- #1

Hall

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- 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*

*A is nonempty**B is nonempty**If ##a\in A## and ##c \lt a## then ##c \in A##**If ##b \in B## and ## c\gt b## then ##c \in B##**If ##b \not\in B## and ## a\lt b##, then ##a \in A##**If ##a \not\in A## and ##b \gt a##, then ##b \in B##**For each ##a \in A## there is some ## b \gt a ## so that ##b \in A##**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.