Does there exists a surjective group homomorphism from $(\mathbb{R},+)$ to $(\mathbb{Z},+)$?

Question

Examine whether there exists any surjective group homomorphism

(1) from $(\mathbb{Q}(\sqrt{2}),+)$ to $(\mathbb{Q},+)$

(2) from $(\mathbb{R},+)$ to $(\mathbb{Z},+)$

My try:

(1) I think $\phi :\mathbb{Q}(\sqrt{2})\rightarrow \mathbb{Q}$ given by $\phi(a+b\sqrt 2)=a$ works well here.

(2) I'm utterly clueless about this one.

I tried many things always hitting a road block:

I tried to figure of if $f:(\mathbb{R},+) \to (\mathbb{Z},+)$ such a homomorphism then how would a rational $\frac{a}{b} $ satisfying $f(\frac{a}{b}) =1$ look like. I eventually realized then it must be $\frac{a}{b}=1$.

Now that we have $f(1)=1$ if we can each $a\in(0,1)$ figure out $f(a)$ then as every $r\in\mathbb{R}$ can be written as $r=n+a$ where $n\in\mathbb{Z}$ and $a\in(0,1)$ and $f(n+a)=f(n)+f(a)=n+f(a)$ we will be done.

I'm stuck here. Please help me out.

Answer 1

Part (1) is good.

For part (2) you're starting from the wrong side. There's no reason why $f(1)=1$. Indeed, if $f\colon G\to G'$ is a homomorphism of (additive) abelian groups, then also $a\mapsto 2f(a)$ defines a homomorphism.

Suppose there exists a surjective (group) homomorphism $f\colon\mathbb{R}\to\mathbb{Z}$. Then there is $r\in \mathbb{R}$ with $f(r)=1$.

Then $1=f(r)=f(2(r/2))=2f(r/2)$.

Answer 2

You have $f(1)=a$ for $n$ integer, $f(n)=an$ thus $f(n(1/n))=nf(1/n)=a$. Impossible if $n$ does not divide $a$.