Proof Verification $(\mathbb{Q},+)$ is not isomorphic to $(\mathbb{R}, +)$.

Question

Let $\gamma\in\mathbb{R}$, $m\in\mathbb{Z}$ and $\varphi:\mathbb{R}\to\mathbb{Q}$. Then $\varphi(\gamma)=\varphi(\pi\cdot \gamma/\pi)=\pi\varphi(\gamma/\pi)$.
But then $\varphi(\gamma)\notin \mathbb{Q}$. Thus, no such isomorphism exists.

My concern is that I can't write $\varphi(\pi\cdot \gamma/\pi)=\pi\varphi(\gamma/\pi)$. I originally thought I could since $\varphi(1)=\varphi(1/q+\cdots 1+q)=q\varphi(1/q)$ for $q\in\mathbb{Z}$. Can this is be generalized to $q\in\mathbb{R}$?

Answer 1

Indeed you cannot do that, because from your notations $\mathbb Q$ and $\mathbb R$ are merely additive groups.

Alternative proofs:

(1) An isomorphism is a bijection. However, $\mathbb Q$ is countable while $\mathbb R$ is not and hence no bijection can exist between them.

(2) Suppose such isomorphism $\varphi$ exists. Let $a=\varphi(1)\neq0,\ b=\varphi(\sqrt2)\neq0$. Then $a,b\in\mathbb Q$ and $$\varphi(b/a)=(b/a)\varphi(1)=b=\varphi(\sqrt2)$$ Apply $\varphi^{-1}$ to both sides: $$b/a=\sqrt2$$ But $b/a$ is rational. A contradiction.

Answer 2

Here’s an argument that doesn't depend on cardinality or anything fancy:

If $\lambda,\mu\in\Bbb Q$, then there are integers $r$ and $s$ such that $r\lambda=s\mu$. (If you’re queasy at the introduction of $\Bbb Z$ here, I’m saying that $\lambda$ added to itself $r$ times is $\mu$ added to itself $s$ times. {modified argument necessary if $\lambda,\mu$ are of different signs})

Of course no such phenomenon holds in $\Bbb R$.

Therefore the two groups are not isomorphic.