Answer to Question #3877 in Real Analysis for junel
If& x, y, z& ∈ R and& x ≤ z, show that& x ≤ y ≤ z& if & |x-y| + |y-z| = |x-z|.& & Interpret this geometrically.
Suppose that x ≤ z, and |x-y| + |y-z| = |x-z|
We have to show that x ≤ y ≤ z
Consider two cases:
1) y < x ≤ z. Then |x-y| = x-y, |y-z| = z-y Hence |x-y| + |y-z| = x-y + z-y = x+z-2y while |x-z| = z-x. So x+z-2y = z-x 2x-2y = 0 x=y which contradicts to the assumption y<x. Thus the case y < x ≤ z is impossible.
2. Similarly, assume that x ≤ z < y Then |x-y| = y-x, |y-z| = y-z Hence |x-y| + |y-z| = y-x + y-z = 2y-x-z while |x-z| = z-x So 2y-x-z = z-x 2y = 2z y=z which again contradicts to the assumption z<y. Thus the case x ≤ z < y is also impossible, so x ≤ y ≤ z.
Geometrically, the statement means that if x ≤ z, then the identity |x-y| + |y-z| = |x-z| is possible if and only if y belongs to the segment [x,z].