As I do not know what to write here, I write only that I tried to prove that $1>0$ using the ordered field axioms (Ordered Field Axioms). I simplified what the site said for shortness only.
Try
Axioms and definitions used
Addition and multiplication
1b) $x+y=y+x$
1c) $x+0=x$
1d) $x+(-x)=0$
2c) $1x=x$ and $1\neq 0$
2e) $x(y+z)=xy+xz$
Inequalities
3a) only one holds: $x=y$ or $x<y$ or $y<x$
3b) if $x<y$ and $y<z$, then $x<z$
3c) if $x<y$, then $x+z<y+z$
3d) if $x<y$ and $0<z$, then $xz < yz$
Definition of greater than
D1) iff $x>y$, then $y<x$
Of that $0x=0$
As logic tell us:
$1x+(-1x)=1x+(-1x)$
$1x+(-1x)=0$ (by 1d)
$x(1+(-1))=0$ (by 2e)
$0x=0$ (by 1d)
which we'll recall by L1.
Of that $x\times -1=-x$
Logic says:
$0+(-x)=0+(-x)$
$0+(-x)=0x+(-x)$ (by L1)
$0+(-x)=0x+(-1x)$ (by 2c)
$0+(-x)=x(0+(-1))$ (by 2e)
$-x=x\times -1$ (by 1c)
which we'll call L2
Of that 1>0
If $1=0$, then we contradict the axiom 2c, therefore either 1 is less than 0 or 0 is less than 1 (3a).
As an assumption, let:
$1<0$
$1+(-1)<0+(-1)$ (by 3c)
$0<-1$ (by 1c and 1d)
As 0 is less than -1 (3d):
$1\times -1<0\times -1$
$-1<0$ (by L1 and L2)
$-1+1<0+1$ (by 3c)
$1+(-1)<1$ (by 1c and 1b)
$0<1$ (by 1d)
which contradicts our assumption that $1<0$. As 1 is not less than 0 nor it is equal to 0, then 0 is less than 1, i.e., (D1) $1>0$.
$\mathcal{Q.E.D.}$
Questions
Is the proof correct? And did I show/write too much? Because I think I did.