How to proof that a field is complete order field?
You have to define an order on $\mathbb{Q}$ and show that it satisfies the required conditions. For this one goes back and recalls how you defined the rational numbers. As far is I know one usually defines it as the field of fractions of $\mathbb{Z}$, i.e. as $(\mathbb{Z} \times \mathbb{Z} )/\sim$ where $(a,b)\sim (c,d)$ iff $ad=bc$. You can define an order on $\mathbb{Q}$ as follows: We say $[(x,y)] \leq [(z,w)]$ if for any two representatives $(a,b)$ and $(c,d)$(of $[(x,y)]$ resp. $[(z,w)]$) s.t. $c,d>0$ we have $ad\leq bc$. Now you can easily verify that this is well defined and defines a total order on $\mathbb{Q}$.
Related videos on Youtube
Jake
Updated on August 01, 2022Comments
-
Jake over 1 year
I know what an ordered field is but how to actually proof that a field, for example $Q$ (rational numbers), is an ordered field?
-
mercio almost 8 yearsAre you saying that (-1,1) is less than (2,-1) ?
-
math635 almost 8 yearsOh you are right. I edited it. No it is a bit ugly but should work.
-
Slade almost 8 yearsIt's probably much easier to just say what the positive elements are, which determines the ordering.
-
DanielWainfleet almost 8 yearsAnd show that the linear order satisfies $(a<b)\to (a+c)<(b+c)$ and that $c>0\to ((a<b)\to (a c< b c))$.