in the set of points on
line segment CA.

If the “second” of A, B, C is between the “first” and the “third,” then the
“second” is accordingly

between the “third” and the “first” (suggested by Prof. Smith).

**B)**

1) If A*B*C, then AC AB U BC (Hyp.)

2) The statement that AB U BC implies that any point x is
included in the set of points in line

segment AB or in line segment BC.

3) Consider the addition of the point x to the line that
includes the distinct, col linear points A, B,

and C, provided that x ≠ A, B, or C and x lies on line segment AC (suggested by
Prof. Smith);

additionally, let a distinct line l pass through x.

a) If B and C are on the opposite side of A from l, then A*x*B (BA-4), and by
the law of

the excluded middle, x lies on AC (Proof I.A, 4-c).

b) If A and B are on the opposite side of C from l, then B*x*C (BA-4), and x

accordingly lies on AC (Proof I.A, 3-c).

4) Therefore, we know that for any point x lying on AC but
not equal to A, B, or C, either

A*x*B or B*x*C, which accordingly implies that A*x*C. AC AB U BC (Conc.)

Q.E.D.

**C)**

1) If A*B*C, then AC = AB U BC and B is the only point
common to segments AB and BC

(Hyp., Prop. 3.5.)

2) For this statement to be true, the fol lowing statements
must both be valid:

a) AC AB U BC holds (Proof I.B).

b) AB U BC AC holds (Proof I.A).

3) Moreover, B must be the only point common to segments
AB and BC (Step 1).

4) Suppose that a fourth point P is included in AB
BC such that B*P*C, but P ≠ B (Suggested

by Prof. Smith).

5) It is not the case that A*P*B, and it is not the case
that P is equal to A or B (Step 4).

6) P lies on line segment BC but not on line segment AB
(Steps 4, 5).

7) Therefore, only B intersects with the set of points
included in both line segments AB and BC

(Conc.)

Q.E.D.

**II. Betweenness Exercise 16**

1) Consider “between” to refer to the midpoint of a line
segment; if A*B*C, then B is the

midpoint of segment AC (AB is congruent to BC).

2) Consider an interpretation in which three distinct,
collinear points A, B, and C exist such that

line segments AB, BC, and AC exist, and such that AB C AC and BC C AC.

3) Segment AB > segment BC.

4) Because segment AB ≠ segments BC, where both segments
are constituents of segment AC,

point B is not the midpoint of AC.

5) By the specified definition of betweenness in this
interpretation, then, it is not the case that

A*B*C.

Q.E.D.

**III. Betweenness Exercise 17**

1) In order to illustrate the failure of the line sepa ration property in an
interpretation, consider a

model including the distinct, collinear points A, P, B, and D, in which the
Euclidean betweenness

relationships A*P*B, A*B*D, and P*B*D hold.

2) Interpret the betweenness of points A, P, and B such
that P*A*B.

3) To show the failure of the line separation property ,
the point P must not lie on either of the

rays formed from the relationship A*B*D (ray BA or ray BD) (Step 1; suggested by
Prof.

Smith).

4) Because the point P does not lie on either ray BA or
ray BD, which exist inherently because

the relationship A*B*D has been given (Step 1; Def. opp. rays), the Line
Separation Property

(Prop. 3.4) fails.

Q.E.D.