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Suppose I have 4 real numbers in $[0,1]$, all different between each other.

Assume that there exists a way of ordering the four numbers such that: $$ \begin{cases} w_1>w_2\\ w_3=1-w_1\\ w_4=1-w_2 \end{cases} $$ where $w_1$ is the first number in the ordered sequence, $w_2$ is the second number in the ordered sequence, $w_3$ is the third number in the ordered sequence, $w_4$ is the fourth number in the ordered sequence.

Is such a way of ordering unique? If yes, could you sketch the proof? In not, could you give a counterexample?

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  • $\begingroup$ What exactly do you mean when you say unique? $\endgroup$ – Ankit Kumar Dec 20 '18 at 13:31
  • $\begingroup$ can you give such an assignment to the following four numbers? $\frac{1}{4},\frac{1}{8},\frac{1}{16},\frac{1}{32}$ $\endgroup$ – user408906 Dec 20 '18 at 13:38
  • $\begingroup$ @Suraj: I'm assuming that such an assignment exists. My assignment requires that you can construct two pairs summing up to 1. In your example, this is not possible. $\endgroup$ – STF Dec 20 '18 at 13:41
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    $\begingroup$ What about $0,\frac{1}{4},\frac{3}{4}, 1$ and $\frac{1}{4},1,0,\frac{3}{4}$? [I choose to do $w_1<w_2$ and swap $w_3$ and $w_4$ it's the same thing.] $\endgroup$ – ancientmathematician Dec 20 '18 at 13:52
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    $\begingroup$ No. Think about it. $\endgroup$ – ancientmathematician Dec 20 '18 at 13:59

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