I have read Ittay Weiss' survey on constructions of the real numbers: https://arxiv.org/abs/1506.03467

He writes that it is basically sufficient to construct the positive real numbers, as inverses (and 0) can be added in a final step. I was wondering how this works exactly. I am aware of the Grothendieck group construction: we can embed a semigroup in an abelian group by considering pairs of elements of the semigroup (representing their difference).

Is this sufficient? That is, can we simply apply this procedure to a totally ordered semifield (namely $\mathbb R_{\ge0}$) and expect to automatically get a totally ordered field? I have been unable to find a reference for this statement, and I am not sure how I would show this.

  • $\begingroup$ Doesn't the author define the positive reals regardless the structure, then extends using $\mathbb R=\mathbb R_{\ge0}\cup-\mathbb R_{\ge0}$ where by definition $\overline r=-r$ so that $\overline r+r=0$, then establishes the group structure and order ? $\endgroup$
    – user65203
    Commented Mar 10, 2020 at 11:00
  • $\begingroup$ He simply writes "typically it makes little difference whether one constructs the positive (or nonnegative) reals $\mathbb R_+$ and then extend to all the reals by formally adding inverses (and a 0 if needed), or constructing all of $\mathbb R$ in one go." No further details are given. $\endgroup$
    – Stefanie
    Commented Mar 10, 2020 at 11:09
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    $\begingroup$ Isn't that compatible with my comment ? $\endgroup$
    – user65203
    Commented Mar 10, 2020 at 11:17
  • $\begingroup$ The answer to your comment is No. He does not. I am getting the impression from his text that constructing $\mathbb R_+$ is enough and that you do NOT need to "establish group structure and order" after extending to $\mathbb R$. $\endgroup$
    – Stefanie
    Commented Mar 10, 2020 at 11:51
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    $\begingroup$ Did you try to "establish group structure and order"? It might not be the best text book style but maybe Ittay Weiss thought that these steps are trivial so he considers them as a little difference. $\endgroup$
    – quarague
    Commented Mar 10, 2020 at 13:30

1 Answer 1


The common approach is to define an operation on $\def\R{\Bbb R^+} \def\x{\times} K=(\R,+)\x(\R,+)$ and then establish an equivalence relation to factor out the "superfluous" pairs resp. the ambiguity of representation. This will be an analogue route to how $\Bbb Q$ can be constructed from $\Bbb Z$, or $\Bbb Z$ can be constructed from $\Bbb N$ just to name a few.

Every Real number can be represented in a non-unique way as difference of two positive Real numbers, so the operation will represent a difference without using operations / notations like $-x$ or $x-y$ which we don't have yet. The non-uniqueness will be addressed by considering equivalence classes, and these classes will later on be called "Real numbers".

Specifically, define a binary operation + as

\begin{align} +:K\x K&\to K \\ (x,y) + (x',y') &\mapsto (x+x',y+y') \end{align} The pairs should function as a difference, i.e. we'd like $(x,y)=x-y$. Notice however that the right side does not mention − in any way, it just uses existing properties of $(\R,+)$.

In order to operate like a difference, i.e. $(x-y)+(x'-y') = (x+x') - (y+y')$, we notice that $x-y = (x+a)-(y+a)$ for any $a$. The equivalence relation will fix the ambiguity and implicitly introduce the properties of −

$$\def\~{\sim} (x,y) \~ (x',y') \quad\iff\quad x+y' = x'+y$$

Again, only features of $(\R,+)$ are used. Then consider the equivalence classes $R=(K,+)/\!\~$ and show:

  1. The definition + is is well defined in $R$, i.e. for any $k,k', q, q'$ in $K$ there is $$ k\~k' \;\land\; q\~q'\quad\implies\quad k+q \~ k'+q' $$ This means the result of $r+q$ is independent of which representative we chose for $r$ or $q$.

  2. $(R,+)$ (and $(K,+)$ for that matter) inherits properties from $(\R,+)$ like: + is closed, associative and kommutative.

  3. For any $x,y,z\in\R$ we have $$(x,y) + (z,z) = (x+z,y+z) \~ (x,y)$$ Thus $(z,z)$ operates as neutral element and we write $0_R:=(1,1)\~(z,z)\in\ R$.

