I don't understand why this axiom makes the difference between a completely ordered field and an ordered...
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I've been reading about ordered fields and completely ordered fields, and I'm stuck on the difference between the rational and real numbers that makes the rational numbers an ordered field and the real numbers a completely ordered field. I see that the reals are complete, and the rationals aren't, but how does that correlate with one being a completely ordered field and the other just an ordered field?
My book, "A Friendly Introduction to Analysis" by Witold A. J. Kosmala, gives the following so-called 'completeness axiom' that separates a completely ordered field from an ordered one:
I understand what the axiom says, but I have a hard time grasping what it really means for the field that it applies to, and what it really means for the difference between a completely ordered and an ordered field. Furthermore, why do the rationals not satisfy this axiom? My thoughts were that a subset of the rationals also contain a supremum, namely the greatest rational number in that subset. What part don't I understand?
All help is appreciated.
field-theory real-numbers rational-numbers ordered-fields
edited Nov 14 at 20:02
Eric Wofsey
175k12202326
asked Nov 14 at 19:46
Benjamin Caris
133
add a comment |
up vote
2
down vote
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I've been reading about ordered fields and completely ordered fields, and I'm stuck on the difference between the rational and real numbers that makes the rational numbers an ordered field and the real numbers a completely ordered field. I see that the reals are complete, and the rationals aren't, but how does that correlate with one being a completely ordered field and the other just an ordered field?
My book, "A Friendly Introduction to Analysis" by Witold A. J. Kosmala, gives the following so-called 'completeness axiom' that separates a completely ordered field from an ordered one:
I understand what the axiom says, but I have a hard time grasping what it really means for the field that it applies to, and what it really means for the difference between a completely ordered and an ordered field. Furthermore, why do the rationals not satisfy this axiom? My thoughts were that a subset of the rationals also contain a supremum, namely the greatest rational number in that subset. What part don't I understand?
All help is appreciated.
field-theory real-numbers rational-numbers ordered-fields
edited Nov 14 at 20:02
Eric Wofsey
175k12202326
asked Nov 14 at 19:46
Benjamin Caris
133
the set of rationals ${x in mathbb{Q} | x^2<2}$ has no supremum. There is no greatest rational number in this set, nor outside the set neither.
– ziggurism
Nov 14 at 19:50
When you say "I see that the reals are complete, and the rationals aren't", what is your understanding of "complete"? Is there some different definition you have? Or do you just mean "complete" in some vague intuitive sense?
– Eric Wofsey
Nov 14 at 20:04
@EricWofsey I understand it now, but just for clarification: yes, I meant that in the most intuitive way you could think of - probably not the best idea, but I had a hard time explaining my question properly. I meant it as it's written down in the book I'm using: "Intuitively, if we were to line up the rational numbers, there would be plenty of "holes" in them. Real numbers, on the other hand, would not have that property."
– Benjamin Caris
Nov 14 at 20:28
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I've been reading about ordered fields and completely ordered fields, and I'm stuck on the difference between the rational and real numbers that makes the rational numbers an ordered field and the real numbers a completely ordered field. I see that the reals are complete, and the rationals aren't, but how does that correlate with one being a completely ordered field and the other just an ordered field?
My book, "A Friendly Introduction to Analysis" by Witold A. J. Kosmala, gives the following so-called 'completeness axiom' that separates a completely ordered field from an ordered one:
I understand what the axiom says, but I have a hard time grasping what it really means for the field that it applies to, and what it really means for the difference between a completely ordered and an ordered field. Furthermore, why do the rationals not satisfy this axiom? My thoughts were that a subset of the rationals also contain a supremum, namely the greatest rational number in that subset. What part don't I understand?
All help is appreciated.
field-theory real-numbers rational-numbers ordered-fields
edited Nov 14 at 20:02
Eric Wofsey
175k12202326
asked Nov 14 at 19:46
Benjamin Caris
133
I've been reading about ordered fields and completely ordered fields, and I'm stuck on the difference between the rational and real numbers that makes the rational numbers an ordered field and the real numbers a completely ordered field. I see that the reals are complete, and the rationals aren't, but how does that correlate with one being a completely ordered field and the other just an ordered field?
My book, "A Friendly Introduction to Analysis" by Witold A. J. Kosmala, gives the following so-called 'completeness axiom' that separates a completely ordered field from an ordered one:
I understand what the axiom says, but I have a hard time grasping what it really means for the field that it applies to, and what it really means for the difference between a completely ordered and an ordered field. Furthermore, why do the rationals not satisfy this axiom? My thoughts were that a subset of the rationals also contain a supremum, namely the greatest rational number in that subset. What part don't I understand?
