In an ordered field, S, with positive elements, P, for all x,y elements of S define x<y if y-x is an element of P,

prove that only one of the following can be true:

x<y,x=y,y<x

Also prove that if x,y,z are arbitrary and x<y and y<z then x<z.

- Sep 15th 2009, 06:41 PMamm345Transitivity and Ordered Fields. How do you prove this? Please help.
In an ordered field, S, with positive elements, P, for all x,y elements of S define x<y if y-x is an element of P,

prove that only one of the following can be true:

x<y,x=y,y<x

Also prove that if x,y,z are arbitrary and x<y and y<z then x<z. - Sep 15th 2009, 07:11 PMNonCommAlg
- Sep 15th 2009, 07:14 PMamm345
To define a total order S.

- Sep 15th 2009, 07:19 PMThePerfectHacker
The set $\displaystyle P$ satisfied that given $\displaystyle a\in S$ we exactly one of following: $\displaystyle a\in P \text{ or }-a\not \in P\text{ or } a=0$. Also given $\displaystyle a,b\in P$ we have $\displaystyle a+b\in P \text{ and }ab\in P$. Given, $\displaystyle x,y\in S$ consider $\displaystyle x-y$. If $\displaystyle x-y=0$ then we are done. Otherwise $\displaystyle x-y\in P$ xor $\displaystyle -(x-y)\in P$. Thus, $\displaystyle x-y\in P$ xor $\displaystyle y-x\in P$ i.e. $\displaystyle x<y$ xor $\displaystyle y<x$.

To prove the other statement, notice that if $\displaystyle y-x\in P$ and $\displaystyle z-y\in P$ then $\displaystyle (y-x)+(z-y) = z-x\in P$ so $\displaystyle x<z$.