$({\mathbb{Q}},+)$ is not finitely generated I'm trying to prove that $G = ({\mathbb{Q}},+)$ is not finitely generated. I have come up with the following, and would like to check it is correct:

$G$ is generated by $\{1/n | n \in \mathbb{N}\}$ Take $x \in G$. Then
  $x = 1/n_1 + ... + 1/n_k$, where $k$ is a positive integer.
Then $x = (n_1+n_2+...+n_k)/(n_1*n_2*...*n_k)$ which implies that y of
  the form:
$y = (n_1+n_2+...+n_k)/(2n_1*n_2*...*n_k)$ is not a rational number,
  contradiction.

I am fully aware of the more standard approach, I would just like to know if this is correct, and if so, how I can make it more 'tidy'.
 A: HINT
You may want to assume that $G$ is finitely generated by $\{g_k\}_{k=1}^N \subset G$ with $N < \infty$ and find an example of some $x \in G$ that is not generated by these $g_k$, reaching a contradiction.
A: Why don't you use the fact that $G = ({\mathbb{Q}},+)$ is abelian group, so if you assume that $G$ is finitely generated, then $G$ is generated by a finite number of rationals, so it is cyclic which is contradictory.
We know that if $G$ is generated by finite number of rationals, then it is cyclic.
A: Suppose on contrary that a finite subset $S$ of $\mathbb{Q}$ generates the group ($\mathbb{Q}, +$). Let $P$ be the set of primes appearing in the denominators of the members of $S$. Then $P$ is finite. Take a rational number $r$ of which denominator is a prime not in $P$. Obviously, $r$ does not in the group $<S>$ generated by S. This yield a contradiction of the assumption that S generates $\mathbb{Q}$. Therefore no finite set of $\mathbb{Q}$ can generate $\mathbb{Q}$ Therefore $\mathbb{Q}$ is not finitely generated. 