All help is appreciated.
field-theory real-numbers rational-numbers ordered-fields
field-theory real-numbers rational-numbers ordered-fields
edited Nov 14 at 20:02
Eric Wofsey
175k12202326
asked Nov 14 at 19:46
Benjamin Caris
133
edited Nov 14 at 20:02
Eric Wofsey
175k12202326
asked Nov 14 at 19:46
Benjamin Caris
133
edited Nov 14 at 20:02
Eric Wofsey
175k12202326
edited Nov 14 at 20:02
Eric Wofsey
175k12202326
edited Nov 14 at 20:02
Eric Wofsey
175k12202326
175k12202326
asked Nov 14 at 19:46
Benjamin Caris
133
asked Nov 14 at 19:46
Benjamin Caris
133
asked Nov 14 at 19:46
Benjamin Caris
133
133
the set of rationals ${x in mathbb{Q} | x^2<2}$ has no supremum. There is no greatest rational number in this set, nor outside the set neither.
– ziggurism
Nov 14 at 19:50
When you say "I see that the reals are complete, and the rationals aren't", what is your understanding of "complete"? Is there some different definition you have? Or do you just mean "complete" in some vague intuitive sense?
– Eric Wofsey
Nov 14 at 20:04
@EricWofsey I understand it now, but just for clarification: yes, I meant that in the most intuitive way you could think of - probably not the best idea, but I had a hard time explaining my question properly. I meant it as it's written down in the book I'm using: "Intuitively, if we were to line up the rational numbers, there would be plenty of "holes" in them. Real numbers, on the other hand, would not have that property."
– Benjamin Caris
Nov 14 at 20:28
add a comment |
the set of rationals ${x in mathbb{Q} | x^2<2}$ has no supremum. There is no greatest rational number in this set, nor outside the set neither.
– ziggurism
Nov 14 at 19:50
When you say "I see that the reals are complete, and the rationals aren't", what is your understanding of "complete"? Is there some different definition you have? Or do you just mean "complete" in some vague intuitive sense?
– Eric Wofsey
Nov 14 at 20:04
@EricWofsey I understand it now, but just for clarification: yes, I meant that in the most intuitive way you could think of - probably not the best idea, but I had a hard time explaining my question properly. I meant it as it's written down in the book I'm using: "Intuitively, if we were to line up the rational numbers, there would be plenty of "holes" in them. Real numbers, on the other hand, would not have that property."
– Benjamin Caris
Nov 14 at 20:28
the set of rationals ${x in mathbb{Q} | x^2<2}$ has no supremum. There is no greatest rational number in this set, nor outside the set neither.
– ziggurism
Nov 14 at 19:50
the set of rationals ${x in mathbb{Q} | x^2<2}$ has no supremum. There is no greatest rational number in this set, nor outside the set neither.
– ziggurism
Nov 14 at 19:50
When you say "I see that the reals are complete, and the rationals aren't", what is your understanding of "complete"? Is there some different definition you have? Or do you just mean "complete" in some vague intuitive sense?
– Eric Wofsey
Nov 14 at 20:04
When you say "I see that the reals are complete, and the rationals aren't", what is your understanding of "complete"? Is there some different definition you have? Or do you just mean "complete" in some vague intuitive sense?
– Eric Wofsey
Nov 14 at 20:04
@EricWofsey I understand it now, but just for clarification: yes, I meant that in the most intuitive way you could think of - probably not the best idea, but I had a hard time explaining my question properly. I meant it as it's written down in the book I'm using: "Intuitively, if we were to line up the rational numbers, there would be plenty of "holes" in them. Real numbers, on the other hand, would not have that property."
– Benjamin Caris
Nov 14 at 20:28
@EricWofsey I understand it now, but just for clarification: yes, I meant that in the most intuitive way you could think of - probably not the best idea, but I had a hard time explaining my question properly. I meant it as it's written down in the book I'm using: "Intuitively, if we were to line up the rational numbers, there would be plenty of "holes" in them. Real numbers, on the other hand, would not have that property."
– Benjamin Caris
Nov 14 at 20:28
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
In the real numbers, if $(a_n)$ is a bounded, monotone increasing sequence, i.e. $a_1 le a_2 le a_3 le dots$ then
$$ lim_{n} a_n = sup{a_n : n in mathbb N} tag{$*$} $$
Thus completeness of an ordered field implies that every bounded, monotone sequence has a limit.
