In an early math class, I was shown how all Reals could be constructed from Rationals using a 2-D representation (ex. Real numbers are represented by (a + b \sqrt{2} ) where a & b are Rational). While using the 'lesser' system of Rationals requires a 2-D representation, we can also represent Reals using decimals as single values (a 1-D representation).

If this is the case, then might our current expression for complex numbers behave the same, such that we use the 'lesser' system of Reals to represent complex numbers as 2-D (x + yi), however there could exist a numeric system that represents complex numbers as single values (1-D representation)?

  • $\begingroup$ Check out Piano's curve if you don't mind non-unique representation. With this you can represent 2 reals number using just 1. And of course, generalizable too. $\endgroup$ – Gina Jun 12 '14 at 11:18
  • $\begingroup$ This question is not precise enough that it could be answered. What do you mean by "1-dimensional"? Any field is 1-dimensional when considered as a vector space over itself. There's nothing special about that. $\endgroup$ – Brusko651 Jun 12 '14 at 11:21
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    $\begingroup$ You might want to have another look at the notes from your math class. Numbers of the form $a+b\sqrt{2}$, a,b rational are a part of the reals, but not all reals can be represented that way. For example $\pi$. In fact you catch nearly none of the reals that way. $\endgroup$ – mlk Jun 12 '14 at 11:22
  • $\begingroup$ Not all reals can be written as $a+b\sqrt{2}$ for $a,b\in\mathbb{Q}$. The set of those which can, is denoted $\mathbb{Q}\left[\sqrt{2}\right]$. It has a different structure than $\mathbb{Q}\left[\sqrt{-1}\right]$. The full set of complex numbers is bigger, $\mathbb{R}\left[\sqrt{-1}\right]$. $\endgroup$ – Jeppe Stig Nielsen Jun 12 '14 at 11:23

Not really. It can be shown that the reals form the only complete totally ordered field, up to isomorphism.

EDIT: In simpler terms: if you map the complex numbers to a line, you are forced to discard some structure.

  • $\begingroup$ Where could I find what is discarded? $\endgroup$ – DGPIS Jun 12 '14 at 11:23
  • $\begingroup$ A complete totally ordered field is a field that is complete with respect to some topology. The complex numbers are complete, so this is not a problem. The complex numbers are also a field. The issue is with the total ordering: there is no way to order the complex numbers while preserving the field structure (i.e. addition, subtraction, multiplication, etc.). This has to do with the fact that squares in an ordered field must necessarily be nonnegative (and $i^2=-1$, so $\mathbb{C}$ fails to satisfy this property). $\endgroup$ – mval Jun 12 '14 at 11:27

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