Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For every integer $n$ with $i$ prime factors associate a unique tile in $\mathbb{R}^m$ with $m \ge i$ as such, for every prime factor $p_j$ of $n$, the tile is a cuboid of dimension $m$ with a sidelength of $p_j$, the rest $m-j$ sides have length 1. E.g. if $m=5$ and $n=6$ the tile is a $2 \times 3 \times 1 \times 1 \times 1$ hypercuboid.

Is it possible to tile every $\mathbb{R}^m$ using each tile of prime factors less than $m$ exactly once?

Is it possible to tile $\mathbb{R}^\infty$ using the unique cuboid associated to every natural integer exactly once?

Is there any tiling of $\mathbb{R}^m$ which do not consist of infinite columns of $\mathbb{R}^{m-1} \times 1$, $\mathbb{R}^{m-1} \times 2$, $\mathbb{R}^{m-1} \times 3 \dots$ ?

share|cite|improve this question
Intuitively, the answer to the first question ought to be "yes". For the second question, what exactly do you mean by R^inf? It would seem to need to be $\mathbb R^{\mathbb N}$, the space of all infinite sequences (rather than, say, the space of sequences with finite support), since you speak of "cuboids". But in that case the answer would have to be "no", because $\mathbb R^{\mathbb N}$ contains uncountably many unit cubes. – Henning Makholm Oct 7 '11 at 20:29
You can delete the question yourself (using the "delete" link underneath it). Also, if you've found a solution, you could write it down and accept it instead of deleting the question. – joriki Oct 7 '11 at 21:25
@joriki: Unregistered users can't delete their contributions themselves (there is no delete link for them). So, in order to delete something, they must either register or flag for moderator attention. Of course, I agree that writing down the solution would be preferable to deletion. – t.b. Oct 8 '11 at 10:24
@t.b.: I see, thanks, sorry. – joriki Oct 8 '11 at 11:07
We cannot flag either – Optimus Prime Oct 8 '11 at 13:42

Here is the trivial tiling:

In R^2 put all nx1 tiles on a line for all n, all mx2 tiles one a line for m>=2 (to form a (inf)x2 column), all bx3 on a line for b>=3 etc. In R^3 take the tiling for R^2 which is R^2x1, then stack on top of it R^2x2, then under it R^2x3, then over it R^2x4 etc.

For any R^m use the same process with R^m-1.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.