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Identifying $\mathbb{C}$ with $\mathbb{R} \times \mathbb{R}$, we can consider the lexicographic order: we define $x_1 + iy_1 < x_2 +iy_2$ if either

A) $x_1 < x_2$ or B) $x_1 = x_2$ and $y_1 < y_2$.

Prove that this relation does not satisfy Axiom 8 of Apostles chapter.

where Axiom 8 states If $x > 0$ and $y > 0$, then $xy > 0$

I don't know how to go about this question, yeah I understand what we were defined on the complex line. But I don't know how to relate this to axiom 8. Please help out, thank you.

Zev Chonoles
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MathGeek
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1 Answers1

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HINT: $0<i$; what about $i^2$?

Brian M. Scott
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  • and $i^2$ > 0 but how would i apply a and b into this question – MathGeek Mar 07 '13 at 22:50
  • @MathGeek: $i^2=-1$ is not greater than $0$. – Brian M. Scott Mar 07 '13 at 23:07
  • sorry that is what I meant i forgot to read over my comment – MathGeek Mar 07 '13 at 23:12
  • so are you saying let x=i and y=$i^2$ ? then xy < 0 ? and this relation wont satisfy axiom 8? – MathGeek Mar 07 '13 at 23:13
  • @MathGeek: No, I’m suggesting that you take $x=y=i$. – Brian M. Scott Mar 07 '13 at 23:16
  • Yes I understand that, but notice we are given conditions A and then B where does that come into play. – MathGeek Mar 07 '13 at 23:19
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    @MathGeek: Those aren’t conditions; they are the two cases defining the linear order on $\Bbb C$. The comparison of $0=0+0i$ and $i=0+1\cdot i$ falls under case (B): $0=0$ in the real part, and $0<1$ in the imaginary part, so $0<i$. The comparison of $i^2=-1$ and $0$, on the other hand, falls under case (A): the real part of $-1$ is $-1$, which is smaller than $0$, the real part of $0$. – Brian M. Scott Mar 07 '13 at 23:24