Set Builder Notation Negation

Question

How would you negate the following set? I am confused if you leave a as element of the set of real numbers or if you would put a there exists sign infront of it.

Answer 1

The negation would be all of the real numbers $a$, for which the property $$ \exists p\in\mathbb{Z} \ \exists q\in\mathbb{Z}^{\not= 0}:a=\frac{p}{q} $$ does not hold. I.e., the set $$ A=\left\{a\in\mathbb{R}\mid\forall p\in\mathbb{Z} \ \forall q\in\mathbb{Z^{\not= 0}}: a\not=\frac{p}{q}\right\}. $$

Answer 2

$q \neq 0$ is there so that $a$ is defined. Negation is just the set of real numbers which cannot be written as $p/q.$

Answer 3

You can only negate logical propositions, not sets. And the negation of the above proposition is just $Q\ne \{ a\in R \cdots$

In this case, either $Q$ would contain at least one element that is not a rational number, or be missing at least one rational number. It could even be empty.