Help with proof concerning ordered fields and smallest elements I have been trying to prove that every ordered field has no smallest positive element for a while now, and I think I have it worked out. However, I feel like something is missing in my proof. Here is what I have so far:
Let $F$ be an ordered field
Let $x$ exist in $F$ such that $0$ is less than $x$
Then $x/2$ is less than $x$
Therefore, $F$ can have no smallest element
This does seem to prove that the field has no smallest element, but it seems like there are some holes. Am I correct in thinking this way? I saw a similar proof involving the positive real numbers, but $x/2$ exists in the positive reals.  How do I know that $x/2$ exists in $F$?  I am just concerned that this is not enough.  Any helpful pushes would be most appreciated. Thank you!
 A: The proof you showed is really a proof by contradition.
Suppose there is a smallest positive number in $F$, call it $x$.
1) Since it is positive by assumption, $0<x$.
2) Now we show $0<x/2<x$, that is $x/2$ is a positive number that is smaller than $x$.
3) Thus $x$ cannot be the smallest positive number, since we found some even smaller positive number, a contradiction.
A: A lot of this proof is spent taking care of preliminaries so that we can smoothly apply an argument of the form you gave. You asked whether $x/2$ exists, so first we give a proof of that; then, we prove it is positive and less than $x$. Finally, we argue that there is no smallest positive element.
Every ordered field $F$ has characteristic zero. In fact, a proof by induction shows that $n>0$ for each $n=\sum_{i=1}^n1$. In particular, $2$ is greater than $0$ and it is invertible. Thus, for any $x\in F$, $x/2$ exists.
Now, suppose $x>0$ but $x/2<0$, then $-x/2>0$. Hence, $-x=2(-x/2)>0$. However, this is a contradiction, since $0=x+(-x)>0$. Therefore, $x/2>0$. On the other hand, $0<x/2$ implies $x/2<x/2+x/2=x$. Thus, for any positive $x$, we are guaranteed a positive $y<x$.
We proceed to prove that there is no smallest positive element. Suppose for the sake of contradiction that $x$ is the smallest positive element of $F$. Since $x$ is positive there is some $y\in F$ such that $0<y<x$. This is a contradiction.
