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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.

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    I think the axiomatic system will still be consistent, however later on there will be results that are no longer as useful or elegant as they used to be - for example the limit $\frac{\sin x }{x}$ as x tends to zero will still be 1, but at zero it'll suddenly jump down to zero - that's just my view though – 2017-02-01
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    @Cato: Why is this any less elegant than the current situation where at zero it is suddenly undefined? However, this is beside the point, and is a matter for a separate discussion. – 2017-02-01
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    The linked axioms also include axioms for order $O1,O2,O3$. Do you intend to include them, or only axioms $F1-F7,\,$ i.e. are you considering field or *ordered* field axioms? – 2017-02-01
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    @BillDubuque: Ordered fields. Specifically, the real number system. – 2017-02-01
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    It will remain consistent, but not in Euclidean plane geometry. – 2017-02-01
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    @BillDubuque: I have re-worded my question. Would you please consider casting a re-open vote for it? – 2017-02-02
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    My pleasure (I don't think it should have been closed anyhow). – 2017-02-02
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    @BillDubuque: Thank you. – 2017-02-02
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    https://en.wikipedia.org/wiki/Projectively_extended_real_line – 2017-02-03
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    @CountIblis: Thanks. Nice try. However, the projected real line is not a field, so it can't be a model for the axiomatic system I described, which includes the field axioms. – 2017-02-03

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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.)

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 ...".

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    I disagree about the usefulness of this definition, but this is a matter for a separate discussion. Thanks. – 2017-02-01
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    @EvanAad What usefulness could it possible have, outside of distinguishing models of the theory by having a slightly different language? – 2017-02-01
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    @StellaBiderman: This is a matter for a separate discussion, but I'll give just one example. Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be some function, and consider the function $g:\mathbb{R}\rightarrow\mathbb{R}$ that satisfies that $g(x) = 1/f(x)$ whenever $f(x)\neq 0$, and $g(x) = 0$ otherwise. Using the newly added axiom, this may be written succinctly as: $g:=f^{-1}\mathbb{1}_{\{f\neq0\}}$. – 2017-02-01
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    @StellaBiderman: Didn't they say the same when negative numbers were introduced to the number system? – 2017-02-01
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    @EvanAad No, the negative numbers are not syntactic sugar. Adding them to the natural numbers changes it's mathematical content. It makes them into a ring. – 2017-02-01
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    Thanks. This is a great answer. I think about reciprocation as a primitive function available in the axiomatic system. The only reason why I haven't marked your answer as THE answer, is that you haven't explained why the addition of this axiom leaves the system consistent; you simply *stated* that it does. – 2017-02-01
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    @EvanAad: I've added the straightforward argument for that. – 2017-02-01
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    @HenningMakholm: I have reworded my question. Would you consider casting a reopen vote for it? – 2017-02-02
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    @EvanAad: Already did; I don't think it should have been closed in the first place. – 2017-02-02
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    @HenningMakholm: Thank you. – 2017-02-02
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    @StellaBiderman: I have reworded my question. Would you consider casting a reopen vote for it? – 2017-02-02
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You would have $$\begin{align} 0\times0^{-1}&=1\\ 0\times0^{-1}&=0\times1=0 \end{align}$$

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    I challenge you to formally derive $0\times0^{-1}=1$ based on the system I described. – 2017-02-01
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    I interpret $0^{-1}$ as the multiplicative inverse of $0$, and $0\times0^{-1}=1$ is a consequence of the definition of multiplicative inverse (F5). – 2017-02-01
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    why does $0\times0^{-1}=0\times1$ hold? – 2017-02-01
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    I did not ask for personal interpretations. – 2017-02-01
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    I guess the axiom about reciprocals still excludes zero – 2017-02-01
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    What other interpretation does $0^{-1}$ have? – 2017-02-01
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    If you are not denoting as $0^{-1}$ the inverse of $0$, then you can denote it with any other symbol like $\alpha$. – 2017-02-01
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    Only those interpretations originating in the axioms. – 2017-02-01
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    @EvanAad You should add your axioms – 2017-02-01
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    @EvanAad Then enlighten us as to what $0^{-1}$ means, if not "the unique number, $a$, such that $0a=a0=1$," as that is the standard meaning of the exponent negative $1$. – 2017-02-01
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    @StellaBiderman: Read my original post and be enlightened, as it is there that I *define* it. – 2017-02-01
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    Well, you have added a new axiom F5': $0^{-1}=1$, but you have not defined how the symbol $0^{-1}$ is related to the number $0$. Axiom F5 defines $a^{-1}$ for $a\ne0$ as the only number, whose existence is guaranteed by the axiom, such that $a\times a^{-1}=1$. – 2017-02-01
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    @EvanAad There's a difference between asserting that $x=$ and defining the symbol $x$. Is $0^{-1}$ some fancy symbol you've made up for $1$, or does it have a particular meaning, perhaps one connected to the meaning of $0$? – 2017-02-01
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    @miracle173 Because the new axiom is $0^{-1}=1$. – 2017-02-01
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    @JuliánAguirre: Exactly: Axiom F5 does not handle the case $a\neq0$, so no inconsistency there with the new definition. – 2017-02-01
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    Axiom F5 gives the relation between the $a$ and $a^{-1}$. What is the relation between the symbol $0^{-1}$ and the number $0$? – 2017-02-01
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    @JuliánAguirre: No, axiom F5 gives the relation between $a$ and $a^{-1}$ _when $a$ is nonzero_. It does not begin saying more than that simply because we add a different axiom that says something about $a^{-1}$ when $a$ _is_ zero. – 2017-02-01
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    @HenningMakholm Yes of course, I included $a\ne0$ in a previous comment, but forgot it in the last one. – 2017-02-01
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    @miracle173: I have reworded my question. Would you consider casting a reopen vote for it? Four people have already voted. Just one more vote is required. – 2017-02-02
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    It's already one. – 2017-02-02
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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.

