What happens when a set has elements that are impossible? eg, $\left\{\frac{m}{n} \mid m \in \Bbb{Z} \wedge n \in \Bbb{Z}\right\}$, which allows $n=0$ I am currently learning naive set theory, and we learned that the set of all rational numbers is defined as:
$$
\mathbb Q = \left\{\frac{m}{n} \mid m \in \Bbb{Z} \wedge n \in \Bbb{Z} \wedge n \neq 0\right\}
$$
I wonder what happens when we allow n to be equal to $0$:
$$
\mathbb Q' = \left\{\frac{m}{n} \mid m \in \Bbb{Z} \wedge n \in \Bbb{Z}\right\}
$$
Are all the impossible numbers not in the set? are they in the set but they are not numbers? is it not a set? or is this some sort of not well defined set?
Thanks!
 A: The arithmetic expression "$1/0$" makes no sense. It's not a number, and not an "impossible number". It's as nonsensical as "$2 + + 2$".
So your second definition of the rational numbers includes a mathematical expression that makes no sense. That means the definition as a whole makes no sense.
If you see it somewhere it is probably intended to represent the rational numbers as a subset of the real numbers, but the person who wrote it down forgot to specify that the denominator can't be $0$.
A: $\newcommand{\Reals}{\mathbf{R}}$Strictly speaking, that definition of the rationals itself is informal. "Set-builder" notation specifies subsets of an existing set $U$ whose elements $x$ satisfy (or not) a condition $P(x)$.
If we already know about the set of real numbers $\Reals$, for example, we can define the set of rationals to be
$$
\{x \in \Reals : \text{$x = \frac{m}{n}$ for some integers $m$ and $n$ with $n \neq 0$}\}.
$$
If not, we could define a rational number to be an equivalence class of ordered pairs $(m, n)$ of integers with $n \neq 0$, in which $(m, n) \sim (m', n')$ if and only if $mn' = nm'$, or some such.
Writing a set by "constructing" its elements can be a useful shorthand informally, but it's sloppy if logical rigor is the aim.
