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If $(\mathbb{Q},+)$ is homomorphic to $( \mathbb{Q^+}, \times)$, then $f(x)=1$ for all $x\in \mathbb{Q}$.

This is one of questions from my assignment. I've been working on it for 2 days, but still don't have a clue. Can somebody give me a hint?

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    One reason you may have been stuck on this is the garbled language of the question itself. I'm surprised that others have offered detailed and useful advice without pointing this out. "If $f$ is a homomorphism from $(\mathbb{Q},+)$ to $(\mathbb{Q}^{+}, \times)$, then $f(x) = 1$ for all $x \in \mathbb{Q}$" is a sensible question. Note that this phrasing *explains what $f$ is* before asking you to prove something about it. (FWIW, the phrase "$A$ is homomorphic to $B$" is not used. It's rather common to talk about isomorphisms without specifying them, but this *isn't done* with homomorphisms.)2012-02-16

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HINT. Given $x\in \mathbb{Q}$ and $n\gt 0$, there always exist $y\in\mathbb{Q}$ such that $ny = x$. What will happen to this equation when you apply $f$? What does that tell you?


Note. The original problem had $(\mathbb{Q},+)$ and $(\mathbb{Q},\times)$. Hence the following:

As stated, the claim is incorrect. Since $(\mathbb{Q},\times)$ is not a group, the only reasonable interpretation is that we are considering these objects as either semigroups, or monoids. But if we consider them as semigroups, then $f(x)=0$ for all $x$ is a perfectly fine semigroup homomorphism from the additive semigroup of rationals to the multiplicative semigroup of rationals, so the claim is false.

If we consider them as monoids (so that $f(0)=1$ necessarily is true), then we do have that any $f\colon\mathbb{Q}\to\mathbb{Q}$ such that $f(a+b) = f(a)f(b)$ is necessarily of the given form.

(If these are meant to be groups, then the second one should be either $(\mathbb{Q}-\{0\},\times)$, or $(\mathbb{Q}_{\gt 0},\times)$. )

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    Yes, this is what I want to say. I tried to split$x$but not this way. Thank you so much.2012-02-17