showing $\mathbb{Q}$ is isomorphic to $\mathbb{R}$ or not as a group under addition

Question

I want to see whether $\mathbb{Q}$ is isomorphic to $\mathbb{R}$ or not as a group under addition.

$(\mathbb{Q}, +) \cong (\mathbb{R},+)$?

I know they are not isomorphic as a ring. The prove is basically done as follows First assume there exist isomorphism $f: \mathbb{R} \rightarrow \mathbb{Q}$, then from homomorphism property $f(1)=1$, and $f(2)=2=f(\sqrt{2}\times \sqrt{2}) = f(\sqrt{2}) f(\sqrt{2})$, this states that $f(\sqrt{2}) = \pm \sqrt{2}$ but this does not belong to $\mathbb{Q}$, so it contradicts.

But how about group under addition, i tried to do the same thing but here $\phi(0)=0$ is all i have.

Answer 1

Apart from the question of cardinality, I'd like to point out this property of $(\mathbf{Q},+)$ that is not shared with $(\mathbf{R},+)$.

Given any $x, y \in \mathbf{Q}$, there exist integers $m$ and $n$, not both zero, such that $mx + ny = 0$. (Multiplication by an integer is definable solely in terms of addition.)

This is not true in $\mathbf{R}$. For example, take $x = 1$ and $y = \sqrt{2}$.

Answer 2

Can there be a bijection between the sets $\mathbb Q$ and $\mathbb R$? (No)