Spheres cause contradictions in dimensions $10$ and more?

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According to this Numberphile video, if you tightly pack hyper-spheres into a hyper-box and then find the radius of the largest hyper-sphere that could possibly fit in the remaining space, the resulting hyper-sphere would somehow exceed the confines of the box that contained all of the hyper-spheres (where the number of dimensions are greater or equal to 10).



Isn't a logical contradiction generally considered a disproof of something?



Wouldn't this disprove the generalised formula being used to find the radius of the resulting sphere on n dimensions?



Is it possible that mathematicians simply do not understand extra dimensional geometry and its inherent rules?










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  • 3




    Sorry, where is the logical contradiction here?
    – Rahul
    Sep 25 at 10:12






  • 1




    This is not a logical contradiction, only a counterintuitive result.
    – Santana Afton
    Sep 25 at 10:14






  • 7




    This is a consequence of the pythagorean theorem extended to higher dimensions. For a 100-dimensional hypercube, a diagonal is ten times the length of any edge; but the hyperspheres you're packing in (excluding this weird center one) stay the same radius in every dimension if your cube's edge length stays the same. So it shouldn't be hard to imagine that the distance between a sphere in one "corner" gets farther from the one in the opposite corner of the cube as the dimension goes up.
    – Malice Vidrine
    Sep 25 at 11:39






  • 6




    That is a bit of a weird description. If anything my intuition has always been that it's cubes that get "spiky".
    – Malice Vidrine
    Sep 25 at 12:03






  • 3




    Another interesting consequence of high dimensional geometry is that the vast majority of the volume of a hypersolid is concentrated in an extremely thin shell near the surface. This seems counterintuitive, but think about it this way. Suppose you have 100 random numbers each from 0 to 1. That defines a point in a 100-d hypercube. What is the probability that this point is extremely close to the center? For that to happen, all the 100 numbers have to be close to 0.5, and that is unlikely! If a randomly chosen point is almost never near the center, the center must have low volume.
    – Eric Lippert
    Sep 25 at 16:49















up vote
14
down vote

favorite
1












According to this Numberphile video, if you tightly pack hyper-spheres into a hyper-box and then find the radius of the largest hyper-sphere that could possibly fit in the remaining space, the resulting hyper-sphere would somehow exceed the confines of the box that contained all of the hyper-spheres (where the number of dimensions are greater or equal to 10).



Isn't a logical contradiction generally considered a disproof of something?



Wouldn't this disprove the generalised formula being used to find the radius of the resulting sphere on n dimensions?



Is it possible that mathematicians simply do not understand extra dimensional geometry and its inherent rules?










share|cite|improve this question



















  • 3




    Sorry, where is the logical contradiction here?
    – Rahul
    Sep 25 at 10:12






  • 1




    This is not a logical contradiction, only a counterintuitive result.
    – Santana Afton
    Sep 25 at 10:14






  • 7




    This is a consequence of the pythagorean theorem extended to higher dimensions. For a 100-dimensional hypercube, a diagonal is ten times the length of any edge; but the hyperspheres you're packing in (excluding this weird center one) stay the same radius in every dimension if your cube's edge length stays the same. So it shouldn't be hard to imagine that the distance between a sphere in one "corner" gets farther from the one in the opposite corner of the cube as the dimension goes up.
    – Malice Vidrine
    Sep 25 at 11:39






  • 6




    That is a bit of a weird description. If anything my intuition has always been that it's cubes that get "spiky".
    – Malice Vidrine
    Sep 25 at 12:03






  • 3




    Another interesting consequence of high dimensional geometry is that the vast majority of the volume of a hypersolid is concentrated in an extremely thin shell near the surface. This seems counterintuitive, but think about it this way. Suppose you have 100 random numbers each from 0 to 1. That defines a point in a 100-d hypercube. What is the probability that this point is extremely close to the center? For that to happen, all the 100 numbers have to be close to 0.5, and that is unlikely! If a randomly chosen point is almost never near the center, the center must have low volume.
    – Eric Lippert
    Sep 25 at 16:49













up vote
14
down vote

favorite
1









up vote
14
down vote

favorite
1






1





According to this Numberphile video, if you tightly pack hyper-spheres into a hyper-box and then find the radius of the largest hyper-sphere that could possibly fit in the remaining space, the resulting hyper-sphere would somehow exceed the confines of the box that contained all of the hyper-spheres (where the number of dimensions are greater or equal to 10).



