I wanted to ask whether I did the task correctly:
Task: Let M be a set satisfying the following property P: ”For every element x of M there is an element y of M such that x + y is an element of M.” Express the statement P using logical quantor, negate P and find a set which fulfills P and one that does not.
My solution:
$$M := \{x: \exists y\in M \ \ \text{such that} \ \ x+y∈M\}$$
$$P: ∀x\in M ∃y\in M(x+y \in M)$$
Negation of $P$: $∃x∈M∀y∈M(x+y∉M)$
A set that would fulfill such a property would be ℕ.
What would be a set of the important number sets that do not fulfill the property above?
Is it correct what I did?
Aha! Now the situation is much clearer; thanks for that edit.
Now, with regards to your answer... For a normal mathematician, your translation of the statement $P$ with logical symbols would be correct. However, the pedantic mathematician out there (or the logician) would argue that you could express $P$ in the following way. You consider a formal language consisting of a binary function symbol $+$ and a unary predicate (also called relation) symbol $M$; for a variable $x$, $M(x)$ would mean that "$x$ is in the set $M$". Then this same pedantic mathematician would argue that a correct translation of the statement $P$ to logical symbols (in the language $\{+, R\}$) would be given by $$\forall x \exists y (M(x) \to (M(y) \wedge M(x+y))).$$ If these concepts sound too alien and weird, you can safely ignore them; I'm assuming your professor wants to see the answer as the one you've given. Also, the negation of the statement is correct.
Finally, sets of numbers which would not satisfy such a property are given by singletons, for example. Indeed, consider the case of $M = \{1\}$; then clearly $M$ doesn't satisfy $P$, since $1+1 = 2 \notin M$. You can also consider finite sets like $\{1, 2, \dots, 665, 666\}$ or $\{\pi, \pi^2, \pi^3\}$ and these would not satisfy $P$ if we interpret $+$ as sum of integers in the first case or sum of real numbers in the second. However, if we consider the set $\{0, 1\}$ and we interpret $+$ as addition of elements modulo $2$, then $\{0, 1\}$ would indeed satisfy $P$.