How to know when to use $P(x)\wedge Q(x)$ and when to use $P(x)\to Q(x)$? When translating English phrases to mathematical statements using logical quantifiers, I find that I'm having trouble knowing the difference between $P(x)\wedge Q(x)$ and $P(x)\to Q(x)$.
For example:
Translate the following sentences using logical quantifiers:


*

*All rationals are real

*No rationals are real

*Some rationals are real

*Some reals are rational


My notes say the following:


*

*$Q(x)$: $x$ is rational

*$R(x)$: $x$ is real


*

*All rationals are real: $(\forall x)(Q(x)\to R(x))$

*No rationals re rel: $(\forall x)(Q(x)\to\neg R(x))$

*Some rationals are real: $(\exists x)(Q(x)\wedge R(x))$

*Some reals are rational: $(\exists x)(R(x)\wedge Q(x))$



So I have to wonder: Why, for number 1 and 2 is it not right to say $(\forall x)(Q(x)\wedge R(x))$? I've gotten many questions like this wrong because I said $A\to B$ when it was supposed to be $A\wedge B$.
 A: Note that it is typically the case that $\forall$ is accompanied by $\rightarrow$, and $\exists$ by $\land$.
For all $x$, if $x$ is $P$ then $x$ is $Q$.
Exists some $x$ such that $x$ is $P$ and $x$ is $Q$.
Exercise:
Negate $\forall x(P(x) \rightarrow Q(x))$ 


*

*$\lnot \forall x(P(x) \rightarrow Q(x)) \equiv \exists x\lnot(P(x) \rightarrow Q(x)) \equiv \exists x \lnot(\lnot P(x) \lor Q(x)) \equiv \exists x (P(x) \land \lnot Q(x))$


Negate $\exists x(P(x) \land Q(x))$


*

*$\lnot \exists x(P(x) \land Q(x)) \equiv \forall x \lnot(P(x) \land Q(x)), \equiv \forall x (\lnot P(x) \lor \lnot Q(x)) \equiv \forall x(P(x) \rightarrow Q(x))$


Moving the negation inward, and see the form you arrive at in each case.
A: Suppose that $P(x)$ is something like "$x$ is rational". When you say "for all rational $x$, $Q(x)$ holds", what you're saying is "for all $x$, if $x$ is rational then $Q(x)$". That is:
$$\forall x(P(x) \to Q(x))$$
When you say "there exists a rational $x$ such that $Q(x)$", what you're saying is "there exists an $x$, which is rational and also $Q(x)$". That is:
$$\exists x(P(x) \wedge Q(x))$$
So the general rule is:


*

*$\forall$ uses $\to$

*$\exists$ uses $\wedge$


There's a more fundamental reason for this. $A \to B$ is the same as $\neg A \vee B$, and $\exists$ is the same as $\neg \forall \neg$. ("There exists a red car" is the same as saying "it's not the case that all cars aren't red".) So...
$$\begin{align}
&\exists x (P(x) \wedge Q(x))\\
\text{is equivalent to}\ &\neg \forall x \neg (P(x) \wedge Q(x))\\
\text{is equivalent to}\ &\neg \forall x (\neg P(x) \vee \neg Q(x))\\
\text{is equivalent to}\ &\neg \forall x (P(x) \to \neg Q(x))\end{align}$$
So really the $\wedge$ in the $\exists$ case does, in some sense, 'come from' a $\to$. The converse also holds.
