Define $f\colon\mathbb R\times \mathbb R\to\mathbb R$ by $f(a,b)=a-b$. Is $f$ associative? I have 3 questions about this question.
First of all I'm confused about the notation.
My understanding is that the first part( $f\colon\mathbb R\times \mathbb R\to\mathbb R$)  means that for any combination of 2 real number $(a,b)$, it will map to one single Real number.  The second part ($f(a,b)=a-b$) means that the mapping of $(a,b)$ will be ($a-b$).  Am I correct?
Second of all, I'm confuse about what this question is even asking.  What does define mean? How do you define a function? Isn't the function already defined in the question
Lastly, I know for example that $a-b-c$ is not associative since $(a-b)-c$ is not the same thing as $a-(b-c)$, however when we only have 1 subtraction, wouldn't it be associative by default since there is only one way to add parentheses to the function (ex.// $(a-b)$).
 A: 
My understanding is that the first part( $f\colon\mathbb R\times \mathbb R\to\mathbb R$)  means that for any combination of 2 real number $(a,b)$, it will map to one single Real number.  The second part ($f(a,b)=a-b$) means that the mapping of $(a,b)$ will be ($a-b$).  Am I correct?

This is indeed correct. You could instead write the mapping as $(a,b) \mapsto (a-b)$.

Second of all, I'm confuse about what this question is even asking. What does define mean? How do you define a function? Isn't the function already defined in the question

To define a function, you need to specify a domain, codomain, and definition for the mapping (what from the domain goes where in the codomain). In this case, your domain is $\Bbb R^2$, your codomain is $\Bbb R$, and your mapping is as described (ordered pairs $(a,b)$ go to $(a-b)$).
"Defining a function" in this sense basically means to describe the function in all these ways, and often attribute it a name for quick reference, e.g. $f,g,h,$ whatever.

Lastly, I know for example that $a-b-c$ is not associative since $(a-b)-c$ is not the same thing as $a-(b-c)$, however when we only have 1 subtraction, wouldn't it be associative by default since there is only one way to add parentheses to the function (ex.// $(a-b)$).

Association is only a concern when three elements are involved. For an operation $\ast$, it is associative over the set of concern if
$$a \ast (b \ast c) = (a \ast b) \ast c$$
for all $a,b,c$ in the set of concern. This requires us to work with three elements, not simply two (as you might do in the case of showing commutativity, $a \ast b = b \ast a$).
