$[\frac{1}{x} = 1$ for all real numbers $x]$ is not true or false

I have a follow-up to a previous question: True or false or not-defined statements.

(In the following I might use the word statement incorrectly.)

In that question/answer I learned that the "statement" that $\frac{1}{0} =1$ is not true or false because the expression isn't a well-formed formula (i.e. the left have side is not well defined). I do understand that one might make the "symbol" $\frac{1}{0}$ in certain areas of mathematics. For this question, though, I am just interested in the math we teach in say an undergraduate calculus class.

My follow-up question is what about the statement

$\frac{1}{x} = 1$ for all real numbers $x$

Here the statement makes fine sense if we had said "for all values of $x\neq 0$".

But would this statement also be not-true and not-false because the expression is not defined/well-formed since $x=0$ is part of the statement?

(Even though I am interested in how one might answer this in a calculus class, I would like the strict logic answer. I know that we often say and write things that technically/strictly speaking not true, but when say teaching about true and false statements, I would like to keep things as precise as possible. Even though we can sometimes overlook some things in an undergraduate class, I also think it is wrong to teach stuff that is outright wrong.)

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Even $1 / 0 = 1$ is a false statement. You are claiming that a nonsense expression is equal to $1$, which is false. –  Austin Mohr Sep 24 '12 at 0:25
@AustinMohr: Did you look at the previous question and the answers given there? I think the conclusion was that if part of an statement is non-sense, then the statement is non-sense and so therefore neither true nor false. –  Thomas Sep 24 '12 at 0:32

The general framework to handle logical systems in which terms can be undefined is called "free logic". However, free logic has been studied mostly in the context of philosophy, and is conspicuous only in its absence in mathematical logic textbooks. This is because the way we handle things in mathematics is just to rephrase the question to avoid the undefined values, and to use vacuous quantification in other settings to work around them.

In free logic there are several semantics, but all of the ones that people usually consider do give a truth value to "$(\forall x \in \mathbb{R})[ 1/x = 1]$". All the usual semantics would say that it is false because it is false for $x = 2$. I spent a while looking into this a little bit ago, and what I learned is that the usual trend in free logic is to assign a truth value to as many quantified sentences as possible. Of course one could define a semantics in which a quantified statement has no truth value if there is a substitution instance that has no truth value (e.g. "1/0 = 1" has no truth value in free logic because $1/0$ is an undefined term; but $(\forall x \in \mathbb{R})[1/x = 1]$ is nonetheless false in all the usual semantics).

My interest in this came from looking at statements such as "$(\forall X \subseteq \mathbb{R})(\forall z \in X) [ z \leq \sup X]$". My opinion is that this statement is erroneous, and has no truth value, because it has no truth value when we take $X = \emptyset$. However, by looking at some references I realized that all of the usual semantics for free logic make this statement true, because they make $(\forall z \in \emptyset)[z \leq \sup \emptyset]$ true as a vacuous quantification. Nevertheless my personal opinion, as yours may be, is that the formula at the beginning of this paragraph cannot be asserted in ordinary mathematics because to do so implies that $\sup\emptyset$ has to be defined. It would be possible, I believe, to write a short paper developing a semantics for free logic that mirrors this, although I have not spent any significant time on it.

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Interesting! I was wondering if there's some formal work going that way. –  tomasz Sep 24 '12 at 1:34
Dr. Mummert: Thank you very much for this answer. That actually helped my understanding. I know that M.SE doesn't encourage follow-up questions in comments, but just to make sure that I understand: You are saying that in "classical logic" the statement in my question does not have a truth value, but that one can do this "Free logic" and thereby assign truth values to statements like the one above? Also, would it be fair to say that the math that is taught in say undergraduate calculus class is based on "classical logic"? –  Thomas Sep 24 '12 at 1:52
@Thomas: in classical first-order logic (CFOL) a function such as the "$/$" in "$a/b$" must be defined for all inputs. This is usually achieved by letting the function have arbitrary values when it is "supposed" to be undefined, and then adding hypotheses to the theorems about the function that mention the cases where it is "supposed" to be defined, e.g. $(\forall x,y)[(x \not = 0) \to (x \cdot (y/x) = y]$. There simply are no undefined values in CFOL. In this particular sense free logic is closer to mathematical practice, but standard techniques let us use CFOL with no ill effects. –  Carl Mummert Sep 24 '12 at 2:03
@CarlMummert: Ok, just one more: What about the statement that $x^0 = 1$ for all real numbers $x$. Would this then be true in free logic? (and so one would never say that this statement would be false) –  Thomas Sep 24 '12 at 2:07
@Thomas: it depends on exactly how $x^0$ is defined; some authors simply define $x^0$ to be $1$. But if $x^y$ is defined as $\exp(y\ln x)$ and the domain is $\mathbb{R}$ and $\ln 0$ is undefined then $(\forall z)[0^z = 0]$ would be false (not undefined) in both the "positive" and "negative" semantics of the article I linked in my answer, because every quantified statement is true or false (it never has an undefined truth value). I believe this is typical in the various free logic semantics that are usually studied. However I am not an expert in free logic, you should look at the literature. –  Carl Mummert Sep 24 '12 at 2:36

