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Doing one problem on my own I found that split-complex numbers might be useful to solve it but couldn't find any information about comparing two split-complex numbers. Can you always find a minimal between two different split-complex numbers without contradiction unlike with complex numbers? If so, how is it done?

epsilonQM
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    The split-complex numbers are just isomorphic to $\mathbb{R} \times \mathbb{R}$, with the isomorphism sending $x + yj$ to the pair $(x + y, x - y)$, so they can be partially ordered pointwise. Whether this ordering is useful for anything depends on what you're trying to do. – Qiaochu Yuan Nov 02 '22 at 19:30
  • Here is a link. – Dietrich Burde Nov 02 '22 at 19:35
  • I think reasonably you can have two different orders: $a+bj\preceq_-c+dj$ iff $a-b\le c-d$, and $a+bj\preceq_+c+dj$ iff $a+b\le c+d$. These are the two coordinate-wise comparisons from @Qiaochu's isomorphism. – Akiva Weinberger Nov 02 '22 at 20:16
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    This is incidental to your question, but it always seemed to me that we should just use the notation $a\pm b$ instead of $a+bj$. Algebraically, using $(\pm1)^2=1$, it behaves identically, and it makes the isomorphism to $\Bbb R\times\Bbb R$ more clear. (And instead of writing$$e^{xj}=\cosh x+j\sinh x$$we'd write $$e^{\pm x}=\cosh x\pm\sinh x,$$which makes a lot of sense.) – Akiva Weinberger Nov 02 '22 at 20:18

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