Show x<=y<=z IFF |x-y|+|y-z|=|x-z|

  • Thread starter Kinetica
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  • #1
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Homework Statement


Hi! Please check my solution.

If x, y, z in R and x<=z, show that x<=y<=z IFF |x-y|+|y-z|=|x-z|

Interpret geometrically.


The Attempt at a Solution



1) Assume that x<=y<=z. We need to prove that |x-y|+|y-z|=|x-z|.

Knowing that x<=y<=z,

-(x-y)+(-(y-z))= -(x-z)
y-x-y+z=-x+z
-x+z=-x+z

2) Assume that |x-y|+|y-z|=|x-z|. We need to prove that x<=y<=z.

a) Assume y<=z<=x.
(x-y)-(y-z)=(x-z)
x-2y+z=x-z
z-2y=z
contradiction

b) Assume z<=x<=y
-(x-y)+(y-z)=(x-z)
-x-z=x-z
contradiction

c) Assume x<=y<=z
-(x-y)+(-(y-z))= -(x-z)
y-x-y+z=-x+z
-x+z=-x+z
Correcto!

Thanks!
 

Answers and Replies

  • #2
22,129
3,297

Homework Statement


Hi! Please check my solution.

If x, y, z in R and x<=z, show that x<=y<=z IFF |x-y|+|y-z|=|x-z|

Interpret geometrically.


The Attempt at a Solution



1) Assume that x<=y<=z. We need to prove that |x-y|+|y-z|=|x-z|.

Knowing that x<=y<=z,

-(x-y)+(-(y-z))= -(x-z)
y-x-y+z=-x+z
-x+z=-x+z

Seems good. However, I have some problems with your second part:

2) Assume that |x-y|+|y-z|=|x-z|. We need to prove that x<=y<=z.

a) Assume y<=z<=x.
(x-y)-(y-z)=(x-z)
x-2y+z=x-z
z-2y=z
contradiction

you know that x<= z, so why split into the case y<= z<= x? Isn't this supposed to be y<=x<= z?

b) Assume z<=x<=y
-(x-y)+(y-z)=(x-z)
-x-z=x-z
contradiction

Same remark as above. Isn't this supposed to be x<=z<=y?

c) Assume x<=y<=z
-(x-y)+(-(y-z))= -(x-z)
y-x-y+z=-x+z
-x+z=-x+z
Correcto!

Technically, this is not wrong, but it is superfluous. You need to show: "IF blabla THEN x<=y<=z". But what you did here is actually proving the converse...
 
  • #3
88
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Hi! Thanks for the feedback. I tried fixing the last part according to your comments. What I did, I squared the both sides of

|x-y|+|y-z|=|x-z|,

then, I solved it and squared |xxxx| again in order to get rid of the absolute value. The answer was short but strange. And probably, wrong.

Any help please?
 
  • #4
22,129
3,297
Why in earth did you do that? :confused:

You just need to use the technique you used this far.
So for y<=x<=z, you just need to simplify...

|x-y|+|y-z|=|x-z|

and show that this does hold.

In don't see why you start squaring things. :confused:
 

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