What fields between the rationals and the reals allow a good notion of 2D distance?
$begingroup$
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
asked Mar 21 at 10:20
Mees de VriesMees de Vries
17.5k12958
$endgroup$
add a comment |
$begingroup$
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
asked Mar 21 at 10:20
Mees de VriesMees de Vries
17.5k12958
$endgroup$
2
$begingroup$
Just so you know, there are other distances besides euclidean distance.
$endgroup$
– PyRulez
Mar 21 at 23:47
add a comment |
$begingroup$
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
asked Mar 21 at 10:20
Mees de VriesMees de Vries
17.5k12958
$endgroup$
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
asked Mar 21 at 10:20
Mees de VriesMees de Vries
17.5k12958
asked Mar 21 at 10:20
Mees de VriesMees de Vries
17.5k12958
edited Mar 21 at 10:27
Mees de Vries
asked Mar 21 at 10:20
Mees de VriesMees de Vries
17.5k12958
asked Mar 21 at 10:20
Mees de VriesMees de Vries
17.5k12958
asked Mar 21 at 10:20
Mees de VriesMees de Vries
17.5k12958
17.5k12958
2
$begingroup$
Just so you know, there are other distances besides euclidean distance.
$endgroup$
– PyRulez
Mar 21 at 23:47
add a comment |
2
$begingroup$
Just so you know, there are other distances besides euclidean distance.
$endgroup$
– PyRulez
Mar 21 at 23:47
2
2
$begingroup$
Just so you know, there are other distances besides euclidean distance.
$endgroup$
– PyRulez
Mar 21 at 23:47
$begingroup$
Just so you know, there are other distances besides euclidean distance.
$endgroup$
– PyRulez
Mar 21 at 23:47
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Mar 21 at 12:54
Jose BroxJose Brox
3,33211129
$endgroup$
$begingroup$
"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
Mar 21 at 13:04
1
$begingroup$
@MeesdeVries Not obvious, it is a consequence of Theorems 5.10-5.13 in Martin's book.
$endgroup$
– Jose Brox
Mar 21 at 13:06
$begingroup$
$K$ is countable, right?
$endgroup$
– PyRulez
Mar 21 at 20:00
$begingroup$
@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
Mar 21 at 21:52
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
Mar 21 at 22:57
add a comment |
$begingroup$
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.
answered Mar 21 at 10:44
DirkDirk
4,418218
$endgroup$
$begingroup$
Why must a field closed under $d$ contain $sqrt{3}$?
$endgroup$
– FredH
Mar 21 at 10:51
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
Mar 21 at 10:53
1
$begingroup$
See en.wikipedia.org/wiki/Spiral_of_Theodorus
$endgroup$
– lhf
Mar 21 at 11:16
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
Mar 21 at 12:05
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
$begingroup$
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.
answered Mar 21 at 12:54
Jose BroxJose Brox
3,33211129
$endgroup$
$begingroup$
"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
Mar 21 at 13:04
1
$begingroup$
@MeesdeVries Not obvious, it is a consequence of Theorems 5.10-5.13 in Martin's book.
$endgroup$
– Jose Brox
Mar 21 at 13:06
$begingroup$
$K$ is countable, right?
$endgroup$
– PyRulez
Mar 21 at 20:00
$begingroup$
@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
Mar 21 at 21:52
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
Mar 21 at 22:57
add a comment |
$begingroup$
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.
answered Mar 21 at 12:54
Jose BroxJose Brox
3,33211129
$endgroup$
$begingroup$
"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
Mar 21 at 13:04
1
$begingroup$
@MeesdeVries Not obvious, it is a consequence of Theorems 5.10-5.13 in Martin's book.
$endgroup$
– Jose Brox
Mar 21 at 13:06
$begingroup$
$K$ is countable, right?
$endgroup$
– PyRulez
Mar 21 at 20:00
$begingroup$
@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
Mar 21 at 21:52
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
Mar 21 at 22:57
add a comment |
$begingroup$
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.
answered Mar 21 at 12:54
Jose BroxJose Brox
3,33211129
$endgroup$
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.
answered Mar 21 at 12:54
Jose BroxJose Brox
3,33211129
edited Mar 21 at 13:08
answered Mar 21 at 12:54
Jose BroxJose Brox
3,33211129
answered Mar 21 at 12:54
Jose BroxJose Brox
3,33211129
answered Mar 21 at 12:54
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
Mar 21 at 13:04
1
$begingroup$
@MeesdeVries Not obvious, it is a consequence of Theorems 5.10-5.13 in Martin's book.
$endgroup$
– Jose Brox
Mar 21 at 13:06
$begingroup$
$K$ is countable, right?
