How do I interpret the formating of these given structures and prove they are not isomorphic? $(\Bbb{Z}^+,\cdot)$ and $(\Bbb{Q}^+,\cdot)$
First and foremost, I am not looking for a direct answer just a method of interpreting what these "structures" are, and some techniques I can use to prove they are not isomorphic. I have recently started Abstract Algebra I and keep hearing Cardinality and the form $f(ab)=f(a)\cdot f(b)$ while looking throughout the internet, but I haven't heard this covered in class.
An attempt at interpretation:
I assume the first structure is a set of all number that can be represented as a multiplication of two positive integers (meaning all positive integers), and the second is the same with positive rational numbers (meaning all positive rational numbers).
An attempt at proving they are not isomorphic:
Because isomorphic means "to have the same shape" I assume this means that the two sets need to cover the same range of numbers. Because the second set can output non-integer values while the first set can't this would prove their isomorphic.
If I am misinterpreting what isomorphic means or if there is a better way to go about proving it I would greatly appreciate any input. Also, I'd like to apologize if I messed up the tags or formatting, I tried to follow the guidelines to the best of my ability; a first-time poster.
 A: To say that two structures are or are not isomorphic, you need to specify what you are saying they are(n't) isomorphic as. In this case, asking about isomorphism of monoids would be reasonable, since both $\mathbb{Z}^+$ and $\mathbb{Q}^+$ are monoids under multiplication.
Fortunately in this case, any reasonable notion of homomorphism $(\mathbb{Q}^+, {\cdot}) \to (\mathbb{Z}^+, {\cdot})$ must satisfy $f(ab) = f(a)f(b)$ for all $a,b \in \mathbb{Q}^+$. We can prove that no such homomorphism is an isomorphism (be it of monoids or otherwise), since it is not even bijective.
Indeed, first note that we must have
$$f(1) = f(1 \cdot 1) = f(1) \cdot f(1) \quad \Rightarrow \quad f(1) = 1$$
We were able to cancel the $f(1)$ since $f(1) > 0$. But then we also have
$$1 = f(1) = f\left( \frac{1}{2} \cdot 2 \right) = f\left(\frac{1}{2}\right) \cdot f(2)$$
and hence $f(2) = 1$ since that is the only positive integral factor of $1$.
But then $f(1)=f(2)$, so $f$ is not a bijection, so $f$ is not an isomorphism.

Footnote: what we actually proved is that $\mathbb{Z}^+$ and $\mathbb{Q}^+$ are not isomorphic as magmas under multiplication. Since every monoid, or even semigroup, has an underlying magma, it follows that they are not isomorphic as semigroups or as magmas.
