The axiom, and where Q breaks
The completeness axiom — the least upper bound property — states: every nonempty set of reals that is bounded above has a supremum in R. This single line, added to the ordered-field axioms, characterizes R completely. There is, up to relabeling, exactly one complete ordered field, and it is the continuum we call the real line.
Here is the counterexample that shows Q lacks this property. Let S = { x in Q : x > 0 and x*x < 2 }. This set is nonempty (1 is in it) and bounded above (by 2) *within the rationals*. Yet S has no rational supremum: any rational upper bound can be nudged smaller and still bound S, and any rational below the bound can be nudged larger and still lie in S. The candidate, the square root of 2, is simply not in Q. The least upper bound is missing — there is a literal hole in the rational line.
Two consequences: Archimedes and density
Completeness immediately yields the Archimedean property: the natural numbers are unbounded above in R, so for any real x there is an integer n with n > x. Equivalently, for any epsilon > 0 there is n with 1/n < epsilon. This is the rigorous version of the intuition that you can make 1/n as small as you like.
Theorem (Archimedean property): N is not bounded above in R. Proof by contradiction. Suppose N is bounded above. Then by COMPLETENESS, s = sup N exists in R. Since s is the LEAST upper bound, s - 1 is NOT an upper bound, so there is a natural number n with n > s - 1. But then n + 1 > s, and n + 1 is a natural number. This contradicts s being an upper bound of N. Hence N is unbounded above. QED Corollary: for every epsilon > 0 there is n in N with 1/n < epsilon.
From the Archimedean property comes the density of the rationals: between any two reals a < b there is a rational number. Pick n with 1/n < b - a so the steps of size 1/n are finer than the gap; the first multiple m/n that exceeds a must land below b. So Q is dense in R — the rationals are everywhere, yet they still leave the holes that completeness fills.