Monotone sequences are important to the real numbers because if you take your favourite real number and look at it's decimal expansions. E.g.
$$ 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, dots $$
you get a monotone increasing sequence of rational numbers. So having every infinite decimal expansion exist (in this case $pi$) follows from having limits for monotone sequences.
For a non-example in the rationals, consider the sequence
$$ frac{1}{1}, frac{3}{2}, frac{8}{5}, frac{21}{13} dots, frac{F_{2n}}{F_{2n-1}},dots $$
where the numerator and denominator are Fibbonacci numbers. This is monotone increasing and converges to the golden ratio, $frac{1 + sqrt 5}{2}$.
The proof of this fact is not important for this discussion so either accept that it's true or imagine some other increasing sequence of rational numbers that converge to an irrational number.
Anyways, with $a_n = F_{2n}/F_{2n-1}$ as above, the set ${a_1,a_2,a_3,dots}$ has no maximum, because $a_n < a_{n + 1}$ so if we said the maximum was $a_n$ we'd have a problem because $a_{n + 1}$ is bigger. By $(*)$ (which you should try to prove) we have
$$ sup{a_1,a_2,a_3,dots} = frac{1 + sqrt 5}{2} $$
So the set has no rational supremum but it does have a supremum in the reals.
answered Nov 14 at 20:04
Trevor Gunn
13.8k32045
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
In the real numbers, if $(a_n)$ is a bounded, monotone increasing sequence, i.e. $a_1 le a_2 le a_3 le dots$ then
$$ lim_{n} a_n = sup{a_n : n in mathbb N} tag{$*$} $$
Thus completeness of an ordered field implies that every bounded, monotone sequence has a limit.
Monotone sequences are important to the real numbers because if you take your favourite real number and look at it's decimal expansions. E.g.
$$ 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, dots $$
you get a monotone increasing sequence of rational numbers. So having every infinite decimal expansion exist (in this case $pi$) follows from having limits for monotone sequences.
For a non-example in the rationals, consider the sequence
$$ frac{1}{1}, frac{3}{2}, frac{8}{5}, frac{21}{13} dots, frac{F_{2n}}{F_{2n-1}},dots $$
where the numerator and denominator are Fibbonacci numbers. This is monotone increasing and converges to the golden ratio, $frac{1 + sqrt 5}{2}$.
The proof of this fact is not important for this discussion so either accept that it's true or imagine some other increasing sequence of rational numbers that converge to an irrational number.
Anyways, with $a_n = F_{2n}/F_{2n-1}$ as above, the set ${a_1,a_2,a_3,dots}$ has no maximum, because $a_n < a_{n + 1}$ so if we said the maximum was $a_n$ we'd have a problem because $a_{n + 1}$ is bigger. By $(*)$ (which you should try to prove) we have
$$ sup{a_1,a_2,a_3,dots} = frac{1 + sqrt 5}{2} $$
So the set has no rational supremum but it does have a supremum in the reals.
answered Nov 14 at 20:04
Trevor Gunn
13.8k32045
add a comment |
up vote
4
down vote
accepted
In the real numbers, if $(a_n)$ is a bounded, monotone increasing sequence, i.e. $a_1 le a_2 le a_3 le dots$ then
$$ lim_{n} a_n = sup{a_n : n in mathbb N} tag{$*$} $$
Thus completeness of an ordered field implies that every bounded, monotone sequence has a limit.
Monotone sequences are important to the real numbers because if you take your favourite real number and look at it's decimal expansions. E.g.
$$ 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, dots $$
you get a monotone increasing sequence of rational numbers. So having every infinite decimal expansion exist (in this case $pi$) follows from having limits for monotone sequences.
For a non-example in the rationals, consider the sequence
$$ frac{1}{1}, frac{3}{2}, frac{8}{5}, frac{21}{13} dots, frac{F_{2n}}{F_{2n-1}},dots $$
where the numerator and denominator are Fibbonacci numbers. This is monotone increasing and converges to the golden ratio, $frac{1 + sqrt 5}{2}$.
The proof of this fact is not important for this discussion so either accept that it's true or imagine some other increasing sequence of rational numbers that converge to an irrational number.
