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Consider the ordered n-tuple $\{a_1,a_2,..., a_n\}$ with $a_i\in \mathbb{R}$ $\forall i=1,...,n$.

$\pi$-operator: Let $\pi$ be an operator that tells me

1) The position in the original n-tuple of $\{a_1,a_2,..., a_n\}$ when ordered from smallest to largest. When two elements of $\{a_1,a_2,..., a_n\}$ are equal we assume the convention that we put firstly the element coming first in $\{a_1,a_2,..., a_n\}$.

2) The relational operator (<,=) between the elements of $\{a_1,a_2,..., a_n\}$ when ordered from smallest to largest.

Examples:

  • take $n=3$, $a_1=0$, $a_2=-100$, $a_3=4$; we have $a_2<a_1<a_3$; hence, $\pi(\{0,-2,4\})=\{\{2,1,3\}, \{<,<\}\}$

  • take $n=3$, $a_1=0$, $a_2=100$, $a_3=0$; we have $a_1=a_3<a_2$; hence, $\pi(\{0,100,0\})=\{\{1,3,2\}, \{=,<\}\}$

Question: Suppose that I know $\pi(\{0,a,b\})$. Does this imply that I know $\pi(\{0,a,b,b-a\})$?

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No, you don't nessecarily know $\pi(\{0,a,b,b-a\}) $.

Consider $a=1$ and $b=3$.$$ \pi(\{0,a,b\})=\{\{1,2,3\},\{<,<\}\} \\ \pi(\{0,a,b,b-a\})=\{\{1,2,4,3\},\{<,<,<\}\} $$

Consider $a=2$ and $b=3$.$$ \pi(\{0,a,b\})=\{\{1,2,3\},\{<,<\}\} \\ \pi(\{0,a,b,b-a\})=\{\{1,3,4,2\},\{<,<,<\}\} $$

These have the same values for $\pi(\{0,a,b\})$, but different values for $\pi(\{0,a,b,b-a\})$. Therefore $\pi(\{0,a,b,b-a\})$ cannot be determined from $\pi(\{0,a,b\})$.

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  • $\begingroup$ Thanks. Just to be sure I understood: would your answer be "YES"if my question was "Suppose that I know $\pi(\{0,a,b\})$. Does this imply that I know $\pi(\{0,-a,-b,b-a\})$"? $\endgroup$ – STF Nov 6 '18 at 16:08
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    $\begingroup$ @STF Note that knowing $\pi(\{0, -a, -b, b-a\})$ is the same as knowing $\pi(\{0, a, b, a-b\})$. Knowing $\pi(\{0, a, b, a-b\})$ is the same as knowing $\pi(\{0, a, b, b-a\})$, so no you cannot know $\pi(\{0, -a, -b, b-a\})$ either. $\endgroup$ – Joey Kilpatrick Nov 6 '18 at 16:12

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