Well, its negation

There exists a real $x$ such that $\frac{1}{x} \ne 1$

is a true statement... despite $1/x$ not being an everywhere-defined formula...

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Wouldn't the negation of you statement be: For all real number $x$ in the domain of $\frac{1}{x}$, $\frac{1}{x} = 1$? –  Thomas Sep 25 '12 at 16:50

Whether you consider the statement true, false, or nonsense really depends on what you expect from the truth value.

In the intuitive sense, it is clearly false (because there are real numbers for which the quantified statement is clearly false).

As you have noted, for $x=0$ the expression $1/x$ may be undefined (unless defined as $\infty$, in which case there's no problem at all), so the statement "for all real numbers" should raise an eyebrow.

In analysis and elsewhere, it is often the case that quantifiers are tacitly assumed to only range over the sets in which the quantified expressions make sense, among some larger set that is obvious from the context. For example, if we have a function defined on a subset of reals (such as a rational function, or $\tan$), when checking for continuity, you might want to use the definition that $f$ is continuous iff $\forall x (\lim_{x'\to x}f(x')=f(x))$. Here the $x$ and $x'$ are only taken from the domain of $f$ and you don't usually go to great lengths to specify that fact.

For that reason, if the statement in question was $\forall x (\frac 1 x=1)$, you would have to be quite picky to take issue with evaluating it as false. When stated as $\forall x\in {\bf R}(\frac 1 x=1)$, though, it looks like an attempt to circumvent the convention, and as such is rather sloppy.

If asked to evaluate its truth value, out of context, I would say that I understand that to either mean $\forall x\in ({\bf R}\setminus \{0\}) (\frac 1 x=1)$ and then is false, and if not, then the $\frac 1 x$ is understood as a function into real projective line, and in that case the sentence is well-formed and false, or otherwise it is just ill-formed.

A "purely logical" answer would require the context to be pointed out exactly, so question (from strictly formal viewpoint) has similar shortcomings as the "sentence": it is not well-formed.

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By the same argument as in the answer to your other question, this statement is not well defined and hence cannot be assigned a meaningful truth value.

I consider your statement in the following way. Consider the family of statements $\mathcal{S}$ of the form $\{1/x=1\}$ in which $x$ is any real number. Assuming (as in your question) the standard definitions of division, these statements are well defined for all real numbers except $0$. Because your statement is effectively the intersection of all such elements of $\mathcal{S}$, your statement is only well defined if every statement in $\mathcal{S}$ is well defined. Because this is not the case, your statement is not well defined.

In abstract math classes, often our first question about any new mathematical concept is to ensure that it is well defined and does not have any internal contradictions.

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It depends.

If you're strict about objects having types, then if $x$ is a variable of type "real number", then no well-formed expression can involve $1/x$.

OTOH, students are implicitly taught to think in terms of partial functions, not functions. In this fashion, if $x$ is of type "real number", then $1/x$ expresses a partial function of $x$ whose domain is the non-zero reals.

In this view, $1/0$ is actually meaningful: it is the "empty" real number. (it helps to understand this if you think of a constant as being a function from the one-point set. So the "empty" constant would be the partial function whose domain is empty)

The fine details of what $1/x=1$ means can get messy at this point. Is it a partial predicate whose domain is the non-zero reals? If so, can we make $\forall x : 1/x=1$ meaningful? Or, we could consider $1/x=1$ identically false because no element of the domain satisfies it. But under this interpretation we have the messy problem that $1/x \neq 1$ is not the negation of $1/x = 1$: both are false for $x=0$.

Alas, I don't know the right way to deal with these issues.

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