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Betweenness Exercise

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.)


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).


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.)



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



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


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.

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.

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