What fields between the rationals and the reals allow a good notion of 2D distance?












20












$begingroup$


Consider a field $K$, let's say $K subseteq mathbb R$. We can consider the 'plane' $K times K$. I am wondering in which cases the distance function $d: K times K to mathbb R$, defined as is normal by $d(x, y) = sqrt{x^2 + y^2}$, takes values in $K$.



Certainly this is not true for $mathbb Q$: we have $d(1, 1) = sqrt{2} notin mathbb Q$. If we take any $K$ which is closed under taking square roots of non-negative numbers, then certainly $d$ will take values in $K$.



However, a priori it might still be true that $a in K$ positive has no square root, yet this does not provide an obstruction because there is no way to write $a = x^2 + y^2$. Thus I am wondering:




Are there fields $K subseteq mathbb R$ which do not have all square roots of positive numbers, yet are closed under $d$?







abstract-algebra field-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Just so you know, there are other distances besides euclidean distance.
    $endgroup$
    – PyRulez
    yesterday
















20












$begingroup$


Consider a field $K$, let's say $K subseteq mathbb R$. We can consider the 'plane' $K times K$. I am wondering in which cases the distance function $d: K times K to mathbb R$, defined as is normal by $d(x, y) = sqrt{x^2 + y^2}$, takes values in $K$.



Certainly this is not true for $mathbb Q$: we have $d(1, 1) = sqrt{2} notin mathbb Q$. If we take any $K$ which is closed under taking square roots of non-negative numbers, then certainly $d$ will take values in $K$.



However, a priori it might still be true that $a in K$ positive has no square root, yet this does not provide an obstruction because there is no way to write $a = x^2 + y^2$. Thus I am wondering:




Are there fields $K subseteq mathbb R$ which do not have all square roots of positive numbers, yet are closed under $d$?







abstract-algebra field-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Just so you know, there are other distances besides euclidean distance.
    $endgroup$
    – PyRulez
    yesterday














20












20








20


4



$begingroup$


Consider a field $K$, let's say $K subseteq mathbb R$. We can consider the 'plane' $K times K$. I am wondering in which cases the distance function $d: K times K to mathbb R$, defined as is normal by $d(x, y) = sqrt{x^2 + y^2}$, takes values in $K$.



Certainly this is not true for $mathbb Q$: we have $d(1, 1) = sqrt{2} notin mathbb Q$. If we take any $K$ which is closed under taking square roots of non-negative numbers, then certainly $d$ will take values in $K$.



However, a priori it might still be true that $a in K$ positive has no square root, yet this does not provide an obstruction because there is no way to write $a = x^2 + y^2$. Thus I am wondering:




Are there fields $K subseteq mathbb R$ which do not have all square roots of positive numbers, yet are closed under $d$?







abstract-algebra field-theory







share|cite|improve this question











$endgroup$




Consider a field $K$, let's say $K subseteq mathbb R$. We can consider the 'plane' $K times K$. I am wondering in which cases the distance function $d: K times K to mathbb R$, defined as is normal by $d(x, y) = sqrt{x^2 + y^2}$, takes values in $K$.



Certainly this is not true for $mathbb Q$: we have $d(1, 1) = sqrt{2} notin mathbb Q$. If we take any $K$ which is closed under taking square roots of non-negative numbers, then certainly $d$ will take values in $K$.



However, a priori it might still be true that $a in K$ positive has no square root, yet this does not provide an obstruction because there is no way to write $a = x^2 + y^2$. Thus I am wondering:




Are there fields $K subseteq mathbb R$ which do not have all square roots of positive numbers, yet are closed under $d$?







abstract-algebra field-theory




abstract-algebra field-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday







Mees de Vries

















asked yesterday









Mees de VriesMees de Vries

17.5k12958




17.5k12958








  • 2




    $begingroup$
    Just so you know, there are other distances besides euclidean distance.
    $endgroup$
    – PyRulez
    yesterday














  • 2




    $begingroup$
    Just so you know, there are other distances besides euclidean distance.
    $endgroup$
    – PyRulez
    yesterday








2




2




$begingroup$
Just so you know, there are other distances besides euclidean distance.
$endgroup$
– PyRulez
yesterday




$begingroup$
Just so you know, there are other distances besides euclidean distance.
$endgroup$
– PyRulez
yesterday










2 Answers
2






active

oldest

votes


















16












$begingroup$

Consider the tower of fields



$K_0:=mathbb{Q}$,



$K_{i+1}:=K_i(sqrt{x^2+y^2}| x,yin K_i)$,



$K:=bigcup_i K_i$.



