Why Does the Cut of $\mathbb{Q}$ Correspond to $\sqrt{2}$?

[Mr Davis 97]

Why does the cut $(\{x \in \mathbb{Q} | x^2 \le 2 ~ \text{or} ~x < 0 \}, \{x \in \mathbb{Q} | x^2 \ge 2 ~ \text{and} ~x > 0 \})$ correspond to $\sqrt{2}$, and not just $(\{x \in \mathbb{Q} | x^2 < 2\}, \{x \in \mathbb{Q} | x^2 \ge 2\})$? Why are the additional inequalities involving 0 necessary?

 

[Someone2841]

Consider the proposed cut you have given and rational numbers p=1 and q=-2. While q<p, p is in the lower set ($x^2<2$) and q is in the upper ($x^2≥2$). Therefore it is not a cut at all, since there are numbers in the upper set (-2) that are less than that of the lower set (1).

 

[Mr Davis 97]

I guess to put the question more simply, why can't we represent $\sqrt{2}$ by the cut $\{q \in\mathbb{Q} ~ | ~ q^2 < 2 \}$? Why does it have to be $\{q \in\mathbb{Q} ~ | ~ q^2 < 2 ~\text{or} ~x <0\}$

 

[Someone2841]

A cut is not merely a set or pair of sets. Sure, $\{q \in \mathbb{Q}:q^2<2\}$ is a set whose l.u.b. is $\sqrt{2}$, but it does not meet the criteria for cut. Usually, a cut is an pair of sets A, B that are disjoint and partition $\mathbb{Q}$; furthermore, A is downward closed and contains no greatest element and B is upward closed. All that is really necessary to define a cut, however, is a downward closed set with no greatest element (since that set's complement would always be our other set).

The problem with your set is it is not downward closed. $q^2<2$ is the same as $-\sqrt{2}<q<\sqrt{2}$, but a cut would have to look like $q<\sqrt{2}$ without a lower bound.

If you want to know why it needs to be downward closed, beyond the fact that that's just the definition of a cut, I would say it has to do with uniqueness more than anything. Many sets have a least upper bound of $\sqrt{2}$, but they are all subsets of $\{q \in \mathbb{Q}:q^2<2 \vee q<0\}$. There is no other downward closed set with this l.u.b.