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Is $2 + i > 1 - i$ true?

First off, what does it even mean to be "bigger" or "smaller" as a number?

If we say that $a>b$ means that $a$ is on the right of $b$ on the number line, could we also say that $a$ V $b$ means that $a$ is below $b$? Would this mean that we could compare "sizes" of complex numbers with an angle?

So I think that $2 + i > 1 - i$ and $2 + i$ ᴧ $1 - i$. But this compares only one part at a time, the real and then the imaginary.

So maybe $2 + i \angle_{_{angle}} 1 - i$ where $\angle_{_0}$ means $>$, $\angle_{_\pi}$ means $<$ and so on clockwise? Would that be a correct way to compare complex numbers?

Peri
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  • See this https://math.stackexchange.com/questions/487997/total-ordering-on-complex-numbers. – Roby5 Mar 07 '19 at 21:01
  • The complex numbers are not ordered. The statement $2 + i > 1 - i$ is meaningless. You could make some convention by comparing first the real parts and then the imaginary parts, say, but this would not have the usual properties of $>$. As for angles, there's something called the argument of a complex number, but since you can rotate around the origin as many times as you like, it's only defined up to a multiple of $2\pi.$ – saulspatz Mar 07 '19 at 21:02
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    First off, what does it even mean to be "bigger" or "smaller" as a number? It means the same thing for "numbers" (whatever those are) as it means for an arbitrary set: it means we have decided upon a total order relation (or perhaps merely even a partial order) for the set. – rschwieb Mar 07 '19 at 21:13
  • It is true under the dictionary order. – John Douma Mar 07 '19 at 21:37
  • @John Douma not the lexicographic (dictionnary) order but the "product order" $(a,b) \leq (c,d) \ \iff a \leq c \ \text{and} \ b \leq d$ which is a partial order whereas the lexicographic order in a total order. – Jean Marie Mar 07 '19 at 21:41

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Complex numbers aren't ordered in any interesting way. (Since bijections of domain $\Bbb C$ and range $\Bbb R$ exist you can order $\Bbb C$, but not in any way that satisfies useful rules.) In particular no set $P\subseteq\Bbb C$ of "positive" complex numbers exists satisfying $$0\notin P,\,\forall z\in\Bbb C\backslash\{0\}(z\in P\iff -z\notin P),\,a,\,b\in P\implies a+b,\,ab\in P,$$ because thanks to $a\in P\implies a^2\in P$ some sign $\pm i$ would satisfy $\pm i\in P\implies -1\in P\implies 1\in P$. Thus we cannot define $a>b$ as $a-b\in P$ as with real numbers viz. $P:=\Bbb R^+$.

J.G.
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