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Well ordering principle for rationals

Is there a known well ordering of the reals?Well ordered set…confusingIs it possible to extend well ordering principle/induction to all well ordered sets?Using the well ordering principle to prove a certain property of an integerPrinciple of mathematical induction to prove well ordering principle for set of rationals.Well ordering theorem, partial orderingdetermining if a set is well ordered setWell ordering of $N^N$Well ordering principleWell ordering principle question

0

$begingroup$

Why can positive rationals be not well ordered ? I we define the relation to be greater than(>) then every subset will have a least element . Or why are positive or even integers not well ordered . By the same logic we can always find a least element in any subset . I know I am wrong at some very fundamental point but please explain me ?

elementary-set-theory well-orders

share|cite|improve this question

asked Sep 21 '18 at 7:02

mathnormiemathnormie

184

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Positive integers equipped with usual order are well ordered. Negative integers are not. E.g. the set ${-nmid n=1,2,3,dots}$ has no least element.
  $endgroup$
  – drhab
  Sep 21 '18 at 7:52
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Why can positive rationals be not well ordered ? I we define the relation to be greater than(>) then every subset will have a least element . Or why are positive or even integers not well ordered . By the same logic we can always find a least element in any subset . I know I am wrong at some very fundamental point but please explain me ?

elementary-set-theory well-orders

share|cite|improve this question

asked Sep 21 '18 at 7:02

mathnormiemathnormie

184

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Positive integers equipped with usual order are well ordered. Negative integers are not. E.g. the set ${-nmid n=1,2,3,dots}$ has no least element.
  $endgroup$
  – drhab
  Sep 21 '18 at 7:52
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Why can positive rationals be not well ordered ? I we define the relation to be greater than(>) then every subset will have a least element . Or why are positive or even integers not well ordered . By the same logic we can always find a least element in any subset . I know I am wrong at some very fundamental point but please explain me ?

elementary-set-theory well-orders

share|cite|improve this question

asked Sep 21 '18 at 7:02

mathnormiemathnormie

184

$endgroup$

share|cite|improve this question

asked Sep 21 '18 at 7:02

mathnormiemathnormie

184

share|cite|improve this question

asked Sep 21 '18 at 7:02

mathnormiemathnormie

184

asked Sep 21 '18 at 7:02

mathnormiemathnormie

184

asked Sep 21 '18 at 7:02

mathnormiemathnormie

184

3 Answers
3

active

oldest

votes

2

$begingroup$

Your claim isn't true.

The positive rationals can be well-ordered

Since $mathbb{Q}$ bijects with $mathbb{N}$, the well-ordering on $mathbb{N}$ will induce a well-ordering on $mathbb{Q}$ and hence on the positive rationals.

However,

The usual ordering of positive rationals is not a well-ordering

The usual ordering is, of course, $frac{a}{b}>frac{c}{d}$ if and only if $ad>bc$ (where $a,b,c,d$ are positive integers).

If it is a well-ordering, then there is a least positive rational $p/q$. But halving it gives a smaller positive rational $p/(2q)$, so $p/q$ can't be the least, contradiction.

share|cite|improve this answer

answered Sep 21 '18 at 7:28

user10354138user10354138

7,4322925

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To clarify your last paragraph: the set of rationals ${qmid qinmathbb Q, q>0}$ has no smallest element.
  $endgroup$
  – Jack M
  Sep 21 '18 at 7:37
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please clarify if the subset is having fixed elements with least element p/q then how is p/2q the least when it was not in the subset initially , are you taking open interval , what about even integers
  $endgroup$
  – mathnormie
  Sep 21 '18 at 7:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The even integers $2mathbb{Z}$ under the usual ordering is not well-ordered. Indeed, the subset ${min 2mathbb{Z}:text{$m$ is negative}}$ has no least element. That is to say, there is no even integer $m_0$ smaller than every element of the chosen subset.
  $endgroup$
  – Alberto Takase
  Oct 5 '18 at 19:52
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Positive integers are well ordered but positive rationals are not because for well ordered, every non empty subset must have least element( least element must belong to subset and there is difference between least element and greatest lower bound). There are many subsets which have no least point in positive rationals like the subset ${1, 1/2, 1/4, 1/8, 1/16, ...}$ has no least element or the set of all positive rationals greater than any irrational number.

share|cite|improve this answer

answered Sep 21 '18 at 7:09

Sumit MittalSumit Mittal

16812

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In the example you mean that the subset was an open interval so we cannot find least element , what about positive integers
  $endgroup$
  – mathnormie
  Sep 21 '18 at 7:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathnormie: Read the first five words in the answer again.
  $endgroup$
  – Hans Lundmark
  Sep 21 '18 at 8:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I meant even integers , my teacher say its not well ordered
  $endgroup$
  – mathnormie
  Sep 21 '18 at 8:15
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathnormie "my teacher say" is neither a good argument nor helpful. Maybe you should post the definition used by your teacher and someone can point a mistake either on the definition or on the argument.
  $endgroup$
  – Mefitico
  Oct 5 '18 at 16:26
  