Isn't a logical contradiction generally considered a disproof of something?



Wouldn't this disprove the generalised formula being used to find the radius of the resulting sphere on n dimensions?



Is it possible that mathematicians simply do not understand extra dimensional geometry and its inherent rules?










share|cite|improve this question















According to this Numberphile video, if you tightly pack hyper-spheres into a hyper-box and then find the radius of the largest hyper-sphere that could possibly fit in the remaining space, the resulting hyper-sphere would somehow exceed the confines of the box that contained all of the hyper-spheres (where the number of dimensions are greater or equal to 10).



Isn't a logical contradiction generally considered a disproof of something?



Wouldn't this disprove the generalised formula being used to find the radius of the resulting sphere on n dimensions?



Is it possible that mathematicians simply do not understand extra dimensional geometry and its inherent rules?







geometry euclidean-geometry spheres paradoxes






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 26 at 3:33









user21820

36.7k441143




36.7k441143










asked Sep 25 at 10:06









Lorry Laurence mcLarry

5221513




5221513







  • 3




    Sorry, where is the logical contradiction here?
    – Rahul
    Sep 25 at 10:12






  • 1




    This is not a logical contradiction, only a counterintuitive result.
    – Santana Afton
    Sep 25 at 10:14






  • 7




    This is a consequence of the pythagorean theorem extended to higher dimensions. For a 100-dimensional hypercube, a diagonal is ten times the length of any edge; but the hyperspheres you're packing in (excluding this weird center one) stay the same radius in every dimension if your cube's edge length stays the same. So it shouldn't be hard to imagine that the distance between a sphere in one "corner" gets farther from the one in the opposite corner of the cube as the dimension goes up.
    – Malice Vidrine
    Sep 25 at 11:39






  • 6




    That is a bit of a weird description. If anything my intuition has always been that it's cubes that get "spiky".
    – Malice Vidrine
    Sep 25 at 12:03






  • 3




    Another interesting consequence of high dimensional geometry is that the vast majority of the volume of a hypersolid is concentrated in an extremely thin shell near the surface. This seems counterintuitive, but think about it this way. Suppose you have 100 random numbers each from 0 to 1. That defines a point in a 100-d hypercube. What is the probability that this point is extremely close to the center? For that to happen, all the 100 numbers have to be close to 0.5, and that is unlikely! If a randomly chosen point is almost never near the center, the center must have low volume.
    – Eric Lippert
    Sep 25 at 16:49













  • 3




    Sorry, where is the logical contradiction here?
    – Rahul
    Sep 25 at 10:12






  • 1




    This is not a logical contradiction, only a counterintuitive result.
    – Santana Afton
    Sep 25 at 10:14






  • 7




    This is a consequence of the pythagorean theorem extended to higher dimensions. For a 100-dimensional hypercube, a diagonal is ten times the length of any edge; but the hyperspheres you're packing in (excluding this weird center one) stay the same radius in every dimension if your cube's edge length stays the same. So it shouldn't be hard to imagine that the distance between a sphere in one "corner" gets farther from the one in the opposite corner of the cube as the dimension goes up.
    – Malice Vidrine
    Sep 25 at 11:39






  • 6




    That is a bit of a weird description. If anything my intuition has always been that it's cubes that get "spiky".
    – Malice Vidrine
    Sep 25 at 12:03






  • 3




    Another interesting consequence of high dimensional geometry is that the vast majority of the volume of a hypersolid is concentrated in an extremely thin shell near the surface. This seems counterintuitive, but think about it this way. Suppose you have 100 random numbers each from 0 to 1. That defines a point in a 100-d hypercube. What is the probability that this point is extremely close to the center? For that to happen, all the 100 numbers have to be close to 0.5, and that is unlikely! If a randomly chosen point is almost never near the center, the center must have low volume.
    – Eric Lippert
    Sep 25 at 16:49








3




3




Sorry, where is the logical contradiction here?
– Rahul
Sep 25 at 10:12




Sorry, where is the logical contradiction here?
– Rahul
Sep 25 at 10:12




1




1




This is not a logical contradiction, only a counterintuitive result.
– Santana Afton
Sep 25 at 10:14




This is not a logical contradiction, only a counterintuitive result.
– Santana Afton
Sep 25 at 10:14




