1

I understand that complex numbers can be neither ordered nor compared by 'size', but if mapped one for one by a transformation, then they can be. Latter point aside, can I say that $2-xi < 2 < 2+xi$?

Thanks

robjohn
  • 345,667
MarvinMaths
  • 11
  • See complex numbers as vectors .Can you say :$\vec a> \vec b$? – ABC Apr 09 '13 at 08:22
  • 1
    Yip duplicate will read other post please delete and reduce my reputation to -2i mu ha ha ha – MarvinMaths Apr 09 '13 at 08:25
  • Actually. Its not really the same. In the other post there is ever only talk of a zero imaginary part which has no effect on comparisons because real numbers are in effect the only things being compared. My question says is 3+3i > 2 ? – MarvinMaths Apr 09 '13 at 08:41
  • You could define that $x+iy \leq x'+iy'$ iff $x \leq x'$. This relation is reflexive and transitive but not antisymmetric. We also get the following implications: (1) for all complex $a,b,z$ it holds that $a \leq b \Rightarrow a+z \leq b+z$, and (2) for all complex $a,b$ and all non-negative reals $x$ it holds that $a \leq b \Rightarrow ax \leq bx$. However you can't generalize Theorem 2 by allowing $x$ to be complex. Could be an interesting idea, but I'm not sure either way. Maybe toy with it a bit and see if any major problems crop up. – goblin GONE Apr 09 '13 at 08:54

0 Answers0