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


  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.

Source : Link , Question Author : Community , Answer Author : vadim123

Leave a Comment