What can be learned for number theory from geometrical constructions (and vice versa)?
Clash Royale CLAN TAG#URR8PPP
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Even though this question of mine was not so well received at MO I'd like to pick two examples and make a question out of them here.
Consider these two pairs of geometrical constructions which yield the same arithmetical results:
Constructing the half $x/2$ for one given positive real $x$ in two different ways.
Constructing the product $mn$ for two given positive integers $m, n$ in two different ways.
It's noteworthy that in each of these pairs one of the constructions does make use of circles while the other one doesn't.
Example 1: Constructing the half $x/2$
You can create the half $mathsfX/2$ of a positive real $mathsfX$ by two different Euclidean constructions:
Example 2: Constructing the product $mn$
You can create the product $nm$ of two positive integers $n, m$ by two different Euclidean constructions:
creating a rectangle
and counting the number of unit squares that fit into the rectanglecreating a line segment
and counting the number of unit lengths that fit into the line segment
In both cases it's not obvious that the two constructions always yield the same result, but for the sake of the theories (i.e. Euclid's â later Descartes' â geometry and number theory, resp. arithmetic geometry) it's essential.
My questions are:
How did (possibly) Euclid formulate the two statements above, i.e. that the two pairs of constructions always yield the same results?
How did (possibly) Euclid prove these statements?
What are the deep insights which we gain from understanding why these two pairs of constructions always yield the same result? (The
same point in Example 1, the same positive integer in Example 2.)
geometry number-theory euclidean-geometry big-picture
add a comment |Â
up vote
1
down vote
favorite
Even though this question of mine was not so well received at MO I'd like to pick two examples and make a question out of them here.
Consider these two pairs of geometrical constructions which yield the same arithmetical results:
Constructing the half $x/2$ for one given positive real $x$ in two different ways.
Constructing the product $mn$ for two given positive integers $m, n$ in two different ways.
It's noteworthy that in each of these pairs one of the constructions does make use of circles while the other one doesn't.
Example 1: Constructing the half $x/2$
You can create the half $mathsfX/2$ of a positive real $mathsfX$ by two different Euclidean constructions:
Example 2: Constructing the product $mn$
You can create the product $nm$ of two positive integers $n, m$ by two different Euclidean constructions:
creating a rectangle
and counting the number of unit squares that fit into the rectanglecreating a line segment
and counting the number of unit lengths that fit into the line segment
In both cases it's not obvious that the two constructions always yield the same result, but for the sake of the theories (i.e. Euclid's â later Descartes' â geometry and number theory, resp. arithmetic geometry) it's essential.
My questions are:
How did (possibly) Euclid formulate the two statements above, i.e. that the two pairs of constructions always yield the same results?
How did (possibly) Euclid prove these statements?
What are the deep insights which we gain from understanding why these two pairs of constructions always yield the same result? (The
same point in Example 1, the same positive integer in Example 2.)
geometry number-theory euclidean-geometry big-picture
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Even though this question of mine was not so well received at MO I'd like to pick two examples and make a question out of them here.
Consider these two pairs of geometrical constructions which yield the same arithmetical results:
Constructing the half $x/2$ for one given positive real $x$ in two different ways.
Constructing the product $mn$ for two given positive integers $m, n$ in two different ways.
It's noteworthy that in each of these pairs one of the constructions does make use of circles while the other one doesn't.
Example 1: Constructing the half $x/2$
You can create the half $mathsfX/2$ of a positive real $mathsfX$ by two different Euclidean constructions:
Example 2: Constructing the product $mn$
You can create the product $nm$ of two positive integers $n, m$ by two different Euclidean constructions:
creating a rectangle
and counting the number of unit squares that fit into the rectanglecreating a line segment
and counting the number of unit lengths that fit into the line segment
In both cases it's not obvious that the two constructions always yield the same result, but for the sake of the theories (i.e. Euclid's â later Descartes' â geometry and number theory, resp. arithmetic geometry) it's essential.
My questions are:
How did (possibly) Euclid formulate the two statements above, i.e. that the two pairs of constructions always yield the same results?
How did (possibly) Euclid prove these statements?
What are the deep insights which we gain from understanding why these two pairs of constructions always yield the same result? (The
same point in Example 1, the same positive integer in Example 2.)
geometry number-theory euclidean-geometry big-picture
Even though this question of mine was not so well received at MO I'd like to pick two examples and make a question out of them here.
