How to show $\mathbf{Q} $ is not free

For the non-free part:

Take any two nonzero elements $x, y ∈ ℚ$ and show they satisfy $λx + μy = 0$ for some nonzero $λ, μ ∈ ℤ$, hence any two elements are linearly dependent. Thus, since $ℚ$ is not cyclic, it cannot have a basis.

For the non-finitely-generated part:

If $ℚ$ was finitely generated then without loss of generality (by finding the common denominator), there were a denominator $s ∈ ℤ$ and numerators $r_1, …, r_n ∈ ℤ$ such that $$ℚ = ℤ\frac{r_1}{s} + \cdots + ℤ\frac{r_n}{s} = (ℤr_1 + \cdots+ ℤr_n)\frac{1}{s} = ℤ\frac{r}{s}$$ for $r = \gcd (r_1, …, r_n)$. Thus, since $ℚ$ isn’t cyclic, it isn’t finitely generated.

Show any finitely generated subgroup of $\mathbf Q$ is cyclic. Since $\mathbf Q$ is not cyclic, it cannot be finitely generated.

$(1)$ It cannot be free: it is not cyclic, so any putative basis has at least two elements. But if $x,y$ are elements of the basis, we know $\mathbf Z^2\simeq \langle x,y\rangle=\langle x'\rangle\simeq\bf Z$ which is impossible. So no basis exists.

$(2)$ It cannot be free: if it were, say $\mathbf Q\simeq\mathbf Z^{(I)}$, then $\mathbf Q\simeq\mathbf Q\otimes_{\mathbf Z}\mathbf Q\simeq \mathbf Q\otimes_{\mathbf Z}\mathbf Z^{(I)}\simeq \mathbf Q^{(I)}$. This forces $|I|=1$, which is absurd.

If an abelian group $G\ne\{0\}$ is free, then it is isomorphic to $\mathbb{Z}^{(X)}$ (direct sum of copies of $\mathbb{Z}$), for some set non empty set $X$.

Then $\mathbb{Z}$ is an epimorphic image of $G$. Since $\mathbb{Z}$ is not divisible, $G$ can't be divisible. But $\mathbb{Q}$ is divisible.

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The group $\mathbb{Q}$ is not finitely generated: if $a_1/b_1, a_2/b_2,\dots,a_n/b_n$ are elements in $\mathbb{Q}$, then we can assume $b_1=b_2=\dots=b_n=b$ (the $a_i$ and $b_i$ are integers, of course); then $$ \frac{1}{2b}=\sum_{i=1}^nc_i\frac{a_i}{b} \qquad(c_1,\dots,c_n\in\mathbb{Z}) $$ implies $\frac{1}{2}\in\mathbb{Z}$, which is false; so the set $\{a_1/b,\dots,a_n/b\}$ doesn't generate $\mathbb{Q}$.