Showing ##\sqrt{2}\in\Bbb{R}## using Dedekind cuts

[Terrell]

1. The problem statement, all variables and given
Prove that $\sqrt{2}\in\Bbb{R}$ by showing $x\cdot x=2$ where $x=A\vert B$ is the cut in $\Bbb{Q}$ such that $A=\{r\in\Bbb{Q}\quad \vert \quad r\leq 0 \quad\lor\quad r^2\lt 2\}$.

I believe that I have to show $A^2=L$ however, it seems that $L\nsubseteq A^2$. What am I doing wrong?

Homework Equations

Let $r^*$ denote the cut of $r\in \Bbb{R}$.

The Attempt at a Solution

By definition of multiplying cuts, if $x=A\vert B$, then $x\cdot x$ is $A^2\vert F$ such that $A^2=\{r\in\Bbb{Q}\quad \vert \quad r\leq 0\quad \lor \quad \exists a,a'\in A(r=aa' \land a,a'\gt 0)\}$. Note that $2^*=L \vert U=\{q\in\Bbb{Q}\vert q\lt 2\}\vert\{q\in\Bbb{Q}\vert 2\leq q\}$. If $x\in A^2$, then either $x\leq 0$ or $x\geq 0$. Suppose $x\geq 0$, then $\exists ]alpha,\alpha'\in A$ s.t. $x=\alpha\alpha'$ and $\alpha,\alpha'\gt 0$. Keep in mind that $\alpha,\alpha'\in A$ means that $\alpha^2,(\alpha')^2<2$ so $\alpha^2(\alpha')^2<4 \Longrightarrow \alpha\alpha'\lt 2$. Thus, $x^2 \lt 2$ which implies $x\in L$. On the other hand, if $x\leq 0$, then clearly, $x\in L$. In all cases, $A^2\subseteq L$.

 

[verty]

I'm thinking you have to show that every member of $A^2 < 2$. Are you able to do that? I would give a hint but I think this already is.

 

[mathwonk]

if you want what i think is a nicer approach, the cauchy sequences approach due to cantor, it is in fundamentals of real analysis by sterling berberian. if you want to stick with the dedekind approach, i suggest you make things easier on yourself and consider only cuts of the positive rationals, where you can just assume the lower part of the square of a (positive) cut consists of the squares of the elements in the lower part of the original cut.

 

[Terrell]

I'm thinking you have to show that every member of $A^2 < 2$. Are you able to do that? I would give a hint but I think this already is.

I believe I did, then concluded with $A^2\subseteq L$. I'm having trouble with $L\subseteq A^2$. Since defining $L$ as the cut of $2$, $L=\{q\in\Bbb{Q}:q<2\}$ meant $3/2\in L$, but $3/2\notin A^2$. Thanks!

 

[Terrell]

i suggest you make things easier on yourself and consider only cuts of the positive rationals, where you can just assume the lower part of the square of a (positive) cut consists of the squares of the elements in the lower part of the original cut.

I believe this is the approach I took. Thank you.

 

[Terrell]

Is it correct that the l.u.b. of $A^2$ is $4$? Which would let me argue that $\{a\in\Bbb{Q}:q\lt 2\}\subseteq A^2$?

 

[verty]

I believe I did, then concluded with $A^2\subseteq L$. I'm having trouble with $L\subseteq A^2$. Since defining $L$ as the cut of $2$, $L=\{q\in\Bbb{Q}:q<2\}$ meant $3/2\in L$, but $3/2\notin A^2$. Thanks!

Sorry, I think this is more complicated than I expected. But regarding $3 \over 2$, ${3 \over 2} = {9 \over 8} \times {4 \over 3}$.

 

[Terrell]

Sorry, I think this is more complicated than I expected. But regarding $3 \over 2$, ${3 \over 2} = {9 \over 8} \times {4 \over 3}$.

But $3/2$ cannot be in $A^2$, right?

 

[verty]

But $3/2$ cannot be in $A^2$, right?

${9 \over 8}$ is in A and ${4 \over 3}$ is in A, therefore ${3 \over 2} = {9 \over 8} \times {4 \over 3}$ is in $A^2$. I don't yet know how to turn this idea into a proof. It seems difficult to select the right numbers.

PS. It seems you should be able to assume that cuts have the least upper bound property because that is usually proven before multiplication is defined.

 

[Terrell]

9898{9 \over 8} is in A and 4343{4 \over 3} is in A, therefore 32=98×4332=98×43{3 \over 2} = {9 \over 8} \times {4 \over 3} is in A2

Right. I should have tried with a calculator.

It seems difficult to select the right numbers.

It's where I am having difficulty too.
Also, the book hinted that I can use the following to solve the problem. Though, I haven't figured how to use it.