• Sep 9th 2009, 07:25 PM
tigergirl
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].
• Sep 10th 2009, 03:48 AM
Taluivren
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, $P$ and $N$, such that $P$ contains $x$ iff $N$ contains $-x$, and $P$ is closed under addition and multiplication.
The ordering $<$ can be then defined as follows: $x < y \mbox{ iff } y-x \in P$.

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

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

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

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

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

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