Is there a way to find out whether complex numbers are greater then or less than another complex number?

For numbers like $i$ and $-i$, it is clear than one is bigger than the other. However, what about numbers like $2+3i$ and $5+2i$?

Here's what I think. In the real numbers, you measured how big numbers are with their magnitude and direction. If two real numbers pointed in the same direction, the one with the bigger magnitude is bigger, and if two real numbers were pointed in opposite directions, the one facing the positive direction is bigger.

So, I tried the same thing for complex numbers. For a complex number $a+bi$, the positive directions are along the north of $a$ and east of $b$ axes. So, $2+3i \gt -2-3i$ because $2+3i$ goes in both positive directions, while $-2-3i$ goes in both negative directions. Also, $2+3i \gt -2+3i$ because $2+3i$ goes in both positive directions, while $-2-3i$ goes positive in only one direction. You have to compare numbers like $3-5i$ and $-3+6i$ using their magnitudes because they both go in a positive and negative direction.

Also, for numbers lying on the axes themselves, you can't compare them unless they are on the same axis.

So, that is my suggestion, does it hold up?

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    $\begingroup$ inequalities are not well-defined in the complex plane... You can compare only absolute value of complex number for example $|1+i|<|5+i|$ . $|a+bi|=\sqrt {a^2+b^2}$ $\endgroup$ – lone student Jan 23 at 21:55
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    $\begingroup$ What would be your purpose of such an ordering: for example, how would you like your ordering to behave with respect to algebraic operations like addition and multiplication? $\endgroup$ – Catalin Zara Jan 23 at 21:55
  • $\begingroup$ shaileshk.medium.com/an-order-on-complex-numbers-6357c8b4aec2 $\endgroup$ – Buraian Jan 23 at 21:56
  • $\begingroup$ Website is blocked on my computer :/ $\endgroup$ – Some Guy Jan 23 at 21:56
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    $\begingroup$ Check this: math.stackexchange.com/q/1184547/42969 $\endgroup$ – Martin R Jan 23 at 22:16

You can define $<$ (nearly) anyway you like.

If wanted two I could define $x < y$ if when you spell them out in english words $x$ occurs lower in alphabetical order than $y$.

so $8 < 18 <805 < 85 < 82< 11< 10 < 3 < 0 < 0.5 < 0.7 < 0.6 < 0.1 < 0.2$ and so on

Because eight < eighte < eighth < eightyf < eightyt < el < te < thr < zero < zero point f < zero point se < zero point si < zero point o < zero point t.

Kinda dumb though.

And we could define an order of compairing $a+bi$ to $c + di$ as $a+bi < c+di$ if $a < c$ or if $a=c$ and $b < d$. But $a + bi = c+di$ if $a=c$ and $b=d$. And $a +bi > c+di$ if $a > c$ or if $a=c$ and $b > d$.

That's called the lexigraphical order and it looks okay by it's actually dumb too.

The thing is if we want to define an order we want it to obey in certain ways.

One thing we want is we want if $a < b$ then $a+c < b+c$ for all $c$

The lexigraphical order does do this but what you, incorrectly, claim is the order on the reals does not.

You claim, incorrectly, $-5 > -2$ because $|-5| > |-2|$ and $-5, -2$ both point in the same direction. If this were true we would like $-5 + 10 > -2 + 10$ and that would mean $5 > 8$ which it isn't.

We also want that if $a < b$ and $m > 0$ then $am < bm$.

This fails on the lexigraphical order. $i = 0 + i > 0 + 0i = 0$. so $i*i$ should be greater than $0*i = 0$. but $i*i=-1 < 0$.

And it is because of this requirement we can not have a reasonable order on the complex. That just isn't possible.

Consider: Is $0 < 1$ or is $0 > 1$?

If $0 > 1$ then $0 + (-1) > 1 +(-1)$ and $-1 > 0$. Therefor $(-1)*(-1) > (-1)*0$ so $1 > 0$. But that contradicts what we assumed. So we can't have $0 > 1$ so we must have $0 < 1$.

And therefor $0 + (-1) < 1 + (-1)$ and $- 1 < 0$.

[Note, this means also that if $5 > 2$ the $5-5 > 2-5$ and $0 > -3$ and $0 + (-2) > -3+(-2)$ and so $-2 > -5$. this is why your claim that $-5 > -2$ was wrong. If they point in the negative direction the one with a bigger magnitude must be less than the one with the smaller magnitude.]

