Assume { F, +,*,, <} is an ordered field. If a<b in F and c=(a+b)/2, prove
a) a<c<b
b) b - c = c - a = (b-a)/2. Thus the distance from c to either a or b is 1/2 the distance from a to b. C is called the mid point of [a,b].

2. Hi,
I don't know what your favourite definition of an ordered field is so i'll use this one: A field is an ordered field if its nonzero elements can be split into two sets, $\displaystyle P$ and $\displaystyle N$, such that $\displaystyle P$ contains $\displaystyle x$ iff $\displaystyle N$ contains $\displaystyle -x$, and $\displaystyle P$ is closed under addition and multiplication.
The ordering $\displaystyle <$ can be then defined as follows: $\displaystyle x < y \mbox{ iff } y-x \in P$.

$\displaystyle 1 \neq 0$ so either $\displaystyle 1\in P$ or $\displaystyle -1\in P$, but if $\displaystyle -1\in P$ then $\displaystyle 1=(-1)(-1)\in P$ (since $\displaystyle P$ is closed under multiplication) which is a contradiction, so $\displaystyle -1\in N$ and $\displaystyle 1\in P$.

Since $\displaystyle P$ is closed under addition, $\displaystyle 2=1+1\in P$.
Furthermore, $\displaystyle \frac{1}{2}\in P$, because if $\displaystyle \frac{1}{2}\in N$ then $\displaystyle -\frac{1}{2}\in P$ and since $\displaystyle 2\in P$ we'd get $\displaystyle -1 =2\left(-\frac{1}{2}\right) \in P$ (since $\displaystyle P$ is closed under multiplication), a contradiction.

Now we're ready to prove a), b).

$\displaystyle c-a=\frac{a+b}{2}-a=\frac{1}{2}(a+b)-\frac{1}{2}2a=\frac{1}{2}(a+b-2a)=\frac{1}{2}(b-a)$.

Similarly, $\displaystyle b-c=b- \frac{a+b}{2}=\frac{1}{2}2b-\frac{1}{2}(a+b)=\frac{1}{2}(2b-a-b)=\frac{1}{2}(b-a)$.

Since $\displaystyle b-a\in P$ was given and we know $\displaystyle \frac{1}{2} \in P$, we get $\displaystyle \frac{1}{2}(b-a)\in P$. Thus $\displaystyle a<c$ and $\displaystyle c<b$.