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    It is not necessarily just a symbol. It can be viewed as the value that the function $f:\mathbb{R}\rightarrow\mathbb{R},\ \ f(x):=x^{-1}$ assumes at the point $x=0$. – 2017-02-01
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    Am I allowed to assume that $f$ satisfies the usual properties, such as $xf(x)=1$ and $f(f(x))=x$ or are you not using that notation in the usual sense? – 2017-02-01
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    @EvanAad: "Just a symbol" in this context means only that you will be moving away from the (intended?) intuition that $(\cdot)^{-1}$ expresses the arithmetical concept of reciprocal. – 2017-02-01
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    As per usual, you may assume those results that can be derived from the axioms, and nothing more. – 2017-02-01
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    @EvanAad Can you stop repeating that mantra? It is not helpful or illuminating, and is only serving as a way for you to avoid actually answering questions. Not all results are derivable solely from axioms. Definition are extremely important to derive results and we are asking about your definitions. – 2017-02-01
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    @EvanAad You used the word "division" in the title of this post. It's not unreasonable for people to think you mean for $0^{-1}$ to have something to do with dividing by zero. Perhaps it even means $1/0!$ But hostility and obstruction to people who are trying to help you will just make no one want to talk to you. – 2017-02-01
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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.

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    where is $0^0$ being derived from? Normally it is undefined - I see it is another issue - I've heard $0^0 = 1$ discussed before - i – 2017-02-01
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    I have fixed it. I thought you were axiomatizing $0^0=1$ but you were in fact demanding $0^{-1}=1$. But you can easily derive the former from the latter. – 2017-02-01
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    This reasoning doesn't work. You can't conclude that $0\times 0^{-1}=1$, because the axioms that tell you something about $x^{-1}$ only do so for $x$ that are not zero. The question did not specify that the $x\ne 0$ conditions were to be removed from the existing axioms. – 2017-02-01
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    Yes, I can because otherwise his inclusion of $0^{-1}$ is a completely empty statement with absolutely zero implications. What he asked in the title of the question was about the consistency of *division* by zero. Division is literally the name of the inversion of multiplication. If he didn't mean $0^{-1} = \frac{1}{0}$ and thus by direct extension $0 * 0^{-1} = 1$ then the question is completely meaningless. It's like asking what happens if we include a number $k=1$ in the axioms with no apparent relation of $k$ to any other number. The answer is nothing happens because $1$ already exists. – 2017-02-01
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    It is not " a completely empty statement with absolutely zero implications" -- it prevents an interpretation with $0^{-1}=42$ from being a model; this interpretation _was_ a model of the original set of axioms. – 2017-02-01
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    When I say that you can't conclude $0\times 0^{-1}=1$ I mean "conclude" in the sense of writing down a valid formal proof, starting from the given axioms, that conlcudes $0\times 0^{-1}=1$. You seem to be engaging in some extra-logical intentional reasoning, but that is irrelevant from a logical point of view. – 2017-02-01
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    It *is* a completely empty statement. You are massively abusing notation by calling some random additional object which just turns out to equal 1 again and is completely unrelated to $0$ in any way $0^{-1}$. It's like me asking what would happen if I set $0^2=42$ and then when you respond it would collapse to the zero ring, I'd say no I didn't mean $0 \times 0$ with $0^2$ I just meant a random other object that I could've given any other name. If you see nothing wrong with this on the side of OP for asking the question in this manner, then I don't know what more to say. – 2017-02-01
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    @TimonG.: Logic does not care about "abusing notation". If you claim that the extended axiom system is inconistent, then please provide a proof of that fact instead of just throwing a tantrum. (You cannot set "$0^2=42$" without first extending the language of the theory, because the language here does not include any function symbol $-^2$. If you do add such a symbol, you can of course add the axiom $0^{42}$ without that in itself causing any inconsistency -- except of you _also_ add some additional axioms about your new $-^2$ function that conflicts with it). – 2017-02-01
