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Let a function exist such that $f(a+b)=f(a)+f(b)$. We have already shown that for any integer n, $f(nx)=n f(x)$. Now we must show that for any rational number $n/m$, $f(n/m)=n/m f(1)$.

The problem is that showing the equation for integers was easy, as multiplication is repeated addition. However, the same can't be done for division.

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Hint: $m f(n/m)=?$ – Thomas Andrews Dec 13 '12 at 19:38

Hint: First deal with $f(1/m)$, where $m$ is positive. Use the fact that $$\frac{1}{m}+\frac{1}{m}+\cdots+\frac{1}{m}=1.$$ (We used $m$ copies of $\dfrac{1}{m}$.)

You could alternately use $m$ copies of $\dfrac{n}{m}$.

For completeness of your proof for $f(k)$, where $k$ is an integer, make sure you have also dealt with negative $k$.

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Thank you! I can't believe I didn't think of that. Also, I have already proven that it is true for both negative and positive integers. – Hayley Dec 13 '12 at 19:53

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