How can one show/prove that $0<1$?
$1$ is not actually defined to be greater than zero, but I think it can be proven. I already know that the real numbers are an ordered field and I am familiar with the field axioms and the definition of a total ordering. However I do not know how to start such a proof.
Any kind of help or advice will be really appreciated.
On
I'll add some details/more information to @lhf answer:
In an ordered field $K$, there is a distinguished subset $P$ such that:
(i) $\{ P,\{0\},-P\}$ is a partition of $K$,
(ii) $\forall x,y\in P$, $\; x+y\in P\;$ and $\;xy\in P$.
In particular, as $\;1=1\cdot 1=(-1)\cdot(-1)$, we see $1\in P$.
We then define a strict order relation on $K$ by setting $x>y$ if $x-y\in P$. In particular, $\;x\in P\iff x>0$.
$1 = 1^2$ and $x^2>0$ if $x\ne 0$.
For the last claim, the main point is that by definition of positive $x,y>0$ implies $xy>0$. This is used as follows:
If $x>0$ then $x^2 = x \cdot x > 0$.
If $x<0$ then $-x > 0 $ and so $x^2=(-x)^2 > 0$.
In turn, this follows from $(-x)y=-(xy)$, which follows from distributivity: $0 = (-x + x)y = (-x)y+(xy)$.