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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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I think $1 = 1^2$ is helpful. – Dylan Moreland Sep 9 '12 at 2:02
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@GerryMyerson I assume the OP means "ordered field axioms". – Alex Becker♦ Sep 9 '12 at 2:28
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You need not only the axioms of fields and the axioms of linearly ordered sets, but also the axioms that say the linear order is compatible with the algebraic operations, i.e. if $a,b>0$ then $ab>0$, etc..... – Michael Hardy Sep 9 '12 at 2:53
In general assume the most minimal set of axioms. I think a lot of textbooks and people might have different terminology. – CodeKingPlusPlus Sep 10 '12 at 1:52

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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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Can't you do the same with $0$ < $1$? Subtract 1 from each side then we have $-1$ < $0$. Now square both sides. We have $1$ < $0$ another contradiction. Thus, this method does not work. – CodeKingPlusPlus Sep 10 '12 at 1:51
@CodeKingPlusPlus There is some missing detail that I believe Marek meant for you to fill in. Of course it is incorrect to say that if $a, b$ are elements of an ordered field then $a < b$ implies $a^2 < b^2$ — one remembers the graph of $ y = x^2$ over $\mathbb R$. But what is true is that if $0 < a < b$ then $0 < a^2 < b^2$. – Dylan Moreland Sep 10 '12 at 3:34
@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. – Marek Sep 10 '12 at 6:29

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