Consider these two pairs of geometrical constructions which yield the same arithmetical results:
Constructing the half $x/2$ for one given positive real $x$ in two different ways.
Constructing the product $mn$ for two given positive integers $m, n$ in two different ways.
It's noteworthy that in each of these pairs one of the constructions does make use of circles while the other one doesn't.
Example 1: Constructing the half $x/2$
You can create the half $mathsfX/2$ of a positive real $mathsfX$ by two different Euclidean constructions:
Example 2: Constructing the product $mn$
You can create the product $nm$ of two positive integers $n, m$ by two different Euclidean constructions:
creating a rectangle
and counting the number of unit squares that fit into the rectanglecreating a line segment
and counting the number of unit lengths that fit into the line segment
In both cases it's not obvious that the two constructions always yield the same result, but for the sake of the theories (i.e. Euclid's â later Descartes' â geometry and number theory, resp. arithmetic geometry) it's essential.
My questions are:
How did (possibly) Euclid formulate the two statements above, i.e. that the two pairs of constructions always yield the same results?
How did (possibly) Euclid prove these statements?
What are the deep insights which we gain from understanding why these two pairs of constructions always yield the same result? (The
same point in Example 1, the same positive integer in Example 2.)
geometry number-theory euclidean-geometry big-picture
edited Aug 29 at 13:26
Mathmo123
17.5k32962
17.5k32962
asked Aug 27 at 18:16
Hans Stricker
4,39613674
4,39613674
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3 Answers
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This is only a partial answer for example 1 part 2. Intersecting circles of the same diameter have a line of symmetry through the two points of intersection and could be loosely considered deep insight into a method of dividing x (in this case the distance between circle centers) into two equal parts.
add a comment |Â
up vote
1
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Addressing only the title of your Q.
Farey-sequences are a useful tool in elementary Number Theory. Let $a,b,c,din Bbb N$ such that $a/b$ and $c/d$ are in lowest terms and $a/b<c/d.$ We say $a/b,c/d $ are Farey-adjacent iff for all $e,fin Bbb N$ such that $a/b<e/f<c/d,$ we have $f>max (b,d).$
An important property is that if $a/b,c/d$ are Farey-adjacent then $|ad-bc|=1.$ This can be proven algebraically. In "Introduction To Geometry" by Coxeter, it is shown how to prove it geometrically:
If $a/b,c/d$ are Farey-adjacent then we can easily show there is a finite sequence $T_1,..., T_n$ of affine area-preserving maps from $Bbb R^2$ to $Bbb R^2$ such that $T=prod_i=1^nT_i$ maps the triangle with vertices $(0,0),(a,b),(c,d)$ onto the triangle with vertices $(0,0), (1,0),(0,1). $ The area of the $triangle$ with vertices $(0,0)(a,b),(c,d)$ is $frac 12|ad-bc|$ while the area of the $triangle$ with vertices $(0,0),(1,0)(0,1)$ is $frac 12.$
add a comment |Â
up vote
0
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Concerning the product $nm$
The "essence" of the first construction is
$$ntimes m = m + m + dots + m (n text times)$$
The "essence" of the second construction is
$$n cdot m = n/(1/m)$$
That they are equivalent â i.e. $n times m = n cdot m$ â means
$$n/(1/m) = m + m + dots + m (n text times)$$
Because they are equivalent one may drop the multiplication sign:
$$nm := n times m = n cdot m$$
For symmetry reasons the first construction also shows that
$$m + m + dots + m (n text times) = n + n + dots + n (m text times)$$
For symmetry reasons the second construction also shows that
$$n/(1/m) = m/(1/n)$$
By the second construction it can be nicely shown that
$$(1 + 1) cdot 1/2 = 1$$
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
This is only a partial answer for example 1 part 2. Intersecting circles of the same diameter have a line of symmetry through the two points of intersection and could be loosely considered deep insight into a method of dividing x (in this case the distance between circle centers) into two equal parts.
add a comment |Â
up vote
1
down vote
This is only a partial answer for example 1 part 2. Intersecting circles of the same diameter have a line of symmetry through the two points of intersection and could be loosely considered deep insight into a method of dividing x (in this case the distance between circle centers) into two equal parts.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This is only a partial answer for example 1 part 2. Intersecting circles of the same diameter have a line of symmetry through the two points of intersection and could be loosely considered deep insight into a method of dividing x (in this case the distance between circle centers) into two equal parts.
