This question already has an answer here:

How do I show that the reals under addition are not isomorphic to the rationals under addition, and that the rationals under addition are not isomorphic to the integers under addition?

I've tried finding elements with different orders, but for each group, the elements have infinite order. I know all three groups are Abelian. Is it because they won't form a bijection?


marked as duplicate by Dietrich Burde, Watson, Stefan4024, Shailesh, Leucippus Sep 23 '16 at 0:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 3
    $\begingroup$ If the groups are isomorphic they are bijective. But $\mathbb{Q}$ and $\mathbb{R}$ are not bijective by the well known Cantor diagonal argument. As for $\mathbb{Q}$ and $\mathbb{Z}$, it's clear that $\mathbb{Z}$ is cyclic, i.e. generated by one element. Show that $\mathbb{Q}$ is not. $\endgroup$ – Mathematician 42 Sep 22 '16 at 13:48

There is no bijection from the reals to neither the rationals nor to the integers. Specifically, there can be no isomorphisms.

As for the integers vs. the rationals, say there is an isomorphism $\phi:\Bbb Q\to \Bbb Z$. Let $p\in \Bbb Q$ be such that $\phi(p)=1$. What property does $\phi(p/2)\in \Bbb Z$ have? Is there such an integer?


Since $\mathbb{Z}$ is cyclic with generator $1$, you can look at the image of $1$ under your supposed isomorphism. No matter what it is, it can't generate $\mathbb{Q}$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.