- #1
MalayInd
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If a cube is cut into finite number of smaller cubes, prove that at least two of them must be of same size.
Can a Cube Be Cut into Smaller Cubes of the Same Size?
- Context: Undergrad
- Thread starter MalayInd
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Discussion Overview
The discussion revolves around the question of whether a cube can be cut into a finite number of smaller cubes of the same size. Participants explore this concept through mathematical reasoning and comparisons with similar problems in two dimensions, particularly focusing on the implications of partitions and the properties of shapes.
Discussion Character
- Exploratory
- Mathematical reasoning
- Debate/contested
Main Points Raised
- One participant suggests that if a cube is cut into a finite number of smaller cubes, at least two of them must be of the same size.
- Another participant draws a parallel to the two-dimensional case of partitioning a square, questioning how the argument for the cube can be applied similarly.
- A different participant expresses uncertainty about formalizing the argument but asserts that the last cut made to create two cubes must result in cubes of the same size.
- One participant challenges the assertion regarding cubes by stating that it is possible to partition a square into smaller squares of different sizes, referencing 'squared squares' as a counterexample.
- Another participant attempts to apply the reasoning from the two-dimensional case to the cube, suggesting that the argument can still hold if certain conditions about the smallest square are met.
- A later reply points to external resources for further clarification on the topic.
Areas of Agreement / Disagreement
Participants do not reach a consensus, as there are competing views regarding the possibility of cutting a cube into smaller cubes of the same size, with some arguing for the necessity of equal sizes and others providing counterexamples from two-dimensional cases.
Contextual Notes
The discussion includes assumptions about the nature of partitions and the properties of shapes, which may not be universally applicable. The arguments presented rely on specific conditions that may not hold in all scenarios.
If a cube is cut into finite number of smaller cubes, prove that at least two of them must be of same size.
- #2
0rthodontist
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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
Integral
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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.
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
rhj23
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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 ]
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 ]
- #5
StatusX
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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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- #6
BSMSMSTMSPHD
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The answer is in rhj23's second link.
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