$({\mathbb{Q}},+)$ is not finitely generated

Question

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'.

Answer 1

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.

Answer 2

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.