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Don't spend a lot of time on this. I'm certain I could bang it out myself; but maybe there's an answer out there that someone already knows.

Say we use Dedekind cuts to construct the reals. Addition is fine and the proof that a set bounded above has a supremum is just lovely. Then we get to defining multiplication and all hell breaks loose - not that there's anything deep or really difficult about it, but at the very least the elegance is lost in a mass of special cases.

So I say to myself this is just because of the historical accident that people figured out how to extend the positive rationals to the rationals before considering completeness; why not get completeness first?

So the plan is to use Dedekind cuts of positive rationals to construct the positive reals, and then go from there to the reals. (In all of this "positive" may mean strictly positive or non-negative, whatever works.)

This leads to at least one question: For what values of whatever and the construction can one prove the following theorem?

Theorem If $F$ is an ordered field then the set of positive elements is a whatever. Conversely, given a whatever $P$, the construction gives an ordered field $F$ such that $P$ is (isomorphic to) the set of positive elements of $F$.

Or if there is no such general theorem we could be more specific. Say $P$ is the set of positive reals. I can think of at least two constructions that do in fact lead from $P$ to $\Bbb R$; the question is what do I need to prove about $P$ in order to prove that the result is in fact $\Bbb R$.

One construction would be to say that a real is just an element of $P$ with a plus or minus sign attached. This seems likely to lead to the sort of special-casing that we're trying to avoid (already when we think about $0$ we see we need to add a special clause to the effect that $+0=-0$; blech).

Another construction is to regard $(a,b)\in P^2$ as representing the real $a-b$. So we'd define $(a,b)\sim(c,d)$ if $a+d=c+b$, we'd define the sum and product of elements of $P^2$ in the obvious way (in particular $(a,b)(c,d)=(ac+bd,ad+bc)$), show these operations lift to the quotient $P^2/\sim$ and be on our way.

I kind of like the second construction. What do I need to prove about $P$ to show that $P^2/\sim$ is an ordered field with positive cone $P$?

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    $\begingroup$ ??? Would be nice if people commented on the reason for downvotes... $\endgroup$ – David C. Ullrich Feb 6 '16 at 17:37
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This falls out from the the definition of a field and the requirement that $P$ embeds in the "group of differences" (your $P^2/\sim$). It will amount to saying that $P$ must be a semiring whose additive semigroup satisfies the cancellation law, if $x + y = x + z$, then $y = z$, and whose multiplicative semigroup is a group.

By the way, did you know that any densely and completely ordered commutative group is isomorphic to the reals? This leads to a little-known but very neat way of defining multiplication by studying the order-preserving homomorphisms of the additive group. I find this a far more satisfactory way of avoiding all those case splits.

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  • $\begingroup$ You need some additional hypothesis (maybe densely ordered) in your second paragraph, to rule out $\mathbb{Z}$. To classify homomorphisms, it also seems to me that you would need a stronger result (e.g., that the isomorphism in question is unique once you determine its value on a single element). $\endgroup$ – Eric Wofsey Oct 18 '15 at 16:06
  • $\begingroup$ $P$ here is the set of positive reals, and I am assuming David has already shown it's the positive cone of a completely ordered commutative group. I agree that to get David's propose theorem to stand without that assumption you need to say more. $\endgroup$ – Rob Arthan Oct 18 '15 at 16:09
  • $\begingroup$ I'm just saying that it certainly isn't true that any completely ordered commutative group is isomorphic to $\mathbb{R}$, because $\mathbb{Z}$ is a counterexample. $\endgroup$ – Eric Wofsey Oct 18 '15 at 16:13
  • $\begingroup$ @RobArthan Very interesting. The first paragraph is the sort of thing I was asking for, but yes the second paragraph may well be much nicer. If we just want to construct $\Bbb R$ we don't need the general theorem about ordered groups. We've constructed the reals as Dedekind cuts, defined $+$ and $<$ and shown that $(\Bbb R, +, <)$ is a complete ordered group. Now say $M$ is the class of homomorphisms from $\Bbb R$ to itself which are either increasing, decreasing or constant. There's a natural map from $M$ to $\Bbb R$ given by $\phi\mapsto\phi(1)$. I suspect the tricky part is [continued] $\endgroup$ – David C. Ullrich Oct 18 '15 at 16:16
  • $\begingroup$ @RobArthan [continued] showing this map is a bijection. Then $M=\{\phi_x:x\in\Bbb R\}$, where $\phi_x(1)=x$. We define multiplication by saying $\phi_{xy}=\phi_x\phi_y$ and yes it seems likely that things then work out very nicely. Thanks. $\endgroup$ – David C. Ullrich Oct 18 '15 at 16:18
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Rather that define dedekind cuts on the set of rational numbers, in a natural way we can modify the definition and define cuts on the positive rational numbers. In this new setting, we construct the non-negative real numbers, $\mathbb R^{\ge0}$, and we can show it forms a commutative semifield, and we don't get bogged down in checking, ad nauseum, the case-by-case breakdown for multiplication.

In exactly the same way as we the integers are constructed from the natural numbers found here, we can get the negative real numbers from $\mathbb R^{\ge0}$, giving us the set of all real numbers. Using the $\mathbb N_0 \to \mathbb Z$ theory, it will be a ring.

Proposition: Every nonzero real number $x$ has a multiplicative inverse.
Proof
If the pair $(u,v)$ with $u,v \in \mathbb R^{\ge0}$ represents $x$, then $u \ne v$.
Recall the definition of multiplication:

$\quad {\displaystyle [(a,b)]\cdot [(c,d)]:=[(ac+bd,ad+bc)]}$

If $u \lt v$, then the number (block) represented by $(0, \frac{1}{v-u})$ is the inverse of $x$.

If $u \gt v$, then the number (block) represented by $(\frac{1}{u-v},0)$ is the inverse of $x$.

So every nonzero element of $\mathbb R$ has a multiplicative inverse and so we have actually constructed a field of numbers. $\quad \blacksquare$

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