# A Simple Maths Question-1

1. Jun 1, 2006

### MalayInd

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

2. Jun 1, 2006

### 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?

3. Jun 1, 2006

### Integral

Staff Emeritus
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.

4. Jun 2, 2006

5. Jun 2, 2006

### 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.

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• ###### square.GIF
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Last edited: Jun 2, 2006
6. Jun 2, 2006

### BSMSMSTMSPHD

The answer is in rhj23's second link.