Okay, but what about $i$. Is $i > 0$. If so then $i*i > 0*i$ and $-1 > 0$. But we just showed we had to have $-1 < 0$.

Okay, so $i < 0$. But that means $i + (-i) < 0 + (-i)$ so $0 < -i$. So $0*(-i) < (-i)(-i)$. So $0 < (-i)^2 = (-1)^2(i)^2 = 1*(-1) = -1$. So $0 < -1$.... but we just said....

So it's impossible.

If we want $a < b \implies a+c < b+c$ and we also want $a<b; m > 1 \implies am < bm$ then it's just not going to be possible to compare two complex numbers. It just can not be done.


It's not at all clear that one of $\mathrm i$ or $-\mathrm i$ is bigger than the other! What makes you think so?

That's a rhetorical question. I know that you think so because of the sign. But there's a big difference between imaginary and real numbers when it comes to signs: $-1$ behaves in a way that is distinctly different from $1$. For instance, if we square $1$, we get back the same number, but if we square $-1$, we obtain a different number. So if I give you a mystery number that is either $1$ or $-1$, you can find out which one it is by asking me what its square is. If you get back the same number it's $1$, otherwise it's $-1$. Such things don't work to differentiate $\mathrm i$ from $-\mathrm i$. If I give you a mystery number that is either $\mathrm i$ or $-\mathrm i$, it will always remain a mystery which one it is, no matter how much you know about its behavior. For instance, if you square the mystery number, you're guaranteed to get $-1$, no matter which of the two it is. If you take the mystery number to the power of $3$, you'll get minus the mystery number, no matter which number it is. Specifically, there is no algebraic way (that is, using only multiplication and addition) to tell the difference. For this reason, there is no algebraic way to tell which of the two should be larger. And when ordering fields, we always want to be able to do so algebraically, since algebra is the only thing we can natively do with fields.

  • $\begingroup$ "So if I give you a mystery number that is either 1 or −1, you can find out which one it is by squaring it. If you get back the same number it's 1, otherwise it's −1." If it's a mystery number, how can you square it? And how can you decide if the result is the same number? $\endgroup$ – Catalin Zara Jan 23 at 22:19
  • $\begingroup$ @CatalinZara I think that the point Vercassivelaunos is making is specifically when you know if the mystery number has changed or not. I'm not clear about your question about not being able to square a mystery number. Why can;t you? $\endgroup$ – A-Level Student Jan 23 at 22:20
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    $\begingroup$ @CatalinZara One number squares to itself. The other number doesn't. If I tell you that $x \in \{-1,1\},$ then $x$ is a "mystery" number, but if I promise to answer truthfully when you ask whether $x = x^2,$ you can solve the mystery. $\endgroup$ – David K Jan 23 at 22:20
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    $\begingroup$ @CatalinZara - We have a field, so we have a machine labeled "$\times$" that takes in two numbers and returns a number. But the numbers are always wrapped in identical grey paper. We also have machines labeled "$+$", "$-$", and "$1/x$", and finally one labeled "$=$", which lights up either the True or False light. You can use these to tell if the grey paper wrapped number someone gave you was $1$ or $-1$. $\endgroup$ – JonathanZ supports MonicaC Jan 23 at 22:32
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    $\begingroup$ Mystery numbers - the best way to explain the concept of field automorphisms :) $\endgroup$ – Hagen von Eitzen Jan 24 at 6:51

You could define inequalities for complex numbers. For example, we could make the following definition:

We say that $z$ is 'greater than' $w$ if $|z|>|w|$.

The only question that remains is whether this definition is helpful. And the answer to that is: not really. The real numbers are closed under multiplication and addition; that is, if $a$ and $b$ are positive, then $a \cdot b$ is positive, and $a + b$ is positive. It turns out that these two properties are crucially important for defining inequalities in a meaningful way. And neither of these properties hold for complex numbers, meaning that writing $z>w$ just doesn't make that much sense.

