Question #3877

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

Expert's answer

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].

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].

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