Answer to Question #3877 in Real Analysis for junel

Question

If& x, y, z& ∈ R and& x ≤ z, show that& x ≤ y ≤ z& if & |x-y| + |y-z| = |x-z|.& & Interpret this geometrically.

Answer 1

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].