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

• Terrell
In summary: 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
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##.

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.

Terrell
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 and fresh_42
verty said:
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!

mathwonk said:
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.

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##?

Terrell said:
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
verty said:
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?

Terrell said:
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.

Last edited:
Terrell
verty said:
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.
verty said:
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.

#### Attachments

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## What is a Dedekind cut?

A Dedekind cut is a mathematical concept used to define the real numbers. It is a partition of the rational numbers into two non-empty sets, such that all elements in the first set are less than any element in the second set. This partition represents the real number that is being defined.

## How does a Dedekind cut show that ##\sqrt{2}\in\Bbb{R}##?

A Dedekind cut can be used to show that ##\sqrt{2}\in\Bbb{R}## by creating a partition of the rational numbers into two sets - one set containing all rational numbers less than ##\sqrt{2}## and one set containing all rational numbers greater than ##\sqrt{2}##. This partition represents the real number ##\sqrt{2}##, which can be shown to satisfy all the properties of a Dedekind cut.

## What are the properties of a Dedekind cut?

A Dedekind cut must satisfy three properties: it must be non-empty, it must have no largest element, and it must have a smallest upper bound. These properties ensure that the cut represents a unique real number and that it is consistent with the properties of the real numbers.

## Can Dedekind cuts be used to define other irrational numbers?

Yes, Dedekind cuts can be used to define any real number, including other irrational numbers such as ##\pi## and ##e##. The process is the same as for ##\sqrt{2}## - creating a partition of the rational numbers that represents the desired real number.

## What is the significance of using Dedekind cuts to define real numbers?

Dedekind cuts provide a rigorous and precise way to define real numbers, which are essential in many areas of mathematics and science. They allow for a clear understanding of the properties and relationships between real numbers, and provide a foundation for more advanced mathematical concepts.

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