  4. For any $x,y\in\R$ we have $$ (x+1,1) + (y+1,1) = (x+y+2, 2)\~(x+y+1, 1)$$ This means the elements of the form $(x+1,1)$ operate under addition exactly the same like the elements of $\R$ do; "$(x+1,1)$" or "$(x+a,a)$" are just fancy ways to write "$x$". This means the subset is isomorphic to $(\R,+)$: $$\left(\{(x+y,y)\in \R\!\x\R \}{\large/}\!\~,+\right) \;\simeq\; (\R,+)$$ This in turn means

    $R$ is actually an extension of $\R$: It contains an isomorphic copy of $(\R,+)$ and also new elements like $0_R=(1,1)$ that cannot be interpreted as element of $\R$.

  5. Observe that $(x,y)+(y,x) = (x+y,x+y)\~(1,1)=0_R$. This means $R$ has a neutral element $0_R$ and each element $(x,y)$ of $R$ has an additive inverse $(y,x)$.

    Hence $(R,+)$ is a group.

  6. We write: $$-(x,y) := (y,x)\quad\text{ and }\quad(x,y)-(x',y') := (x,y)+(y',x')$$

  7. Proceed in the same way for a multiplication provided we already have $\def\.{\cdot} (\R,+,\.)$: Define, now with $K=(\R,+,\.)\x(\R,+,\.)$ \begin{align} \cdot :K\x K&\to K \\ (x,y) \cdot (x',y') &\mapsto (x\.x'+y\.y', x\.y'+x'\.y) \end{align} Again, show it's well defined when applying ~, notice that it inherits closedness, associativity, kommutativity, distributivity from $K$.

  8. Notice $(x,y)\.(2,1) = (2x+y, 2y+x)\~(x+y)$ and hence we have a One $1_R\~(2,1)$ in $R$. And the same embedding like above still applies because $$(x+1,1)\.(y+1,1)=((x+1)(y+1)+1,x+y+2)\~(xy+1,1)$$ So $(K/\!\~,+,\.)$ is an extension of $(\R,+,\.)$.

You got the idea. You can proceed the same way for division, or if you don't have division yet define it similar and factor out that $x/y = (ax)/(ay)$. The only thing is to keep in mind that zero is special.

Likewise, relations like $=$, $\neq$, $<$, $\leqslant$, $>$, $\geqslant$ can be carried over, some of them with restrictions. For example $$x>y \iff z\.x > z\.y$$ for $x,y,z\in\R$, but monotony in $R$ only holds true if $z>0$.

The final step is then to identify $\R$ with it's isomorphic embedding in $R$ and use the notation $\Bbb R := R$ calling the equivalence classes Real numbers (again).

Note: The notation above is somewhat sloppy as it uses the same notation $(x,y)$ for elements of $\R\x\R$ and equivalence classes modulo ~ for brevity. More strict notation would be using $$R\ni\overline{(x,y)}:=\{(x',y')\in K \mid (x,y)\~(x',y')\}\subseteq K$$

  • $\begingroup$ Thank you very much for your extensive reply! Do I get it right that this construction does not depend on how the positive real numbers themselves are constructed? It seems that you only need that this has been done and that we have established that they form a semifield. $\endgroup$
    – Stefanie
    Commented Mar 12, 2020 at 8:01
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    $\begingroup$ @Stefanie : All you need are the used properties of $\Bbb R^+$, how you got these properties does not matter. However if your construction yields already all of $\Bbb R$, then you don't need constructions like above in the first place. $\endgroup$ Commented Mar 12, 2020 at 10:49
  • $\begingroup$ Of course. But it seems much easier to construct $\mathbb R^+$ first. Avoiding negative numbers just makes everything so much easier! Think conditionally convergent series etc. Thanks again! $\endgroup$
    – Stefanie
    Commented Mar 12, 2020 at 14:02
  • $\begingroup$ BTW, do you have a (textbook) reference for this? The only constructions that I have seen in some detail construct all of $\mathbb R$ in one go. $\endgroup$
    – Stefanie
    Commented Mar 12, 2020 at 14:57

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