Only homomorphism from rationals to integers

[vikas92]

Kindly see the attached image.i can't understand the step where he write f(x)=f(1+1+...+1)=xf(1).But the homomorphism is from <Q,.> to <Z,+>

 

[oli4]

Sorry, but this is unreadable
Cheers...

 

[vikas92]

Sorry for bad image quality.
Show that only homomorphism from the group <Q^*,.> to <Z,+> is the zero homomorphism where Q^*=Q-{0}
Solution
Let 0≠f be a homomorphism from <Q^*,.> to <Z,+>
Let f(1)=n$\in$Z. Suppose f(1)=0
Then f(x)=f(1+1+...+1) if x>0,x$\in$Z
=x f(1)=0
This is the step I couldn't understand as the homomorphism is from <Q^*,.> to <Z,+> not from <Q^*,+> to <Z,+> .
Also,0=f(x)=f(-1χ(-x))=f(-1)+f(-x)
$\Rightarrow$f(-1)=-1f(-x)
Thus f(-x)=f(-1χx)=f(-1)+f(x)=f(-1)+0=-f(-x) $\Rightarrow$ f(-x)=0

Further he proved that f(r/s)=0

 

[oli4]

Hi,
Indeed something is very fishy about this proof.
Furthermore I think he is trying to prove something which is wrong.
Unless I miss something,
define for any x of Q being a multiple of a power of 2 f(x)= said power
for instance, f(-1/2)=f(15/2)=-1
f(4)=f(-4/17)=2
f(0)=1
and fx)=0 otherwise, than it looks to me that this is a counter example, but I could miss something...

Cheers...

 

[vikas92]

Thanks for your reply.Perhaps it is serving as counterexample to what the question was given.In some other book the same question was there with <Q,+> instead of <Q,.>
.Maybe the solution provided by the author is wrong

 

[oli4]

Well i,n fact Q+ sounds like a much better candidate to do the job, so my counter example would be even simpler, but I probably miss something.
Anyway, the proof you linked to is completely bogus as far as I can tell, and, just think about exp(x) ln(x); if we were to talk about R+ -> R (or the other way around, pick one) you would have a perfect example of the same idea, so I think the proof should be a lot more involved than that and.

Cheers..