Prove that the additive groups $\mathbb{R}$ and $\mathbb{Q}$ are not isomorphic. 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
 A: 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. 
A: The quotient of $\mathbb{Q}$ by a cyclic subgroup (namely $\mathbb{Z}$) is torsion. $\mathbb{R}$ has no such subgroup.
A: 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.
A: Let $F : Q → R$ be isomorphism of additive groups.
$n^\frac{1}{2}$ belongs to $\Bbb R$ for all $n$ natural number
Since $F$ is isomorphism, there exist unique $\frac{p}{q}$ in $Q$ for unique $n$ such that $F(\frac{p}{q}) =  n^\frac{1}{2}$
$$
F\left(\frac{p}{q}\right) = F\left(\frac{1}{q} + \frac{1}{q} + \ldots + \frac{1}{q}\right) =  p\cdot F\left(\frac{1}{q}\right) = n^\frac{1}{2}\qquad\qquad\mathbf{(1)}
$$
$$F(1) = F\left(\frac{q}{q}\right) = q \cdot F\left(\frac{1}{q}\right) = q \cdot \frac{n^\frac{1}{2}}{p}\qquad\qquad\mathbf{(2)}$$
Hence,
$$F(1) = \frac{q_1 \cdot (1)^\frac{1}{2}}{p_1} = \frac{q_2 \cdot (2)^\frac{1}{2}}{p_2} = \frac{q_3 \cdot (3)^\frac{1}{2}}{p_3} =\ldots$$
So, $\frac{q_1 \cdot p_2}{p_1 \cdot q_2} = (2)^\frac{1}{2}$ where LHS is rational and RHS irrational
This imply our assumption is incorrect. Therefore, there doesn't exist an isomorphism $F$.
