|
|
Post by Trichotomy on Jul 2, 2015 2:29:32 GMT
Hi guys, have any ideas about proving trichotomy? especially the property that exactly one holds.
BTW, I can't use the definition given in class that x=y <=> (x>y and x<y) since this implies two can hold at the same time.
|
|
|
|
Post by KBEAL on Jul 2, 2015 2:52:07 GMT
For clarity: if you use the symbol "<" to mean "is subset of", then you can have x"<y" and "y<x". (In this case, x and y denote sets).
I'm not sure what I wrote for the 3 order axioms, but they should be
for all x,y \in F,
(i) exactly one of the following is true: x<y, y<x, or x=y
(ii) x<y \iff 0<(x-y)
(iii) 0<x, 0<y \implies 0<(x+y) and 0<(xy).
Use (ii) to help show (i)
|
|
|
|
Post by Guest on Jul 2, 2015 19:36:17 GMT
I think this works
ab>0 ==> p/q-f/g>0 and so p/q>f/g
or
ab<0 ==> -ab>0 ==> f/g-p/q>0 and so f/g>p/q
or
(not ab>0) and (not ab<0) ==> (not p/q>f/g) and (not f/g>p/q) i.e. ab=0 implies f/g=p/q
I still don't understand what the Archimedean property would mean for F
|
|
|
|
Post by KBeal on Jul 4, 2015 0:02:23 GMT
tip 1: first, Explain why b_k is one (it makes computations for the rest of the problems a lot easier).
Re part (i): second order axiom is an \iff statement
and \mathbb{R} is ordered and is in F and the ordering on F is the same as the ordering on R for elements in R
Let me know if this helps.
|
|
