Say we have two functions $f(x)$ and $g(x)$ where $\frac{df(x)}{dx}=\frac{dg(x)}{dx}$. This surely means that we can write $f(x)=g(x)+C$. Now let's say we are graphing these two functions and we want to edit one of them, say $f(x)$ in such a way that the graph of $f(x)$ lies perfectly on the graph of $g(x)$. This would be possible to do by adding $C$ to $f(x)$ however only if $C$ is a rational constant; if it is an irrational constant then we would need to use infinitely many digits of $C$ to accomplish our goal since if we are using a finite number of digits of $C$ it would be possible to "zoom in" the graph and see that the two functions don't overlap. This is only the context from which the question I want to ask emerged.
What makes this constant irrational? Is the rationality/irrationality of the constant somehow dependent on the nature of the derivative of $f(x)$ and $g(x)$ ?
Let's think about the simplest possible case.
Let's suppose that $f'(x) = g'(x)$ for all $x$ but further, suppose that these are both equal to $0$. Then both $f$ and $g$ are constant functions.
So $f(x) = A$ and $g(x) = B$ for some $A$ and $B$, and we see that $$ f(x) = g(x) + C $$ for $C = A-B$.
You can imagine that you discover $f$ by integration, and I discover $g$ the same way. One possibility is that you find $A = 0$, and that I find $B = 1$. In this case, $C = -1$ is a rational.
But another possibility is that you come up with $A = \pi$ and I come up with $B = 2$, in which case $C = \pi - 2$, which is irrational.
There's nothing in the nature of the derivative -- the constant function $0$ -- that says anything about what the values $A$ and $B$ can be (indeed, any real number is a possible value).