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A Conjecture

Last post 14/02/12 at 18:55 by Andrew Jeffrey, 16 replies
Post started by Andrew Jeffrey on 06/02/12 at 13:26

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  • Andrew Jeffrey avatar
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    Posted by: Andrew Jeffrey 06/02/2012 at 13:26
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    Any rectangle can be cut into squares if the squares don't have to be the same size.

    Or it can't.

    Now I have not seen a really good proof either way (I have thrown it about on Twitter with Adam Creen and James Grime who both made interesting contributions to my thinking).

    Any thoughts? Adam and James both mentioned irrationals though I am not yet convinced that even a rectangle in the ratio1:root 2 could not be cut.

    So what do you think? Sorry 'AJs' Conjecture' isn't curriculum maths, it just interests me, and I am curious that I cannot find this as a theorem out there anywhere. It seems such a basic thing to want to find out!

    And could it be used with A-level students or top-set GCSE ones?

  • sedemihcrA avatar
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    Posted by: sedemihcrA 06/02/2012 at 14:54
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    The following page may be of interest:

    http://www.squaring.net/sq/tws.html

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    Posted by: Andrew Jeffrey 06/02/2012 at 15:07
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    It is indeed, thank you, but having read it twice now I still cannot find anywhere that says it is or is not always possible to square a rectangle. Am I missing something very obvious indeed?
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    Posted by: bombaysapphire 06/02/2012 at 16:20
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    If you have a rectangle with lengths x, y where x > y then you could cut off a square size y by y.

    Now your square is x-y, y.  Decide which is lesser and cut off a square with those lengths.

    If your sides are ever equal then you are done but otherwise the remaining rectangle will keep on decreasing in size.  Not sure where that gets you.  I'll have to try it out with some numbers!

    I would hypothesise that it will definitely work for x & y rational but not sure about irrational.
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    Posted by: PaulDG 06/02/2012 at 17:43
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    For it not to be possible for a rectangle to be "squared", there'd have to be some number z ( = xy), which cannot be made from addition of a set square numbers.

    I can't see why such a number should exist, but proving that seems like a re run of the 4 colour problem to me.

    (One thing I can see may be confusing with the talk of irrational numbers is that, while it's not possible to measure an irrational length, constructing one is quite easy. So I believe the concern about irrational numbers is a red herring.)
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    Posted by: Andrew Jeffrey 06/02/2012 at 19:19
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    Thanks both. I have been thinking along similar lines. But I still have a hunch that there is something we are missing! My first attempt was to cut a square from the left hand end, then from the left again if possible, then from the top, then back to the left, then the top, and so on.

    This throws up whether it is acceptable as it will throw up all squares other than an infinitely small rectangle at the end, so this proves that this particular algorithm will not work unless the width is an exact factor of the length. Or multiple...back to drawing board...

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    Posted by: pipipi 06/02/2012 at 19:32
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    hmm. A good one to think about thanks!
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    Posted by: brookes 06/02/2012 at 20:34
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    It is a good question! I found myself wondering what kind of rectangles it *would* easily be possible with, and I think answering that would be a good way to pursue this with a KS4 class (and not just the most able).
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    Posted by: Piranha 06/02/2012 at 20:47
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    My hunch is that it would not be possible for most rectangles. E.g. Root 2 by 1. Clearly, rational sided squares won't do, as they can't add up to root 2 (assuming a finite number of them). Multiples of root 2 won't do either as then the side length 1 can't be made. Off the top of my head, I can't think of a proof for more complicated irrational numbers. 

    Easy to do is rational sides. If they are of the form a/b and c/d, a-d integers, then squres of side 1/(ab) will do it, even using the same sized squares throughout. Similalry, sides of rational numbers times the same irrational number will also do it. E.g. root 8 and root 18.
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    Posted by: algebraist 06/02/2012 at 21:03
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    Interesting question! However, as I belatedly found out, it turns out it was solved in 1940 by Tutte and others, and the reference is in the Wikipedia article on rectangles :-) The answer is that you can tile a rectangle with finitely many squares iff the sides of the rectangle are commensurable, i.e., rational multiples of the same number. The paper is

    http://dx.doi.org/10.1215%2FS0012-7094-40-00718-9

    - which looks, shall we say, non-trivial. 

    I think it's obvious that if you're allowed infinitely many squares, then it is always possible: the greedy procedure in which you put in/cut off a square whose side is the smallest side of the remaining rectangle,and then repeat, converges by completeness of the reals, surely...

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