7




7




This is a consequence of the pythagorean theorem extended to higher dimensions. For a 100-dimensional hypercube, a diagonal is ten times the length of any edge; but the hyperspheres you're packing in (excluding this weird center one) stay the same radius in every dimension if your cube's edge length stays the same. So it shouldn't be hard to imagine that the distance between a sphere in one "corner" gets farther from the one in the opposite corner of the cube as the dimension goes up.
– Malice Vidrine
Sep 25 at 11:39




This is a consequence of the pythagorean theorem extended to higher dimensions. For a 100-dimensional hypercube, a diagonal is ten times the length of any edge; but the hyperspheres you're packing in (excluding this weird center one) stay the same radius in every dimension if your cube's edge length stays the same. So it shouldn't be hard to imagine that the distance between a sphere in one "corner" gets farther from the one in the opposite corner of the cube as the dimension goes up.
– Malice Vidrine
Sep 25 at 11:39




6




6




That is a bit of a weird description. If anything my intuition has always been that it's cubes that get "spiky".
– Malice Vidrine
Sep 25 at 12:03




That is a bit of a weird description. If anything my intuition has always been that it's cubes that get "spiky".
– Malice Vidrine
Sep 25 at 12:03




3




3




Another interesting consequence of high dimensional geometry is that the vast majority of the volume of a hypersolid is concentrated in an extremely thin shell near the surface. This seems counterintuitive, but think about it this way. Suppose you have 100 random numbers each from 0 to 1. That defines a point in a 100-d hypercube. What is the probability that this point is extremely close to the center? For that to happen, all the 100 numbers have to be close to 0.5, and that is unlikely! If a randomly chosen point is almost never near the center, the center must have low volume.
– Eric Lippert
Sep 25 at 16:49





Another interesting consequence of high dimensional geometry is that the vast majority of the volume of a hypersolid is concentrated in an extremely thin shell near the surface. This seems counterintuitive, but think about it this way. Suppose you have 100 random numbers each from 0 to 1. That defines a point in a 100-d hypercube. What is the probability that this point is extremely close to the center? For that to happen, all the 100 numbers have to be close to 0.5, and that is unlikely! If a randomly chosen point is almost never near the center, the center must have low volume.
– Eric Lippert
Sep 25 at 16:49











4 Answers
4






active

oldest

votes

















up vote
19
down vote













No. It just means that (hyper)cubic lattice sphere packing (where the centers of the spheres are placed in a cubic grid, say spheres with radius $frac12$ centered at each point with integer cartesian coordinates) is very inefficient in higher dimensions, and the room between the spheres become large enough to fit even larger spheres.



Counterintuitive? Yes, but mostly because we are relatively low-dimensional beings with limited imagination. Paradox or contradiction? No.






share|cite|improve this answer





























    up vote
    13
    down vote













    In the video, they get a sphere too large to fit in the cube because they did not look for a sphere that fit inside the cube, they just looked for a sphere that fit between the other spheres.



    If you require that the central sphere also has to fit inside the original cube, as well as fit between the spheres in the corners, then once you get past dimension $4$ the central sphere will no longer touch the spheres in the corners.
    There will be gaps between the central sphere and the others.



    That is, in $5$ or more dimensions, if the central sphere is small enough to fit inside the box it is too small to touch all the corner spheres at once;
    if you make it large enough to touch all the corner spheres at once then of course it is now too large to fit between the sides of the box.



    It's all a result of the (possibly counterintuitive) fact that in $5$ or more dimensions, if you start at the center of a hypercube you have to travel farther to reach one of the "corner spheres" than you do to reach one of the sides of the hypercube.






    share|cite|improve this answer





























      up vote
      4
      down vote













      Think of it another way. The hyperspheres are the size you expect, but the hyperbox is much, much bigger than you'd expect.



      A 1m diameter circle in a square has 4 pyramids (well, triangles), each with a height of about 0.207m.



      A 1m sphere in a cube has 8 pyramids, each with a height of about 0.366m



      A 1m 10-sphere in a 10-cube has 1024 pyramids, each with a height of 1.081m - that's now longer than the cube's edge.



