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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 $q\in\mathbb{R}$?

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

2 Answers 2

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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.

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

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  • $\begingroup$ That’s clever. This seems to relate to the idea that $\mathbb{R}$ and $\mathbb{Q}$ do not have the same generators? $\endgroup$
    – smileemote
    Sep 5, 2019 at 15:45
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    $\begingroup$ 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 $\Bbb Z$. $\endgroup$
    – Lubin
    Sep 5, 2019 at 18:46
  • $\begingroup$ 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. $\endgroup$
    – smileemote
    Sep 5, 2019 at 20:21
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    $\begingroup$ It’s almost the same as dimension, for vector spaces. $\endgroup$
    – Lubin
    Sep 5, 2019 at 22:52

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