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I'm having trouble with this problem:

Show that on $\mathbb{C}$ there is no total order relation $\preceq$ such that:

(i) $\forall x,y,z\in\mathbb{C},~~x\preceq y\Rightarrow x+z\preceq y+z$

(ii) $z\preceq 0,~~x\preceq y\Rightarrow xz\preceq yz$

JMoravitz
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The Physics Student
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  • I believe there was a typo in your original image, that statement (ii) should have read $0\preceq z$ instead, as that corresponds with our usual intuition of how orders should work using the real numbers as an example. Remember that orders should be anti-symmetric. I.e. if $x\preceq y$ and $y\preceq x$, then they must be equal. – JMoravitz Sep 07 '15 at 04:32

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If there is a total order relation would $ i<0$ or $i>0 $ Your contradiction will be shown that none of them can hold.

JonesY
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  • Sorry, Im not understanding. Can you explain a little more? – The Physics Student Sep 08 '15 at 16:37
  • There are certain functions that keep the order of a set of numbers. Such as $ x^2 $ will keep the order in this case: $ 0\leq x_1 \leq x_2 \rightarrow 0\leq x_1 ^2\leq x_2 ^2$ – JonesY Sep 08 '15 at 19:42