Are imaginary numbers really incomparable?
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
Solution 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.
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.
Solution 2:
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 $i > -i$.