IOW, to approximate an irrational number to 10 decimal places... use 10 decimal places? Geen, use 11 decimal places just to be double-sure and not have to worry.
What am I missing?
A different measure of goodness takes into consideration not only the distance between the real number and its rational approximation, but how big the denominator of the approximation is. Note that, in the scheme you describe, a denominator of q guarantees an error of 1/q. But if you are allowed to choose both the numerator and the denominator of the rational approximation, then perhaps you can make the denominator smaller?
It turns out you can! Dirichlet showed that for every irrational alpha there are infinitely many "good approximations" of the form p/q where |alpha - p/q| < 1/q^2. It turns out the golden ratio shows that you can't do better than 1/(sqrt(5) * q^2) if you want the result to hold for every irrational number. This leads to the problem of identifying "interesting" sets of irrational numbers that allow approximations substantially better than Dirichlet's bound (this is the problem of Diophantine approximation [0]). Klaus Roth got his Fields Medal primarily for a result in this area [1].
[0] https://en.wikipedia.org/wiki/Diophantine_approximation [1] https://en.wikipedia.org/wiki/Roth%27s_theorem
Getting an approximation p/q of x within 1/q is easy. Getting it within 1/q^2 takes work. Within 1/(10 q^2) is only sometimes possible.
For any q?
Yes.
The hardest real number to approximate with rationals is (1+sqrt(5))/2. Its approximations that come closest are of the form fib(n+1)/fib(n) where fib(n) is the n'th fibonacci number. The error converges to 1/(sqrt(5) fib(n)^2). None are within 1/(10 q^2).
To learn more about this, read up on continued fractions.