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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
$begingroup$
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
asked Sep 21 '18 at 7:02
mathnormiemathnormie
184
$endgroup$
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
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
asked Sep 21 '18 at 7:02
mathnormiemathnormie
184
$endgroup$
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
elementary-set-theory well-orders
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
asked Sep 21 '18 at 7:02
mathnormiemathnormie
184
asked Sep 21 '18 at 7:02
mathnormiemathnormie
184
184
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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 |
3 Answers
3
active
oldest
votes
$begingroup$
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.
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 |
$begingroup$
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.
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 |
$begingroup$
As a supplement, the definition can be written as
Every nonempty set of nonnegative integers absolutely has a smallest element.
Some sets of nonnegative rationals do not have a smallese element. (note that we only say some sets do not posess this property)
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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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
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.
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.
answered Sep 21 '18 at 7:28
user10354138user10354138
7,4322925
$endgroup$
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.
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
answered Sep 21 '18 at 7:28
user10354138user10354138
7,4322925
7,4322925
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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 |
$begingroup$
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.
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.
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.
answered Sep 21 '18 at 7:09
Sumit MittalSumit Mittal
16812
$endgroup$
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.
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
answered Sep 21 '18 at 7:09
Sumit MittalSumit Mittal
16812
16812
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
@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
$begingroup$
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
$begingroup$
@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 |
$begingroup$
As a supplement, the definition can be written as
Every nonempty set of nonnegative integers absolutely has a smallest element.
Some sets of nonnegative rationals do not have a smallese element. (note that we only say some sets do not posess this property)
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
Every nonempty set of nonnegative integers absolutely has a smallest element.
Some sets of nonnegative rationals do not have a smallese element. (note that we only say some sets do not posess this property)
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
Every nonempty set of nonnegative integers absolutely has a smallest element.
Some sets of nonnegative rationals do not have a smallese element. (note that we only say some sets do not posess this property)
answered Mar 12 at 3:37
王文军 or Wenjun Wang王文军 or Wenjun Wang
534
$endgroup$
As a supplement, the definition can be written as
Every nonempty set of nonnegative integers absolutely has a smallest element.
Some sets of nonnegative rationals do not have a smallese element. (note that we only say some sets do not posess this property)
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
answered Mar 12 at 3:37
王文军 or Wenjun Wang王文军 or Wenjun Wang
534
534
add a comment |
add a comment |
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$begingroup$
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