Closed under an operation which is not well-defined?

Question

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?

Answer 1

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...