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

How did Captain America manage to do this?

Image
2 How does Captain America lift Thor's hammer? In Avengers: Age of Ultron , he is not worthy but how he is worthy in this film. plot-explanation marvel-cinematic-universe avengers-age-of-ultron avengers-endgame share | improve this question edited 19 mins ago A J ♦ 42.7k 16 229 246 asked 39 mins ago Andi AR Andi AR 325 1 4 14
Read more

迪纳利

Image
迪纳利峰 位于迪纳利国家公园和保留区的迪纳利峰（旧名麦金利峰） 最高点 海拔 20,237英尺（6,168米）   NGVD 29 . [1] [2] 相對高度 20,073英尺（6,118米） 列表 各洲最高峰列表 特独立山峰 各国最高点列表 美国最高点  排名第一 [3] 坐标 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley ) 坐标： 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley )   地理 位置   美國阿拉斯加州迪纳利自治市镇 迪纳利国家公园和保留区 山脈 阿拉斯加山脉 攀山 首次登頂 （ 英语 ： First ascent ） 1913年6月7日 哈德逊·史塔克 哈里·卡斯坦斯 沃尔特·哈珀 罗伯特·塔特姆 最簡路線 西坡路线（攀登冰川/雪） 迪纳利 （北阿萨巴斯卡语支、 英语： Denali ， / d ᵻ ˈ n ɑː l i / ，直译“高山”），1917年至2015年官方名称为 麦金利山 （另译作 麦金利峰 ， 英语： Mount McKinley ），位于阿拉斯加州东南部、阿拉斯加山脉中段，海拔6194米，是北美洲最高峰，也是美国的最高峰。它的俄语名称是 Большая Гора ， 罗马化：Bolshaya Gora ，意思是“大山”。 迪纳利山地区拥有变幻莫测的高山风、典型的北极植被以及野生动植物。这里大部分地区终年积雪，山间经常浓雾不断，雾气在皑皑白雪中缭绕弥漫时，几百米之外的景物便不可见。夏季，迪纳利山的青青山坡上鲜花盛开，紫色的杜鹃和精巧的铃状石南花随处可见。迪纳利山还是世界登山爱好者的汇集之地。每年5月到7月有数百人登山，只有一半不到的人能登顶成功。一次登山约花3个星期。 2015年8月31日，美國政府宣佈將山名由“麦金利山”改回原名“迪纳利山”。美國內政部發表聲
Read more

南乌拉尔铁路局

Image
南乌拉尔铁路局 Южно-Уральская железная дорога 成立 1934年1月4日 国家   蘇聯（1934年—1991年）   俄羅斯（1991年—） 总部 车里雅宾斯克 隶属 俄罗斯铁路股份公司 路局代码 80 铁路里程 4,934.9公里 奖项 网址 http://yuzd.rzd.ru/ 查 论 编 南乌拉尔铁路局 （俄语： Южно-Уральская железная дорога ， 羅馬化： Yuzhno-Uralskaya zheleznaya doroga ，简称 ЮУЖД ）是俄罗斯铁路股份公司属下的铁路管理局之一，总部位于车里雅宾斯克，负责管辖俄罗斯联邦乌拉尔南部地区的铁路系统，范围包括奥伦堡州、车里雅宾斯克州、库尔干州的大部分铁路，以及萨马拉州、斯维尔德洛夫斯克州、巴什科尔托斯坦共和国的少部分铁路，并有部分铁路延伸至或穿过哈萨克斯坦境内。南乌拉尔铁路局的营业里程为4,934.9公里，与伏尔加铁路局、古比雪夫铁路局、斯维尔德洛夫斯克铁路局、西西伯利亚铁路局相邻，并与南方的哈萨克斯坦铁路连接。 外部链接 （俄文） Южно-Уральская железная дорога — филиал ОАО РЖД （俄文） 南乌拉尔铁路局管辖路线图（一） （俄文） 南乌拉尔铁路局管辖路线图（二） 查 论 编 俄罗斯铁路（集团）股份公司 十月铁路局 · 莫斯科铁路局 · 北高加索铁路局 · 高尔基铁路局 · 古比雪夫铁路局 · 北方铁路局 · 伏尔加铁路局 · 东南铁路局 · 斯维尔德洛夫斯克铁路局 · 南乌拉尔铁路局 · 西西伯利亚铁路局 · 克拉斯诺亚尔斯克铁路局 · 东西伯利亚铁路局 · 后贝加尔铁路局 · 远东铁路局 · 加里宁格勒铁路局 This page is only for reference, If you need detailed information, please check here
Read more