I've had this idea in my head for a while, but I've never told anybody because... well, I really don't know. I just never thought that it might even be remotely correct, but here goes. Here is just an example that, I think, makes a good job of explaining this principle, but there are other scenarios, this is just the simplest one I can think of: We can prove that the derivative of a function $f(x)=ax^n$ is equal to $f'(x)=anx^{n-1}$, for all rational numbers $\frac{n}{m}$ where $n$ and $m$ are integers, but that doesn't prove that this property holds for irrational/transcendental numbers, right? But I say it does, just from the fact that it works for all rational numbers, and here is why: say you have the expression $\frac{d}{dx}x^\pi$. Now we all know that $\pi$ is transcendental, but why can't we just "pretend" that we have a two immensely big numbers $n$ and $m$, such that $\frac{n}{m}$ is such a close approximation to $\pi$ that there is no way to distinguish them. This number is so close to $\pi$ that it differs on the googolplexth digit. But it is still a repeating decimal. So Why can't we just do that, and say that the power rule applies to irrational/transcendental numbers too?

I apologize if this questions is completely ludicrous.


2 Answers 2


The type of argument you describe is a big motivation for the definition of a continuous function. A basic result which captures the spirit of what you want is this: a continuous function $f : X \to Y$ between topological spaces is determined by its values on a dense subset of $X$. In particular, a continuous function $f : \mathbb{R} \to Y$ is determined by its values on the rationals. This gives rise to the following intuition:

Any "sufficiently continuous" statement you can prove for the rationals should be true of all real numbers.

The above result doesn't directly apply to this situation, at least not in a straightforward way: for one thing, knowing that the function $x^q$ exists for $q$ rational is not enough to immediately conclude that a function called $x^r$ even exists, let alone that it is continuous or differentiable. But it is still an important motivation for suspecting that such a thing ought to be true anyway.

For $x \ge 0$ it is a standard exercise in real analysis to prove that we can define

$$x^r = \lim_{n \to \infty} x^{q_n}$$

where $q_n$ is any sequence of rational numbers tending to $r$. Furthermore this function satisfies all of the expected algebraic identities, and so forth. But it is still not completely trivial to see that $x^r$ is differentiable for arbitrary real $r$ until you prove the basic properties of logarithms and exponentials, after which you then know that

$$x^r = e^{r \log x}.$$

Once you know this, you're in the money: the composite of differentiable functions is differentiable, and everything works out exactly the way you expect it to by the chain rule.

One might be tempted to argue as follows: surely if the function $x^r$ is a limit of a sequence of functions $x^{q_n}$, then the derivative of the function $x^r$ is a limit of the sequence of derivatives $q_n x^{q_n-1}$. But this argument is not so straightforward to formalize. First, there are various notions of convergence of a sequence of functions: there is pointwise convergence but also convergence with respect to various norms, and one must decide what notion is sensible here. Second, even if one takes a fairly restrictive definition of convergence (say uniform convergence), it is still false in general that a uniform limit of differentiable functions is differentiable (see Weierstrass function).

One needs the additional constraint that the derivatives themselves also converge uniformly. Now, the derivatives $q_n x^{q_n-1}$ do not converge uniformly to $rx^{r-1}$ on $\mathbb{R}_{\ge 0}$ (let's provisionally assume we only care about $r > 1$), but they do converge to $rx^{r-1}$ uniformly on the intervals $\left[ \frac{1}{k}, k \right]$ for all $k \in \mathbb{N}$ (say). I think that is enough to get this result without going through the stuff above about logarithms and exponentials, but I haven't checked it closely.

  • 1
    $\begingroup$ "tending to zero" should be "tending to $r$", right? $\endgroup$ Jul 14, 2011 at 6:10
  • $\begingroup$ @Zev: ha. Yes, it should. $\endgroup$ Jul 14, 2011 at 6:29

The rationals are dense in the reals, and differentiation is, in some sense, continuous, and this can used to turn your argument into a rigorous proof, with some well-timed application of limits. One has to be careful - there are other properties of rationals that don't apply to reals in general, and that's why we have the apparatus of limits.

  • $\begingroup$ For example, all rationals are rational, but almost all reals are not, because being rational is not a continuous property. $\endgroup$
    – user21820
    Apr 19, 2016 at 1:41

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