Then $K$ is closed under $d$ and contains $1+sqrt 5$ but not $sqrt{1+sqrt 5}$, as I have found by following the Pythagorean fields Wikipedia link given by @Dirk in his answer: If $sqrt{1+sqrt 5}$ were in $K$ then $1+sqrt 5$ would be a sum of two squares in some extension $K_i$, and then it would be so in an extension of $mathbb{Q}(sqrt 5)$, which implies that it is a sum of squares in $mathbb{Q}(sqrt 5)$, which is impossible because that would entail that $1-sqrt 5$, which is negative, is also a sum of squares in $mathbb{Q}(sqrt 5)$.



The details can be found in Chapter 5 of the book Geometric constructions by Martin. The relevant results are Theorems 5.10-5.15.



Similarly, $sqrt 2in K$ but $sqrt[4]2notin K$, and more in general, this is true for any positive number which is not a sum of squares in the first extension in which it appears.



Geometrically, numbers in $K$ correspond to constructible points by ruler and dividers. Hence $sqrt[4]2$ is constructible by rule and compass but not by rule and dividers.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    "which implies that it is a sum of squares in $mathbb Q(sqrt{5})$" -- I do not see directly how this follows from the previous sentence, but I could very well be missing something obvious.
    $endgroup$
    – Mees de Vries
    yesterday






  • 1




    $begingroup$
    @MeesdeVries Not obvious, it is a consequence of Theorems 5.10-5.13 in Martin's book.
    $endgroup$
    – Jose Brox
    yesterday










  • $begingroup$
    $K$ is countable, right?
    $endgroup$
    – PyRulez
    yesterday












  • $begingroup$
    @PyRulez yeah the same construction which makes $mathbb Q$ countable from $mathbb N^2$ plus skipping over duplicates should work to imply that $K_{i+1}$ is countable given that $K_{i}$ is countable, skipping over duplicates; by induction therefore all $K_i$ are countable; then we should be able to repeat the same construction again with $K_m(n)$, again skipping over duplicates, to find that $K$ is countable.
    $endgroup$
    – CR Drost
    yesterday








  • 1




    $begingroup$
    @PyRulez Yes: clearly, all elements of $K$ are algebraic, and algebraic numbers are countable (there is a countable number of rational polynomials, with a finite number of roots each)
    $endgroup$
    – Jose Brox
    yesterday





















5












$begingroup$

edit: Look what I found:
Wiki




The field
$$mathbb{Q}(sqrt{p} mid p in mathbb{P})$$
might be a good candidate.

At least, all fields closed under $d$ must contain this field.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why must a field closed under $d$ contain $sqrt{3}$?
    $endgroup$
    – FredH
    yesterday






  • 1




    $begingroup$
    @FredH, it must contain $sqrt{2} = d(1, 1)$, and thus it must contain $sqrt{3} = d(1, sqrt{2})$.
    $endgroup$
    – Mees de Vries
    yesterday






  • 1




    $begingroup$
    See en.wikipedia.org/wiki/Spiral_of_Theodorus
    $endgroup$
    – lhf
    yesterday






  • 2




    $begingroup$
    I don't think this works: $d(sqrt 2 + 1, 1) = sqrt{2sqrt2 + 4}$, but that doesn't look like a sum of square roots of rational numbers.
    $endgroup$
    – Arthur
    yesterday













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3156636%2fwhat-fields-between-the-rationals-and-the-reals-allow-a-good-notion-of-2d-distan%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









16












$begingroup$

Consider the tower of fields



$K_0:=mathbb{Q}$,



$K_{i+1}:=K_i(sqrt{x^2+y^2}| x,yin K_i)$,



$K:=bigcup_i K_i$.



