My Math Forum  

Go Back   My Math Forum > College Math Forum > Number Theory

Number Theory Number Theory Math Forum


Thanks Tree4Thanks
  • 1 Post By Maschke
  • 2 Post By Petek
  • 1 Post By complicatemodulus
Reply
 
LinkBack Thread Tools Display Modes
July 11th, 2017, 02:42 PM   #1
Newbie
 
Joined: Jul 2017
From: Wahkiacus, WA, USA

Posts: 2
Thanks: 0

Question about Dedekind's paper "Continuity and Irrational Numbers"

Hi all,

I've been working through this paper (available at https://www.gutenberg.org/files/21016/21016-pdf.pdf) and am confused about a couple points. I was hoping someone here could clarify things for me.

I use his terminology, which might be different from the current terminology.

In section 6 (at the bottom of page 10 in the pdf), Dedekind is reducing addition with real numbers to addition with rationals (if I understand him correctly). At one point in the proof, he states that α + β = c2 + p, where α and β are real numbers and c2 and p are rational numbers. But since α and β can be irrational numbers and as far as I understand an irrational number cannot be the sum of two rationals, why isn't this equation invalid? What am I missing?

Similarly, a couple lines later he states that α - 1/2p is a number in A1 and β - 1/2p is a number in B2. But if I understand his notation correctly, A1 and B1 are classes in the domain of rational numbers and α - 1/2p and β - 1/2p can be irrational numbers. Again, I can't figure out what I'm missing.

Thanks!
Yosef is offline  
 
July 11th, 2017, 03:25 PM   #2
Senior Member
 
Joined: Aug 2012

Posts: 1,521
Thanks: 364

Quote:
Originally Posted by Yosef View Post
At one point in the proof, he states that α + β = c2 + p, where α and β are real numbers and c2 and p are rational numbers. But since α and β can be irrational numbers and as far as I understand an irrational number cannot be the sum of two rationals, why isn't this equation invalid? What am I missing?
Didn't look at the paper but the sum of two irrationals can be rational, which seems to be the case you're describing. What's the context of Dedekind's argument?
Thanks from Yosef
Maschke is online now  
July 11th, 2017, 11:21 PM   #3
Senior Member
 
Joined: Nov 2010
From: Berkeley, CA

Posts: 174
Thanks: 35

Math Focus: Elementary Number Theory, Algebraic NT, Analytic NT
Dedekind defines a real number to be what he calls a cut (now called a Dedekind cut). A cut is a subset of the rational numbers. In the section you cite, Dedekind is defining the sum of two cuts. This sum also will be a cut and hence will be a subset of the rationals. Leaving out some details, that's why the sum of two reals can be expressed in terms of rationals.
Thanks from topsquark and Yosef
Petek is offline  
July 12th, 2017, 03:40 AM   #4
Banned Camp
 
Joined: Dec 2012

Posts: 1,028
Thanks: 24

$(\sqrt(2)-1)+(2-\sqrt(2))=?$
Thanks from Yosef
complicatemodulus is offline  
July 12th, 2017, 10:46 AM   #5
Newbie
 
Joined: Jul 2017
From: Wahkiacus, WA, USA

Posts: 2
Thanks: 0

Thanks for the replies!

Rereading the section with your points in mind, I think what he's doing is defining the operation of adding two reals as producing a certain cut from two other cuts. Then the equations are only dealing with rational numbers, to show that this newly defined operation doesn't contradict any of the definitions regarding rational numbers.
Yosef is offline  
Reply

  My Math Forum > College Math Forum > Number Theory

Tags
dedekind, irrationals, paper, question, rationals



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
How to prove that "i-(ioota)" is irrational? muhammadmasood Algebra 11 September 6th, 2012 02:07 PM
A "simple" application of dirac delta "shift theorem"...help SedaKhold Calculus 0 February 13th, 2012 11:45 AM
sample exeriment-need help finding "statistic" and "result" katie0127 Advanced Statistics 0 December 3rd, 2008 01:54 PM
How to prove that "i-(ioota)" is irrational? muhammadmasood Abstract Algebra 6 December 31st, 1969 04:00 PM





Copyright © 2017 My Math Forum. All rights reserved.