To explain why closure under addition and multiplication is relevant, we must first define what we actually mean by $a>b$. One common way to do this is to take the notion of a positive number for granted. Let's denote the collection of positive numbers as $\mathbf{P}$. One of the most important properties of $\mathbf{P}$ is the trichotomy law:

For every number $x$, one and only one of the following holds:

  • $x=0$.
  • $x$ is in the collection $\mathbf{P}$.
  • $-x$ is in the collection $\mathbf{P}$

While this may seem obvious, it is necessary for the following definitions:

  • $a>b$ if $a-b$ is in $\mathbf{P}$
  • $a<b$ if $b>a$
  • $a \geq b$ if $a>b$ or $a=b$
  • $a \leq b$ if $b \geq a$

If it were possible for $x$ to be in the collection $\mathbf{P}$, and $-x$ also be in the collection, then we could have $a>b$ and $b>a$ ! Hence, the trichotomy law is necessary if we are to maintain our sanity. Another important property is closure under multiplication and addition, as mentioned earlier. If the real numbers weren't closed under multiplication, this would also cause headaches. For example, we wouldn't be able to say that if $a>0$, and $b>0$, then $ab>0$.

Fortunately, the real numbers do contain a collection $\mathbf{P}$ with all these properties, but the complex numbers do not. This can even be proven fairly easily. Suppose that the complex numbers did satisfy the trichotomy law and closure under multiplication and addition. Let $x$ be an arbitrary complex number. If $x$ is positive, then $xx = x^2$ would also be positive. If $x$ is negative, then $-x$ would be positive, and so $(-x)(-x) = (-x)^2=x^2$ would be positive. It follows that for all non-zero $x$, $x^2$ is positive. In particular, if $x=i$, then $x^2 = -1$ would be positive. But $1=1^2$ would also be positive. This violates the trichotomy law: $1$ and $-1$ can't both be positive.

Hence, the idea of positivity is incoherent for complex numbers, and so too is the idea of inequalities. The best we can do is say that two complex numbers are either equal, or unequal.


I think that to define an order on a field, you would want a positive cone $\Bbb P$in that field, behaving like the set of positive reals within $\Bbb R$. So you would want, for complex numbers $z$ and $w$, satisfaction of the three conditions \begin{align} \Bbb P+(-\Bbb P)=\Bbb C\\ z,w\in\Bbb P&\Longrightarrow z+w\in\Bbb P\\ z,w\in\Bbb P&\Longrightarrow zw\in\Bbb P\,. \end{align} The first condition means that every complex number is to be the sum of a “positive” number and the negative of some positive number. I think you’ll persuade yourself, no matter how you try, that with the first condition in place, the third condition is unsatisfiable.


All of the answers are great and all, but really the issue is that the greater than and less than cannot exist within the complex numbers. It is actually provable that such a comparison does not exist in any way comparable to greater than and less than for real numbers.

Suppose $0 < i$. Since any number greater than $0$ squared is still greater than $0$ we have that $0 < -1$. This is a contradiction. So therefore it must be that $i < 0$. However subtracting $i$ from both sides yields $0 < -i$. Once again squaring we get that $0 < -1$, which is also a contradiction. Therefore $i$ is neither greater nor less than $0$ nor equal to $0$.

Simply put, the complex numbers do not satisfy the algebraic rules governing $<$. Therefore, defining a concept of less than and greater than would be nothing more than defining a two input function on the complex numbers whose value is true or false. It would be meaningless.

The way I think of this is the complex numbers form a complex plane rather than a real number line. This is all just a way of thinking of things, of course, but it makes sense from this that a plane cannot be ordered in an algebraic fashion without it just becoming equivalent to the real numbers. The real numbers are the penultimate set of comparable numbers. They were meant to be the densest set of numbers possible that obey the laws of comparison and arithmetic. Complex numbers are useful and interesting because they are a set that breaks some of those laws. And so they are not comparable.

The closest is comparison by absolute value, but this is not equivalent to the real number idea of comparison. It’s really just a comparison of the results of computing the norm of a complex number.

  • $\begingroup$ This is just simply wrong! There are models of exactly the same ordered field axioms as the real numbers with every infinite cardinality, including countable and the cardinality of $ℝ$ and every possible cardinality between or beyond. So it is totally false to claim that the reals are "the penultimate set of comparable numbers" or "the densest set of numbers possible that obey the laws of comparison and arithmetic". $\endgroup$ – user21820 Jan 24 at 8:09

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