      Now imagine a 2m side 10-cube packed with 1024 of those 10-spheres. In between the 10-spheres there is a 2.162m wide void that can fit a 2.162m diameter hypersphere. It feels wrong, but it follows naturally as you increase in dimensions.






      share|cite|improve this answer



























        up vote
        3
        down vote













        Just take an $n$-dimensional hypercube of edge size 2 centered at the origin. Take furthermore hyperballs with unit radius centered at each of its vertices. Then those evidently will be touching by construction.



        By Pythagoras the distance from the origin, i.e. from the body center of that hypercube, towards either of its vertices happens to be
        $$sqrt1^2+1^2+...+1^2=sqrtn$$



        Accordingly an hyperball, which is centered at the origin and is touching those corner-centered unit hyperballs, should have a radius of
        $$sqrtn-1$$
        But that number clearly gets larger than $1$, which in turn clearly is the radius of the in-hyperball of that hypercube, as soon as $n>4$. - That is all of that counterintuitive magic.



        --- rk






        share|cite|improve this answer




















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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          19
          down vote













          No. It just means that (hyper)cubic lattice sphere packing (where the centers of the spheres are placed in a cubic grid, say spheres with radius $frac12$ centered at each point with integer cartesian coordinates) is very inefficient in higher dimensions, and the room between the spheres become large enough to fit even larger spheres.



          Counterintuitive? Yes, but mostly because we are relatively low-dimensional beings with limited imagination. Paradox or contradiction? No.






          share|cite|improve this answer


























            up vote
            19
            down vote













            No. It just means that (hyper)cubic lattice sphere packing (where the centers of the spheres are placed in a cubic grid, say spheres with radius $frac12$ centered at each point with integer cartesian coordinates) is very inefficient in higher dimensions, and the room between the spheres become large enough to fit even larger spheres.



            Counterintuitive? Yes, but mostly because we are relatively low-dimensional beings with limited imagination. Paradox or contradiction? No.






            share|cite|improve this answer
























              up vote
              19
              down vote










              up vote
              19
              down vote









              No. It just means that (hyper)cubic lattice sphere packing (where the centers of the spheres are placed in a cubic grid, say spheres with radius $frac12$ centered at each point with integer cartesian coordinates) is very inefficient in higher dimensions, and the room between the spheres become large enough to fit even larger spheres.



              Counterintuitive? Yes, but mostly because we are relatively low-dimensional beings with limited imagination. Paradox or contradiction? No.






              share|cite|improve this answer














              No. It just means that (hyper)cubic lattice sphere packing (where the centers of the spheres are placed in a cubic grid, say spheres with radius $frac12$ centered at each point with integer cartesian coordinates) is very inefficient in higher dimensions, and the room between the spheres become large enough to fit even larger spheres.



              Counterintuitive? Yes, but mostly because we are relatively low-dimensional beings with limited imagination. Paradox or contradiction? No.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Sep 25 at 10:21

























              answered Sep 25 at 10:12









              Arthur

              103k798179




              103k798179




















                  up vote
                  13
                  down vote













                  In the video, they get a sphere too large to fit in the cube because they did not look for a sphere that fit inside the cube, they just looked for a sphere that fit between the other spheres.



                  If you require that the central sphere also has to fit inside the original cube, as well as fit between the spheres in the corners, then once you get past dimension $4$ the central sphere will no longer touch the spheres in the corners.
                  There will be gaps between the central sphere and the others.



                  That is, in $5$ or more dimensions, if the central sphere is small enough to fit inside the box it is too small to touch all the corner spheres at once;
                  if you make it large enough to touch all the corner spheres at once then of course it is now too large to fit between the sides of the box.



                  It's all a result of the (possibly counterintuitive) fact that in $5$ or more dimensions, if you start at the center of a hypercube you have to travel farther to reach one of the "corner spheres" than you do to reach one of the sides of the hypercube.






                  share|cite|improve this answer


























                    up vote
                    13
                    down vote













                    In the video, they get a sphere too large to fit in the cube because they did not look for a sphere that fit inside the cube, they just looked for a sphere that fit between the other spheres.



                    If you require that the central sphere also has to fit inside the original cube, as well as fit between the spheres in the corners, then once you get past dimension $4$ the central sphere will no longer touch the spheres in the corners.
                    There will be gaps between the central sphere and the others.



                    That is, in $5$ or more dimensions, if the central sphere is small enough to fit inside the box it is too small to touch all the corner spheres at once;
                    if you make it large enough to touch all the corner spheres at once then of course it is now too large to fit between the sides of the box.



