# Study of functions from $\mathbb{Q}$ to $\mathbb{Q}$

Is it possible to study functions from $\mathbb{Q}$ to $\mathbb{Q}$ with ordinary calculus ? Obviously with the limitation that $\mathbb{Q}$ is not complete. So much less limits, derivatives and integrals exist; but does it make sense a tangent in $\mathbb{Q^2}$ ?

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One difficulty is that if $f'(x) = 0$ for all $x \in \mathbb{Q}$, you couldn't conclude that $f$ is constant. –  littleO Oct 31 '12 at 11:44
Also, any intuition on continuity goes out the window. Take the function which is constantly $0$ if $x^2 < 2$ and $1$ otherwise. It has a clear jump at what would be $x = \pm\sqrt{2}$, but is at no rational point discontinuous. –  Arthur Oct 31 '12 at 11:52

On the other hand, in $\mathbb Q$ (and, indeed, in other fields) use of the formal derivative for a polynomial is useful for purely algebraic purposes. Establishing when a polynomial has multiple roots, for example.

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It could make some sense, but it will be quite pathological. $\bf Q$ is not a differentiable manifold, so differentiation in the usual sense doesn't make much sense, furthermore, there is no nontrivial continuous measure on $\bf Q$ (because $\bf Q$ is countable), so Lebesgue integral will be pretty much completely useless in studying it.

If course, you can still calculate limits, derivatives and Riemann integrals as if you were working in $\bf R$. After all, limits make sense in any topological space, including $\bf Q$, and all these objects are defined as limits, so at worst, they may just fail to exist, which can lead to rather pathological examples.

For instance, the function $1/x$ will still be continuous except in $0$, but it will be very non-integrable (as the logarithm of a positive rational number distinct from $1$ is irrational). It is not hard to imagine a function $f:{\bf Q}\to{\bf Q}$ and three rational numbers $a<b<c$ such that the “integral” from $a$ to $c$ is well-defined, but not from $a$ to $b$ (because the integral from $a$ to $b$ converges to an irrational number, but from $a$ to $c$ to a rational one).

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Adding to the above comments and appreciating the idea of complete spaces, one can see that the intermediate value theorem, which holds on reals, does not hold on rationals. For instance,

$$f(x) = \begin{cases} -1 & \mbox{if } x^2 < 2 \\ 1 & otherwise. \end{cases} \,.$$

$f(x)$ is continuous with $f(0)=-1,$ $f(2)=1 ,$ yet there does not exist a $c$ with $f(c)=0.$

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