Is there a better (or other) way(s) to prove the following statement? Also, the same argument works for multiplicative groups $\mathbb{R}-\{0\}$ and $\mathbb{Q}-\{0\}$, right?
Problem Prove that the additive groups $\mathbb{R}$ and $\mathbb{Q}$ are not isomorphic.
Solution By cantor's diagonal argument, there is no possible bijection between $\mathbb{Q}$ and $\mathbb{R}$. Since an isomorphism needs to be a bijection, there is no possible isomorphism between the additive groups $\mathbb{R}$ and $\mathbb{Q}$.
Thanks
Best Answer
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.