Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $k$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.