# Cubing the cube



## robert@fm (Sep 15, 2018)

It is possible (never mind how, as that's not relevant to this puzzle; do a web search if you're interested) to cut a square into a finite number of smaller squares, no two of which are the same size.

Is it possible to divide a cube into a finite number of smaller cubes of all-different sizes? Prove your answer. (It's a simple inductive proof.)

Given a tesseract (the analogue of a cube in four-dimensional space), is it possible to slice it into smaller tesseracts in the same way? (This answer follows on quite simple from the previous one.)


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## C&E Guy (Sep 20, 2018)

I would say No. Because mini square #1 would have an edge touching another square inside the big one. Consequently, if it is then a perfect square, it must be the same dimensions as mini square #1.


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## chaoticcar (Sep 20, 2018)

Ehh ?
   Carol


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## C&E Guy (Sep 20, 2018)

chaoticcar said:


> Ehh ?
> Carol



Clear as mud!


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## robert@fm (Sep 21, 2018)

I think you've got the first part of the solution (each of the six faces of a perfect cubed cube would have to be a perfect squared square), but you seem to have lost your way after that. 

A little more thought and explanation is needed...


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## robert@fm (Jan 14, 2019)

OK then...

If a perfect cubed cube were possible, the smallest cube on any face would be surrounded on all sides by taller cubes, thus could have only smaller cubes on top of it. But the smallest of those could likewise be topped only by still smaller cubes... and so on to infinity.

If a perfect tesseracted tesseract were possible, each of its eight sides would be a perfect cubed cube — which we know to be impossible. Hence no perfect hypercubed hypercube in any higher dimension is possible either.


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