Can a Cube Be Cut into Smaller Cubes of the Same Size?

[MalayInd]

If a cube is cut into finite number of smaller cubes, prove that at least two of them must be of same size.

 

[0rthodontist]

Consider the 2 dimensional case. If this is true in 3 dimensions then it works in 2 dimensions too for partitions of a square (a cross section of the cube will be a partition of the square with the same property). Now (under the assumption that a partition exists with no two squares the same size) consider the left side of the square and the smallest square of the partition that touches that side. You know that an edge of this square must be less than half the size of the whole square, or it must be equal to the size of the whole square (why?). Discarding that second case for now because it is a trivial partition, consider the rightmost edge of this smallest square. How can you re-apply the same argument to this edge? Where does re-applying the argument a large number of times lead you?

 

[Integral]

I am not sure how to expess this formally.

You must make a last cut to create 2 cubes. The cube faces you create with this last cut must be the same side length. Therefore the cubes are the same size.

 

[rhj23]

i don't know about the case for cubes but it's definitely not true for squares.

it's possible to partition a square into smaller squares with integer sides such that none have the same side length. they are known as 'squared squares'

[See: http://mathworld.wolfram.com/PerfectSquareDissection.html

http://en.wikipedia.org/wiki/Squaring_the_square ]

 

[StatusX]

The same type of argument Orthodontist suggested can be applied directly to the cube case. While this argument fails in the case of the square due to what might happen if the smallest square lies on an edge, as shown in the attached picture, it can be shown to still work in the cube case if you show that the smallest square of a tiled square (with all tiles being squares of different sizes) cannot lie on the edge of the square. This can be shown by exhausting a few possibilities.

 

[BSMSMSTMSPHD]

The answer is in rhj23's second link.