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If $a+ib$, $c+id$, $e+if$ are three complex numbers, than can we tell which one is greater or smaller between them? If yes, then how and if no then why not?

Can somebody give explanation on this.... I will be grateful to him.

Amzoti
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Sachin Sharmaa
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  • What do you mean by "greater"/"smaller"? – Cameron Buie Feb 22 '13 at 05:32
  • a+ib > c+id > e+if – Sachin Sharmaa Feb 22 '13 at 05:33
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    And what exactly do you mean by x>y (With both of them complex). – Ishan Banerjee Feb 22 '13 at 05:34
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    I think the question might be about whether $\mathbb C$ is an ordered field, in which case it is not: we only have magnitude to compare, but there is no strict order on the numbers themselves, at least in the case of comparing numbers of the form $a + bi$, $b\neq 0$ – amWhy Feb 22 '13 at 05:37
  • @amWhy: That isn't (quite) true. There are, in fact, at least continuum-many strict orders on the complex numbers (perhaps more than that, but I'll have to think on it). We certainly can't make $\Bbb C$ an ordered field with the usual operations, though. – Cameron Buie Feb 22 '13 at 05:59
  • Thanks for the correction, @Cameron I should have left it at "not an ordered field"...with the usual operations :-) – amWhy Feb 22 '13 at 06:00

3 Answers3

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We can define a partial order on $\Bbb C$ by $z_1\prec z_2$ if and only if $|z_1|<|z_2|$.

We can define a total order on $\Bbb C$ in various ways--I'll give you a few if you're interested.

We cannot give an order that is compatible with the operations on $\Bbb C$ so that $\Bbb C$ is an ordered field. If we could, would $i$ be positive or negative?

Basically, it depends on how you want to define "bigger/smaller" in this instance. Give us more detail, and we can better answer your question.

Cameron Buie
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You can't define a total order $\le$ so that $\mathbb{C}$ is an ordered field:

$(1)\space\forall a, b, c \in \mathbb{C}, a ≤ b \Rightarrow a + c ≤ b + c$

$(2)\space\forall a, b \in \mathbb{C}, 0 ≤ a \land 0 ≤ b \Rightarrow 0 ≤ a b$

Either $0 \le 1$ or $1 \le 0$ in which case $0\le-1$. Let $\varepsilon \in\{-1,1\}$ so that $0 \le \varepsilon$

Either $0\le i$ or $i \le 0$ in which case $0\le-i$. Let $\delta \in \{-i,i\}$ so that $0\le\delta$

Now you can just derive something absurd:

You have $0\le \varepsilon$ and $0\le \delta$

So by $(2)$, $0\le\varepsilon\delta$

By $(2)$ again, $0\le\varepsilon\delta\delta = \varepsilon(-1) = -\varepsilon$

So $\varepsilon \le 0$

Now we have $0\le\varepsilon$ and $\varepsilon\le 0$ so $\varepsilon=0$

But we have $\varepsilon\in\{-1,1\}$

Absurd.

xavierm02
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I'm thinking that you'd want to find their magnitude. Given a complex number $z = a +ib, b \ne0$, the magnitude is given by $|z| = \sqrt{z\cdot z^*} = \sqrt{(a+ib)\cdot (a-ib)} = \sqrt{a^2+b^2}$, where $z^*$ is the complex number's conjugate. Then you can order them from greatest to least in this way.

MITjanitor
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  • Although if you sort them in this way, there will be more than one number of a given "size" (in fact, there will be infinitely many, unless you're talking about $\left|z\right| = 0$). – Stahl Feb 22 '13 at 05:43
  • True, I didn't think about that. – MITjanitor Feb 22 '13 at 05:44
  • Then, you can sort it lexically - the greater the real part, the greater the number. – m93a Apr 18 '14 at 18:16