Total order on complex numbers [closed]

Note that the proof does not prove that there is no total ordering of the complex numbers. It is easy to order the complex numbers totally.

What it does prove is that there is no total ordering of the complex numbers which is compatible with their algebraic structure.

You say you want to have both $i<0$ and $-i<0$, but that is not compatible with the algebraic structure. Namely, if we have $i<0$ then adding $-i$ to both sides gives $0<-i$. Since you're asserting $-i<0$ which contradicts this, your ordering doesn't obey the rule $$ a<b \implies a+c<b+c $$ which is one of the axioms for an ordered field.

$n\in N, a_{n+1} = a_n (1 - a_n)$ , $0 < a_0 < 1$. Prove that $\lim\limits_{n\to\infty} a_n\cdot n = 1$.

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