This is only a partial answer for example 1 part 2. Intersecting circles of the same diameter have a line of symmetry through the two points of intersection and could be loosely considered deep insight into a method of dividing x (in this case the distance between circle centers) into two equal parts.
answered Aug 27 at 19:24
Phil H
1,9772311
1,9772311
add a comment |Â
add a comment |Â
up vote
1
down vote
Addressing only the title of your Q.
Farey-sequences are a useful tool in elementary Number Theory. Let $a,b,c,din Bbb N$ such that $a/b$ and $c/d$ are in lowest terms and $a/b<c/d.$ We say $a/b,c/d $ are Farey-adjacent iff for all $e,fin Bbb N$ such that $a/b<e/f<c/d,$ we have $f>max (b,d).$
An important property is that if $a/b,c/d$ are Farey-adjacent then $|ad-bc|=1.$ This can be proven algebraically. In "Introduction To Geometry" by Coxeter, it is shown how to prove it geometrically:
If $a/b,c/d$ are Farey-adjacent then we can easily show there is a finite sequence $T_1,..., T_n$ of affine area-preserving maps from $Bbb R^2$ to $Bbb R^2$ such that $T=prod_i=1^nT_i$ maps the triangle with vertices $(0,0),(a,b),(c,d)$ onto the triangle with vertices $(0,0), (1,0),(0,1). $ The area of the $triangle$ with vertices $(0,0)(a,b),(c,d)$ is $frac 12|ad-bc|$ while the area of the $triangle$ with vertices $(0,0),(1,0)(0,1)$ is $frac 12.$
add a comment |Â
up vote
1
down vote
Addressing only the title of your Q.
Farey-sequences are a useful tool in elementary Number Theory. Let $a,b,c,din Bbb N$ such that $a/b$ and $c/d$ are in lowest terms and $a/b<c/d.$ We say $a/b,c/d $ are Farey-adjacent iff for all $e,fin Bbb N$ such that $a/b<e/f<c/d,$ we have $f>max (b,d).$
An important property is that if $a/b,c/d$ are Farey-adjacent then $|ad-bc|=1.$ This can be proven algebraically. In "Introduction To Geometry" by Coxeter, it is shown how to prove it geometrically:
If $a/b,c/d$ are Farey-adjacent then we can easily show there is a finite sequence $T_1,..., T_n$ of affine area-preserving maps from $Bbb R^2$ to $Bbb R^2$ such that $T=prod_i=1^nT_i$ maps the triangle with vertices $(0,0),(a,b),(c,d)$ onto the triangle with vertices $(0,0), (1,0),(0,1). $ The area of the $triangle$ with vertices $(0,0)(a,b),(c,d)$ is $frac 12|ad-bc|$ while the area of the $triangle$ with vertices $(0,0),(1,0)(0,1)$ is $frac 12.$
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Addressing only the title of your Q.
Farey-sequences are a useful tool in elementary Number Theory. Let $a,b,c,din Bbb N$ such that $a/b$ and $c/d$ are in lowest terms and $a/b<c/d.$ We say $a/b,c/d $ are Farey-adjacent iff for all $e,fin Bbb N$ such that $a/b<e/f<c/d,$ we have $f>max (b,d).$
An important property is that if $a/b,c/d$ are Farey-adjacent then $|ad-bc|=1.$ This can be proven algebraically. In "Introduction To Geometry" by Coxeter, it is shown how to prove it geometrically:
If $a/b,c/d$ are Farey-adjacent then we can easily show there is a finite sequence $T_1,..., T_n$ of affine area-preserving maps from $Bbb R^2$ to $Bbb R^2$ such that $T=prod_i=1^nT_i$ maps the triangle with vertices $(0,0),(a,b),(c,d)$ onto the triangle with vertices $(0,0), (1,0),(0,1). $ The area of the $triangle$ with vertices $(0,0)(a,b),(c,d)$ is $frac 12|ad-bc|$ while the area of the $triangle$ with vertices $(0,0),(1,0)(0,1)$ is $frac 12.$
Addressing only the title of your Q.
