This appears to be a new question on MSE. The only post on here after a search using the string irrational "packing" does not mention (explicitly) what I have in mind. A Google search doesn't return anything specific worth noting to an amateur like me. Finally, MO is empty in this regard too (but that's without looking into the minutiae of the posts that show up).

The Question:

Is it possible to pack $\Bbb R^n$ for some/each $n$, with copies of a regular shape and no gaps, where each copy is of a distinct, positive irrational defining length?


I'm guessing the answer is "no". Why? I don't know: pure speculation. It just seems typical of the behaviour of irrational numbers.

Having said that, though, it's just occurred to me that $\Bbb R$ can be packed with line segments, each of a distinct, (positive) irrational length. This seems obvious. List the line segments (assuming countably infinitely many is sufficient); spiral out from zero and leave no gaps.

I'm reminded of squared squares for when $n=2$. They give me confidence that the $n=2$ case is possible. Multiply a squared square by an irrational number. Build a bigger squared square from it by keeping the ratios between the square side lengths the same as the original squared square. Carry on ad infinitum.

Please help :)

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    $\begingroup$ Getting the lengths irrational doesn't feel like a challenge at all if you use an absolute scale (as opposed to, say, requiring the ratios of the sizes to be irrational). After all, I would guess that "regular shape" implies that there will be only countably infinitely many tiles. A cardinality argument then tells that you can scale the whole thing to consist of irrational sized tiles only. $\endgroup$ – Jyrki Lahtonen Jan 26 at 12:57
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    $\begingroup$ One of the 110x110 made out of 22 smaller squares uses a 1x1 tile. So you can use that squared square as the smallest unit, and blow up the rest by a factor of 110. Rinse, repeat. Should give you the whole plane covered by distinct sized squares , no? $\endgroup$ – Jyrki Lahtonen Jan 26 at 13:17
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    $\begingroup$ But, what exactly you require from a shape to call it regular? $\endgroup$ – Jyrki Lahtonen Jan 26 at 13:20

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