# Implications of knowing how to order the elements of an $n$-tuple.

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 $$a_2; hence, $$\pi(\{0,-2,4\})=\{\{2,1,3\}, \{<,<\}\}$$

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

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

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 $$\pi(\{0,a,b\})$$.
• 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 $\pi(\{0,-a,-b,b-a\})$"? – STF Nov 6 '18 at 16:08
• @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. – Joey Kilpatrick Nov 6 '18 at 16:12