As normed division algebras, when we go from the complex numbers to the quaternions, we lose commutativity. Moving on to the octonions, we lose associativity. Is there some analogous property that we lose moving from the reals to the complex numbers?
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J. M. ain't a mathematician
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Tudwell
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2(This is posted after the very good "linear order is lost" solution appeared.) It's known that commutativity and associativity have geometric interpretations. I'm curious now whether or not that linear order fits this pattern, or if there's another property that does fit the pattern that is lost... – rschwieb Nov 20 '12 at 02:12
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15In passing from the reals to the complex, we lose a few people who can't wrap their head around sqrt(-1). :) – Kaz Nov 20 '12 at 03:45
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Also, not to belittle the very apt answers, but I was wondering basically what rschwieb is asking when I posed this question: is there something "more analogous" to commutativity and associativity that we lose? – Tudwell Nov 20 '12 at 20:13
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Does losing ordering define the complex numbers? Ie is there only one field (strictly) containing the reals that can not be ordered but is complete? – Michaela Light Nov 20 '12 at 22:32
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What property do quaternions have that complex numbers don't, again? – Joe Z. Jan 31 '14 at 03:04
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2Possible duplicate of What is lost when we move from reals to complex numbers? – Autar Apr 09 '18 at 13:30
2 Answers
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The most obvious property that we lose is the linear (or total) ordering of the real line.
Per Erik Manne
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Thanks. I guess I never really associated that with a typical property of algebras. Probably because it's so unusual. – Tudwell Nov 20 '12 at 00:13
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4i.e. you can't use $<$, $>$, $\le$, $\ge$ to (directly) compare complex numbers... – J. M. ain't a mathematician Nov 20 '12 at 00:19
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1@Tudwell One possible definition of the real numbers is as a totally ordered field which satisfies the least upper bound property, see for instance Rudin's Principles of Analysis. – Per Erik Manne Nov 20 '12 at 00:27
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You cannot use <, >, ≤, ≥ in a way which is backward compatible with how these are defined for the reals, yet makes sense in the complex plane. – Kaz Nov 20 '12 at 03:46
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We lose equality of the complex conjugate and total order.
So:
$x+i y \ne x-i y$ for complex numbers which are not also reals.
And you can't say wether $ i > -i $ or $i < -i$, etc.
All you have is the magnitude which, in the given example, is equal.
kram1032
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3@TheChaz The magnitude of the difference of two complex numbers is their distance. If that's what you mean, then yes. But that doesn't help you at all with finding an order. You still can't say wether i>-i or vice-versa. – kram1032 Nov 20 '12 at 01:33
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2Magnitude does not map into the reals in a way that is "backward compatible" with how reals are compared; i.e. that -2 is less than 1. – Kaz Nov 20 '12 at 03:47