Why is the rational number system inadequate for analysis?

Solution 1:

The problem with working in just the rationals is that they aren't complete. This means that there are Cauchy sequences in $\mathbb{Q}$ that don't converge or equivalently that given some bounded set, the $\sup$ or $\inf$ of that set don't necessarily lie in $\mathbb{Q}$.

This means that lots of key results in real analysis fail to be true when working over $\mathbb{Q}$ instead of $\mathbb{R}$. For example, functions that are continuous in $\mathbb{Q}$ don't necessarily have the intermediate value property. Consider $$f(x) = \begin{cases} -1, & x^2 < 2 \\ 1, & x^2 > 2 \end{cases}$$ Since there isn't a rational such that $x^2 = 2$ we can prove that this function is continuous in $\mathbb{Q}$ but there isn't a $y$ such that $f(y) = 0$. The problem is that proofs of the intermediate value theorem typically either generate a Cauchy sequence using nested intervals and claim it converges or look at something like $\sup\{x: f(x)<0\}$. However both of these claims use completeness of the reals and so don't hold over the rationals. Other results fail in similar ways (or because they rely on IVT).

Solution 2:

Many important theorems of calculus fail if you consider $\Bbb Q$ in place of $\Bbb R$. To name a few examples:

Intermediate value theorem: Function $f(x)=x^2$ is continuous and $f(1)=1,f(2)=4$ but for no rational number $p$ we have $f(p)=2$ (this is pretty much the example mentioned)

Bolzano-Weierstrass theorem: If you take any sequence of rational numbers which converges in $\Bbb R$ to an irrational (e.g. $3,3.1,3.14,3.141,3.1415,...$) then this is a bounded sequence with no subsequence convergent to a rational number.

Cauchy completeness, monotone convergence theorem: The above example also shows that a Cauchy sequence doesn't necessarily have a limit, and same goes for monotone sequences.

Extreme value theorem: The function $f(x)=x-x^3$ is clearly continuous on interval $[0,1]$, but doesn't achieve a maximum on this interval - indeed, as we go closer to $\frac{1}{\sqrt{3}}$ (which is not in $\Bbb Q$) we get values closer and closer to the supremum $\frac{2}{3\sqrt{3}}$ but we never achieve it.

Boundedness theorem: The function $f(x)=\frac{1}{x^2-2}$ is defined and continuous for every rational number, but it isn't bounded on a closed interval $[1,2]$.

Mean value theorem: The derivative of the function $f(x)=x-x^3$ is never zero for rational $x$, so it's impossible to find rational $\xi$ such that $f(1)-f(0)=f'(\xi)(1-0)$.