What is the difference between ∀x(p(x) →q(x)) and ∀x(p(x) ∧q(x))? Textbook(Discrete Mathematics with Applications by Susanna S. Epp) says that
$\forall x$ in $D$, $p(x)$ can be written as $\forall x$ in $D$$\to p(x)$,
$\exists x$ in $D$, $p(x)$ can be written as $\exists x$ in $D$$\land p(x)$.
but I am wonder, does it make no sense to say that 
$\forall x$ in $D$, $p(x)$ can be written as $\forall x$ in $D$$\land p(x)$,
$\exists x$ in $D$, $p(x)$ can be written as $\exists x$ in $D$$\to p(x)$?
I cannot understand why can $\forall x$ in $D$, $p(x)$ / $\exists x$ in $D$, $p(x)$ be written as  
$\forall x$ in $D$$\to p(x)$ / $\exists x$ in $D$$\land p(x)$ 
but $\forall x$ in $D$$\land p(x)$ / $\exists x$ in $D$$\to p(x)$.
 A: Saying $$\forall x(x\in D\wedge p(x))$$ is saying that all $x$ is in $D$ and has property $p$. Instead, $$\forall x(x\in D\to p(x))$$ says that if $x$ is in $D$, $x$ has property $p$. The first claim is most likely incorrect, while the second is more profound.
On the other hand, $$\exists x(x\in D\to p(x))$$ is almost meaningless, because it is saying that there is at least one $x$ that either is not in $D$, or is in $D$ and has property $p$. Rather,
$$\exists x(x\in D\wedge p(x))$$ says that there is a $x$ that is both in $D$ and has property $p$, a stronger statement.
Do you see the differences between $\exists$ and $\forall$?
A: Let's take a simple concrete example:
Let the domain of discourse (i.e. the set of objects that the quantifiers range over) be the natural numbers
Let $E$ be the set of even numbers, $O$ the set of odd numbers, and $P(x)$ be the property of being twice some natural number, which we can express as $\exists y \ x = 2*y$
Now compare and contrast the following two statements:
$$\forall x (x \in E \rightarrow \exists y \ x = 2*y)$$
This statement is saying that 'If you are an even number, then you are twice some natural number' ... which we know is true for the natural numbers
On the other hand:
$$\forall x (x \in E \land \exists y \ x = 2*y)$$
ends up saying: 'All natural numbers are even and are equal to twice some natural number' ... which of course is false!
So, we see the two statements have different truth-values, and are therefore different statements.  OK, so which one is the right translation of '$\forall x \in E: P(x)$'?  It is the first one: "For all even numbers, it is true that they are twice some natural number"
Now let's look at the existentials. 
$$\exists x (x \in O \land \exists y \ x = 2*y)$$
This statement is saying that some odd numbers are twice some natural number .. False!
However:
$$\exists x (x \in O \rightarrow \exists y \ x = 2*y)$$
ends up saying: "there is a natural number such that if it is odd, then it is twice some natural number' ... which is not only a very unnatural statement .. but is in fact true:  you can pick the number $2$, which is part of the domain of discourse, and for the number two it is false that it is odd, meaning that the conditional 'if it is odd, then it is twice some natural number' becomes true (any conditional with a false antecedent is true).
So again, we see the two statements have different truth-values, so they are different. And which one is actually correctly capturing '$\exists x \in O: P(x)$'?  It should be clear that it is the first: 'There is some odd number that is twice some natural number' (which is indeed a false statement)
