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How do I prove $1 > 0$ using only field axioms and order axioms? I have tried using the cancellation law, with the identities in a field and I cannot get anywhere. Does anybody have any suggestions?

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    In general assume the most minimal set of axioms. I think a lot of textbooks and people might have different terminology.2012-09-10

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Suppose $1 < 0$. Adding $(-1)$ to both sides we'd also have $0 < -1$ (addition axiom). But if $0 < a$ then it must also hold that $0 < a^2$ (multiplication axiom). For $a = -1$ this means $0 < (-1)^2 = 1$, a contradiction.

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    @CodeKingPlusPlus: I was not squaring. I was multiplying by (a > 0) -- a positive number by assumption (since$0$< -1). In your argument (-1 < 0) is a negative number, so the multiplication axiom cannot be applied.2012-09-10