SOLUTION: r,s,t are collinear with coordinates r,s, and t respectively.which of r,s, and t is between the other two if /r-s/+/t-r/=/t-s/

|r-s| + |t-r| = |t-s|

Suppose r < s < t

Then there exists positive a and b such that

s = r+a and t=r+a+b 

|r-s| + |t-r| = |t-s|

|r-(r+a)| + |(r+a+b)-r| = |(r+a+b)-(r+a)|

|r-r-a| + |r+a+b-r| = |r+a+b-r-a|

|-a| + {a+b| = |b|

     a + a+b = b
      2a + b = b
          2a = 0
           a = 0

That's impossible since a and b are positive

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Suppose r < t < s

Then there exists positive a and b such that

t = r+a and s=r+a+b 

|r-s| + |t-r| = |t-s|

|r-(r+a+b)| + |(r+a)-r| = |(r+a)-(r+a+b)|

|r-r-a-b| + |r+a-r| = |r+a-r-a-b|

|-a-b| + |a| = |-b|

  a+b + a = b
   2a + b = b
       2a = 0
        a = 0

That's impossible since a and b are positive

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Suppose s < r < t

Then there exists positive a and b such that

r = s+a and t=s+a+b 

|r-s| + |t-r| = |t-s|

|(s+a)-s| + |(s+a+b)-(s+a)| = |(s+a+b)-s|

|s+a-s| + |s+a+b-s-a| = |s+a+b-s|

|a| + {b| = |a+b|

That's true a and b are positive

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Suppose s < t < r

Then there exists positive a and b such that

t = s+a and r=s+a+b 

|r-s| + |t-r| = |t-s|

|(s+a+b)-s| + |(s+a)-(s+a+b)| = |(s+a)-s|

|s+a+b-s| + |s+a-s-a-b| = |s+a-s|

|a+b| + {-b| = |a|

     a+b + b = a
          2b = 0
           b = 0

That's impossible since a and b are positive

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Suppose t < r < s

Then there exists positive a and b such that

r = t+a and s=t+a+b 

|r-s| + |t-r| = |t-s|

|(t+a)-(t+a+b)| + |t-(t+a)| = |t-(t+a+b)|

|t+a-t-a-b| + |t-t-a| = |t-t-a-b|

|-b| + {-a| = |-a-b|

b + a = a + b

That's true since a and b are positive

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Suppose t < s < r

Then there exists positive a and b such that

s = t+a and r=t+a+b 

|r-s| + |t-r| = |t-s|

|(t+a+b)-(t+a)| + |t-(t+a+b)| = |t-(t+a)|

|r-r-a| + |r+a+b-r| = |r+a+b-r-a|

|-a| + {a+b| = |b|

a + a+b = b
 2a + b = b
     2a = 0
      a = 0

That's impossible since a and b are positive

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So the only possible cases are:

s < r < t  and t < r < s

and in both cases r is between the other two.

Edwin