# Closed under an operation which is not well-defined?

Consider the operation defined on $\mathbb Q$ as follows:

$a*b=\frac{\mathrm {num}(a)}{\mathrm {den} (b)}$, where $\mathrm {num}(a)$ and $\mathrm {den}(b)$ are numerator of $a$ and denominator of $b$, respectively.

This operation is of course not well-defined, such as the case of $a=\frac 13 = \frac 26, b = \frac 12 =\frac 36$, it is obvious that the two different representations of $a$ and $b$ gives different outcomes.

However, the operation is closed because it is the quotient of two integers, where $\mathrm {den}(b)$ is not zero, right?

I am not sure that I can say this operation is closed since I heard that we cannot discuss closedness of operation which is not well-defined.

How should I say about the closedness of this operation? Closed? or Can't discuss?

• I would say you can't discuss anything about an operation that is ill-defined. However there is nothing unusual about an ill-defined operator being closed. If the output is rational (and what else would it be) then it is closed. unless one is to argue that the function has not output or has output of set classes. Which I won't argue against. Commented Feb 27, 2017 at 2:20
• Sometimes there is an assumption of irreducibility of the fractions as in the case of the Farey Series. I'm not saying this is your case, but you could make the assumption and see what happens. Commented Feb 27, 2017 at 2:24
• @fleablood I also agree, but you mean I can say either, right? Commented Feb 27, 2017 at 2:24
• @FabioSomenzi There is no such assumption. Commented Feb 27, 2017 at 2:24
• I'm saying the discussion is not worth having. @FabioSomenzi such an assumption would be a very simple thing and it'd make the operation quite well defined. Commented Feb 27, 2017 at 2:30

Let $a=\frac{n_a}{d_a}$ with $\gcd(n_a,d_a)=1$ and $b=\frac{n_b}{d_b}$ with $\gcd(n_b,d_b)=1$

$a\star b=\{\frac{n\times n_a}{m\times d_b},(m,n)\in\mathbb N^{*2}\}$

By this definition it is a well defined operation from $\mathbb Q^2\to \mathcal P(\mathbb Q)\quad$ but

for $a\neq 0$, $n=p\,d_b\in\mathbb N$ and $m=q\,n_a\in\mathbb N$ then $\frac{n\times n_a}{m\times d_b}=\frac{p}{q}$ for arbitrary $(p,q)\in\mathbb N^{*2}$.

So, if you don't work on irreductible fractions, not only is the operation ill-defined as an operation from $\mathbb Q^2\to\mathbb Q$ but also completely useless as a set operation from $\mathbb Q^2\to \mathcal P(\mathbb Q)$ since it has only two values : $a\star b=\mathbb Q^*$ and $\{0\}$.

Given that, I think it's better to define $a\star b=\frac{n_a}{d_b}$ as a closed operation from $\mathbb Q^2\to\mathbb Q$.

Note: you operation is also closed from $\mathcal P(\mathbb Q)^2\to\mathcal P(\mathbb Q)$, this times it has $3$ values $\{0\},\mathbb Q^*,\mathbb Q$ depending wether the subset $a$ is equal to/contains $0$ or not, which is not very interesting either...