Plane Geometry - Bruce E. Shapiro

SECTION 14. BETWEENNESS 73andx > 2a 2 = agivinga < x < bHence eitherf(A) < f(M) < f(B)orf(A) > f(M) > f(B).By theorem 14.25 A ∗ M ∗ B.To verify uniqueness, let M ′ ∈ AB, where M ′ ≠ M and AM ′ = M ′ B.Suppose that f(A) < f(B). Then both the following hold:f(A) < f(M) < f(B)f(A) < f(M ′ ) < f(B)Furthermore, since M and M ′ are midpoints|f(A) − f(M)| = AM= 1 2 AB= AM ′= |f(A) − f(M ′ )|Since f(A) < f(M) and f(A) < f(M ′ ), this givesf(M) − f(A) = f(M ′ ) − f(A)f(M) = f(M ′ )Since f is 1-1 then M = M ′ , which proves uniqueness when f(A) < f(B).A similar argument holds to prove uniqueness when f(A) > f(B).Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.