  
  
  
  

add a comment | 

0

$begingroup$

As a supplement, the definition can be written as

1. Every nonempty set of nonnegative integers absolutely has a smallest element.




2. Some sets of nonnegative rationals do not have a smallese element. (note that we only say some sets do not posess this property)

share|cite|improve this answer

answered Mar 12 at 3:37

王文军 or Wenjun Wang王文军 or Wenjun Wang

534

$endgroup$

add a comment | 

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2

$begingroup$

Your claim isn't true.

The positive rationals can be well-ordered

Since $mathbb{Q}$ bijects with $mathbb{N}$, the well-ordering on $mathbb{N}$ will induce a well-ordering on $mathbb{Q}$ and hence on the positive rationals.

However,

The usual ordering of positive rationals is not a well-ordering

The usual ordering is, of course, $frac{a}{b}>frac{c}{d}$ if and only if $ad>bc$ (where $a,b,c,d$ are positive integers).

If it is a well-ordering, then there is a least positive rational $p/q$. But halving it gives a smaller positive rational $p/(2q)$, so $p/q$ can't be the least, contradiction.

share|cite|improve this answer

answered Sep 21 '18 at 7:28

user10354138user10354138

7,4322925

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To clarify your last paragraph: the set of rationals ${qmid qinmathbb Q, q>0}$ has no smallest element.
  $endgroup$
  – Jack M
  Sep 21 '18 at 7:37
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please clarify if the subset is having fixed elements with least element p/q then how is p/2q the least when it was not in the subset initially , are you taking open interval , what about even integers
  $endgroup$
  – mathnormie
  Sep 21 '18 at 7:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The even integers $2mathbb{Z}$ under the usual ordering is not well-ordered. Indeed, the subset ${min 2mathbb{Z}:text{$m$ is negative}}$ has no least element. That is to say, there is no even integer $m_0$ smaller than every element of the chosen subset.
  $endgroup$
  – Alberto Takase
  Oct 5 '18 at 19:52
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Your claim isn't true.

The positive rationals can be well-ordered

Since $mathbb{Q}$ bijects with $mathbb{N}$, the well-ordering on $mathbb{N}$ will induce a well-ordering on $mathbb{Q}$ and hence on the positive rationals.

However,

The usual ordering of positive rationals is not a well-ordering

The usual ordering is, of course, $frac{a}{b}>frac{c}{d}$ if and only if $ad>bc$ (where $a,b,c,d$ are positive integers).

If it is a well-ordering, then there is a least positive rational $p/q$. But halving it gives a smaller positive rational $p/(2q)$, so $p/q$ can't be the least, contradiction.

share|cite|improve this answer

answered Sep 21 '18 at 7:28

user10354138user10354138

7,4322925

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To clarify your last paragraph: the set of rationals ${qmid qinmathbb Q, q>0}$ has no smallest element.
  $endgroup$
  – Jack M
  Sep 21 '18 at 7:37
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please clarify if the subset is having fixed elements with least element p/q then how is p/2q the least when it was not in the subset initially , are you taking open interval , what about even integers
  $endgroup$
  – mathnormie
  Sep 21 '18 at 7:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The even integers $2mathbb{Z}$ under the usual ordering is not well-ordered. Indeed, the subset ${min 2mathbb{Z}:text{$m$ is negative}}$ has no least element. That is to say, there is no even integer $m_0$ smaller than every element of the chosen subset.
  $endgroup$
  – Alberto Takase
  Oct 5 '18 at 19:52
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Your claim isn't true.

The positive rationals can be well-ordered

Since $mathbb{Q}$ bijects with $mathbb{N}$, the well-ordering on $mathbb{N}$ will induce a well-ordering on $mathbb{Q}$ and hence on the positive rationals.

However,

The usual ordering of positive rationals is not a well-ordering

The usual ordering is, of course, $frac{a}{b}>frac{c}{d}$ if and only if $ad>bc$ (where $a,b,c,d$ are positive integers).