$endgroup$
– PyRulez
Mar 21 at 20:00
$begingroup$
@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
Mar 21 at 21:52
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
Mar 21 at 22:57
add a comment |
$begingroup$
"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
Mar 21 at 13:04
1
$begingroup$
@MeesdeVries Not obvious, it is a consequence of Theorems 5.10-5.13 in Martin's book.
$endgroup$
– Jose Brox
Mar 21 at 13:06
$begingroup$
$K$ is countable, right?
$endgroup$
– PyRulez
Mar 21 at 20:00
$begingroup$
@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
Mar 21 at 21:52
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
Mar 21 at 22:57
$begingroup$
"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
Mar 21 at 13:04
$begingroup$
"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
Mar 21 at 13:04
1
1
$begingroup$
@MeesdeVries Not obvious, it is a consequence of Theorems 5.10-5.13 in Martin's book.
$endgroup$
– Jose Brox
Mar 21 at 13:06
$begingroup$
@MeesdeVries Not obvious, it is a consequence of Theorems 5.10-5.13 in Martin's book.
$endgroup$
– Jose Brox
Mar 21 at 13:06
$begingroup$
$K$ is countable, right?
$endgroup$
– PyRulez
Mar 21 at 20:00
$begingroup$
$K$ is countable, right?
$endgroup$
– PyRulez
Mar 21 at 20:00
$begingroup$
@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
Mar 21 at 21:52
$begingroup$
@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
Mar 21 at 21:52
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
Mar 21 at 22:57
$begingroup$
@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
Mar 21 at 22:57
add a comment |
$begingroup$
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.
answered Mar 21 at 10:44
DirkDirk
4,418218
$endgroup$
$begingroup$
Why must a field closed under $d$ contain $sqrt{3}$?
$endgroup$
– FredH
Mar 21 at 10:51
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
Mar 21 at 10:53
1
$begingroup$
See en.wikipedia.org/wiki/Spiral_of_Theodorus
$endgroup$
– lhf
Mar 21 at 11:16
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
Mar 21 at 12:05
add a comment |
$begingroup$
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.
answered Mar 21 at 10:44
DirkDirk
4,418218
$endgroup$
$begingroup$
Why must a field closed under $d$ contain $sqrt{3}$?
$endgroup$
– FredH
Mar 21 at 10:51
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
Mar 21 at 10:53
1
$begingroup$
See en.wikipedia.org/wiki/Spiral_of_Theodorus
$endgroup$
– lhf
Mar 21 at 11:16
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
Mar 21 at 12:05
add a comment |
$begingroup$
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.
answered Mar 21 at 10:44
DirkDirk
4,418218
$endgroup$
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.
answered Mar 21 at 10:44
DirkDirk
4,418218
edited Mar 21 at 12:19
answered Mar 21 at 10:44
DirkDirk
4,418218
answered Mar 21 at 10:44
DirkDirk
4,418218
answered Mar 21 at 10:44
DirkDirk
4,418218
4,418218
$begingroup$
Why must a field closed under $d$ contain $sqrt{3}$?
$endgroup$
– FredH
Mar 21 at 10:51
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
Mar 21 at 10:53
1
$begingroup$
See en.wikipedia.org/wiki/Spiral_of_Theodorus
$endgroup$
– lhf
Mar 21 at 11:16
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
Mar 21 at 12:05
add a comment |
$begingroup$
Why must a field closed under $d$ contain $sqrt{3}$?
$endgroup$
– FredH
Mar 21 at 10:51
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
Mar 21 at 10:53
1
$begingroup$
See en.wikipedia.org/wiki/Spiral_of_Theodorus
$endgroup$
– lhf
Mar 21 at 11:16
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
Mar 21 at 12:05
$begingroup$
Why must a field closed under $d$ contain $sqrt{3}$?
$endgroup$
– FredH
Mar 21 at 10:51
$begingroup$
Why must a field closed under $d$ contain $sqrt{3}$?
$endgroup$
– FredH
Mar 21 at 10:51
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
Mar 21 at 10:53
$begingroup$
@FredH, it must contain $sqrt{2} = d(1, 1)$, and thus it must contain $sqrt{3} = d(1, sqrt{2})$.
$endgroup$
– Mees de Vries
Mar 21 at 10:53
1
1
$begingroup$
See en.wikipedia.org/wiki/Spiral_of_Theodorus
$endgroup$
– lhf
Mar 21 at 11:16
$begingroup$
See en.wikipedia.org/wiki/Spiral_of_Theodorus
$endgroup$
– lhf
Mar 21 at 11:16
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
Mar 21 at 12:05
$begingroup$
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
Mar 21 at 12:05
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
2
$begingroup$
Just so you know, there are other distances besides euclidean distance.
$endgroup$
– PyRulez
Mar 21 at 23:47