One of the questions in my exercise asks:
We have made several logical connectives using the propositions p and q. For instance p ∧ q, p ∨ q and p ↑ q . Can this go on indefinitely; that is, making an unlimited number of connectives with p and q ? Or is there a finite number of distinct connectives? If so, can you determine how many there can be?
I was trying to come up with an answer but got stuck. Does it mean an unlimited number of connectives with p and q such that $$p \land q$$ $$p \land (p \land q)$$ $$p \land (p \land (p \land q))$$
In that way you definitely can keep going forever, or for another example: $$ \neg p $$ $$ \neg (\neg p)$$ $$ \neg (\neg (\neg p))$$
So does it mean
Can you create infinite connectives using the same operation?
or
How many connectives can be made using unique operations?
Also, for the last question
If so, can you determine how many there can be?
Wouldn't it be the number of conjunctions available? (That mean something different to each other)
I would really appreciate some help and a better explanation of the question at hand, maybe even hints so that I can complete it on my own.
Thank you in advance.
It makes a difference whether you talk about syntactically or semantically distinct connectives.
Syntactically, $$p \land q$$ $$p \land (p \land q)$$ $$p \land (p \land (p \land q))$$
are different: The formulas obviously have a different form, they have a different number of atoms and connectives and thus they can syntactically not be identical.
Now you could make up a connective defined as $p \land q$ and one that is shorthand for $p \land (p \land q)$ and have two connectives. By their form, they are indeed distinct entities, so yes, there is theoretically an infinite number of connectives. This observation is already given by simple recursion (as you exemplified).
Semantically, however, they are equivalent: All of the formulas yield the same truth conditions, namely $1$ if both $p$ and $p$ are true and $0$ else, so you can not get any new logical meaning by creating $p \land (p \land q)$ because the truth value of this statement is already defined by another connective (namely the simple $\land$).
If you understand a connective by its semantics, i.e. the meaning of a connective is exactly the conditions under which the formla becomes true and "a connective" corresponds to "a certain permutation of truth values in a truth table" (when you write the formulas out in a truth table, the last column would always look the same for any of the connectives you used), then the number of possible connectives is limited, since there are only finitely many possible permutations of truth values for a given number of atoms there are and the number of truth values a statement can have.
More precisely, the number of possible semantically distinct $n$-ary connectives in an $m$ valued logic is given by $m^{m^{n}}$, so in the case of a classical two-valued logic and binary connectives, there are $2^{2^{2}} = 16$ possible semantically distinct connectives.
For unary connectives in a two-valued logic, (like negation), there are $2^{2^{1}} = 4$ different possibilities to assign a truth value to the combination of connective + variable, and so on.
Hence, the number of syntactically differently defined connectives is theoretically infinite, but if you understand a distinct connective by its semantics, then the number of possible connectives is restricted; in the case of binary connectives for a two-valued logic to 16.
I assume the textbook refers to semantically distinct connectives because any logical element is normally defined by its truth conditions, thus the answer would be yes, there is a finite number of distinct connectives, given by $m^{m^{n}}$ for $n$-ary connectives in an $m$-valued logic.