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Given a closed rectangle $R$ are there closed squares $(S_{i})_{1\leq i\leq n}$ such that $R=\underset{1\leq i\leq n}{\cup}S_{i}$ and $S_{i}^{\circ}\cap S_{j}^{\circ}=\varnothing$ for $i\neq j$ ?

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If your rectangle has incommensurable sides, I don't see how it can be a finite union of almost disjoint squares. –  Gerry Myerson Apr 27 '12 at 3:32
    
I don't either. But I really don't see how to show it can't. –  Bill Apr 27 '12 at 3:41
    
A proof that this isn't possible in the case of incommensurable side lengths is in 'Proofs from the book' - Google Books has a preview (books.google.com.au/…) which is unfortunately (or perhaps fortunately) missing a page of the proof. I'd post this as an answer, but I haven't figured out the rest of the proof for myself yet! –  Alex Amenta Apr 27 '12 at 5:13
    
Another Google Books-able proof is available in "Problems and theorems in classical set theory" by Komjath and Totik (search within the book for "union of squares"). AFAIK the result is originally due to Max Dehn, who used similar ideas to solve Hilbert's third problem. (This brings to mind a good exercise in proof writing: give a short, rigorous proof that if a rectangle is a finite and essentially disjoint union of other rectangles, then every edge of each of these other rectangles is either parallel to (or perpendicular to, or contained in) an edge of the original rectangle.) –  leslie townes Apr 27 '12 at 5:24
    
See also Tiling a Rectangle by Squares, which is item 12 (pages 39-40) of Jiri Matousek, Thirty-three Miniatures. This is a book of attractive applications of Linear Algebra. –  Gerry Myerson Apr 27 '12 at 6:15
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