Answer on Real Analysis Question for junel
We have to show that x ≤ y ≤ z
Consider two cases:
1) y < x ≤ z.
|x-y| = x-y,
|y-z| = z-y
|x-y| + |y-z| = x-y + z-y = x+z-2y
|x-z| = z-x.
x+z-2y = z-x
2x-2y = 0
which contradicts to the assumption y<x.
Thus the case y < x ≤ z is impossible.
2. Similarly, assume that x ≤ z < y
|x-y| = y-x,
|y-z| = y-z
|x-y| + |y-z| = y-x + y-z = 2y-x-z
|x-z| = z-x
2y-x-z = z-x
2y = 2z
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].
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