Mathematical Difference between "there is one" and "there is EXACTLY one" I know that I can say ∃x(P(x)) which means there is at least one x for P(x), but how do I express for exactly one?

Here's the questions:

(a) Not everyone in your class has an internet connection.
(b) Everyone except one student in your class has an internet connection.

So for the first one I wrote:

(a) ∀x∃x(¬I(x))

"For all x there exists an x (or more) such that an x does not have an internet connection" (where I is the state of having an internet connection)

(b) Don't know how to express

I could be wrong please correct me since i'm pretty new to expressing this all mathematically

Thanks for help
You are correct that "There exists ..." means that there exists at least one. To say that there is exactly one you need to say the following:

$$\exists x(\varphi(x)\land\forall z(\varphi(z)\rightarrow z=x)).$$

Namely, there exists $x$ satisfying whatever, and whenever $z$ satisfies whatever, $z$ has to be equal to $x$. Also note the scope of the existential quantifier is over the entire statement.

(As a mathematical example, There exists a natural number which is larger than $1$; but there exists exactly one natural number which is smaller than $1$ (here we take $0$ to be a natural number))
Umm... when I studied math logic back there in my uni, we had exact quantification, expressed as $∃!$

So, let $I(x)$ === "x has internet connection", $P$ === "pupils". Then, a) is represented as

$$
\exists x(x \in P \land \neg I(x))
$$

and b) is represented as
$$
\exists!x(x \in P \land \neg I(x))
$$

Difference between them is only in exactness of second statement: it's not just "not everyone", it's "exactly one" who does not have internet.

We, however, used slightly different notation:

$$
x \in P
$$
$$
\exists ! x : I(x) = false
$$

which, arguably, reads more naturally.