Then $K$ is closed under $d$ and contains $1+sqrt 5$ but not $sqrt{1+sqrt 5}$, as I have found by following the Pythagorean fields Wikipedia link given by @Dirk in his answer: If $sqrt{1+sqrt 5}$ were in $K$ then $1+sqrt 5$ would be a sum of two squares in some extension $K_i$, and then it would be so in an extension of $mathbb{Q}(sqrt 5)$, which implies that it is a sum of squares in $mathbb{Q}(sqrt 5)$, which is impossible because that would entail that $1-sqrt 5$, which is negative, is also a sum of squares in $mathbb{Q}(sqrt 5)$.



The details can be found in Chapter 5 of the book Geometric constructions by Martin. The relevant results are Theorems 5.10-5.15.



Similarly, $sqrt 2in K$ but $sqrt[4]2notin K$, and more in general, this is true for any positive number which is not a sum of squares in the first extension in which it appears.



Geometrically, numbers in $K$ correspond to constructible points by ruler and dividers. Hence $sqrt[4]2$ is constructible by rule and compass but not by rule and dividers.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    "which implies that it is a sum of squares in $mathbb Q(sqrt{5})$" -- I do not see directly how this follows from the previous sentence, but I could very well be missing something obvious.
    $endgroup$
    – Mees de Vries
    yesterday






  • 1




    $begingroup$
    @MeesdeVries Not obvious, it is a consequence of Theorems 5.10-5.13 in Martin's book.
    $endgroup$
    – Jose Brox
    yesterday










  • $begingroup$
    $K$ is countable, right?
    $endgroup$
    – PyRulez
    yesterday












  • $begingroup$
    @PyRulez yeah the same construction which makes $mathbb Q$ countable from $mathbb N^2$ plus skipping over duplicates should work to imply that $K_{i+1}$ is countable given that $K_{i}$ is countable, skipping over duplicates; by induction therefore all $K_i$ are countable; then we should be able to repeat the same construction again with $K_m(n)$, again skipping over duplicates, to find that $K$ is countable.
    $endgroup$
    – CR Drost
    yesterday








  • 1




    $begingroup$
    @PyRulez Yes: clearly, all elements of $K$ are algebraic, and algebraic numbers are countable (there is a countable number of rational polynomials, with a finite number of roots each)
    $endgroup$
    – Jose Brox
    yesterday


















16












$begingroup$

Consider the tower of fields



$K_0:=mathbb{Q}$,



$K_{i+1}:=K_i(sqrt{x^2+y^2}| x,yin K_i)$,



$K:=bigcup_i K_i$.



Then $K$ is closed under $d$ and contains $1+sqrt 5$ but not $sqrt{1+sqrt 5}$, as I have found by following the Pythagorean fields Wikipedia link given by @Dirk in his answer: If $sqrt{1+sqrt 5}$ were in $K$ then $1+sqrt 5$ would be a sum of two squares in some extension $K_i$, and then it would be so in an extension of $mathbb{Q}(sqrt 5)$, which implies that it is a sum of squares in $mathbb{Q}(sqrt 5)$, which is impossible because that would entail that $1-sqrt 5$, which is negative, is also a sum of squares in $mathbb{Q}(sqrt 5)$.



The details can be found in Chapter 5 of the book Geometric constructions by Martin. The relevant results are Theorems 5.10-5.15.



Similarly, $sqrt 2in K$ but $sqrt[4]2notin K$, and more in general, this is true for any positive number which is not a sum of squares in the first extension in which it appears.



