Are there any natural (or at least pretty) ways to tile a cube with an infinite number of distinct cubes? Jim On Tue, Feb 7, 2017 at 8:50 PM, David Wilson <davidwwilson@comcast.net> wrote:
Ok, so stack yet smaller cubes on top of it to climb the curb... But then the small cube's top face is dissected into yet smaller squares, and the cube sitting on that square has a curb. Infinite descent, ergo no smallest cube. Hence a cube cannot be tiled with a finite number of distinct cubes.
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Allan Wechsler Sent: Tuesday, February 07, 2017 9:16 AM To: math-fun Subject: Re: [math-fun] Lifted from Littlewood, hacked from Hacking
It then continues: The *smallest* square of that dissection is the bottom of a cube, whose top is a square platform with a nonzero "curb" all the way around its edges.
On Tue, Feb 7, 2017 at 9:04 AM, James Propp <jamespropp@gmail.com> wrote:
I remember how the proof starts: Look at the bottom face of the cube and how it's dissected into squares.
Jim Propp
On Tuesday, February 7, 2017, Fred Lunnon <fred.lunnon@gmail.com> wrote:
I had seen this before, long ago, but then forgotten it.
A square can be dissected into finitely many unequal squares, but a cube cannot be dissected into finitely many unequal cubes.
Why not?
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