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$$\begin{array}{c|ccc} + & 0 & 1 & 2 \\ \hline 0 & 0 & 1 & 2 \\ 1 & 1 & 2 & 0 \\ 2 & 2 & 0 & 1 \\ \end{array}$$ $$\begin{array}{c|ccc} \times &0&1&2\\ \hline 0&0&0&0\\ 1&0&1&2\\ 2&0&2&1\\ \end{array}$$ Using a proof by contradiction, show that it is impossible to define an operation $<$ that satisfies the order axioms.

I'm really stuck here and I don't know how to prove this. The two tables are $3\times3$ addition and multiplication tables.

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    $\begingroup$ First, write down the order axioms. And actually, these are 3×3 tables. $\endgroup$ – k.stm Sep 11 '17 at 19:36
  • $\begingroup$ But how do I use the axioms to prove this. A hint was given: If x /= 0 then x^2 > 0. In particular, 1>0 $\endgroup$ – Chance Gordon Sep 11 '17 at 19:54
  • $\begingroup$ Have a look at the duplicate: $0<1<1+1<\ldots<\underbrace{1+\ldots+1}_{p\text{ times}} = 0$. Here $p=3$. $\endgroup$ – Dietrich Burde Sep 11 '17 at 20:36
  • $\begingroup$ @Dietrich Burde In fact, you need only the addition table to obtain this contradiction. $\endgroup$ – Jean Marie Sep 11 '17 at 22:05
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We know that an ordered field cannot be finite, see this duplicate. Since we have the finite field $\mathbb{F}_3$ here with addition and multiplication, it cannot be ordered.

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