Geometrically, numbers in $K$ correspond to constructible points by ruler and dividers. Hence $sqrt[4]2$ is constructible by rule and compass but not by rule and dividers.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    "which implies that it is a sum of squares in $mathbb Q(sqrt{5})$" -- I do not see directly how this follows from the previous sentence, but I could very well be missing something obvious.
    $endgroup$
    – Mees de Vries
    yesterday






  • 1




    $begingroup$
    @MeesdeVries Not obvious, it is a consequence of Theorems 5.10-5.13 in Martin's book.
    $endgroup$
    – Jose Brox
    yesterday










  • $begingroup$
    $K$ is countable, right?
    $endgroup$
    – PyRulez
    yesterday












  • $begingroup$
    @PyRulez yeah the same construction which makes $mathbb Q$ countable from $mathbb N^2$ plus skipping over duplicates should work to imply that $K_{i+1}$ is countable given that $K_{i}$ is countable, skipping over duplicates; by induction therefore all $K_i$ are countable; then we should be able to repeat the same construction again with $K_m(n)$, again skipping over duplicates, to find that $K$ is countable.
    $endgroup$
    – CR Drost
    yesterday








  • 1




    $begingroup$
    @PyRulez Yes: clearly, all elements of $K$ are algebraic, and algebraic numbers are countable (there is a countable number of rational polynomials, with a finite number of roots each)
    $endgroup$
    – Jose Brox
    yesterday
















16












16








16





$begingroup$

Consider the tower of fields



$K_0:=mathbb{Q}$,



$K_{i+1}:=K_i(sqrt{x^2+y^2}| x,yin K_i)$,



$K:=bigcup_i K_i$.



Then $K$ is closed under $d$ and contains $1+sqrt 5$ but not $sqrt{1+sqrt 5}$, as I have found by following the Pythagorean fields Wikipedia link given by @Dirk in his answer: If $sqrt{1+sqrt 5}$ were in $K$ then $1+sqrt 5$ would be a sum of two squares in some extension $K_i$, and then it would be so in an extension of $mathbb{Q}(sqrt 5)$, which implies that it is a sum of squares in $mathbb{Q}(sqrt 5)$, which is impossible because that would entail that $1-sqrt 5$, which is negative, is also a sum of squares in $mathbb{Q}(sqrt 5)$.



The details can be found in Chapter 5 of the book Geometric constructions by Martin. The relevant results are Theorems 5.10-5.15.



Similarly, $sqrt 2in K$ but $sqrt[4]2notin K$, and more in general, this is true for any positive number which is not a sum of squares in the first extension in which it appears.



Geometrically, numbers in $K$ correspond to constructible points by ruler and dividers. Hence $sqrt[4]2$ is constructible by rule and compass but not by rule and dividers.






share|cite|improve this answer











$endgroup$



Consider the tower of fields



$K_0:=mathbb{Q}$,



$K_{i+1}:=K_i(sqrt{x^2+y^2}| x,yin K_i)$,



$K:=bigcup_i K_i$.



Then $K$ is closed under $d$ and contains $1+sqrt 5$ but not $sqrt{1+sqrt 5}$, as I have found by following the Pythagorean fields Wikipedia link given by @Dirk in his answer: If $sqrt{1+sqrt 5}$ were in $K$ then $1+sqrt 5$ would be a sum of two squares in some extension $K_i$, and then it would be so in an extension of $mathbb{Q}(sqrt 5)$, which implies that it is a sum of squares in $mathbb{Q}(sqrt 5)$, which is impossible because that would entail that $1-sqrt 5$, which is negative, is also a sum of squares in $mathbb{Q}(sqrt 5)$.



The details can be found in Chapter 5 of the book Geometric constructions by Martin. The relevant results are Theorems 5.10-5.15.



Similarly, $sqrt 2in K$ but $sqrt[4]2notin K$, and more in general, this is true for any positive number which is not a sum of squares in the first extension in which it appears.



Geometrically, numbers in $K$ correspond to constructible points by ruler and dividers. Hence $sqrt[4]2$ is constructible by rule and compass but not by rule and dividers.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday

























answered yesterday









Jose BroxJose Brox

3,33211129




3,33211129












  • $begingroup$
    "which implies that it is a sum of squares in $mathbb Q(sqrt{5})$" -- I do not see directly how this follows from the previous sentence, but I could very well be missing something obvious.
    $endgroup$
    – Mees de Vries
    yesterday






  • 1




    $begingroup$
    @MeesdeVries Not obvious, it is a consequence of Theorems 5.10-5.13 in Martin's book.
    $endgroup$
    – Jose Brox
    yesterday










  • $begingroup$
    $K$ is countable, right?
    $endgroup$
    – PyRulez
    yesterday