Anyways, with $a_n = F_{2n}/F_{2n-1}$ as above, the set ${a_1,a_2,a_3,dots}$ has no maximum, because $a_n < a_{n + 1}$ so if we said the maximum was $a_n$ we'd have a problem because $a_{n + 1}$ is bigger. By $(*)$ (which you should try to prove) we have
$$ sup{a_1,a_2,a_3,dots} = frac{1 + sqrt 5}{2} $$
So the set has no rational supremum but it does have a supremum in the reals.
answered Nov 14 at 20:04
Trevor Gunn
13.8k32045
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
In the real numbers, if $(a_n)$ is a bounded, monotone increasing sequence, i.e. $a_1 le a_2 le a_3 le dots$ then
$$ lim_{n} a_n = sup{a_n : n in mathbb N} tag{$*$} $$
Thus completeness of an ordered field implies that every bounded, monotone sequence has a limit.
Monotone sequences are important to the real numbers because if you take your favourite real number and look at it's decimal expansions. E.g.
$$ 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, dots $$
you get a monotone increasing sequence of rational numbers. So having every infinite decimal expansion exist (in this case $pi$) follows from having limits for monotone sequences.
For a non-example in the rationals, consider the sequence
$$ frac{1}{1}, frac{3}{2}, frac{8}{5}, frac{21}{13} dots, frac{F_{2n}}{F_{2n-1}},dots $$
where the numerator and denominator are Fibbonacci numbers. This is monotone increasing and converges to the golden ratio, $frac{1 + sqrt 5}{2}$.
The proof of this fact is not important for this discussion so either accept that it's true or imagine some other increasing sequence of rational numbers that converge to an irrational number.
Anyways, with $a_n = F_{2n}/F_{2n-1}$ as above, the set ${a_1,a_2,a_3,dots}$ has no maximum, because $a_n < a_{n + 1}$ so if we said the maximum was $a_n$ we'd have a problem because $a_{n + 1}$ is bigger. By $(*)$ (which you should try to prove) we have
$$ sup{a_1,a_2,a_3,dots} = frac{1 + sqrt 5}{2} $$
So the set has no rational supremum but it does have a supremum in the reals.
answered Nov 14 at 20:04
Trevor Gunn
13.8k32045
In the real numbers, if $(a_n)$ is a bounded, monotone increasing sequence, i.e. $a_1 le a_2 le a_3 le dots$ then
$$ lim_{n} a_n = sup{a_n : n in mathbb N} tag{$*$} $$
Thus completeness of an ordered field implies that every bounded, monotone sequence has a limit.
Monotone sequences are important to the real numbers because if you take your favourite real number and look at it's decimal expansions. E.g.
$$ 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, dots $$
you get a monotone increasing sequence of rational numbers. So having every infinite decimal expansion exist (in this case $pi$) follows from having limits for monotone sequences.
For a non-example in the rationals, consider the sequence
$$ frac{1}{1}, frac{3}{2}, frac{8}{5}, frac{21}{13} dots, frac{F_{2n}}{F_{2n-1}},dots $$
where the numerator and denominator are Fibbonacci numbers. This is monotone increasing and converges to the golden ratio, $frac{1 + sqrt 5}{2}$.
The proof of this fact is not important for this discussion so either accept that it's true or imagine some other increasing sequence of rational numbers that converge to an irrational number.
Anyways, with $a_n = F_{2n}/F_{2n-1}$ as above, the set ${a_1,a_2,a_3,dots}$ has no maximum, because $a_n < a_{n + 1}$ so if we said the maximum was $a_n$ we'd have a problem because $a_{n + 1}$ is bigger. By $(*)$ (which you should try to prove) we have
$$ sup{a_1,a_2,a_3,dots} = frac{1 + sqrt 5}{2} $$
So the set has no rational supremum but it does have a supremum in the reals.
answered Nov 14 at 20:04
Trevor Gunn
13.8k32045
answered Nov 14 at 20:04
Trevor Gunn
13.8k32045
answered Nov 14 at 20:04
Trevor Gunn
13.8k32045
answered Nov 14 at 20:04
Trevor Gunn
13.8k32045
13.8k32045
add a comment |
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the set of rationals ${x in mathbb{Q} | x^2<2}$ has no supremum. There is no greatest rational number in this set, nor outside the set neither.
– ziggurism
Nov 14 at 19:50
When you say "I see that the reals are complete, and the rationals aren't", what is your understanding of "complete"? Is there some different definition you have? Or do you just mean "complete" in some vague intuitive sense?
– Eric Wofsey
Nov 14 at 20:04
@EricWofsey I understand it now, but just for clarification: yes, I meant that in the most intuitive way you could think of - probably not the best idea, but I had a hard time explaining my question properly. I meant it as it's written down in the book I'm using: "Intuitively, if we were to line up the rational numbers, there would be plenty of "holes" in them. Real numbers, on the other hand, would not have that property."
– Benjamin Caris
Nov 14 at 20:28