Understanding Dedekind Cuts: Exploring Addition and Negative Numbers

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In summary, the conversation discussed the definition of addition for Dedekind cuts and how it is defined as the sum of elements from two lower sets. It also touched on the neutral cut, the proof for A + S0 = A, and the relationship between Dedekind cuts and hyperrationals. The conversation revealed a mistake in mental arithmetic and led to the discovery of a consistency condition for defining hyperrationals.
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This is a really basic stupid question but has me beat. I am almost certainly missing something but for the life of me cat see what. If A and B are are the lower sets of two Dedekind cuts addition is defined as the sum of the elements (r+s) with r in A and s in B. Is that correct or have I missed something? What I cant see is why if x is a negative number x + 0 = x. As I said I am likely missing something basic but for the life of me can't see what it is. Working on a paper defining the numbers but using a different approach based on hyperrationals and hyperreals. This came up while writing it but like I said has me beat. I will probably feel terrible when told where I went wrong about such a basic question.

Thanks
Bill
 
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Thanks. I will quote what one reference says and spell out my issue. By the definition of addition of Dedekind cuts above we have f(a)+f(b) is the Dedekind cut determined by the set{x+y|x∈A,y∈B}. Well, the set of rational numbers of the form x+y where x<a and y<b and x,y∈Q is exactly the set of rational numbers less than a+b. Therefore, {x+y|x∈A,y∈B}= C, so f(a)+f(b)=f(a+b).

Take x as -1. The Dedekind cut is A1 = {Q| Q < -1}. Take y as 0. The Dedekind cut is A2 = {Q| Q < 0}. Let the element from A1 be -1.000001. Let the element from A2 be -.000001. Then -1.000001 - .000001 = -1.000002 < -1.

I see my error. Dumb mental arithmetic.

Teaches me a lesson. Don't do stuff in your head - when necessary write it out. Yes and I do feel stupid, but will leave it up anyway so others can learn from my mistake.

Thanks
Bill
 
  • #4
As a service to other readers and for us to have common ground, I'll quote the definitions in my book.

##A\subset \mathbb{Q}## is called a (Dedekind) cut if
  1. ##\emptyset \neq A \neq \mathbb{Q}##
  2. ##\alpha \in A\, , \,\beta \ge \alpha \Longrightarrow \beta \in A##
  3. ##A## does not contain a minimal element.

bhobba said:
This is a really basic stupid question but has me beat. I am almost certainly missing something but for the life of me cat see what. If A and B are are the lower sets of two Dedekind cuts addition is defined as the sum of the elements (r+s) with r in A and s in B. Is that correct or have I missed something?
My book says: If ##A## and ##B## are cuts then ##A+B=\{r+s | r \in A, s \in B\},## i.e. the author (Christian Blatter) doesn't refer explicitly to "lower sets" but to cuts instead.

bhobba said:
What I cant see is why if x is a negative number x + 0 = x.
I can't see where negative should matter. The neutral cut is defined as ##S_0=\{\xi\in \mathbb{Q}\,|\,\xi>0\}.##
bhobba said:
As I said I am likely missing something basic but for the life of me can't see what it is. Working on a paper defining the numbers but using a different approach based on hyperrationals and hyperreals. This came up while writing it but like I said has me beat. I will probably feel terrible when told where I went wrong about such a basic question.

Thanks
Bill
The proof for ##A+S_0=A## goes as follows:

We get from the second condition that ##A+S_0\subseteq A.## For a given ##\alpha\in A## there is always an ##\alpha' < \alpha## which is still in ##A## by the third condition. Hence
$$
\alpha=\alpha' +(\alpha - \alpha') \in A+ S_0
$$
so ##A\subseteq A+S_0## and ##A=A+S_0.##
 
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  • #5
if you want a good source to have that walks you through, look at Bloch : Real Numbers and Real Analysis.
 
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Perhaps the Dedekind cuts is not the best way to introduce reals. There is a general way to complete a metric space.

Df: We shall say that ##\{x_n\}\subset\mathbb{Q}## is a Cauchy sequence iff for any (rational) ##\varepsilon>0## there is a number ##N## such that
$$n,m>N\Longrightarrow |x_n-x_m|<\varepsilon.$$

Let ##M## be a set of Cauchy sequences.

Th: The following relation in ##M## is an equivalence relation:
$$\{x_n\}\sim\{y_k\}\Longleftrightarrow |x_n-y_n|\to 0.$$

Df: ##\mathbb{R}:=M/\sim##

UPDATE
Let ##p:M\to\mathbb{R}## be the projection. Define a set ##U_r\subset \mathbb{R},\quad r>0## as follows.
$$p(\{x_k\})\in U_r$$ iff there exist ##K,\varepsilon>0## such that
$$k>K\Longrightarrow |x_k|<r-\varepsilon.$$ The sets ##U_r## form a base of neighbourhoods of the origin.

etc
 
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  • #7
Thanks, Wrrebel.

I am exploring all this in an insights article on what numbers are.

My error was dumb - relying on mental arithmetic instead of doing it on paper. I am so embarrassed.

While writing the article, which is nearly finished, I discovered something very interesting - it's all interrelated to a consistency condition needed to define the hyperrationals well. Each hyperrational must be <,=, > a rational. That forces the finite hyperrationals to be a Cauchy Sequence and a Dedekind cut.

It was a surprising result that emerged while writing the article.

I hope people find reading as interesting as I did in writing it.

Thanks
Bill
 

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