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June 15

Comparing complex numbers

We all know that 2>1, and you can tell because 2 is farther to the right on the number line than 1. But since 1 and i can only be viewed together on a plane, how can they be compared? Would it even be accurate to say i is greater than or less than one? --65.31.80.100 22:06, 15 June 2007 (UTC)[reply]

Unlike the reals, we can't give the complex numbers a total ordering. But we can compare their magnitudes -- in which case 1 and i have the same magnitude. iames 22:12, 15 June 2007 (UTC)[reply]

It's not an Ordered field because squared elements can be negative. However, it should be noted that you could give the complex numbers an ordering (e.g. dictionary ordering), but that the ordering won't be compatible with the field axioms in the way we're accustomed. nadav (talk) 22:21, 15 June 2007 (UTC)[reply]

Is the lexical ordering a total ordering for this case? Twma 00:44, 16 June 2007 (UTC)[reply]

I was about to argue that certainly you could give them a total ordering but then I just realized nadav just argued the same thing. It's quite like sorting two fields in a Table (database); the two fields to sort by would be the real component and the imaginary component. However, you'd never want to sort by potential decimal representation (or any base for that matter) unless you could constrain numbers not to end with things like 0.999... Root⁴(one) 03:56, 16 June 2007 (UTC)[reply]

BTW a more direct answer is here: Total order#Orders on the Cartesian product of totally ordered sets.Root⁴(one) 03:59, 16 June 2007 (UTC)[reply]