# Analysis Question Involving Real Numbers

So I’ve been sick and I’ve missed a couple of lectures of my Analysis class. But I didn’t want to be too behind in lecture tomorrow, so I was trying to catch up by reading my textbook and solving some problems.

Here’s one I was having some trouble with:

Prove $\sqrt2\in\mathbb R$ by showing $x^2 = 2$ where $x = A|B$ is a cut w/ $A = \{r \in \mathbb Q : r \leqslant 0 \text{ or } r^2 < 2\}$.

Any help would be appreciated!

Thank you! 🙂

#### Solutions Collecting From Web of "Analysis Question Involving Real Numbers"

Take $A$ and $B=A^c$. We have to show first that $(A|B)$ is a cut. For this we need to show that for all $x\in A$ and $y\in B$ we have $x\leq y$.

Assume $x\in A$ and $y\in B$. Then, $y>0$ and $y^2\geq2$. If $x>y$ then $x>0$ and it follows that $x^2>xy>y^2\geq2$. Contradiction. Therefore $x\leq y$. Therefore $A,B$ is a cut.

Now consider $(A_1|B_1)=(A|B)^2$, i.e. where $B_1=\{z=y_1y_2:\ y_1,y_2\in B\}$ and $A_1=B_1^c$.

Observe that $B_1\supset \{y>0:\ y\geq2\}$.

Conversely, if $y\in B_1$ then $y=y_1y_2$ with $y_1,y_2\in B$. Therefore $y_1,y_2>0$ and $y_1^2,y_2^2\geq2$. Then $(y_1y_2)^2\geq4$ and $y_1y_2>0$. Therefore $y_1y_2\geq2$. Hence $B_1=\{y>0:\ y\geq2\}$. This is, $(A_1|B_1)=2$.

Hint

I think that you have to apply the definition of Dedekind cut, which must be “something like” :

Let $D \subseteq \mathbb Q$ be a set. The set $D$ is a Dedekind cut if the following
three properties hold :

a. $D \ne \emptyset$ and $D \ne \mathbb Q$.

b. Let $r \in D$; if $y \in \mathbb Q$ and $y \ge r$, then $y \in D$.

c. Let $r \in D$; then there is some $y \in D$ such that $y < r$.

With your symbolism, we have to consider $A|B$ where $B=A^C$ and apply the above definition with $B$ in place of $D$.

See :

• Ethan Bloch, The real numbers and real analysis (2011), Ch.1.6 Dedekind Cuts, page 33-on.