least upper bound for rational numbers In proofs for incompleteness of rational numbers
In demonstrating the rational numbers $\Bbb Q$ are incomplete the following example is most often used.
Let $S$ is the set of all rational numbers $p$ such that $p^2<2$.
The proof goes something like this.
The least upper bound is $\sqrt 2$ and you can show that $\sqrt 2$ is not rational, therefore the rationals $\Bbb Q$ for this set does not exist.
I’ve seen this explanation dozens of times.
My problem is this:
A supremum does not have to be in the set $S$. So, any rational number $> \
\sqrt2~$ is an upper bound e.g. $ 3, 50$ and $1000$. Since we assume no rational numbers exist (yet), the set of upper bounds consists of only rational numbers.  So, we must prove it is impossible to find a least upper bound that is a rational number.  In effect Suppose $A$ is the set of all rational numbers $p$ such that $p^2<2$ and $B$ is the set of all rational numbers $p$ such that $p^2>2$. We want to show that $B$ contains no smallest rational number.
 A: I think I have a solution to my own question.
Some of the commenters may have misunderstood the question. It's not about proving that $\sqrt{2}$  is irrational. That's easy enough. We know $\sqrt{p}$ is irrational for any p that is prime. Since $2$ is prime then $\sqrt{2}$ is irrational.
However, the question is how to show the rationals do not satisfy the Axiom of Completeness (AoC). That is, how to show that for any bounded finite set of rationals S there is no least upper bound sup S in the rationals.
Solution
Try to find a counter example. Consider $\sqrt{2}$ as a cut that splits the rational numbers Q into two sets A and B. A is the set of rationals to the left and B is the set of rationals to the right. This is a Dedekind cut.
Consider set A as consisting of the decimal expansion of $\sqrt{2}$,
A = {a1 , a2, a3 ... } = {1.4, 1.41, 1.414, 1.4142, 1.41421, ... }.
These are strictly rational numbers $a_i\in Q$. The terms $a_i$ get as close to $\sqrt{2}$ as we like but we can never reach it, there is always another rational closer. This is consistent with the rules for the Dedekind cut.
Next is the part I am trying to prove. Here is my attempt:
We need to show that the set $B=\{b_1,b_2,b_3,...\}$ does not have a least element or LUB. To do this, let's take set A and add to each element a positive rational number $e_i$ so that each $b_i$ = $a_i+e_i$ is greater than $\sqrt{2}$. In other words $B=\{b1,b2,b3,...\} = \{a1+e1, a2+e2 ... \}$.
The terms $a_i$ get as close to $\sqrt{2}$ as we like. Choose $ei$ such that each $b_i$ is greater than $\sqrt{2}$. Then the magnitude of $e_i$ required to keep $b_i=a_i+e_i > \sqrt{2}$ gets arbirarily small. Now we need to ask, is there a least element in B they would prevent us from making $e_i$ as small as we like.
The answer is no since the rationals are densely packed. That is, no matter how small we make $e_i$, you can always find a smaller rational number $e_i'$ between $0$ and $e_i$. This means there is no lower bound for the set B. QED
Not sure if I proved it correctly?
