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Some texts define the field of real numbers as a set $\mathbb{R}$, on which there are two operations defined (called Addition $+$ and Multiplication $\cdot$, respectively), such that:

  1. $(\mathbb{R},+)$ is an abelian group with neutral element $0$;
  2. $(\mathbb{R}\setminus\{0\},\cdot)$ is an abelian group with neutral element $1$;
  3. For all $a,b,c\in\mathbb{R}$ we have (distributivity laws): \begin{align*} a\cdot (b+c)=(a\cdot b)+(a\cdot c) \\ (b+c)\cdot a=(b\cdot a)+(c\cdot a). \end{align*}

At first this seemed to me like a nice way to define the real numbers. However, on second thought, from 2. we have associativity, commutativity, and the existence of a neutral element for multiplication for all $a,b,c\in\mathbb{R}\setminus\{0\}$ instead of for all $a,b,c\in\mathbb{R}$. Did I miss something or is the short definition given above indeed incomplete?

PAT
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2 Answers2

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In fact the axioms given do determine how $0$ multiplies - you can show that we must have $0\times r=r\times 0=0$ for every $r\in\mathbb{R}$. This is a good exercise (see here).

That said, this description is still very far from complete. For example, both $\mathbb{Q}$ and $\mathbb{C}$ satisfy all those rules as well, despite being very different from $\mathbb{R}$. At the very least we should add further axioms which ensure real closure.

(In fact there is a precise sense in which there is no "simple definition" of $\mathbb{R}$, although that's a bit technical - this is discussed in various places at this site, e.g. 1 or 2.)

Noah Schweber
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What you wrote cannot possibly a definition of $\Bbb R$ since it applies to every field.

Anyway, nothing is missing. First of all, note that you always have $a.0=0$, since\begin{align}0&=a.0-a.0\\&=a.(0+0)-a.0\\&=(a.0+a.0)-a.0\\&=a.0+(a.0-a.0)\\&=a.0+0\\&=a.0.\end{align}For the same reason, $0.a=0$. So, commutativity holds when one of the elements is $0$. Since $1.0=0.1=0$, $1$ is still the neutral element. And if one of the numbers $a$, $b$, and $c$ is $0$, $(ab)c=a(bc)$ since both numbers are equal to $0$.

José Carlos Santos
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  • Right, I was only talking about the field axioms, actually. Well, if the only axiom we state for multiplication is that $(R\0,\cdot)$ is an abelian group, then to prove $0\cdot a=0$ we have to use the laws of distributivity as noticed in your answer. But all that gives us is $a\cdot 0=a\cdot (0+0)=a\cdot 0+a\cdot 0$, not $a\cdot 0=0$... unless I missed something here.

    I can follow your arguments after establishing $0\cdot a=0$ though.

     – PAT Sep 30 '21 at 18:18
  • I have expanded my justification of the fact that $a.0=0$. What do you think now? – José Carlos Santos Sep 30 '21 at 18:23
  • Ok, that makes sense now :) Thanks. – PAT Sep 30 '21 at 18:25