Difference due to scope of quantifiers I'm trying to understand the difference between the scopes of quantifiers in the following statements and how that impacts taking negations and contrapositives.
Statement 1: $\quad \exists M \in \mathbb{R}^+ \quad \forall x\in \mathbb{R}^+ \quad \big(x\geq M \implies f(x)>0 \big).$
I believe I know how to take the negation and contrapositive of statement 1:

*

*Negation: $\quad \forall M \in \mathbb{R}^+ \quad \exists x\in \mathbb{R}^+ \quad \big(x\geq M \land f(x)\leq0 \big).$

*Contrapositive: $\quad \exists M \in \mathbb{R}^+ \quad \forall x\in \mathbb{R}^+ \quad \big(f(x)\leq0 \implies x<M \big).$
Statement 2: $\quad \big(\exists M \in \mathbb{R}^+ \quad \forall x\in \mathbb{R}^+ \quad x\geq M\big) \implies f(x)>0 .$
Given statement 2 as above, my questions are:

*

*Is there a difference between statement 2 and statement 1, and if so, what is it? An intuitive approach would be appreciated if possible.

*If there is a difference, how does that impact the negation and contrapositive of statement 2?

*Finally, if there are any mistakes in my negation and contrapositive of statememt 1, I would appreciate corrections.

 A: Statement 1 doesn't have a contrapositive.
Only conditionals have contrapositives, and 1 isn't a conditional.
More carefully, in a formal context, only formulae whose main logical operator is the conditional (i.e. which are instances of the schema $A \to B$) have contrapositives (the corresponding instance of $\neg B \to \neg A$). And the main operator of 1 is the initial quantifier, so the notion of contraposition doesn't directly apply.
However, if you have a quantified formula $Q_1Q_2\ldots Q_n(A \to B)$ then that will be equivalent to what you get by contraposing $(A \to B)$, i.e. to $Q_1Q_2\ldots Q_n(\neg B \to \neg A)$ but of course leaving the quantifier prefix untouched!
Statement 2 is simply ill-formed in any sensible syntax.
A: 
Statement 1: $\quad \exists M \in \mathbb{R}^+ \quad \forall x\in \mathbb{R}^+ \quad \big(x\geq M \implies f(x)>0 \big).$
Statement 2: $\quad \big(\exists M \in \mathbb{R}^+ \quad \forall x\in \mathbb{R}^+ \quad x\geq M\big) \implies f(x)>0 .$

*

*Is there a difference between statement 2 and statement 1, and if so, what is it? An intuitive approach would be appreciated if
possible.


Good observation that Formulae 1 & 2 have different negations and different contrapositives, and thus different meanings!
Side note: Formula 2 is more clearly written as $$\quad \big(\exists M \in \mathbb{R}^+ \quad \forall y\in \mathbb{R}^+ \quad y\geq M\big) \implies f(x)>0$$ (notice that its variable $x$ is actually free and thus requires context), so is called an open formula instead of a statement.
Now, consider the logic formula \begin{gather}\Big(∀x\:Px\Big)\to Qy \quad↔\quad ∃x\:\Big(Px\to Qy\Big).\tag1\end{gather} If $(1)$'s RHS is false, then $∀x\:\Big(Px\land \lnot Qy\Big),$ so $Px$ is universally true and $\lnot Qy$ true, so $Px$ is universally true while $Qy$ false, so $(1)$'s LHS is false; by contrapositive and since we have been abstractly inferring, $(1)$'s LHS logically implies $(1)$'s RHS. On the other hand, if $(1)$'s LHS is false, then $Px$ is universally true and $Qy$ false, so $(Px\land \lnot Qy)$ is universally true, so $(Px\to Qy)$ is universally false, so $(1)$'s RHS is false; by contrapositive and since we have been abstractly inferring, $(1)$'s RHS logically implies $(1)$'s LHS. Hence, $(1)$'s LHS and RHS are logically equivalent, that is, \begin{gather}\Big(∀x\:Px\Big)\to Qy \quad\equiv\quad ∃x\:\Big(Px\to Qy\Big).\end{gather} Similarly, \begin{gather}\Big(∃x\:Px\Big)\to Qy \quad\equiv\quad ∀x\:\Big(Px\to Qy\Big).\end{gather}



*If there is a difference, how does that impact the negation and contrapositive of statement 2?

*Finally, if there are any mistakes in my negation and contrapositive of statement 1, I would appreciate corrections.


Your negation and contrapositive of Formula $(1)$ are both correct; just do the same for Formula $(2)$ (the procedure is more straightforward as the main connective ‘${\Rightarrow}$’ is not being quantified).
