Can the function $x\mapsto x^{-1}$ be extended to zero without introducing inconsistencies?

Question

Assume that the function $x\in\mathbb{R}\setminus\{0\}\mapsto x^{-1}$, which assigns to every non-zero real number its unique reciprocal, is a primitive of the field axioms that, together with other axioms, define the real number system. Is it possible to extend this function to all of $\mathbb{R}$ (i.e. to define it at $x=0$) in such a way that the resulting real number system will remain logically consistent? The axioms should otherwise remain unchanged, and no special stipulations on the properties of the extension are to be assumed.

To put it more concretely, suppose I add the following axiom: '$0^{-1}=1$' to the axioms of the real number system as listed here, but with the understanding that the function $x\mapsto x^{-1}$ is a primitive of the system. Will the resulting axiomatic system remain consistent?

My question is purely logical/mathematical. I do not wish to engage in extra-mathematical discussions about the merits of such an extension or about the question of dividing by zero.

Answer 1

As long as the other axioms you have do not claim anything about the behavior of $x^{-1}$ when $x=0$, then it does not have any ill effects to add an axiom requiring $0^{-1}=1$.

This won't buy you much in practice, of course, because the axiom that tells you that $x\cdot x^{-1}=1$ still only tells you that under the condition that $x\ne 0$.

But you can certainly do it, for example, to restrict the possible models of the theory such that you can be sure non-isomorphic models are indeed different. Sometimes that is technically convenient.

(The extended system is consistent because it has a model -- in particular, the structure whose universe is $\mathbb R$, the interpretation of the symbols $0$ and $1$ are the real numbers $0$ and $1$, the interpretation of the symbols $-$, $+$, and $\cdot$ are negation, addition, and multiplication from usual real arithmetic, and the interpretation of ${}^{-1}$ is the function defined by $$ f(x) = \begin{cases} 1/x & \text{when }x\ne 0 \\ 1 & \text{when }x=0\end{cases} $$ is a model.)

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Hmm, let me qualify that a bit. Actually this depends on how one formalizes the English statement of your field axioms in actual first-order logic. The source you link to states the relevant axiom as

For every nonzero real number $x$ there exists a nonzero real number $x^{-1}$ such that $x\cdot x^{-1}=1$.

1. My first interpretation (and the one the above answer is directed at) is that the logical language we're talking about contains a function symbol that we write as $-^{-1}$, and that the axiom means $$ \forall x: x\ne 0 \Rightarrow ( x^{-1}\ne 0 \land x\cdot x^{-1}=1 ) $$

2. However it is also plausible to say that $^{-1}$ is not a function symbol at all, but "$x^{-1}$" is just a suggestively named variable, in which case the formal axiom is $$ \forall x: x\ne 0 \Rightarrow \exists y: ( y \ne 0 \land x \cdot y = 1 ) $$ (where I've renamed the fancy variable $x^{-1}$ to $y$ for clarity). In that case, however, your plan doesn't really make sense: When $^{-1}$ is not a symbol of its own, all you're doing is adding a fresh constant symbol (identified by an ink blob of the shape "$0^{-1}$") and specifying by a new axiom that this new constant symbol must have the same meaning as the constant symbol $1$ you already have. (Or perhaps your "$0^{-1}$" is to count as a variable, in which case your new axiom is just $\exists y:y=1$ which was already provable).

Which of these interpretations the author you're quoting intends is a bit unclear. On one hand he seems to say that the only operations on a field are $+$ and $\cdot$, so his non-logical vocabulary would be $\{{+},{\cdot}\}$. But on the other hand he has one axiom stating "there exist real numbers $0$ and $1$ with such-and-such properties" and then he uses those symbols again in other axioms, which would be nonsense if F3 merely meant $\exists 0:\exists 1:(\cdots)$.

So perhaps the language is $\{0,1,{+},{\cdot}\}$? But $-x$ and $x^{-1}$ are introduced with the same phrase as $0$ and $1$ were, so I think it is most reasonable to say that the language is $\{0,1,{-},{}^{-1},{+},{\cdot}\}$ and we're therefore in case (1) above.

Some of the other answers clearly seem to have (2) in mind, and there are, for some purposes, technical advantages to writing the field axioms that way (for example, the theory of fields then has the same logical language as the theory of rings).

Mostly I think the author is being slightly confused about what his language is, an impression reinforced by the fact that what he calls "field axioms" speak about "there is a real number ..." rather than "the field has an element ...".

Answer 2

You would have $$\begin{align} 0\times0^{-1}&=1\\ 0\times0^{-1}&=0\times1=0 \end{align}$$

Answer 3

You would instantly collapse the number system you are working on to the Zero ring, which is the ring on which the multiplicative identity and the additive identity are the same, meaning that every number is the same, i.e. the only number is $0$.

You can see this rather easily by using $1 = 0^{-1}$ meaning that $0^{1} \times 0^{-1} = 0^{1-1} = 0^0$. Because $a$ and $a^{-1}$ are multiplicative inverses, you just axiomatically set $0^0=1$, add to that, that you know $0$ is the additive identity (i.e. $0 \times a = 0$) and you get $1=0$.

Will it be consistent? Yes, you can do math (technically) on the zero ring. It's just not particularly interesting and you can't avoid this happening.

EDIT: I misread your question originally as asking if the axiom $0^0=1$ would be okay. I've fixed this now.

Answer 4

As said in the comments, if $0^{-1}$ is the inverse of $0$, then you can derive $0=1$. Otherwise, you are adding a generic element $\alpha$ to the real field, and then introducing the axiom $\alpha=1$. The axiom itself ensures that you are actually not adding any element, since $1$ is already included in $\mathbb R$.

If there are any other relations you have in your mind that include $0^{-1}$ in it, then we can discuss about the inconsistency of the axioms.

Answer 5

Assuming you don't give an interpretation of $0^{-1}$ and we treat it as a formal symbol (notably axiom F5 in the linked post only applies to non-zero $a$), then you wind up with the exact same system, because all you've done is introduce an unusual and syntactically confusing symbol for $1$. It makes no mathematical difference, as $1$ is already a real number. You can contrast this with if you had defined a symbol that satisfied $x^2=-1$, in which case you'd obtain the Complex Numbers because your new symbol would be functionally identical to $i$.

However, if you were to give $0^{-1}$ the natural interpertation, i.e. that $0\times 0^{-1}=0^{-1}\times 0=1$ then as Timon noted you wind up with the zero. However, given your aggressive rebuke of that idea in the comments, I'm forced to conclude you're using the symbol in the sense of the first paragraph.