Solution to Exercise 2.24.
The beginning is the half of
every action.
— Greek Proverb
Exercise
2.24. Prove a segment has a unique midpoint.
Proof. Let be a segment. By the Ruler Placement Postulate
there is a coordinate system (ruler)
such that f(A) = 0 and f(B) > 0. Let
We show that M is a
midpoint of the segment. Since M is defined from the ruler f
for the line through A and B, the points A, B, and M are
collinear. Further,
Hence, M is a midpoint of the segment
We need to show that M
is the unique midpoint of the segment. Suppose N is a midpoint of segment Then
|f(A) – f(N)| = d(A, N) = d(N, B) = |f(N) – f(B)|.
That is, |f(N)| = |f(N) –
f(B)|. One of three cases must be true: or
Case 1. Assume Then |f(N)| =
|f(N) – f(B)| implies
–f(N)
= –(f(N) – f(B)). Hence, f(B)
= 0, but this contradicts that f(B) > 0.
Case 2. Assume Then |f(N)| =
|f(N) – f(B)| implies f(N)
= f(N) – f(B). Hence, f(B)
= 0, but this contradicts that f(B) > 0.
Case 3. Assume Then |f(N)| =
|f(N) – f(B)| implies f(N)
= –(f(N) – f(B)). Hence
Thus
Therefore, the point M is the unique midpoint of segment
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