Confusion regarding the completeness axiom Why would we use square root of 2 in our example to show that the rationals don't have a supremum when square root of 2 is not an element of the rationals? Wouldnt the supremum be the element of the rationals that is greater square root 2 or less than square root 2? 
 A: We can still talk about the set of rationals less than $\sqrt 2$ without making direct reference to $\sqrt 2.$ Write it as $$\{q\in\mathbb Q: \mbox{$q^2<2$ or $q<0$}\}.$$
As to your second sentence, I'm not sure I understand... any element of the rationals is greater than or less than $\sqrt{2}.$ Maybe you meant greater than or equal to and less than or equal to  $\sqrt 2$? But this just means it is equal to $\sqrt{2},$ and the usual proof that $\sqrt{2}$ is irrational shows that this is impossible. 
Showing the set has no least upper bound in the rationals requires one more step than showing there is no rational $q$ with $q^2 =2.$ We also use the fact that the rationals are dense. We know that any upper bound must have $q^2 > 2,$ and then we can show that there is a slightly smaller rational that still squares to greater than two.
A: You don't use the square root of $2$ directly, rather you look at the set of rational numbers $q$ such that $q^2 < 2$, so the definition of that set does not refer to anything other than rational numbers and multiplication of them.
