Show that $({\mathbb{Q}},+)$ is not finitely generated using the Fundamental Theorem of Finitely Generated Abelian Groups.

If $\mathbb Q$ wants to be finitely generated, then it can't be divisible group. But it is.

Assume that $(\mathbb Q,+)$ is finitely generated. Then, by the fundamental theorem there exist $m,n\ge 0$ and $d_1\mid\cdots\mid d_m$ with $d_i>1$ such that $\mathbb Q\simeq \mathbb Z/d_1\mathbb Z\oplus\cdots\oplus\mathbb Z/d_m\mathbb Z\oplus \mathbb Z^n$. If $m\ge 1$, then there exists $x\in\mathbb Q$, $x\neq 0$, such that $d_1x=0$, a contradiction. Thus we get $m=0$. Then $\mathbb Q\simeq \mathbb Z^n$. If $n\ge 2,$ then there exist $x_1,x_2\in\mathbb Q$ which are linearly independent over $\mathbb Z$. But $x_1=a_1/b_1$ and $x_2=a_2/b_2$ give $(b_1a_2)x_1+(-b_2a_1)x_2=0$, a contradiction. So we must have $n=1$, that is, $\mathbb Q$ is cyclic. Assume that it is generated by $a/b$ with $b\ge 1$. Then $\frac{1}{b+1}$ can not be written as $\frac{ka}{b}$, and again we reached to a contradiction.

(Of course, there are simpler and much more natural arguments to show that $(\mathbb Q,+)$ is not finitely generated.)

Edit. In particular, this shows that the additive group of any field of characteristic $0$ is not finitely generated. (However the property holds for any infinite field.)

As per the OP's request, here's an explanation of my comment:

I don't see why you need that theorem (FToFGAG). It couldn't be finitely generated since, letting p be any prime greater than the product of the denominators of the generators in lowest form, one could not generate 1/p.

Suppose $(\mathbb{Q}, +)$ were finitely generated. Let $$\left\{\frac{n_1}{d_1}, \dots, \frac{n_k}{d_k}\right\}$$ be a generating set. Let $p$ be any prime that doesn't divide $\prod_{i=1}^k d_i$. Then clearly one cannot generate $\frac 1p$ by adding and subtracting (whole number multiples of) the elements of the generating set, contradicting the fact that it's supposed to be a generating set.

To be explicit, let's consider an arbitrary ($\mathbb{Z}$-linear) combination of the elements in the generating set:

$$m_1\frac{n_1}{d_1}+ \cdots + m_k\frac{n_k}{d_k}=\frac{\mathrm{long\ expression}}{\prod_i^kd_i}$$

There's no way to reduce such a fraction to $\frac 1p$ if $p$ doesn't divide the denominator.