Is $(\mathbb{Z}, +)$ isomorphic to $(\mathbb{R}, +)$

Question

Is $(\mathbb{Z}, +)$ isomorphic to $(\mathbb{R}, +)$

my idea:

$1_{\mathbb{Z}} = \phi(r) = \phi\big(2(r/2)\big) = 2\phi(r/2$)

i.e., $\phi(r/2) = \frac{1}{2}$

which contradiction so no isomorphism

Answer 1

There are several ways of doing this. Perhaps the easiest are:

Cardinality: A isomorphism is a bijection. Can you see a problem with this?

$(\mathbb{Z},+)$ is cyclic generated by $1$ or $-1$. Is $(\mathbb{R},+)$ cyclic? Think, if there were a generator, would it be rational or irrational? In either case, could I add a rational/irrational to itself a number of times and get the other 'type'?

EDIT. I suppose I have ignored your proof. The idea is sound just be clear. Your map clearly (since you consider $r/2$ which would not be in $\mathbb{Z}$) goes from $(\mathbb{R},+)$ to $(\mathbb{Z},+)$. Is this map an isomorphism? You do not say. You seem to try to use $\phi$ is a homomorphism under multiplication not addition. [Though you could be tacitly using the fact that $2 \cdot r=r+r$.] What is $r$? Why is $\phi(r)=1_\mathbb{Z}$? I'm sure you know all of this, but be clear to a reader.

Suppose that there were an isomorphism $\phi: (\mathbb{R},+) \to (\mathbb{Z},+)$. Then there is a $r \in \mathbb{R}$ such that $\phi(r)=1_\mathbb{Z}$ since $\phi$ must be surjective. Now note $r \in \mathbb{R}$ so that $r/2 \in \mathbb{R}$. But $$ 1_\mathbb{Z}= \phi(r)= \phi\left(2 \cdot \frac{r}{2}\right)= \phi\left(\dfrac{r}{2}+\dfrac{r}{2}\right)= \phi\left(\dfrac{r}{2}\right)+\phi\left(\dfrac{r}{2}\right)=2 \phi\left(\dfrac{r}{2}\right) $$

Now $\phi$ is an isomorphism and hence a bijection. But as $\phi(0)=0$ (homomorphisms take identity to identity, which is $0$ under addition for both rings), so $\phi(r/2) \neq 0$. It cannot be that $\phi(r/2)=1$ as then $2=1+1=\phi(r/2)+\phi(r/2)=\phi(r/2+r/2)=\phi(r)=1$. For the same reason, $\phi(r/2) \neq -1$ as then $-2= (-1)+(-1)=\phi(r/2)+\phi(r/2)=\phi(r/2+r/2)=\phi(r)=1$. Then $\phi(r/2)>1$ or $\phi(r/2)< -1$. But then $\phi(r/2) \in \mathbb{Z}$ so $2\phi(r/2)>1$ or $2\phi(r/2)<1$, a contradiction.

This would be the 'clearest' version of your proof. Though some detail here may be too much and in other places a bit quick. But if I were to write your proof in a form for a person only just seeing groups, this is how I would do it.

Answer 2

No, there is no isomorphism between $(\Bbb Z,+)$ and $(\Bbb R,+)$ because $(\Bbb Z,+)$ is cyclic and $(\Bbb R,+)$ is not cyclic.