Derivatives don't really go wrong, it's antiderivatives. (EDIT: Actually, the more I think about it, this is just a symptom. The underlying cause is that continuity on the rationals is a much weaker notion than continuity on the reals.)
Consider the function f:Q→Q given by
f(x)={01x<πx>π
This function is continuous and differentiable everywhere in its domain. If x<π, then there's a neighborhood of x in which f is a constant 0, and so it's continuous there, and f′(x)=0. But if x>π, there's a neighborhood of x in which f is a constant 1, so it's continuous there too, and f′(x)=0 again.
So the antiderivatives of font-style: inherit; font-variant: inherit; font-weight: inherit; font-stretch: inherit; lin
manpreet![Tuteehub forum best answer]()
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2 years ago
I was once told that one must have a notion of the reals to take limits of functions. I don't see how this is true since it can be written for all functions from the rationals to the rationals, which I will denote ff, that
Since, as far as I know, functions like ∣x∣∣x∣ and relations like << can be defined on the rationals. Is it true you couldn't do calculus on just the rational numbers? At the moment I can't think of any rational functions that differentiate to real functions. If it's true that it isn't formally constructible on the rationals, what about the algebraic numbers?
Edit
Thanks for all the help, but I haven't seen anyone explicitly address whether or not we could construct integrals with only algebraic numbers. Thanks in advance to anyone who explains why or why not this is possible.