Is it OK if a proposition contains $\frac{1}{0}$?

Question

I think the following statement is accepted as a proposition.

$$\forall x\in \mathbb{Q}\left(x\neq 0\Rightarrow \frac{1}{x}\in\mathbb{Q}\right).$$

This means

$$\bigwedge_{x\in \mathbb{Q}}\left(x\neq 0\Rightarrow \frac{1}{x}\in\mathbb{Q}\right).$$

The conjunction includes $$0\neq 0\Rightarrow \frac{1}{0}\in\mathbb{Q}.$$

This sentence appears to be gramarically incorrect because $\frac{1}{0}$ is undefined.

Is $0\neq 0\Rightarrow \frac{1}{0}\in\mathbb{Q}$ a proposition?

If so, is something like $0\neq 0 \Rightarrow \frac{)+>1/}{=\in \subset}\mathbb{Q}$ also a proposition?

Answer 1

As with many questions in logic, the answer to this question is "it depends on how you set things up".

Most questions in math don't depend on the precise details of your foundational system, but the closer to logic you get the more you start having to deal with exactly what you're working with under the hood. In particular, for this question, it's important to ask what the fraction symbol $\frac{\cdot}{\cdot}$ is representing${}^1$! The usual setup for model theory doesn't allow partial functions, so we can't say $\frac{\cdot}{\cdot} : \mathbb{Q} \times (\mathbb{Q} \setminus 0) \to \mathbb{Q}$ is undefined when the denominator is $0$. How do we get around this?

The typical trick is to relationalize things. We add a new ternary relation $\text{"}\frac{x}{y} = z\text{"} \subseteq \mathbb{Q}^3$ so that the proposition "$\frac{x}{y} = z$" is given the value "true" exactly when, well... the relevant equation is true, haha.

There are other options too, for instance some authors${}^2$ really want $\frac{\cdot}{\cdot}$ to be a function, so they're willing to define $\frac{x}{y}$ to be $0$ when $y = 0$! Doing this renders $\frac{\cdot}{\cdot}$ a total operation, so that model theory can talk about it, but it's fairly nonstandard.

Yet a third option is to work with a multi-sorted logic, which explicitly has a sort for $\mathbb{Q}$, a separate sort for $\mathbb{Q}^\times = \mathbb{Q} \setminus \{ 0 \}$, and a function symbol $\frac{\cdot}{\cdot} : \mathbb{Q} \times \mathbb{Q}^\times \to \mathbb{Q}$.

Let's take a look at these three options.

In case we have a ternary relation, then the formula "$\frac{1}{0} \in \mathbb{Q}$" is actually an abbreviation! It's shorthand for

$$ \exists z . \left ( \text{"}\frac{1}{0} = z\text{"} \right ) \land \left ( z \in \mathbb{Q} \right ) $$

This is obviously a well formed expression, where the proposition "$\frac{1}{0} = z$" happens to be false for all choices of $z$, so that this formula is false!

In case we flippantly define $\frac{1}{0} = 0$, then, of course, we have no issues at all (but again, most authors go for the ternary relation approach).

Lastly, in case we work with a multisorted logic, $\frac{1}{0}$ is not a well formed formula (since proper sorting is baked into the definition of a "term" in the logic). So the entire formula $\frac{1}{0} \in \mathbb{Q}$ is itself not well formed.

Obviously when we work with fractions in our day to day life, the precise formalization of $\frac{\cdot}{\cdot} = \cdot$ as a ternary relation symbol, or something given a weird "default value" of $0$, etc. doesn't matter at all. But when you start asking questions like "is this a well formed formula" the answer sort of has to depend on the specifics of however you choose to formalize such things.

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I hope this helps ^_^

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${}^1$: As a cute aside, I'm pretty sure this is where the division symbol $\div$ comes from! It's a fraction with two $\cdot$s as placeholders for the arguments!

${}^2$: See, for example, chapter 1 exercise 7 in Hodge's A Shorter Model Theory

Answer 2

Let's first ask, what is a fraction from the perspective of classical algebra and ring theory.

