# Sorting $i$ amongst integers

In the game Wits & Wagers, players each answer a numerical question and the answers are then sorted in ascending order for scoring, details of which being irrelevant for this question.

In this week's Tabletop, Wil Wheaton answered the imaginary number $i$ to one such question (Around 10:40 in the video). Given that the other answers were positive non-zero integers, where should $i$ have been sorted and why?

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There is no right answer. If the possibility of such an answer had not previously been taken into account in the rules, it would have been better to disqualify the answer than to try to deal with it on the fly. One can come up with ad hoc justifications for various decisions, but allowing this to be done while the game is in progress amounts to letting the person running the game influence the result. E.g., if the imaginary part is ignored, and the ranking is based on the real part, then $i$ ranks as if it were $0$. If the complex numbers are ordered lexicographically, real part first, then $I$ also precedes every positive integer. If the complex numbers are ordered lexicographically, imaginary part first, then $i$ follows every positive integer. If all numbers are ordered by magnitude, then $i$ ties with $1$.
You have to design a custom comparison of a sorts, because in general, comparing complex numbers (equivalent to comparing 2D vectors) with ordinary "<" is only a partial order (i.e. not everything can be compared). In other words, it makes sense to say that $(1,2) < (3,4)$ (or equivalently, $1+2i < 3+4i$), but how can you compare $(1,3)$ and $(4,2)$? Now, $4>1$ but $2<3$?!
One order which is occasionally used is you compare the distance of vectors from the origin, i.e $|(1,3)|^2 = 3^2+1^2 = 10$ and $|(4,2)|^2 = 20$ so $(1,3)<(4,2)$. The problem here is that it renders all points at the same distance from the origin equivalent to each other, e.g. in your example, $1$ and $i$ come out the same...
Or you can use the lexicographic order, where $a+bi \lt c+di \Leftrightarrow (a \lt c) \vee (a=c \wedge b \gt d)$. Then $i \lt 1$. It is a total order, but doesn't respect the operations of the field at all. –  Ross Millikan Sep 20 '12 at 23:33