A proof that $\mathbb{Q}^+$ is not finitely generated. Is it correct? 
Is $\mathbb{Q}^+$ finitely generated? Justify your answer.

This is an abstract algebra exercise I've tried to solve, but I don't know whether my answer (proof) is right or not.


Proof 
    So, let us suppose that $\mathbb{Q}^+$ is finitely generated and $\mathbb{Q}^+=\langle q_1,\dots , q_n \rangle$ , where $q_1, \dots , q_n$ generators and $n\in \mathbb{N}$. The "mean value" of $q_1 \dots , q_n$ is in $\mathbb{Q}$. Thus there exist $k_1, \dots , k_n$ such that $$\frac{q_1+\dots +q_n}{n}=k_1q_1+\dots +k_nq_n$$ or $$\left(k_1-\frac{1}{n} \right)q_1 +\dots +\left(k_n-\frac{1}{n}\right)q_n=0 \in \mathbb{Q}^+$$ Now, every $\left(k_i-\frac{1}{n} \right)$, where $i=1,\dots,n$ is not an integer (something we would have wanted), as $\frac{1}{n} \notin \mathbb{N}$ and that leads us to contradiction. Thus, $\mathbb{Q}^+$ is not finitely generated.


But, here comes my question. Are we sure that any rational number can be written uniquely as a word in $q_1,\dots,q_n$? Where are the possible mistakes in my proof? Thank you in advance?
 A: No, there is no guarantee that the representation is unique. The mean of a sequence of rational numbers can even be an element of the sequence, so nothing is precluding the mean from being expressible as an integer linear combination of generators, just with different coefficients than the ones you give. In a general group, there's no guarantee that an expression of an element in terms of generators is unique, even if it is a minimal generating set.
The way to show it is not finitely generated is to look at the denominators. Every element expressible in terms of a finite number of elements has a denominator that divides the product of all of the denominators of the generators, hence if we pick a prime that does not divide that product as a denominator then the element is not expressible in terms of the generators.
A: As mentioned, you have not justified that every rational number is a unique $\mathbb{Z}$-combination. However, you may assume WLOG that $q_1, \dots, q_n$ are independent over $\mathbb{Z}$ by pruning dependencies if this is not the case. Multiplying by $n$ gives you $nk_i - 1 = 0$ for all $i$ now that you have assumed independency. This means $k_i = 1/n,$ contradiction. You also forgot to dispose of the trivial case $n=1.$
