order relation and numbers I want to prove the following: If $a$ and $b$ are elements of an ordered field such that $a \leq b+c$ for every $c >0$, then $a \leq b$.
So suppose that $a>b$. Then $a-b > 0$ or $b-a \leq 0$. We know that $a \leq b+c$ so that $b+c-a \geq 0$ or $b-a \geq -c$. How would I get the contradiction?

First, note that in ordered field we know that $2$ is positive and in particular, we may divide by $2$ while preserving positivity. Now, suppose $a$ and $b$ have your property, but $b\lt a$. Let $c=(a-b)/2$, which is positive, but also $b+c+c=a$ and so $b+c\lt a$, contradicting your hypothesis. 
Meanwhile, observe that the fact is not necessarily true in ordered rings, such as the integers $\mathbb{Z}$, since $5\leq 4+c$ for any integer $c\gt 0$, but $5\not\leq 4$. So we should expect to divide or otherwise use the field axioms at some point in any proof of your fact. 
