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












19












$begingroup$


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











$endgroup$












  • $begingroup$
    Just so you know, there are other distances besides euclidean distance.
    $endgroup$
    – PyRulez
    2 hours ago
















19












$begingroup$


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











$endgroup$












  • $begingroup$
    Just so you know, there are other distances besides euclidean distance.
    $endgroup$
    – PyRulez
    2 hours ago














19












19








19


3



$begingroup$


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 15 hours ago







Mees de Vries

















asked 15 hours ago









Mees de VriesMees de Vries

17.5k12958




17.5k12958












  • $begingroup$
    Just so you know, there are other distances besides euclidean distance.
    $endgroup$
    – PyRulez
    2 hours ago


















  • $begingroup$
    Just so you know, there are other distances besides euclidean distance.
    $endgroup$
    – PyRulez
    2 hours ago
















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




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










2 Answers
2






active

oldest

votes


















14












$begingroup$

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
    12 hours ago






  • 1




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










  • $begingroup$
    $K$ is countable, right?
    $endgroup$
    – PyRulez
    5 hours ago












  • $begingroup$
    @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
    3 hours ago








  • 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
    2 hours ago





















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
    14 hours ago






  • 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
    14 hours ago






  • 1




    $begingroup$
    See en.wikipedia.org/wiki/Spiral_of_Theodorus
    $endgroup$
    – lhf
    14 hours ago






  • 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
    13 hours ago













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









14












$begingroup$

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
    12 hours ago






  • 1




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










  • $begingroup$
    $K$ is countable, right?
    $endgroup$
    – PyRulez
    5 hours ago












  • $begingroup$
    @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
    3 hours ago








  • 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
    2 hours ago


















14












$begingroup$

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
    12 hours ago






  • 1




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










  • $begingroup$
    $K$ is countable, right?
    $endgroup$
    – PyRulez
    5 hours ago












  • $begingroup$
    @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
    3 hours ago








  • 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
    2 hours ago
















14












14








14





$begingroup$

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 12 hours ago

























answered 12 hours ago









Jose BroxJose Brox

3,31211129




3,31211129












  • $begingroup$
    "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
    12 hours ago






  • 1




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










  • $begingroup$
    $K$ is countable, right?
    $endgroup$
    – PyRulez
    5 hours ago












  • $begingroup$
    @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
    3 hours ago








  • 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
    2 hours ago




















  • $begingroup$
    "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
    12 hours ago






  • 1




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










  • $begingroup$
    $K$ is countable, right?
    $endgroup$
    – PyRulez
    5 hours ago












  • $begingroup$
    @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
    3 hours ago








  • 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
    2 hours ago


















$begingroup$
"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
12 hours ago




$begingroup$
"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
12 hours ago




1




1




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




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












$begingroup$
$K$ is countable, right?
$endgroup$
– PyRulez
5 hours ago






$begingroup$
$K$ is countable, right?
$endgroup$
– PyRulez
5 hours ago














$begingroup$
@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
3 hours ago






$begingroup$
@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
3 hours ago






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
2 hours ago






$begingroup$
@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
2 hours ago













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
    14 hours ago






  • 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
    14 hours ago






  • 1




    $begingroup$
    See en.wikipedia.org/wiki/Spiral_of_Theodorus
    $endgroup$
    – lhf
    14 hours ago






  • 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
    13 hours ago


















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
    14 hours ago






  • 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
    14 hours ago






  • 1




    $begingroup$
    See en.wikipedia.org/wiki/Spiral_of_Theodorus
    $endgroup$
    – lhf
    14 hours ago






  • 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
    13 hours ago
















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 13 hours ago

























answered 15 hours ago









DirkDirk

4,333218




4,333218












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






  • 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
    14 hours ago






  • 1




    $begingroup$
    See en.wikipedia.org/wiki/Spiral_of_Theodorus
    $endgroup$
    – lhf
    14 hours ago






  • 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
    13 hours ago




















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






  • 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
    14 hours ago






  • 1




    $begingroup$
    See en.wikipedia.org/wiki/Spiral_of_Theodorus
    $endgroup$
    – lhf
    14 hours ago






  • 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
    13 hours ago


















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




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




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
14 hours ago




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




1




1




$begingroup$
See en.wikipedia.org/wiki/Spiral_of_Theodorus
$endgroup$
– lhf
14 hours ago




$begingroup$
See en.wikipedia.org/wiki/Spiral_of_Theodorus
$endgroup$
– lhf
14 hours ago




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
13 hours ago






$begingroup$
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
13 hours ago




















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