If it is a well-ordering, then there is a least positive rational $p/q$. But halving it gives a smaller positive rational $p/(2q)$, so $p/q$ can't be the least, contradiction.

share|cite|improve this answer

answered Sep 21 '18 at 7:28

user10354138user10354138

7,4322925

$endgroup$

share|cite|improve this answer

answered Sep 21 '18 at 7:28

user10354138user10354138

7,4322925

answered Sep 21 '18 at 7:28

user10354138user10354138

7,4322925

answered Sep 21 '18 at 7:28

user10354138user10354138

7,4322925

1

$begingroup$

Positive integers are well ordered but positive rationals are not because for well ordered, every non empty subset must have least element( least element must belong to subset and there is difference between least element and greatest lower bound). There are many subsets which have no least point in positive rationals like the subset ${1, 1/2, 1/4, 1/8, 1/16, ...}$ has no least element or the set of all positive rationals greater than any irrational number.

share|cite|improve this answer

answered Sep 21 '18 at 7:09

Sumit MittalSumit Mittal

16812

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In the example you mean that the subset was an open interval so we cannot find least element , what about positive integers
  $endgroup$
  – mathnormie
  Sep 21 '18 at 7:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathnormie: Read the first five words in the answer again.
  $endgroup$
  – Hans Lundmark
  Sep 21 '18 at 8:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I meant even integers , my teacher say its not well ordered
  $endgroup$
  – mathnormie
  Sep 21 '18 at 8:15
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathnormie "my teacher say" is neither a good argument nor helpful. Maybe you should post the definition used by your teacher and someone can point a mistake either on the definition or on the argument.
  $endgroup$
  – Mefitico
  Oct 5 '18 at 16:26
  

  
  
  
  

add a comment | 

1

$begingroup$

Positive integers are well ordered but positive rationals are not because for well ordered, every non empty subset must have least element( least element must belong to subset and there is difference between least element and greatest lower bound). There are many subsets which have no least point in positive rationals like the subset ${1, 1/2, 1/4, 1/8, 1/16, ...}$ has no least element or the set of all positive rationals greater than any irrational number.

share|cite|improve this answer

answered Sep 21 '18 at 7:09

Sumit MittalSumit Mittal

16812

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In the example you mean that the subset was an open interval so we cannot find least element , what about positive integers
  $endgroup$
  – mathnormie
  Sep 21 '18 at 7:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathnormie: Read the first five words in the answer again.
  $endgroup$
  – Hans Lundmark
  Sep 21 '18 at 8:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I meant even integers , my teacher say its not well ordered
  $endgroup$
  – mathnormie
  Sep 21 '18 at 8:15
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathnormie "my teacher say" is neither a good argument nor helpful. Maybe you should post the definition used by your teacher and someone can point a mistake either on the definition or on the argument.
  $endgroup$
  – Mefitico
  Oct 5 '18 at 16:26
  

  
  
  
  

add a comment | 

$begingroup$

Positive integers are well ordered but positive rationals are not because for well ordered, every non empty subset must have least element( least element must belong to subset and there is difference between least element and greatest lower bound). There are many subsets which have no least point in positive rationals like the subset ${1, 1/2, 1/4, 1/8, 1/16, ...}$ has no least element or the set of all positive rationals greater than any irrational number.

share|cite|improve this answer

answered Sep 21 '18 at 7:09

Sumit MittalSumit Mittal

16812

$endgroup$

share|cite|improve this answer

answered Sep 21 '18 at 7:09

Sumit MittalSumit Mittal

16812

answered Sep 21 '18 at 7:09

Sumit MittalSumit Mittal

16812

answered Sep 21 '18 at 7:09

Sumit MittalSumit Mittal

16812

0

$begingroup$

As a supplement, the definition can be written as

1. Every nonempty set of nonnegative integers absolutely has a smallest element.




2. Some sets of nonnegative rationals do not have a smallese element. (note that we only say some sets do not posess this property)

share|cite|improve this answer

answered Mar 12 at 3:37

王文军 or Wenjun Wang王文军 or Wenjun Wang

534

$endgroup$

add a comment | 

0

$begingroup$

As a supplement, the definition can be written as

1. Every nonempty set of nonnegative integers absolutely has a smallest element.




2. Some sets of nonnegative rationals do not have a smallese element. (note that we only say some sets do not posess this property)

share|cite|improve this answer

answered Mar 12 at 3:37

王文军 or Wenjun Wang王文军 or Wenjun Wang

534

$endgroup$

add a comment | 

$begingroup$

As a supplement, the definition can be written as

1. Every nonempty set of nonnegative integers absolutely has a smallest element.




2. Some sets of nonnegative rationals do not have a smallese element. (note that we only say some sets do not posess this property)

share|cite|improve this answer

answered Mar 12 at 3:37

王文军 or Wenjun Wang王文军 or Wenjun Wang

534

$endgroup$

share|cite|improve this answer

answered Mar 12 at 3:37

王文军 or Wenjun Wang王文军 or Wenjun Wang

534

answered Mar 12 at 3:37

王文军 or Wenjun Wang王文军 or Wenjun Wang

534

answered Mar 12 at 3:37

王文军 or Wenjun Wang王文军 or Wenjun Wang

534