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

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 $$q\in\mathbb{R}$$?

• Note: $\Bbb Q$ and $\Bbb R$ have different cardinality Sep 5, 2019 at 3:37

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.

• How can we say that $\varphi(b/a)=(b/a)\varphi(1)$? Sep 5, 2019 at 11:22
• @dallas2019 You have shown $\varphi(1)=q\varphi(1/q)$ for an integer $q$. Multiply by another integer $p$ then $(p/q)\varphi(1)=\varphi(p/q)$. Sep 5, 2019 at 11:54
• Ah of course! Thank you. Sep 5, 2019 at 15:40

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.

• That’s clever. This seems to relate to the idea that $\mathbb{R}$ and $\mathbb{Q}$ do not have the same generators? Sep 5, 2019 at 15:45
• You can look at it as saying that $\Bbb Q$ is of rank one over the integers, so that any two elements are $\Bbb Z$-linearly dependent. But $\Bbb R$ isn’t even of countable rank over $\Bbb Z$. Sep 5, 2019 at 18:46
• I do not think I have covered this concept yet, but I will keep this in the back of my mind for when I do. Sep 5, 2019 at 20:21
• It’s almost the same as dimension, for vector spaces. Sep 5, 2019 at 22:52