TITLE: How far can one get in analysis without leaving $\mathbb{Q}$?
QUESTION [167 upvotes]: Suppose you're trying to teach analysis to a stubborn algebraist who refuses to acknowledge the existence of any characteristic $0$ field other than $\mathbb{Q}$.  How ugly are things going to get for him?
The algebraist argues that the real numbers are a silly construction because any real number can be approximated to arbitrarily high precision by the rational numbers - i.e., given any real number $r$ and any $\epsilon>0$, the set $\left\{x\in\mathbb{Q}:\left|x-r\right|<\epsilon\right\}$ is nonempty, thus sating the mad gods of algebra.
As @J.M. and @75064 pointed out to me in chat, we do start having some topology problems, for example that $f(x)=x^2$ and $g(x)=2$ are nonintersecting functions in $\mathbb{Q}$.  They do, however, come arbitrarily close to intersecting, i.e. given any $\epsilon>0$ there exist rational solutions to $\left|2-x^2\right|<\epsilon$.  The algebraist doesn't find this totally unsatisfying.

Where is this guy really going to start running into trouble?  Are there definitions in analysis which simply can't be reasonably formulated without leaving the rational numbers?  Which concepts would be particularly difficult to understand without the rest of the reals?

REPLY [3 votes]: The fact that the only functions on an interval with derivative zero everywhere are constants is at the heart of applications to science. This fact is equivalent to the completeness axiom. Continuity, differentiation and intervals make sense in any ordered field but if there exists a nonempty set with an upper bound but no least upper bound, then there exists a non-constant function with derivative zero everywhere.