# A Conjecture

Discussion in 'Mathematics' started by Andrew Jeffrey, Feb 6, 2012.

1. ### Andrew JeffreyNew commenter

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?

3. ### Andrew JeffreyNew commenter

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?

4. ### bombaysapphireStar commenter

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.

5. ### PaulDGOccasional commenter

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.)

6. ### Andrew JeffreyNew commenter

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...

7. ### pipipiNew commenter

hmm. A good one to think about thanks!

8. ### brookes

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).

9. ### PiranhaStar commenter

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.

10. ### algebraist

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...

11. ### Andrew JeffreyNew commenter

Thank you very much indeed! I can only read the first page, frustratingly, and yet it throws up more questions than it answers, at least in my head.
I think that there is mileage in this even though someone has proved it. Have they disproved it for non-commensurables, I wonder, as surely ANY two rational numbers are multiples of the same amount, as long as the 'amount' is sufficiently small but finite?So is their proof 'if and only if' or just 'if'?!

12. ### algebraist

Yes, they have disprovedit for non-commensurables - the "iff" in my previous post is standard shorthand for "if and only if" but I guess I should have written that out in full here!
That's right, any two rational numbers are commensurable. So too are, say, 2789 pi / 87235655 and 70 pi / 44447, or 7 root 2 and 19/23 root 2, or... you get the idea. However, root 2 is not commensurable with any rational number, or with pi, or with e, or with root 3, or with... in fact, none of these quantities is commensurable with any of the others. Some of these statements are easy to prove - for example, that root 2 is incommensurable with any rational follows from the fact that it isn't itself rational, which is in turn just a few lines of fraction manipulation. Others are deep. If you like this kind of thing, you might enjoy a good book on fields and Galois theory, such as Ian Stewart's, a lot.

13. ### pbperth

This conjecture is very worthwhile and it can be very useful in teaching. Just keep it as simple as possible by considering only integer-sided squares, and pose the question "What squares can be partitioned into smaller (integer-sided) squares?" and it becomes an interesting prompt for all sorts of good mathematical thinking, even leading to a nice introduction to proof by induction. See my induction assignment, available at www.pbperth.com free downloads section.

Paul

14. ### NazardNew commenter

I can see this being of use when clarifying the concept of area, but I would have thought that pupils would be entirely underwhelmed by any sort of proof or explanation of this because it is so straightforward and, once you have thought about it, obvious.
As has already been pointed out (by Brookes, I think), if you start with an integer-sided rectangle (of which such squares are a subset) you can simply cut it into cm squares - which will be the area, and all of which will be integer-sided squares.
Or am I missing something?

15. ### pbperth

My apologies for not giving the whole story: you would have to see the download to get it all. The question I pursued was whether it was possible to partition into 2 squares, 3 squares, 4 squares... (all integer-sided and not necessarily the same size).

Obviously, partitioning a square into 2 squares is impossible. 3 squares also. And there are some quite challenging cases (is 7 squares possible?) But a lot falls into place after that. You can prove that partitioning a square into n squares is possible for all n above a certain value, and I think this exercise is quite a good way to approach induction techniques more generally.

Do please take a look: it's all a free download, including the solutions, at www.pbperth.com

16. ### PiranhaStar commenter

I've done this one a few times, although not insisting on integer sides. A bright Year 7 came up with a very convincing argument on the value on n.

17. ### Andrew JeffreyNew commenter

Credit where it's due - I have just re-read an article by Jo Tomlinson in January's MT magazine which is obviously what subconsciously sent me off thinking about this interesting idea in the first place. Thanks Jo!