What can be learned for number theory from geometrical constructions (and vice versa)?

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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:




  1. Constructing the half $x/2$ for one given positive real $x$ in two different ways.


  2. 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:




  1. enter image description here



  2. enter image description here


Example 2: Constructing the product $mn$



You can create the product $nm$ of two positive integers $n, m$ by two different Euclidean constructions:



  1. creating a rectangle
    enter image description here
    and counting the number of unit squares that fit into the rectangle


  2. creating a line segment
    enter image description here
    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 – later Descartes' – geometry and number theory, resp. arithmetic geometry) it's essential.




My questions are:




  1. How did (possibly) Euclid formulate the two statements above, i.e. that the two pairs of constructions always yield the same results?


  2. How did (possibly) Euclid prove these statements?


  3. 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.)








share|cite|improve this question


























    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:




    1. Constructing the half $x/2$ for one given positive real $x$ in two different ways.


    2. 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:




    1. enter image description here



    2. enter image description here


    Example 2: Constructing the product $mn$



    You can create the product $nm$ of two positive integers $n, m$ by two different Euclidean constructions:



    1. creating a rectangle
      enter image description here
      and counting the number of unit squares that fit into the rectangle


    2. creating a line segment
      enter image description here
      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 – later Descartes' – geometry and number theory, resp. arithmetic geometry) it's essential.




    My questions are:




    1. How did (possibly) Euclid formulate the two statements above, i.e. that the two pairs of constructions always yield the same results?


    2. How did (possibly) Euclid prove these statements?


    3. 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.)








    share|cite|improve this question
























      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:




      1. Constructing the half $x/2$ for one given positive real $x$ in two different ways.


      2. 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:




      1. enter image description here



      2. enter image description here


      Example 2: Constructing the product $mn$



      You can create the product $nm$ of two positive integers $n, m$ by two different Euclidean constructions:



      1. creating a rectangle
        enter image description here
        and counting the number of unit squares that fit into the rectangle


      2. creating a line segment
        enter image description here
        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 – later Descartes' – geometry and number theory, resp. arithmetic geometry) it's essential.




      My questions are:




      1. How did (possibly) Euclid formulate the two statements above, i.e. that the two pairs of constructions always yield the same results?


      2. How did (possibly) Euclid prove these statements?


      3. 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.)








      share|cite|improve this question














      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:




      1. Constructing the half $x/2$ for one given positive real $x$ in two different ways.


      2. 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:




      1. enter image description here



      2. enter image description here


      Example 2: Constructing the product $mn$



      You can create the product $nm$ of two positive integers $n, m$ by two different Euclidean constructions:



      1. creating a rectangle
        enter image description here
        and counting the number of unit squares that fit into the rectangle


      2. creating a line segment
        enter image description here
        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 – later Descartes' – geometry and number theory, resp. arithmetic geometry) it's essential.




      My questions are:




      1. How did (possibly) Euclid formulate the two statements above, i.e. that the two pairs of constructions always yield the same results?


      2. How did (possibly) Euclid prove these statements?


      3. 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.)










      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 29 at 13:26









      Mathmo123

      17.5k32962




      17.5k32962










      asked Aug 27 at 18:16









      Hans Stricker

      4,39613674




      4,39613674




















          3 Answers
          3






          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.






          share|cite|improve this answer



























            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 $frac 12.$






            share|cite|improve this answer



























              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 – i.e. $n times m = n cdot m$ – 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



              $$(1 + 1) cdot 1/2 = 1$$






              share|cite|improve this answer






















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                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.






                share|cite|improve this answer
























                  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.






                  share|cite|improve this answer






















                    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.






                    share|cite|improve this answer












                    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.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Aug 27 at 19:24









                    Phil H

                    1,9772311




                    1,9772311




















                        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 $frac 12.$






                        share|cite|improve this answer
























                          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 $frac 12.$






                          share|cite|improve this answer






















                            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 $frac 12.$






                            share|cite|improve this answer












                            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 $frac 12.$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Aug 28 at 3:20









                            DanielWainfleet

                            32.1k31644




                            32.1k31644




















                                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 – i.e. $n times m = n cdot m$ – 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



                                $$(1 + 1) cdot 1/2 = 1$$






                                share|cite|improve this answer


























                                  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 – i.e. $n times m = n cdot m$ – 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



                                  $$(1 + 1) cdot 1/2 = 1$$






                                  share|cite|improve this answer
























                                    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 – i.e. $n times m = n cdot m$ – 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



                                    $$(1 + 1) cdot 1/2 = 1$$






                                    share|cite|improve this answer














                                    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 – i.e. $n times m = n cdot m$ – 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



                                    $$(1 + 1) cdot 1/2 = 1$$







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Aug 28 at 7:11

























                                    answered Aug 28 at 7:05









                                    Hans Stricker

                                    4,39613674




                                    4,39613674



























                                         

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