  • $begingroup$
    @PyRulez yeah the same construction which makes $mathbb Q$ countable from $mathbb N^2$ plus skipping over duplicates should work to imply that $K_{i+1}$ is countable given that $K_{i}$ is countable, skipping over duplicates; by induction therefore all $K_i$ are countable; then we should be able to repeat the same construction again with $K_m(n)$, again skipping over duplicates, to find that $K$ is countable.
    $endgroup$
    – CR Drost
    yesterday








  • 1




    $begingroup$
    @PyRulez Yes: clearly, all elements of $K$ are algebraic, and algebraic numbers are countable (there is a countable number of rational polynomials, with a finite number of roots each)
    $endgroup$
    – Jose Brox
    yesterday




















  • $begingroup$
    "which implies that it is a sum of squares in $mathbb Q(sqrt{5})$" -- I do not see directly how this follows from the previous sentence, but I could very well be missing something obvious.
    $endgroup$
    – Mees de Vries
    yesterday






  • 1




    $begingroup$
    @MeesdeVries Not obvious, it is a consequence of Theorems 5.10-5.13 in Martin's book.
    $endgroup$
    – Jose Brox
    yesterday










  • $begingroup$
    $K$ is countable, right?
    $endgroup$
    – PyRulez
    yesterday












  • $begingroup$
    @PyRulez yeah the same construction which makes $mathbb Q$ countable from $mathbb N^2$ plus skipping over duplicates should work to imply that $K_{i+1}$ is countable given that $K_{i}$ is countable, skipping over duplicates; by induction therefore all $K_i$ are countable; then we should be able to repeat the same construction again with $K_m(n)$, again skipping over duplicates, to find that $K$ is countable.
    $endgroup$
    – CR Drost
    yesterday








  • 1




    $begingroup$
    @PyRulez Yes: clearly, all elements of $K$ are algebraic, and algebraic numbers are countable (there is a countable number of rational polynomials, with a finite number of roots each)
    $endgroup$
    – Jose Brox
    yesterday


















$begingroup$
"which implies that it is a sum of squares in $mathbb Q(sqrt{5})$" -- I do not see directly how this follows from the previous sentence, but I could very well be missing something obvious.
$endgroup$
– Mees de Vries
yesterday




$begingroup$
"which implies that it is a sum of squares in $mathbb Q(sqrt{5})$" -- I do not see directly how this follows from the previous sentence, but I could very well be missing something obvious.
$endgroup$
– Mees de Vries
yesterday




1




1




$begingroup$
@MeesdeVries Not obvious, it is a consequence of Theorems 5.10-5.13 in Martin's book.
$endgroup$
– Jose Brox
yesterday




$begingroup$
@MeesdeVries Not obvious, it is a consequence of Theorems 5.10-5.13 in Martin's book.
$endgroup$
– Jose Brox
yesterday












$begingroup$
$K$ is countable, right?
$endgroup$
– PyRulez
yesterday






$begingroup$
$K$ is countable, right?
$endgroup$
– PyRulez
yesterday














$begingroup$
@PyRulez yeah the same construction which makes $mathbb Q$ countable from $mathbb N^2$ plus skipping over duplicates should work to imply that $K_{i+1}$ is countable given that $K_{i}$ is countable, skipping over duplicates; by induction therefore all $K_i$ are countable; then we should be able to repeat the same construction again with $K_m(n)$, again skipping over duplicates, to find that $K$ is countable.
$endgroup$
– CR Drost
yesterday






$begingroup$
@PyRulez yeah the same construction which makes $mathbb Q$ countable from $mathbb N^2$ plus skipping over duplicates should work to imply that $K_{i+1}$ is countable given that $K_{i}$ is countable, skipping over duplicates; by induction therefore all $K_i$ are countable; then we should be able to repeat the same construction again with $K_m(n)$, again skipping over duplicates, to find that $K$ is countable.
$endgroup$
– CR Drost
yesterday






1




1




$begingroup$
@PyRulez Yes: clearly, all elements of $K$ are algebraic, and algebraic numbers are countable (there is a countable number of rational polynomials, with a finite number of roots each)
$endgroup$
– Jose Brox
yesterday






$begingroup$
@PyRulez Yes: clearly, all elements of $K$ are algebraic, and algebraic numbers are countable (there is a countable number of rational polynomials, with a finite number of roots each)
$endgroup$
– Jose Brox
yesterday













5












$begingroup$

edit: Look what I found:
Wiki




The field
$$mathbb{Q}(sqrt{p} mid p in mathbb{P})$$
might be a good candidate.

