Prove that the additive groups $\mathbb{R}$ and $\mathbb{Q}$ are not isomorphic.

Let $\Phi: \mathbb{Q} \rightarrow \mathbb{R}$ homomorphism of additive groups. Then $\Phi$ is already determined by $\Phi(1)$ (as $\Phi(\frac{a}{b})= \Phi(\frac{1}{b})+ ... + \Phi(\frac{1}{b})$ ($a$ summands) and $\Phi(1)= \Phi(\frac{1}{b}) + ... + \Phi(\frac{1}{b}) $, ($b$ summands))

Now, say $\sqrt{2} \cdot \Phi(1)$ doesn't have a preimage.

Edit: I just saw this was remarked by @Robert M in a comment to his answer.

This is another approach: If $(\Bbb{R},+) \cong (\Bbb{Q},+)$ then $\Bbb{R} \cong \Bbb{Q}$ as vector spaces over $\Bbb{Q}$ which is impossible since one is one dimensional and the other is infinite dimensional.