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    But **StackExchange** cares about asking questions in a manner that makes your intention *clear*. This question abuses notation intentionally and assumes a random leap of understanding on the side of person taking his time to formulate an answer. If OP meant specifically that $0^{-1}$ is wholly unrelated to the multiplicative inverse of 0, then he should not have called the question "Consistency of **division** by zero" and then used the symbol $0^{-1}$ without *any* indication of him meaning a whole new number entirely unrelated to $0$. – 2017-02-01
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    I nowhere said it's inconsistent to add a $k=1$ to the axiom system entirely unrelated to $0$. You can add infinitely many of those, since you already have an object that is $1$. Even in my answer, where acc. to you I misunderstood, I suggest the appropriate answer is the zero ring - an entirely *consistent* mathematical object. – 2017-02-01
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    The question is completely clear. It asks **"Will the resulting axiomatic system remain consistent?"** (this is a verbatim quote, not my paraphrase).The only correct answer to that question is "yes, the resulting system will remain consistent". Your comments seem to amount to "but I don't _like_ the resulting axiomatic system", but that is not what was asked. – 2017-02-01
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    The zero ring, is, however **not a model of the proposed theory**, because the field axioms linked to in the question includes (as it is common) an assertion that $0$ and $1$ are distinct. – 2017-02-01
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    It says they are distinct, it doesn't say explicitly $0 \neq 1$. Just like you are apparently saying that without any additional context the object $0^{-1}$ bears no burden to be a multiplicative inverse or even related to zero even though OP *called* it a "*division* by zero", "distinct" can (apparently) just refer to whatever anybody wants even though we all know what "division" and "distinct" actually are intended to mean. If I had the necessary rep, I'd vote close this question as unclear what OP is even asking. He can open a new question and be more clear in formulating his issue. – 2017-02-01
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    I never said I don't "like" the resulting system (whatever that even means, I'd appreciate it if you stuck to math instead of personal jabs), you are making things up about my opinion even though I already explicitly explained I'm not making that case. You can introduce whatever symbol you want to mean whatever you want, however, you cannot consistently refer to it as division in the way division is commonly defined and expect other humans to understand what you mean without any further clarification, which is *exactly* what OP did. I recommend that you read the question title again. – 2017-02-01
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    Let us leave it at that, Henning. Thanks for the discussion, I enjoy having one once in a while. However, as SE is constantly reminding us here, this isn't the place for this kind of talk. I think your answer to OP is more thorough and well thought out than the question deserved given how unintuitively he named and asked it. – 2017-02-01
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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.

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    $0^{-1}$ is exactly what it is defined to be. No more, no less. – 2017-02-01
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    For example, does it follow power rules? Like, $0^{-1} \times 0^2 = 0^{2-1}$? – 2017-02-01
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    Only if these rules can be derived from the axioms. – 2017-02-01
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    Ok, then, as I already explained, your sistem is consistent, and coincides with $\mathbb R$. You are only renaming the constant $1$. – 2017-02-01
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    How do you propose to derive $0=1$ using only the specified axioms. The one that tells you $a\ne 0 \Rightarrow a^{-1}\cdot a = 1$ does not suddenly lose its left-hand side just because we add a new axiom the that specifies the (previously unrestricted) interpretation of $0^{-1}$. – 2017-02-01
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    In fact I'm giving choices in my answer: Like I said, if you want $0^{-1}\times 0=1$, then you obtain $0=1$. If you are not implying this property or any other relations, then you're ok, except for that your notation is a bit more complicated.. – 2017-02-01
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    $0^{-1}\times 0^2=0^{2-1}$ is not even in the language of the theory; it contains no $-^2$ symbol, much less a $-^{2-1}$ symbol. – 2017-02-01
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    @HenningMakholm The question has been settled. As me(and many others) were wondering if the OP had other rules in mind for the new symbols, I tried only to suggest some that could have been plausible. Since the OP itself declared the non-existence of further relations, I drew the only conclusion possible. Thanks – 2017-02-01
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    @Exodd: I have reworded my question. Would you consider casting a reopen vote for it? – 2017-02-02