Proving there is no smallest positive element of any ordered field F

Question

I missed lecture so these notes may be incomplete and I'm trying to fill in the blanks. Really not following this proof though (feel like maybe some stuff was said out loud but not written down)

Appreciate help understanding/filling in the blanks

Answer 1

First, by the axioms, we have $0<1=1+0<1+1=2$ (run through this step-by-step if needed). Now use some more axioms to obtain $0<\frac{1}{2}<\frac{1}{1}=1$ (again, step-by-step as necessary).

Now, to prove the theorem, assume the opposite, i.e., assume there is a smallest positive element. Unraveling what this means, we assume there is some positive element $a$ in $\mathbb F$ (so $a>0$) such that for any positive element $b$ of $\mathbb F$, we have $a\leq b$. Since $0<\frac{1}{2}<1$, we have $0<\frac{1}{2}a<a$. But $\frac{1}{2}a$ is positve, so by assumption we have $a\leq\frac{1}{2}a<a$. This is a contradiction (can you see why?), and thus the theorem is true.