Isomorphic set to the rational numbers I'm looking for examples of isomorphic sets where the sets aren't finite to get a better understanding.
Is it correct to say that the rational numbers are isomorphic to $\Bbb Z/p \Bbb Z$? If not, could you explain why?
 A: I'm going to assume you're talking about group isomorphisms here (although there are other contexts where the concept of isomorphism makes sense). A group isomorphism $\varphi:G\to H$ needs to satisfy two conditions:

*

*$\varphi$ should be a homomorphism: that is, $\varphi(x\cdot y)=\varphi(x)\cdot\varphi(y)$ for each $x,y\in G$;

*$\varphi$ should be a bijection.

The group $\mathbb{Z}/n\mathbb{Z}$ (in one of its constructions) consists of the equivalence classes of integers modulo $n$, with the group operation being addition modulo $n$. In particular, the cardinality of $\mathbb{Z}/n\mathbb{Z}$ is exactly $n$ - if you divide an integer by $n$, there are only $n$ possibilities for the remainder: $0,1,\ldots,n-1$. On the other hand, the rational numbers $\mathbb{Q}$ (as an additive group) are infinite. Therefore there can't be any bijection between $\mathbb{Z}/n\mathbb{Z}$ and $\mathbb{Q}$, since bijections preserve cardinality. This rules out the possibility of a group isomorphism $\varphi:\mathbb{Z}/n\mathbb{Z}\to\mathbb{Q}$ for any $n\in\mathbb{N}$ - property 2 automatically fails. (The case where $n=p$ for $p$ a prime is a special case of this.)
