Show that $(\Bbb Z, +)$ and $(\Bbb R, +)$ are not isomorphic groups.
What I did so far:
To show that these groups are not isomorphic, we need first assume that, there exists an isomorphism $f$ from $(\Bbb Z, +)$ onto $(\Bbb R, +)$. Then,$f(a^n)=f(\underbrace {a.a...a}_{\text{n times}})$(,where $.$ is an arbitary binary operation. But $+$ is the binary operation given, so $f(1^n)=f(\underbrace{1+1+\cdots+1}_{\text{n times}})=f(n)=\underbrace {f(1)+\cdots+f(1)}_{\text{n times}}=nf(1)$. Thus, we have, $f(n)=nf(1)$. So, $f$ is completely known if $f(1)$ is known. If $f(1)=r\in\Bbb R$, then, $\forall x\in\Bbb R$,$\exists n\in \Bbb Z$, such that, $f(n)=rn=x$.
But I dont know how to arrive at a contradiction from here. My general strategies include : trying to show that if one group has an element of a particular order then the other group does not have a corresponding element of that particular order, showing that we can't find a pre-image of the "image" group. But I don’t know what to do here? Also, I am still learning elementary group theory and I need an "elementary" solution for the same(, without using cardinality of sets, as I still don'tknow about countability,uncountablity and other related topics).
Although , the following thread Is $(\mathbb{Z}, +)$ isomorphic to $(\mathbb{R}, +)$ is about this same topic, but the answers there weren't suitable for me, because I couldn't understand the 1st answer , furthermore, it addresses a specific doubt of the user (if I am not mistaken). As for the 2nd answer, I am not quite fond of it . Actually,I was looking for the particular type of answer provided in the comments by lulu.