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This question already has an answer here:

Let $\Bbb Q^+$ be the set of positive rational numbers. Find all solutions $f:\Bbb Q^+ \to \Bbb R$ of the functional equation $$ f(xy)=f(x)f(y), \quad x, y\in \Bbb Q. $$

Is $f(x)=x^a$ the only solution? If not, is it true if we assume that $f$ is continuous?

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marked as duplicate by user61527, Semsem, Claude Leibovici, Shuchang, user63181 Mar 13 '14 at 7:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Your question has an answer at math.stackexchange.com/questions/43964/…. $\endgroup$ – user122283 Mar 13 '14 at 1:59
  • $\begingroup$ The answer is for the equation on real numbers which is very well known, current question is for functions defined in rationals, and continuity assumed. Thank you. $\endgroup$ – Chung. J Mar 13 '14 at 4:32
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Fundamental theorem of Arithmetic can be interpreted as saying that the set of positive rational numbers under multiplication is a free abelian group with prime numbers as basis.

So one can define an endomorphism of the group by specifying it arbitrarily on primes. (An endomorphism is also a function from rationals to real numbers as required by you)

For example define $f(1)=1, f(p) = p$ for primes less than 100, and $f(p) = 1$ for primes $ p > 100$, and extend it multiplicatively. This is in particular not a 1-1 function and satisfies your functional equation and is not of the form $f(x) = x^a$.

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  • $\begingroup$ Thank you very much for your perfect answer. Do you have an idea if we assume $f$ is continuous? $\endgroup$ – Chung. J Mar 13 '14 at 4:12

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