I have not much experience with proofs, so any feedback will be welcome! If you could me with the last step it would be great!
A Dedekind Cut is defined as a set $D \subset \mathbb{Q}$ such that
- $D \neq \mathbb{Q}$ and $D \neq \emptyset$
- Let $x \in D, $ if $y \in \mathbb{Q}$ and $y > x$, then $y \in D$
- Let $x \in D$. Then there is some $y \in D$ such that $x>y$
We seek to prove that the set $T=\{x \in \mathbb{Q} \mid x>0 \text{ and } x^2>2\}$ is a Dedekind cut.
Proof:
Any $y \in \mathbb{Q}, y<0$ is not in $T$, hence $T \neq \mathbb{Q}$. Any $y \in \mathbb{Q}, y>2$ will be in $T$, since $y^2>2\cdot y=y \cdot 2>1\cdot 2=2$. So $T \neq \emptyset $. Hence 1) is fulfilled.
Let $x \in T$, and let $y \in \mathbb{Q} , y>x$. Then since $x,y>0$, $y^2>x \cdot y=y \cdot x>x^2>2$. Hence $y \in T$. Thus 2) is fulfilled.
For 3), I'm struggling to prove that for every $x \in T$, there exists a rational number $y<x$ such that $y^2>2$. If anyone could help me with this step I would be really grateful!
Choose $\delta\in\mathbb{Q}$, $\delta>0$ such that $x^2-\delta>2$. Choose $k\in\mathbb{Z}$, $k>0$ such that $$ \frac{2x}{k}<\delta. $$ Define $$ y=x-\frac{1}{k}. $$ Then we have $x>y>0$, $y\in\mathbb{Q}$ and $$ y^2=\left(x-\frac{1}{k}\right)^2= x^2-\frac{2x}{k}+\frac{1}{k^2}> x^2-\delta>2. $$