Using the axioms of real numbers prove that 0 < 1

Question

These are the axioms that I am allowed to use:
(1) $x + 0 = 0 + x = x$
(2)$x \cdot 1 = 1 \cdot x = x$
(3) $xy = 1 \iff y = \frac{1}{x}$, $x \neq0$
(4) $x+y = 0 \iff y = -x$
(5) $ x(y+z) = xy + xz$
(6) $(x < y) \land (y<z) \Rightarrow x<z \Rightarrow x+c < z+c$

Could you give me any suggestions on how to tackle this problem?

EDIT Would this problem be solvable if we were allowed to use the properties of + and *?

Answer 1

It can't be done from these axioms alone. Consider the $>$ relation. Your sixth axiom (which is the only one dealing with $<$) is valid if we put there $>$ instead of $<$. Therefore, if we could prove that $0<1$ from these axioms alone, then we would also be able to prove that $0>1$. And this is not true.

Answer 2

You're missing axioms. I like to use the axioms of $\mathbb R$ that describes it as a complete ordered field. An ordered field contains many axioms (all the axioms of a field and an ordered set plus the translation invariance of the order and $x > 0 \wedge y>0 \Rightarrow x\cdot y > 0$). The completeness roughly means that the supremum of any non-empty set exists.

Answer 3

With only those axioms you couldn't. To see that you just create an example where this doesn't work. Let's just define $x<y$ as always being false, but otherwise the arithmetics is as ordinary for $\mathbb R$. Then you fulfil the axioms, but still $\neg(0<1)$.

Note that these axioms are quite meagre. They don't even require $\mathbb R$ to be very large at all. You can fulfil them with even a two element set. If you don't require $0$ and $1$ to be distinct members you don't even need that (and if you don't require them to be members you don't even need elements to fulfil those axioms).