We pick a ring $R$ (a set which has addition, substraction, multiplication; think about integers, $\mathbb{Z}$). Then we pick a multiplicative subset $S$ in $R$ (a subset which is closed under multiplication and contains $1$; think about all non-zero integers, $\mathbb{Z}-\{0\}$). Then we consider all the pairs $(r,s)\in R\times S$, which we tend to denote as $\frac{r}{s}$ after we announce the identification of $(r_1,s_1)$ with $(r_2,s_2)$ on $R\times S$ by the same logic as we believe $\frac{r_1}{s_1}=\frac{r_2}{s_2}$ to be the same if and only if $r_1s_2=r_2s_1$ (in many cases, including $R=\mathbb{Z}$, but not always). The point of defining this identification is to get a set of rational numbers, $\mathbb{Q}$.

As we see, from this perspective $\frac10$ doesn't have an established definition yet because $0\notin S=\mathbb{Z}-\{0\}$. So instead of "$\frac10$ is undefined", I prefer "$\frac10$ is not defined yet, but we can always agree on a definition and proceed, as long as it doesn't contradict with the mathematics built so far".

Answer 3

If you wish to simultaneously be extremely rigorous and practical, a way out is simply define $$\frac x 0 = 0$$ because division is just another operation and we can define it as whatever we want, as long as it is convenient. All the usual theorems about division still holds because they all include a premise "the denominator is not zero", and so whatever we define $x/0$, those theorems will not be affected.

An additional way is to modify the syntax for division. Usually we take $x/y$ to be a notation for something like $\textrm{divide}(x, y)$. But instead of that we can do $\textrm{divide}(x, y, p)$, where $p$ is a proof that $y \ne 0$. And we just omit the proof and write $x/y$ in our usual notation. Omitting stuff in notation is extremely common. We write $\mathbb R$ instead of $(\mathbb R, +, 0)$ for the group of real numbers. So omitting $p$ is also quite justifiable.

Answer 4

is something like $$0\neq 0 \Rightarrow \frac{)+>1/}{=\in \subset}\mathbb{Q}\tag1$$ also a proposition?

The string of symbols $(1)$ is well formed according to first-order logic's syntactic rules, so it arguably is a first-order-logic proposition (in the way that $\text“1+g(1)=4\iff g(1)=g(1)\text”$ is generally considered a proposition).

It is meaningful under no common mathematical interpretation, so it is not in general a mathematical statement. (However, it is a statement in some possible interpretation.)

Is $$0\neq 0\Rightarrow \frac{1}{0}\in\mathbb{Q}\tag2$$ a proposition?

Similarly, string $(2)$ is arguably an FOL proposition, but in a given interpretation if $\frac10$ is not defined then $(2)$ is not technically a statement.

Outside of formal logic, I think it is acceptable to call $\forall {\in}\mathbb Q\,\big(x\neq 0\Rightarrow \frac{1}{x}\in\mathbb{Q}\big)$ a true statement.

Answer 5

Let's simplify the problem. Suppose we have at least $0\in X$ and $1\in X$, and $f$ such that:

$~~~~~~\forall a:[a\in X \land a\neq 0 \implies f(a)=1]$

This definition of function $f$ does not give us a truth value for $f(0)=0$. Likewise for $f(0)=1$.

We say that this definition leaves $f(0)$ undefined. A subsequent extension of $f$ may assign a value to $f(0)$, so we cannot say the $f(0)$ is somehow "grammatically incorrect."

Back to the original problem. The definition of division on $Q$ leaves $1/0$ undefined. It cannot give us a truth value for the expression $1/0 \in Q$. The implication $0\neq 0 \to 1/0 \in Q$, however, is actually (vacuously) true since the antecedent $0\neq 0$ is false. What is meant by vacuously true?

The truth table for $P \to Q$

Note that when the antecedent $P$ is false (lines 3-4), the implication $P\to Q$ is true regardless of the truth value of the consequent $Q$. This is the so-called principle of vacuous truth.