Ordering four real numbers such that some conditions are satisfied












1












$begingroup$


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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$endgroup$












  • $begingroup$
    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






  • 2




    $begingroup$
    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








  • 1




    $begingroup$
    No. Think about it.
    $endgroup$
    – ancientmathematician
    Dec 20 '18 at 13:59
















1












$begingroup$


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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    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






  • 2




    $begingroup$
    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








  • 1




    $begingroup$
    No. Think about it.
    $endgroup$
    – ancientmathematician
    Dec 20 '18 at 13:59














1












1








1


1



$begingroup$


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?










share|cite|improve this question











$endgroup$




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?







combinatorics permutations combinations






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share|cite|improve this question













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edited Dec 20 '18 at 13:53







STF

















asked Dec 20 '18 at 13:13









STFSTF

781420




781420












  • $begingroup$
    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






  • 2




    $begingroup$
    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








  • 1




    $begingroup$
    No. Think about it.
    $endgroup$
    – ancientmathematician
    Dec 20 '18 at 13:59


















  • $begingroup$
    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






  • 2




    $begingroup$
    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








  • 1




    $begingroup$
    No. Think about it.
    $endgroup$
    – ancientmathematician
    Dec 20 '18 at 13:59
















$begingroup$
What exactly do you mean when you say unique?
$endgroup$
– Ankit Kumar
Dec 20 '18 at 13:31




$begingroup$
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




$begingroup$
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




$begingroup$
@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




2




2




$begingroup$
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






$begingroup$
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






1




1




$begingroup$
No. Think about it.
$endgroup$
– ancientmathematician
Dec 20 '18 at 13:59




$begingroup$
No. Think about it.
$endgroup$
– ancientmathematician
Dec 20 '18 at 13:59










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