Proof: Zero is less than one 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.

$1 = 1^2$ 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:
$0 = (-x + x)y = (-x)y+(xy)$.
