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Let $(F,+,*,{\preceq})$ be a totally ordered field.

Then ${\forall}f\,{\in}\,F:0_F\,{\preceq}\,f*f$.

It's obvious from the definition of total ordering on a field that this applies to $f:0_F\,{\preceq}\,f$. Since $0_F\,{\preceq}\,a\,{\land}\,0_F\,{\preceq}\,b{\implies}0_F\,{\preceq}\,a*b$, we know that $f^2$ must succeed $0_F$.

What about the cases $f:f\,{\preceq}\,0_F\,{\land}\,f\,{\neq}\,0_F$? How do we prove the theorem for them?

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    That's because you consider $-f$ whose square is the same as $f$. – 2017-01-02
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    Why does $(-f)*(-f)=f*f$ – 2017-01-02
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    @asdasdfsss $(-1)^2 - 1^2 = (-1 -1)(-1 +1) = 0$. – 2017-01-02

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