                    It's all a result of the (possibly counterintuitive) fact that in $5$ or more dimensions, if you start at the center of a hypercube you have to travel farther to reach one of the "corner spheres" than you do to reach one of the sides of the hypercube.






                    share|cite|improve this answer
























                      up vote
                      13
                      down vote










                      up vote
                      13
                      down vote









                      In the video, they get a sphere too large to fit in the cube because they did not look for a sphere that fit inside the cube, they just looked for a sphere that fit between the other spheres.



                      If you require that the central sphere also has to fit inside the original cube, as well as fit between the spheres in the corners, then once you get past dimension $4$ the central sphere will no longer touch the spheres in the corners.
                      There will be gaps between the central sphere and the others.



                      That is, in $5$ or more dimensions, if the central sphere is small enough to fit inside the box it is too small to touch all the corner spheres at once;
                      if you make it large enough to touch all the corner spheres at once then of course it is now too large to fit between the sides of the box.



                      It's all a result of the (possibly counterintuitive) fact that in $5$ or more dimensions, if you start at the center of a hypercube you have to travel farther to reach one of the "corner spheres" than you do to reach one of the sides of the hypercube.






                      share|cite|improve this answer














                      In the video, they get a sphere too large to fit in the cube because they did not look for a sphere that fit inside the cube, they just looked for a sphere that fit between the other spheres.



                      If you require that the central sphere also has to fit inside the original cube, as well as fit between the spheres in the corners, then once you get past dimension $4$ the central sphere will no longer touch the spheres in the corners.
                      There will be gaps between the central sphere and the others.



                      That is, in $5$ or more dimensions, if the central sphere is small enough to fit inside the box it is too small to touch all the corner spheres at once;
                      if you make it large enough to touch all the corner spheres at once then of course it is now too large to fit between the sides of the box.



                      It's all a result of the (possibly counterintuitive) fact that in $5$ or more dimensions, if you start at the center of a hypercube you have to travel farther to reach one of the "corner spheres" than you do to reach one of the sides of the hypercube.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Sep 25 at 17:29

























                      answered Sep 25 at 15:07









                      David K

                      49.7k340111




                      49.7k340111




















                          up vote
                          4
                          down vote













                          Think of it another way. The hyperspheres are the size you expect, but the hyperbox is much, much bigger than you'd expect.



                          A 1m diameter circle in a square has 4 pyramids (well, triangles), each with a height of about 0.207m.



                          A 1m sphere in a cube has 8 pyramids, each with a height of about 0.366m



                          A 1m 10-sphere in a 10-cube has 1024 pyramids, each with a height of 1.081m - that's now longer than the cube's edge.



                          Now imagine a 2m side 10-cube packed with 1024 of those 10-spheres. In between the 10-spheres there is a 2.162m wide void that can fit a 2.162m diameter hypersphere. It feels wrong, but it follows naturally as you increase in dimensions.






                          share|cite|improve this answer
























                            up vote
                            4
                            down vote













                            Think of it another way. The hyperspheres are the size you expect, but the hyperbox is much, much bigger than you'd expect.



                            A 1m diameter circle in a square has 4 pyramids (well, triangles), each with a height of about 0.207m.



                            A 1m sphere in a cube has 8 pyramids, each with a height of about 0.366m



                            A 1m 10-sphere in a 10-cube has 1024 pyramids, each with a height of 1.081m - that's now longer than the cube's edge.



                            Now imagine a 2m side 10-cube packed with 1024 of those 10-spheres. In between the 10-spheres there is a 2.162m wide void that can fit a 2.162m diameter hypersphere. It feels wrong, but it follows naturally as you increase in dimensions.






                            share|cite|improve this answer






















                              up vote
                              4
                              down vote










                              up vote
                              4
                              down vote









                              Think of it another way. The hyperspheres are the size you expect, but the hyperbox is much, much bigger than you'd expect.



                              A 1m diameter circle in a square has 4 pyramids (well, triangles), each with a height of about 0.207m.



                              A 1m sphere in a cube has 8 pyramids, each with a height of about 0.366m



                              A 1m 10-sphere in a 10-cube has 1024 pyramids, each with a height of 1.081m - that's now longer than the cube's edge.



