Are imaginary numbers really incomparable?

Question

If we really don't know which is bigger if $ i $ is greater or $ 2i $ or so on then why do we plot $ i $ first then $ 2i $ and so on, on the imaginary axis of the Argand plane? My teacher said that imaginary numbers are just points and all are dimensionless so they are incomparable and the distance really doesn't matter. I want to get this more clear

Answer 1

1. There is no way to order complex numbers, in a way that preserves operations in a sensible way. The precise term is ordered field; among their properties, $x^2 \ge 0$ for every $x$. Since $i^2=-1$, we would need $-1\ge 0$, which is impossible.

2. However, if you want to measure distance, you can do that with a norm. In the complex numbers, this is calculated as $|a+bi|=\sqrt{a^2+b^2}$. Hence, it is correct to say that $2i$ is twice as far from the origin as $i$ is.

Answer 2

In an ordered field, you can prove that $x^2 \geq 0$. In $\mathbb{C}$, we have $i^2=-1<0$ so you cannot put an order on it.

Answer 3

1. We plot $i$ before $2i$ and so on purely by convention (and maybe that our brains are acquainted to working with numbers from smaller magnitudes to larger magnitudes). On a computer screen, for example, $2i$ would have lower $y$ coordinate than $i$ and thus is plotted first.

2. There is no way of imposing a total ordering on $\mathbb C$ that would give you the same nice properties as one on $\mathbb R$, for example, having $x^2 \geq 0$ for all $x$, is not possible.

3. You can compare the magnitudes of complex numbers. That is, their absolute values. Distance in this case does matter because distance is a real number and is thus comparable. It is just that the complex points themselves are not comparable.

Answer 4

Here is one way to compare them With undermentioned orders we have Example see that $$\color{blue}{i \lesssim 2i~~~~~and~~~~i \lessapprox 2i}$$

With the usual oder on $\Bbb R$ it is not possible since we have the obstruction $i^2 = -1<0$ which directly drive us to contradiction.

However,given to couple $(x,y)$ and $(a,b)$ whose complex representations are $ x+iy$ and $a+ib$ respectively we can define the following lexical oders

We say that $$(x,y)\lesssim (a,b)~~~~\mbox{ iff}~~~x<a ~~~or (~~x=a~~~and ~~~y< b.)$$

$$(x,y)\lessapprox (a,b)~~~~\mbox{ iff}~~~x\le a ~~~or (~~x=a~~~and ~~~y\le b.)$$

you can check these are actually on $\Bbb C$.

Example see that $$\color{blue}{i \lesssim 2i~~~~~and~~~~i \lessapprox 2i}$$

Be aware that they do not preserve classical well known properties of the oders $<$ and $\le $ such as $$Z\lesssim W \not\implies ZZ'\lesssim WZ'$$

Answer 5

For purely imaginary numbers, the ones with zero real part, i.e. $z = bi$ (where $b$ is a real), it's trivial to set an ordering: just consider the imaginary parts and compare those as reals. That's what we do implicitly when drawing the imaginary axis or choosing the "$y$" coordinate when plotting complex numbers on a plane.

For complex numbers in general: $z = a+bi$, where both $a$ and $b$ can be nonzero, there's no clear order. E.g. any ordering between all four of $(\pm 1 \pm i)$ would be arbitrary.

Answer 6

Imaginary numbers are comparable

As you said we can order the imaginary numbers like $-i < 0 < i < 2i < \ldots$ and in general write $ai \le bi$ to mean $a \le b$ for real numbers $a,b$. There is nothing wrong with this.

Some complex numbers are comparable

If you want to order all the complex numbers we can define $a+bi \le c+di$ to mean that $a \le c$ and $b \le d$ as real numbers. This gives what's called a partial order on $\mathbb C$. This order is unlike the order on the real (or imaginary numbers) in that there exist pairs of elements that are incomparable. For example $1+2i$ and $2+1i$ are incomparable.

Edit: We cannot compare the complex numbers $1+2i$ and $2+1i$.

This is because we defined $1+2i \le 2+i$ to mean $1\le 2$ and $2 \le 1$ which is untrue.

And we defined $2+i \le 1+2i$ to mean $2 \le 1$ and $1 \le 2$ which is untrue.

Answer 7

There is a good reason to avoid defining "orderings" like $i > -i$, or any similar comparisons, even if you don't care about ordered fields. That is: there is no way to distinguish between $i$ and $-i$, except for the symbols used to denote them. $i$ is defined as a solution of $x^2 +1 = 0$. But there are two solutions, so how do you decide which one gets the privilege to be bigger than the other?

If you take any sentence in the language of complex numbers in which $i$ occurs, and replace it everywhere with $-i$, then the truth of the sentence doesn't change. That isn't true anymore if you introduce definitions like $i > -i$.

Answer 8

The well-ordering theorem tells us that if we accept the Axiom of Choice then any set, obviously including the complex numbers, can be ordered. But it doesn't tell us how to construct such an ordering, and the ordering won't have the useful properties of the ordering of the real numbers (the properties of an ordered field). So while we can in principle order the complex numbers, in practice it is both difficult and pointless.

Answer 9

I've come to think of i as a rotation operator, so ordering may not make any sense. But these are the thoughts of an old man, and I'm not sure there is any basis to this? This probably ought to be a comment, but you can't comment on a question until you have a reputation level of 50+.