At least, all fields closed under $d$ must contain this field.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why must a field closed under $d$ contain $sqrt{3}$?
    $endgroup$
    – FredH
    yesterday






  • 1




    $begingroup$
    @FredH, it must contain $sqrt{2} = d(1, 1)$, and thus it must contain $sqrt{3} = d(1, sqrt{2})$.
    $endgroup$
    – Mees de Vries
    yesterday






  • 1




    $begingroup$
    See en.wikipedia.org/wiki/Spiral_of_Theodorus
    $endgroup$
    – lhf
    yesterday






  • 2




    $begingroup$
    I don't think this works: $d(sqrt 2 + 1, 1) = sqrt{2sqrt2 + 4}$, but that doesn't look like a sum of square roots of rational numbers.
    $endgroup$
    – Arthur
    yesterday


















5












$begingroup$

edit: Look what I found:
Wiki




The field
$$mathbb{Q}(sqrt{p} mid p in mathbb{P})$$
might be a good candidate.

At least, all fields closed under $d$ must contain this field.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why must a field closed under $d$ contain $sqrt{3}$?
    $endgroup$
    – FredH
    yesterday






  • 1




    $begingroup$
    @FredH, it must contain $sqrt{2} = d(1, 1)$, and thus it must contain $sqrt{3} = d(1, sqrt{2})$.
    $endgroup$
    – Mees de Vries
    yesterday






  • 1




    $begingroup$
    See en.wikipedia.org/wiki/Spiral_of_Theodorus
    $endgroup$
    – lhf
    yesterday






  • 2




    $begingroup$
    I don't think this works: $d(sqrt 2 + 1, 1) = sqrt{2sqrt2 + 4}$, but that doesn't look like a sum of square roots of rational numbers.
    $endgroup$
    – Arthur
    yesterday
















5












5








5





$begingroup$

edit: Look what I found:
Wiki




The field
$$mathbb{Q}(sqrt{p} mid p in mathbb{P})$$
might be a good candidate.

At least, all fields closed under $d$ must contain this field.






share|cite|improve this answer











$endgroup$



edit: Look what I found:
Wiki




The field
$$mathbb{Q}(sqrt{p} mid p in mathbb{P})$$
might be a good candidate.

At least, all fields closed under $d$ must contain this field.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday

























answered yesterday









DirkDirk

4,408218




4,408218












  • $begingroup$
    Why must a field closed under $d$ contain $sqrt{3}$?
    $endgroup$
    – FredH
    yesterday






  • 1




    $begingroup$
    @FredH, it must contain $sqrt{2} = d(1, 1)$, and thus it must contain $sqrt{3} = d(1, sqrt{2})$.
    $endgroup$
    – Mees de Vries
    yesterday






  • 1




    $begingroup$
    See en.wikipedia.org/wiki/Spiral_of_Theodorus
    $endgroup$
    – lhf
    yesterday






  • 2




    $begingroup$
    I don't think this works: $d(sqrt 2 + 1, 1) = sqrt{2sqrt2 + 4}$, but that doesn't look like a sum of square roots of rational numbers.
    $endgroup$
    – Arthur
    yesterday




















  • $begingroup$
    Why must a field closed under $d$ contain $sqrt{3}$?
    $endgroup$
    – FredH
    yesterday






  • 1




    $begingroup$
    @FredH, it must contain $sqrt{2} = d(1, 1)$, and thus it must contain $sqrt{3} = d(1, sqrt{2})$.
    $endgroup$
    – Mees de Vries
    yesterday






  • 1




    $begingroup$
    See en.wikipedia.org/wiki/Spiral_of_Theodorus
    $endgroup$
    – lhf
    yesterday






