Proving a property of a Dedekind cut

[Hall]

Homework Statement

$A = \{x: x\lt 0 or x^2 \lt 2\}$
$B = \{x: x \gt 0 ~and~x^2 \gt 2\}$

Prove that $(A,B)$ is a Dedekind cut.

Relevant Equations

There are some properties which a cut must have.

A Dedekind cut is a pair $(A,B)$, where $A$ and $B$ are both subsets of rationals. This pair has to satisfy the following properties

1. A is nonempty
2. B is nonempty
3. If $a\in A$ and $c \lt a$ then $c \in A$
4. If $b \in B$ and $c\gt b$ then $c \in B$
5. If $b \not\in B$ and $a\lt b$, then $a \in A$
6. If $a \not\in A$ and $b \gt a$, then $b \in B$
7. For each $a \in A$ there is some $b \gt a$ so that $b \in A$
8. For each $b \in B$ there is some $a \lt b$ so that $a \in B$

For the given cut, I tried to prove Property 5. Here is my attempt:

If $b \not \in B$ then we have:

Case 1: $b \leq 0$
Sub-Case (I): $b=0$. If $a \lt b \implies a \lt 0$, well then $a \in A$.
Sub-Case (II): $b \lt 0$. If $a \lt b \implies b \lt 0$, again $a \in A$

Case 2: $b^2 \leq 2$
Sub-Case (I): $b^2 = 2$. If $a \lt b implies a^2 \lt b^2 = 2$, well then $a \in A$
Sub-Case (II): $b^2 \lt 2$. If $a \lt b \implies a^2 \lt b^2 \lt 2$, again $a \in A$

That proves Property 5.

Am I right? I had a hard time in understanding why would $b \not\in B$ and $b \not\in A$ when all $a$ and $b$ in th properties listed above has to be rational numbers.

 

[FactChecker]

It looks ok to me except that your Case 2 hypothesis and proofs should specify that $b$ is positive and consider both positive and negative cases for $a$. (In fact, maybe you should just change Case 1 to the case of $a \le 0$.)