How many 3D cubes can fit inside a hypercube in R^4?

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let suppose we have an hypercube in R^4 then m y question is how many 3-dimensional cubes could we put inside our hypercube?...
 
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A rhetorical question for you to ponder: how many squares are there in a cube?
 
eljose said:
let suppose we have an hypercube in R^4 then m y question is how many 3-dimensional cubes could we put inside our hypercube?...


If you mean "How many cube faces does a hyper-cube have"?
then 8, possibly

otherwise your question doesn't make sense, as there is no number of 3-dimensional cubes that we could put inside a hypercube.
 
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the question is let,s suppose we have a four dimensional space,then could we put inside this four dimensional space our 3-dimensional space?,i think the question has been answered when considering a plane made by an infinite numer of curves or a line made by an infinite numer of points
 
Is the statement \mathbb{R}^3 \subseteq \mathbb{R}^4 true? (Hint: NO!)
 
Good point. (But there is, of course, a subset of R4 that is diffeomorphic to R3.)
 
True. And now that I think about it, my original post isn't anything close to a good answer to the original question at all~
 
matt grime said:
A rhetorical question for you to ponder: how many squares are there in a cube?

there are infinite squares in a cube. so to draw a conclusion: there are infinite cubes in a hypercube.
 
I don't really understand the erm "put inside". What if a squre is bigger than the face of a cube? Wouldn't the cube only be able to contain squares that are smaller than or equal to the size of its faces?
 
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Icebreaker said:
I don't really understand the erm "put inside". What if a squre is bigger than the face of a cube? Wouldn't the cube only be able to contain squares that are smaller than or equal to the size of its faces?

Only if they weren't on curved surfaces.
 

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