Homomorphism between integers and $(-1,1)$ under any operation contains an irrational point I need to prove that 

Any homomorphism $h$ between integers and $(-1,1)$ under any operation contains an irrational point, i.e. there exists a $k\in\mathbb Z$ such that $h(k)$ is irrational.

For example, consider the group $(-1,1)$ with the operation $a\oplus b=\dfrac{a+b}{1+ab}$. It is not hard to see that $h:\mathbb Z\to(-1,1):=k\mapsto \tanh(k)$ is a homomorphism from $\mathbb Z$ under regular addition to $(-1,1)$ under $\oplus$, and, clearly, $h(1)=\tanh(1)$ is irrational, so we have found an irrational point in the homomorphism.
Is it true for any operation and homomorphism?
I feel that the fact that $(-1,1)$ is bounded and the integers not may play a role in the argument, but I cannot see how, any help would be appreciated. Thanks.
 A: It's not true.   There is a one-to-one function $f_1$ from $\mathbb Q \cap (-1,1)$ onto $\mathbb Z$ (because they have the same cardinality).  Similarly there is a one-to-one function $f_2$ from $(-1,1) \backslash \mathbb Q$ onto $\mathbb R \backslash \mathbb Z$.
Put them together to get a one-to-one function $f$ from $(-1,1)$ onto $\mathbb R$ that maps $\mathbb Q \cap (-1,1) \to \mathbb Z$.  Let the operation on $(-1,1)$ be $\oplus$ defined 
by 
$$ x \oplus y = f^{-1}(f(x) + f(y))$$
This makes $(-1,1)$ into a group and $f$ into a homomorphism (in fact an isomorphism).  $h = f_1^{-1}$ is a homomorphism from $\mathbb Z$ into $(-1,1)$ whose image contains only rationals.
A: This is false. For any operation on $(-1,1)$ with a rational identity (like the example operation in your question, which has $0$ as an identity), there exists a homomorphism that sends all of $\mathbb Z$ to that rational identity. Hence, that homomorphism has no irrational points in its image.
A: As we say in mathematics, the answer is trivially NO because you allowed for a total set-theoretical freedom of selecting $h$. You can easily get even a strictly increasing $\ h\ $ which can be interpreted as a homomorhism which has rational values only:
Define bijections $\ H:\Bbb R\to(-1;1)\ $ and
$\ F:(-1;1)\to\Bbb R\ $ -- each inverse to the other
-- as follows,
$$ \forall_{x\in\Bbb R}\quad H(x)\ :=\ \frac x{1+|x|} $$
$$ \forall_{y\in(-1;1)}\quad F(y)\ :=\ \frac y{1-|y|} $$
Finally, define binary operation $\ \#\ $ in $\ (-1;1)\ $
by borrowing it from $\ \Bbb R$,
$$ \forall_{a\ b\,\in(-1;1)}\quad
     a\,\#\,b\,\ :=\,\ H(F(a) + F(b)) $$
Thus $\ \#\ $ is a group operation such that
$\ H:\Bbb R\to(-1;1)\ $ and $\ F:(-1;1)\to\Bbb R\ $ are
isomorphisms, each inverse to the other.
Finally, define $\ h:\Bbb Z\to(-1;1)\ $ as
$$ h\ :=\ H|\Bbb Z $$
Then $\ h\ $ is the required homomorphism (monomorphism)
such that
$$ \forall_{n\in\Bbb Z}\quad h(n)\in\Bbb Q.$$
Great!