Farey-sequences are a useful tool in elementary Number Theory. Let $a,b,c,din Bbb N$ such that $a/b$ and $c/d$ are in lowest terms and $a/b<c/d.$ We say $a/b,c/d $ are Farey-adjacent iff for all $e,fin Bbb N$ such that $a/b<e/f<c/d,$ we have $f>max (b,d).$
An important property is that if $a/b,c/d$ are Farey-adjacent then $|ad-bc|=1.$ This can be proven algebraically. In "Introduction To Geometry" by Coxeter, it is shown how to prove it geometrically:
If $a/b,c/d$ are Farey-adjacent then we can easily show there is a finite sequence $T_1,..., T_n$ of affine area-preserving maps from $Bbb R^2$ to $Bbb R^2$ such that $T=prod_i=1^nT_i$ maps the triangle with vertices $(0,0),(a,b),(c,d)$ onto the triangle with vertices $(0,0), (1,0),(0,1). $ The area of the $triangle$ with vertices $(0,0)(a,b),(c,d)$ is $frac 12|ad-bc|$ while the area of the $triangle$ with vertices $(0,0),(1,0)(0,1)$ is $frac 12.$
answered Aug 28 at 3:20
DanielWainfleet
32.1k31644
32.1k31644
add a comment |Â
add a comment |Â
up vote
0
down vote
Concerning the product $nm$
The "essence" of the first construction is
$$ntimes m = m + m + dots + m (n text times)$$
The "essence" of the second construction is
$$n cdot m = n/(1/m)$$
That they are equivalent â i.e. $n times m = n cdot m$ â means
$$n/(1/m) = m + m + dots + m (n text times)$$
Because they are equivalent one may drop the multiplication sign:
$$nm := n times m = n cdot m$$
For symmetry reasons the first construction also shows that
$$m + m + dots + m (n text times) = n + n + dots + n (m text times)$$
For symmetry reasons the second construction also shows that
$$n/(1/m) = m/(1/n)$$
By the second construction it can be nicely shown that
$$(1 + 1) cdot 1/2 = 1$$
add a comment |Â
up vote
0
down vote
Concerning the product $nm$
The "essence" of the first construction is
$$ntimes m = m + m + dots + m (n text times)$$
The "essence" of the second construction is
$$n cdot m = n/(1/m)$$
That they are equivalent â i.e. $n times m = n cdot m$ â means
$$n/(1/m) = m + m + dots + m (n text times)$$
Because they are equivalent one may drop the multiplication sign:
$$nm := n times m = n cdot m$$
For symmetry reasons the first construction also shows that
$$m + m + dots + m (n text times) = n + n + dots + n (m text times)$$
For symmetry reasons the second construction also shows that
$$n/(1/m) = m/(1/n)$$
By the second construction it can be nicely shown that
$$(1 + 1) cdot 1/2 = 1$$
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Concerning the product $nm$
The "essence" of the first construction is
$$ntimes m = m + m + dots + m (n text times)$$
The "essence" of the second construction is
$$n cdot m = n/(1/m)$$
That they are equivalent â i.e. $n times m = n cdot m$ â means
$$n/(1/m) = m + m + dots + m (n text times)$$
Because they are equivalent one may drop the multiplication sign:
$$nm := n times m = n cdot m$$
For symmetry reasons the first construction also shows that
$$m + m + dots + m (n text times) = n + n + dots + n (m text times)$$
For symmetry reasons the second construction also shows that
$$n/(1/m) = m/(1/n)$$
By the second construction it can be nicely shown that
$$(1 + 1) cdot 1/2 = 1$$
Concerning the product $nm$
The "essence" of the first construction is
$$ntimes m = m + m + dots + m (n text times)$$
The "essence" of the second construction is
$$n cdot m = n/(1/m)$$
That they are equivalent â i.e. $n times m = n cdot m$ â means
$$n/(1/m) = m + m + dots + m (n text times)$$
Because they are equivalent one may drop the multiplication sign:
$$nm := n times m = n cdot m$$
For symmetry reasons the first construction also shows that
$$m + m + dots + m (n text times) = n + n + dots + n (m text times)$$
For symmetry reasons the second construction also shows that
$$n/(1/m) = m/(1/n)$$
By the second construction it can be nicely shown that
$$(1 + 1) cdot 1/2 = 1$$
edited Aug 28 at 7:11
answered Aug 28 at 7:05
Hans Stricker
4,39613674
4,39613674
add a comment |Â
add a comment |Â
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