For example consider $x=e.$ So $0<e=p/q$ where $p,q$ are both integer and positive. I'm trying to show that there is no integer $p,q$ that satisfies the equation $e=p/q$. If we differentiate $e=p/q$ with respect to $p$ we would have $1=0$. So I'm assuming we can never differentiate this equation, even if we create some function $f(p)$, to write $f(p)e=f(p)\:p/q$.
Or integrate, for example: $\int_0^\infty f(p)e\: dp = \int_0^\infty f(p)\: p/q\: dp$ because then $p$ is now a real number and this ruins the proof? I chose the improper $\int_0^\infty dp$ to cover all possible values for $p>0$.
For example, for $f(p)=p$, $\int_0^3 ep\:dp=\int_0^3 p^2/q \: dp \implies 9e/2=9/q \implies e=2/q$. This is invalid because when we integrated $p$ becomes real and this ruins the proof?
I just chose $\int_0^3$ because I couldn't find a convergent $\int_0^\infty$ for $f(p)=p$.
We cannot differentiate functions on the integers at all, as every point of the domain is isolated.