  • 2




    $begingroup$
    I don't think this works: $d(sqrt 2 + 1, 1) = sqrt{2sqrt2 + 4}$, but that doesn't look like a sum of square roots of rational numbers.
    $endgroup$
    – Arthur
    yesterday


















$begingroup$
Why must a field closed under $d$ contain $sqrt{3}$?
$endgroup$
– FredH
yesterday




$begingroup$
Why must a field closed under $d$ contain $sqrt{3}$?
$endgroup$
– FredH
yesterday




1




1




$begingroup$
@FredH, it must contain $sqrt{2} = d(1, 1)$, and thus it must contain $sqrt{3} = d(1, sqrt{2})$.
$endgroup$
– Mees de Vries
yesterday




$begingroup$
@FredH, it must contain $sqrt{2} = d(1, 1)$, and thus it must contain $sqrt{3} = d(1, sqrt{2})$.
$endgroup$
– Mees de Vries
yesterday




1




1




$begingroup$
See en.wikipedia.org/wiki/Spiral_of_Theodorus
$endgroup$
– lhf
yesterday




$begingroup$
See en.wikipedia.org/wiki/Spiral_of_Theodorus
$endgroup$
– lhf
yesterday




2




2




$begingroup$
I don't think this works: $d(sqrt 2 + 1, 1) = sqrt{2sqrt2 + 4}$, but that doesn't look like a sum of square roots of rational numbers.
$endgroup$
– Arthur
yesterday






$begingroup$
I don't think this works: $d(sqrt 2 + 1, 1) = sqrt{2sqrt2 + 4}$, but that doesn't look like a sum of square roots of rational numbers.
$endgroup$
– Arthur
yesterday




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3156636%2fwhat-fields-between-the-rationals-and-the-reals-allow-a-good-notion-of-2d-distan%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

數位音樂下載

Image
數位音樂下載 為非以「實體」方式來販賣音樂的相關產品，而使用數位格式如mp3、AAC等文件格式來進行銷售販賣的一种音乐产业形式。著名的數位音樂下載服務商有iTunes Store、RecoChoku、Google Play音乐和微软Zune等。数字音乐下载在二十一世纪初一度风靡全球， [1] [2] 使得传统唱片产业走向衰落，但其自身也在2010年代流媒体音乐服务如Apple Music、Spotify等的挤压下迅速失去市场份额。 [3] [4] 数字音乐下载有便于携带、传播迅速、制作简单成本低等优点，与流媒体音乐相比也能更好保障音乐人权益； [5] 但它同时很可能带来侵犯版权和滥用的情形， [6] 也给音乐产业带来了诸多不利影响。数字音乐下载为音乐人和乐迷提供了更加便捷直接的交流渠道，也一定程度上推动了网络音乐人的出现和独立音乐的迅速发展。 [7] 目录 1 历史 1.1 技术积淀（1987-2000） 1.2 出现与兴起（2000-2004） 1.3 唱片市场的颠覆者（2004-2007） 1.4 巅峰和衰落（2005-2013） 1.5 摇摇欲坠的现状与不可预知的明天（2014至今） 2 主要文件格式 2.1 PCM编码格式 2.1.1 WAV 2.1.2 APE 2.1.3 FLAC 2.2 苹果推出的无损文件格式 2.2.1 AIFF 2.2.2 ALAC 2.3 有损压缩格式 2.3.1 MP3 2.3.2 AAC 3 主要市场 3.1 iTunes Store 3.2 Google Play音乐商店 3.3 亚马逊音乐商店 3.4 部分其他音乐商店 4 影响与评价 4.1 对音乐产业的影响 4.2 对科技行业的影响 4.3 正面评价 4.4 负面评价 4.4.1 侵权问题 4.4.2 经营模式不明晰 4.4.3 不适应移动时代 4.5 文化现象 5 延伸阅读 6 参考资料 历史 苹果iTunes+iPod的软硬件搭配让数字音...
Read more

When can things happen in Etherscan, such as the picture below?

Image
2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  ...
Read more

格利澤436b

Image
格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
Read more