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There is a persuasive school of thought that argues exactly the opposite, that imaginary numbers are not real http://geometry.mrao.cam.ac.uk/1993/01/imaginary-n

There is a persuasive school of thought that argues exactly the opposite, that imaginary numbers are not real http://geometry.mrao.cam.ac.uk/1993/01/imaginary-numbers-are-not-real-the-geometric-algebra-of-spacetime/ http://geometry.mrao.cam.ac.uk/1993/01/imaginary-numbers-are... .

topaz0 5y ago
I'm familiar with (and partial to) the geometric algebra literature, but I think it's a bit disingenuous/clickbaity to say that it implies imaginary numbers aren't real. I think of it as meaning that the algebra of the imaginary numbers arises as a slice through a more useful, bigger algebra.
- hgomersall 5y ago
  Yes, I think it's rather tongue in cheek. It's probably not defined as clickbait since it was written in '93!
  - topaz0 5y ago
    Fair enough!
wrycoder 5y ago
You can make some progress by thinking of the imaginary numbers as being a two dimensional vector space over the reals, equipped with a very peculiar multiplication operator that is a combination of the usual stretch with a rotation. Addition is the usual addition of components, but the multiplication (using polar coordinates) looks like C = ABexp(T1+T2), where the T are the phases of A and B. To multiply two positive reals, since their phases are zero, it’s only a stretch: AB To multiply two negative reals, since their phases are pi, it’s ABexp(pi+pi). But exp(2pi) is unity, so we have AB again, a positive number. To find the square root of a positive number D, find a positive real (zero phase) that, when multiplied by itself, gives D. AA, for example. So, real number arithmetic is a subset of complex number arithmetic, with zero phase. And therefore zero rotation. Now, find the square root of a negative number, say -4. To do this, we have to step off the real axis for the first time: out into the Argand plane! -4 is represented as 4exp(pi) OK? So we are looking for a number out here on the Argand plane that when multiplied by itself following our rule gives 4exp(pi). And that number is 2exp(pi/2), which is two units out on the “imaginary” axis of our Argand plane. 2exp(pi/2)2exp(pi/2) equals 4exp(pi/2+pi/2) equals 4exp(pi) QED

topaz0 5y ago

I'm familiar with (and partial to) the geometric algebra literature, but I think it's a bit disingenuous/clickbaity to say that it implies imaginary numbers aren't real. I think of it as meaning that the algebra of the imaginary numbers arises as a slice through a more useful, bigger algebra.

hgomersall 5y ago
Yes, I think it's rather tongue in cheek. It's probably not defined as clickbait since it was written in '93!
- topaz0 5y ago
  Fair enough!

wrycoder 5y ago

You can make some progress by thinking of the imaginary numbers as being a two dimensional vector space over the reals, equipped with a very peculiar multiplication operator that is a combination of the usual stretch with a rotation. Addition is the usual addition of components, but the multiplication (using polar coordinates) looks like C = ABexp(T1+T2), where the T are the phases of A and B. To multiply two positive reals, since their phases are zero, it’s only a stretch: AB To multiply two negative reals, since their phases are pi, it’s ABexp(pi+pi). But exp(2pi) is unity, so we have AB again, a positive number. To find the square root of a positive number D, find a positive real (zero phase) that, when multiplied by itself, gives D. AA, for example. So, real number arithmetic is a subset of complex number arithmetic, with zero phase. And therefore zero rotation. Now, find the square root of a negative number, say -4. To do this, we have to step off the real axis for the first time: out into the Argand plane! -4 is represented as 4exp(pi) OK? So we are looking for a number out here on the Argand plane that when multiplied by itself following our rule gives 4exp(pi). And that number is 2exp(pi/2), which is two units out on the “imaginary” axis of our Argand plane. 2exp(pi/2)2exp(pi/2) equals 4exp(pi/2+pi/2) equals 4exp(pi) QED