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**Swlabr** So, $\displaystyle 15=3.5$. Now, $\displaystyle 1$ is coprime to $\displaystyle 15$, $\displaystyle 2$ is coprime, $\displaystyle 3$ is not, $\displaystyle 4=2.2$ is, $\displaystyle 5$ is not, $\displaystyle 6=2.3$ is not (note the $\displaystyle 3$), $\displaystyle 7$ is, $\displaystyle 8=2^3$ is, $\displaystyle 9=3^2$ is not, $\displaystyle 10=2.5$ is not, $\displaystyle 11$ is, $\displaystyle 12=2^2.3$ is not, $\displaystyle 13$ is, and $\displaystyle 14=2.7$ is. Counting all the is-es, we see that there are $\displaystyle 7$ numbers less than $\displaystyle 15$ that are coprime to $\displaystyle 15$. Thus, $\displaystyle f(15)=7$.