Rationals as a direct sum of two proper subgroups Prove that the group $(\mathbb{Q},+)$ is not isomorphic to the direct sum of two proper subgroups.

My approach: Suppose that $\mathbb{Q}=A\oplus B$ where $A,B$ - proper subgroups of $\mathbb{Q}$. By the definition it means that each $q\in \mathbb{Q}$ has unique representation, i.e. if $q=a_1+b_1$ and $q=a_2+b_2$ then $a_1=a_2$, $b_1=b_2$. Since $1\in \mathbb{Q}$ and $0\in A$ then $1=0+1$ where $1\in B$. Since $2=1+1\in B$ hence $1=(-1)+2$ and so $-1\in A$. We get two distinct representations, contradiction.

Remark: Please do not duplicate this question. I have sen some topics with this question but i did not saw the solution similar to above. 

What's wrong with this solution?
You assert $1 \in B$, but there's no reason to think this is true.
Here's one way to do it. 

We claim $(\mathbb{Q}, +)$ can not be the direct product of two non-trivial subgroups.

Suppose we two subgroups $H, K$ of $\mathbb{Q}$ such that $Q \cong H \times K$. Then, there would be a homomorphism $h : \mathbb{Q} \rightarrow K$ with kernel $H$. But since any non-zero homomorphism from $(\mathbb{Q}, +)$ to $(\mathbb{Q}, +)$ is always bijective, we conclude $H = \{0\}$, which is a contradiction.