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Sorry if have incomplete knowledge, as am new to this topic. I.e., even though have chosen 'ordered-fields' as tag; am not clear about it.

It seems (as shown below) that the simple property of ordered fields is not limited to just have a total ordering; but much bigger. I mean that it extends to taking power of a number, and ensuring that still the property (of being in a given class, out of $3$ possible) remains the same. But, this fact is never stated explicitly, and is the cause of my doubts.

I got details about these here: 1.1, 1.2

The proof/answers rely on stating that if there some ordered field $P$, then an element $x$ in that would be following mutually exclusive eventualities :
(1) itself be a member of that, or
(2) $-x$ be a member of that, or
(3) $x = 0$.

I cannot understand why have mutual-exclusiveness, & as shown below it is not meaning that; but wants to state that a value $x$ cannot be at the same time - positive, negative, or zero. Also, it applies for any positive powers of the number $x$. So, actually need a better symbolism than used to express the property.

An alternate way given is to show by contradiction that the square of $i$ does not follow the same value (positive/negative, as $i$). This approach is best shown in 1.2.
This alternate-way seems better, but cannot gel this approach with the (earlier stated) one that states $x$ to be in any one of the $3$ classes, as it does not say that $x^n, n\in \mathbb{Z+}$; and so the two ways do not gel as to my understanding.

jiten
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  • Can you state what your definition of "ordered field" is? Or is the problem that you don't know what the definition is? – Eric Wofsey Jul 07 '18 at 00:05
  • Your alternate way is actually the same proof as the other. It just stated in terms of a different means of specifying the ordering (but both means are equivalent). And as Eric Wofsey is hinting, $P$ is not an "ordered field". That is $\scr F$. $P$ is a subset of $\scr F$ often called the "positive set", since it determines the elements that will be $> 0$ in the induced ordering. – Paul Sinclair Jul 07 '18 at 00:19
  • @EricWofsey Thanks, my defn. is: It is a total ordering in which there is a fixed precedence based on value in continuum; such that there are positive, negative, & a single zero value. In a field with multiple dimensions to consider, with additional property that $i$ is second domain (imaginary numbers) & $i^2$ is real again, that too coupled with a sign change; it becomes difficult to describe ordering without modifications. One such modification can be - consider real part first, then imaginary part. – jiten Jul 07 '18 at 06:38
  • @PaulSinclair Thanks, but $P$ was arbitrarily chosen to describe the complete field (don't know to display $\mathbb{F}$ as given- so please help in that part too). Issue was: second defn. is clear- but is never explicitly stated in terms of powers (hope is +ve ones only, or might be even power - but unclear as even power will make a -ve number +ve) of $x$. The first defn. never states that it means (i) $x$ to simultaneously take $\gt 1$ subset value (+ve, -ve, or $0$). Hence, 1st defn. complicates issue. (ii) There is no description of powers of $x$, that would make 'alternate' defn. explicit. – jiten Jul 07 '18 at 07:04
  • The ordering you describe, consider real then imaginary, is a fine total ordering of the complex numbers. The problem for an ordered field is that it does not respect addition and multiplication. – Ross Millikan Jul 07 '18 at 12:54
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    @jiten - Since you've accepted Ross' post, I'll assume your mathematical questions are settled. $\scr F$ can be obtained as \mathscr F or just \scr F. The difference between the two is that \mathscr affects only the next character, while \scr affects all following characters up the end of the { } block \scr occurs in, of the end of the mathjax expression, More generally, if you want to see how an expression was created, right-click on it and select "Show Math as > TeX Commands". Also:: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Paul Sinclair Jul 07 '18 at 13:20

1 Answers1

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The point of an ordered field is to have a total order that respects the field operations. We would like $a \lt b$ to imply $a+c \lt b+c$ and $ac \lt bc$ as long as $c \gt 0$. It is possible to order the elements of a field in an arbitrary way, but that is not useful for what we want. We demand that the order be total. You can get good intuition thinking of the usual order on the rationals or reals.

I didn't click through to your links, but I think you have a couple confusions. There are a couple ways to define an ordered field. One way is just to say that it is a field equipped with a total order that respects the operations as I mentioned above. That is a lot to check. An alternate way is to say the the field $F$ has a subset $P$, which are intended to be the positive elements. We then demand that all elements of the field are either $0$, a member of $P$, or a negative of a member of $P$, and only one of those. We also demand that $P$ is closed under addition and multiplication. This is less to check about a proposed $P$ than order in the earlier approach. We can then define $a \gt b$ as $a-b \in P$ and prove that the order is total and the operations are respected.

Having defined an ordered field we would like to prove that $\Bbb C$ is not an ordered field. For that we need to prove that there is no order you can impose on the complex numbers that respects the operations. Asking whether $i$ is greater than or less than $0$ is the path to that. We know it has to be one of them because the order is total. We then show that either one leads to a contradiction and conclude that there cannot be a way to make $\Bbb C$ an ordered field.

Ross Millikan
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  • But there is a difference, as normally for reals would not take an even power as that would change the sign to positive. But here, the case is different - only even powers can get a real out of $i$. So, is the power taking a proper way to disprove the ordering of $\mathbb{C}$, is unclear. – jiten Jul 07 '18 at 07:01
  • We required that for all $x\neq 0$, either $x \gt 0$ or $x \lt 0$. We show that either choice leads to a contradiction. It comes because the square of an element has to be positive, but the square of $i$ is not. – Ross Millikan Jul 07 '18 at 12:52