Let us say that we have constructed the positive real numbers in some fashion (possibly including 0). Let us furthermore assume that we have defined addition and multiplication for ordered pairs of such numbers (an ordered pair $(x,y)$ is supposed to represent the difference $x-y$, but you cannot properly define subtraction for positive real numbers), and proved that these operations "work properly".

So we have reached the state where we have a ring $(K/\sim, +,\cdot)$ which extends $(\mathbb R^+,+,\cdot)$. Finally, let us assume that we already have division for $\mathbb R^+$.

Now. How exactly do we show that multiplicative inverses exist, that is, that for any pair $(x,y)$ there exists a pair $(x',y')$ such that their product is equivalent to 1? Using the definition of multiplication for ordered pairs, we need to find $(x',y')$ such that $$(xx'+yy', xy'+x'y) \sim (1,0).$$

I have tried, but all I can get is that this reduces to $$x'-y' = \frac{1}{x-y}.$$ Unfortunately, as we still cannot actually subtract positive real numbers, this does not seem to help.

One possible solution would be to show that having gotten this far, we can now actually subtract positive real numbers by using their isomorphic embedding in the extension. I have the feeling that this is the way to go, but I simply do not see how to write down a proof. It should be based on the following ordering of ordered pairs: $$(x,y)\le(x',y') \Leftrightarrow x+y'\le x'+y.$$ Basically, it seems we have to show somehow that expressions like $\sqrt2-\sqrt3$ (and also $\sqrt3-\sqrt2$!) make sense now.


1 Answer 1


Why not just do it piecewise? For what it's worth, though, this might be a good reason to use an alternative, "sign-magnitude representation", instead of the difference-based representation, which is closer to how we typically work with signed numbers anyways and hence is more intuitive, or at least I think so. That is, if we are already given $\mathbb{R}^{+}$ by some alternative construction procedure, you can define an element of $\mathbb{R}$ to be

$$(\mbox{$+$ or $0$ or $-$}, m)$$

where the first item listed is the sign, to be one of the three symbols given, $m \in \mathbb{R}^{+}$ is the magnitude of the number, and we identify all elements of the form $(0, m)$ as being the same thing. Then just define

$$\frac{1}{x} := \left(s, \frac{1}{m}\right)$$

when $s \ne 0$, where we've done the right-hand reciprocal in the already-existing nonnegative reals. On the other hand, you have to now construct addition piecewise, but at least here it's considerably more obvious and natural how to do that (e.g. "plus plus minus is subtraction", etc.).

I don't think there's any good way to handle the difference-based definition that would not be piecewise - note that with the positive reals you can already subtract $x - y$ when $x > y$, and then just take $x - y$ for $x < y$ as $-(y - x)$. It works more slickly for building the additive group structure of the integers from the natural numbers, but not so much for the multiplicative structure of reals from positive reals. In particular, note that any "differencing" definition you are going to come up with has to mirror some operation on fractions of the form

$$\frac{a}{b} - \frac{c}{d}$$

which equals $\frac{ad - bc}{bd}$, but there's no way to get a difference in the denominator $bd$ without subtracting things there, i.e. using only addition, multiplication, and division of positive reals, and hence we're pretty much back to handling the case of the left operand being less than the right operand piecewisedly.

  • $\begingroup$ OK, so how do I construct addition in this case? It is now clear that additive inverses exist, we can just define $-(+,m) := (-,m)$. But now I want to define $(+,a)-(+,b)$. All I can get is that $$ (+,a) - (+,b) = (+,a) + (-,b) = \begin{cases} (+,a-b) & a\ge b\\ (-,b-a) & a < b \end{cases}$$ which only brings me back to having to subtract positive real numbers, becaus what is $(+,a-b)$? $\endgroup$
    – Stefanie
    Mar 20, 2020 at 11:02
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    $\begingroup$ @Stefanie : Subtraction of positive real numbers, $a - b$, when $a > b$, does not require the existence of negative numbers, and hence also their construction. Are you actually asking for how to construct subtraction of positive real numbers itself where it would otherwise be defined, or how to prove it possible, and not just the extension of positive real numbers with negative numbers (which is how I interpreted the question)? $\endgroup$ Mar 20, 2020 at 11:27
  • $\begingroup$ Thank you, this comment has made me figure it out. Yes, of course it is possible to define a-b as long as a is at least b. At least, I think you prove it by completeness: consider all numbers such that c+b is at most a, and also all numbers such that c+b is at least a. Then the supremum of the first set must be the infimum of the second set, both must exist by completion, and the value is a-b. Right? Then given this and the sign-magnitude notation, you can also define a-b for the case where a is less than b. $\endgroup$
    – Stefanie
    Mar 20, 2020 at 13:48
  • $\begingroup$ @Stefanie : Yes, I believe you can use the completeness property as you suggest, forming a Dedekind cut where the difference should be. In fact, technically multiplication can be considered redundant - see Tarski's very compact axiomatization of the reals. $\endgroup$ Mar 20, 2020 at 13:50

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