Squaring the square

[igam...@univ-aix.fr]

I am looking for an ingenious proof which demonstrates that if a
rectangle is tiled by squares of different sizes, then the sizes of the
tiles and of the rectangle are commensurable (integer multiples of a
single number).
I am also looking for the proof about the impossibility for tiling a
cube using cubes of different sizes.

If you know where I can find those proofs please send me an email
(igam...@univ-aix.fr).

Thanks,

Ian

[James Smith]

igam...@univ-aix.fr wrote:

> I am also looking for the proof about the impossibility for tiling a
> cube using cubes of different sizes.

You mean that are not integer multiples of each other's size. You could
fill a 4x4x4 cube with 7 2x2x2 cubes and 8 1x1x1 cubes.

Jim

[Mark J. Tilford]

Cubes of different sizes means no two cubes have the same size.

--
-----------------------
Mark Jeffrey Tilford
til...@cco.caltech.edu

[Virgil]

In article <375F6195...@cc.newcastle.edu.au>, James Smith
<en...@cc.newcastle.edu.au> wrote:

>> I am also looking for the proof about the impossibility for tiling a
>> cube using cubes of different sizes.

I assume you mean impossible with finitely many smaller cubes of different
sizes.

Then:

Consider the square formng the bottom surface layer of such a tiling. It
is must be tiled by the cubes into different sized squares, and contains
a smallest square.

The height of the cube on this smallest square is less than the heights of
cubes on the of the adjacent squares, so the top surface of this cube is
below the tops of the other cubes.

This smallest square must be tiled in a way similar to the original square
with a similar result.

Clearly the process cannot end.

--
Virgil
vm...@frii.com