# Implications of multiple ways to order eight numbers

Consider two sets $$A,B$$ composed of four real numbers each.

These eight real numbers are in $$[0,1]$$.

Consider other four real numbers $$c,d,e,f$$ each in $$[0,1]$$, all different between each other.

Assume there exists a way of ordering the four numbers in each set $$A,B$$ such that $$\begin{cases} w^A_1+w^B_1=c\\ w^A_2+w^B_2=d\\ w^A_3+w^B_3=e\\ w^A_4+w^B_4=f\\ \end{cases}$$ where

• $$w^A_h$$ denotes the $$h$$th element of set $$A$$ once we have ordered its 4 elements

• $$w^B_h$$ denotes the $$h$$th element of set $$B$$ once we have ordered its 4 elements

Claim: if such an ordering is not unique, then it should be that two numbers in $$A$$ are equal and/or that two numbers in $$B$$ are equal.

Is this claim correct? If yes, how can I prove it? If not, can you provide a counterexample?

(similar question here but with 2 elements per set)

Maybe the claim is wrong? Let $$\{a_1,a_2,a_3,a_4\}$$ be the elements of $$A$$ and $$\{b_1,b_2,b_3,b_4\}$$ be the elements of $$B$$. We could have:

order I $$\begin{cases} a_2+b_3=c\\ a_4+b_4=d\\ a_1+b_1=e\\ a_3+b_2=f \end{cases}$$

and

order II $$\begin{cases} a_1+b_2=c\\ a_2+b_1=d\\ a_3+b_4=e\\ a_4+b_3=f \end{cases}$$

which implies $$\begin{cases} a_2+b_3=a_1+b_2\\ a_4+b_4=a_2+b_1\\ a_1+b_1=a_3+b_4\\ a_3+b_2=a_4+b_3 \end{cases}$$ Does this imply that two numbers in $$A$$ are equal and/or that two numbers in $$B$$ are equal?

The claim is wrong. Pick the following partial solution of the last system: $$(a_1,a_2,a_3,a_4)=(0,a,a’,a+a’)$$ and $$(b_1,b_2,b_3,b_4)=(a’,a+b,b,0)$$. Then $$(c,d,e,f)=(a+b,a+a’,a’,a+a’+b)$$. For instance, we can put $$a=0.1$$, $$a’=0.15$$ and $$b=0.2$$. Then $$(a_1,a_2,a_3,a_4)=(0,0.1,0.15,0.25)$$, $$(b_1,b_2,b_3,b_4)=(0.15,0.3,0.2,0)$$, and $$(c,d,e,f)=(0.3,0.25,0.15,0.45)$$.