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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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    What is $*$? ${}$2012-02-16
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    @Shannon: Please include the question in the body of the post, not just the title. Your post began with "This is one of the question from my assignment" and never stated the question in the post, only in the title. Titles of posts are like titles of books written on the spine: informative when you are looking on the main site, but should not be considered part of the 'narrative.'2012-02-16
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    @ArturoMagidin Thanks for doing revision for me. I was using my phone to post this question which caused some information hidden. Thanks for your suggestion.2012-02-16
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    @KannappanSampath multiplication. Sorry.2012-02-16
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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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    the second one is ( \mathbb{Q^+}, \times). Sorry for my carelessness. I was trying f(x)=f(0+x)= f(0)f(x) ,the only thing I can get is f(0)=1.2012-02-16
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    @Shannon: And did you *now* look at the hint I gave? (P.S.: The fact that $f(0)=1$ follows simply from the fact that you have a group homomorphism: it *must* map the identity of $(\mathbb{Q},+)$, which is $0$, to the identity of $(\mathbb{Q},\times)$, which is $1$).2012-02-16
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    @Shannon: To be more explicit: whatever $f(1)$ is, what is $f(\frac{1}{2})$, in terms of $f(1)$? What is $f(\frac{1}{3})$, in terms of $f(1)$. What is $f(\frac{1}{n})$, in terms of $f(1)$? What does that tell you about $f(1)$?2012-02-16
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    I got f(x)=f(x/n)^n. Is it right?2012-02-16
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    I got f(x)=f(x/n)^n, is it enough to conclude f(x) = 1 since f(x) and f(x/n) are both in Q.2012-02-17
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    What you say I think is not enough, you have to *explain*, not just guess (we aren't cooking pasta; you don't throw things at the wall and hope something sticks). Yes: for any $x$, $f(x)$ must be an $n$th power for every $n$. But the only rational number that is an $n$th power for *every* natural number $n$ is $1$, so $f(x)$ must equal $1$ for any $x$.2012-02-17
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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