• Sep 15th 2009, 06:41 PM
amm345
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 PM
NonCommAlg
Quote:

Originally Posted by amm345
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.

there are a couple of ways to define an ordered field. which one are you using?
• Sep 15th 2009, 07:14 PM
amm345
To define a total order S.
• Sep 15th 2009, 07:19 PM
ThePerfectHacker
Quote:

Originally Posted by amm345
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.

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

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