Why is the rational number system inadequate for analysis?
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).
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 $f(1)-f(0)=f'(\xi)(1-0)$.
Consider the sequence $$ x_{n+1} = \dfrac12\left(x_n+\dfrac{2}{x_n}\right) $$
When $x_0$ is rational, all $x_n$ are rational.
The sequence is a Cauchy sequence but does not converge in the rationals because the limit is $\sqrt 2$ when $x_0>0$.
For further discussion of how the rationals fail to be adequate for analysis, see A Companion to Analysis by Körner. Chapter 1 is freely available and contains some of that discussion.