Proof: To prove this theorem, we need to be able to show that there exists a subset of the rationals that is bounded above but doesn't have a rational lub or that there exists a subset of the rationals that is bounded below but doesn't have a rational glb.

Consider the set `S={n in QQ : n>0 and n^2<2}`. `S` is a subset of the rationals. We will prove that this set does not have a rational lub. Since `sqrt(2)` is clearly an upper bound for this set, but `sqrt(2)` is irrational, a rational lub (call it `p`) would have to be near to, but greater or less than `sqrt(2)`. We'll consider each case.

1. `0 < p^2 < 2`. If `0 < p^2 < 2` then `p in S` and so `p` must be the maximum of `S`. Then, by definition there is no element of `S` greater than `p`. But consider `q = p + (2-p^2)/(p+2) = (2p+2)/(p+2)`. If it can be proven that `q > p` and `q^2 < 2`, then `q in S`, and that would contradict `p` being the lub of `S`. Since `p^2 < 2`, then `2-p^2 > 0`, and since `p > 0`, then `p+2 > 0`. Therefore, `(2-p^2)/(p+2)` is positive, and so `q > p`. It only remains to show that `q^2 < 2`. `q^2-2 = ((2p+2)^2)/((p+2)^2) - 2 = (4p^2+8p+4)/((p+2)^2) - 2 = (4p^2+8p+4-(2p^2+8p+8))/((p+2)^2) = (2p^2-4)/((p+2)^2) = (2(p^2-2))/((p+2)^2)`. Since `p^2 < 2`, `p^2-2 < 0`, so the numerator is negative. And since `p > 0`, the denominator is positive. Thus, `q^2-2 < 0`, so `q^2 < 2`. So `S` does not have a maximum, and if it has a rational lub, it cannot be less than `sqrt(2)`.
2. `p^2 > 2`. If `p^2 > 2` and `p` is the lub of `S`, then there can exist no rational number `q` such that `sqrt(2) < q < p`. But consider the same formula as above, `q = p + (2-p^2)/(p+2) = (2p+2)/(p+2)`. If it can be proven that `q < p` and `q^2 > 2`, then that would contradict `p` being the lub of `S`. Since `p^2 > 2`, then `2-p^2 < 0`, and since `p > 0`, then `p+2 > 0`. Therefore, `(2-p^2)/(p+2)` is negative, so `q < p`. It only remains to show that `q^2 > 2`. `q^2-2 = (2(p^2-2))/((p+2)^2)`, and since `p^2 > 2`, then the numerator is positive, and since `p > 0`, the denominator is positive. Therefore, `q^2-2 > 0`, so `q^2 > 2`. So `S` cannot have a rational lub greater than `sqrt(2)`.

Since `S` cannot have a rational lub that is less than, equal to, or greater than `sqrt(2)`, `S` cannot have a rational lub. Therefore, `QQ` does not have the completeness property.