TITLE: How to show that $\mathbb{Q}_p$ cannot be ordered?
QUESTION [6 upvotes]: I've seen many references on this site and others to the fact that the $p$-adic numbers cannot be ordered, but the closest I've seen to a proof of this is Wikipedia's vague reference to "$\mathbb{Q}_2$ contains a square root of −7 and $\mathbb{Q}_p$ ($p > 2$) contains a square root of $1 − p$" on the page for "Ordered Field". How does this statement imply that $\mathbb{Q}_p$ cannot be ordered? Or is there a more intuitive approach to the proof?

REPLY [10 votes]: It can be ordered, it just won't obey.
An ordered field has a set of positive numbers $P,$ such that $1 \in P,$ the sum or product of any two elements of $P$ is once again in $P.$ Finally, there is trichotomy, given a field element $x,$ exactly one of three is true: $x = 0,$ or $x \in P,$ or $-x \in P.$ 
These numbers $-7$ or $1-p$ would be "negative," as they are $-1$ times sums of copies of $1.$ But, if they are squares, they are squares of something positive, as $(-x)^2 = x^2,$ a contradiction.
The Artin-Schreier Theorem says that a field an be ordered if and only if $-1$ is not the sum of finitely many squares.