                              Now imagine a 2m side 10-cube packed with 1024 of those 10-spheres. In between the 10-spheres there is a 2.162m wide void that can fit a 2.162m diameter hypersphere. It feels wrong, but it follows naturally as you increase in dimensions.






                              share|cite|improve this answer












                              Think of it another way. The hyperspheres are the size you expect, but the hyperbox is much, much bigger than you'd expect.



                              A 1m diameter circle in a square has 4 pyramids (well, triangles), each with a height of about 0.207m.



                              A 1m sphere in a cube has 8 pyramids, each with a height of about 0.366m



                              A 1m 10-sphere in a 10-cube has 1024 pyramids, each with a height of 1.081m - that's now longer than the cube's edge.



                              Now imagine a 2m side 10-cube packed with 1024 of those 10-spheres. In between the 10-spheres there is a 2.162m wide void that can fit a 2.162m diameter hypersphere. It feels wrong, but it follows naturally as you increase in dimensions.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Sep 25 at 14:09









                              HP Williams

                              1412




                              1412




















                                  up vote
                                  3
                                  down vote













                                  Just take an $n$-dimensional hypercube of edge size 2 centered at the origin. Take furthermore hyperballs with unit radius centered at each of its vertices. Then those evidently will be touching by construction.



                                  By Pythagoras the distance from the origin, i.e. from the body center of that hypercube, towards either of its vertices happens to be
                                  $$sqrt1^2+1^2+...+1^2=sqrtn$$



                                  Accordingly an hyperball, which is centered at the origin and is touching those corner-centered unit hyperballs, should have a radius of
                                  $$sqrtn-1$$
                                  But that number clearly gets larger than $1$, which in turn clearly is the radius of the in-hyperball of that hypercube, as soon as $n>4$. - That is all of that counterintuitive magic.



                                  --- rk






                                  share|cite|improve this answer
























                                    up vote
                                    3
                                    down vote













                                    Just take an $n$-dimensional hypercube of edge size 2 centered at the origin. Take furthermore hyperballs with unit radius centered at each of its vertices. Then those evidently will be touching by construction.



                                    By Pythagoras the distance from the origin, i.e. from the body center of that hypercube, towards either of its vertices happens to be
                                    $$sqrt1^2+1^2+...+1^2=sqrtn$$



                                    Accordingly an hyperball, which is centered at the origin and is touching those corner-centered unit hyperballs, should have a radius of
                                    $$sqrtn-1$$
                                    But that number clearly gets larger than $1$, which in turn clearly is the radius of the in-hyperball of that hypercube, as soon as $n>4$. - That is all of that counterintuitive magic.



                                    --- rk






                                    share|cite|improve this answer






















                                      up vote
                                      3
                                      down vote










                                      up vote
                                      3
                                      down vote









                                      Just take an $n$-dimensional hypercube of edge size 2 centered at the origin. Take furthermore hyperballs with unit radius centered at each of its vertices. Then those evidently will be touching by construction.



                                      By Pythagoras the distance from the origin, i.e. from the body center of that hypercube, towards either of its vertices happens to be
                                      $$sqrt1^2+1^2+...+1^2=sqrtn$$



                                      Accordingly an hyperball, which is centered at the origin and is touching those corner-centered unit hyperballs, should have a radius of
                                      $$sqrtn-1$$
                                      But that number clearly gets larger than $1$, which in turn clearly is the radius of the in-hyperball of that hypercube, as soon as $n>4$. - That is all of that counterintuitive magic.



                                      --- rk






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                                      Just take an $n$-dimensional hypercube of edge size 2 centered at the origin. Take furthermore hyperballs with unit radius centered at each of its vertices. Then those evidently will be touching by construction.



                                      By Pythagoras the distance from the origin, i.e. from the body center of that hypercube, towards either of its vertices happens to be
                                      $$sqrt1^2+1^2+...+1^2=sqrtn$$



                                      Accordingly an hyperball, which is centered at the origin and is touching those corner-centered unit hyperballs, should have a radius of
                                      $$sqrtn-1$$
                                      But that number clearly gets larger than $1$, which in turn clearly is the radius of the in-hyperball of that hypercube, as soon as $n>4$. - That is all of that counterintuitive magic.



                                      --- rk







                                      share|cite|improve this answer












                                      share|cite|improve this answer



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                                      answered Sep 25 at 17:56









                                      Dr. Richard Klitzing

                                      8616




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