Consider the case:
Let:
- S(x) = “x is a student”
- F(x) = “x is a faculty member”
- A(x, y) = “x has asked y a question”
- Dx and Dy = Consists of all people associated with your school.
Use quantifiers to express this statement:
Some student has never been asked a question by a faculty member.
Attempted Solution:
Exist x, All y ( S(x) AND F(y) AND (NOT A(y, x)) )
Book Solution:
Exist x ( S(x) AND All y ( F(y) -> NOT A(y, x) ) )
What is the point of using implication here? Would my answer also be correct?
Let's make it slightly simpler. Fix an $x$ and assume that $x$ is a student. We want to say that $x$ has never been asked a question by a faculty member. The relevant part of what you wrote is $\forall y\, (F(y) \mathbin{\And} \neg A(y,x))$. However, this does not express the desired property of $x$. The problem is that it asserts that everyone is a faculty member (even the janitors!)
To express "for all faculty members $y$,..." you should instead write "for all $y$, if $y$ is a faculty member, then...." Admittedly, this is awkward English because first it asks us to consider all $y$ in the domain of discourse (even the janitors,) and then it becomes clear that we only care about faculty members $y$; that is, we only care about $y$ if $y$ is a faculty member. This is where the if comes in.
So to express "$x$ has never been asked a question by a faculty member" we would write
(1) $\forall y\,(F(y) \implies \neg A(y,x))$.
This formula (1) translates to the awkward but correct English phrase "for all $y$, if $y$ is a faculty member, then $y$ has not asked $x$ a question." Slightly less formally, and abusing notation by using $F$ for the set of faculty members rather than for the predicate "is a faculty member," we could write
(2) $\forall y \in F\; \neg A(y,x)$.
This formula (2) translates to the less awkward English phrase "for every faculty member $y$, $y$ has not asked $x$ a question." However, in formal logic (2) is just defined to be an abbreviation for (1), so at its root the implication is still there.