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I have a three-dimensional vector space over integers: $\mathbb{Z}^3$.

I want to split it up into $n$ partitions. For $n=2$ I would simply split it into two sets, one where $x$ just has even numbers,and one where $x$ is odd. But how to do it for $n > 2$? I need some generalization, but lack the math skills. Also just $x$ and $y$ should be used for partitioning, all possibilities for, lets say $(1,1,z)$ should be in one set.

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Since $\mathbb{Z}$ is not a field what you have is not a vector space, it is a module –  Belgi Oct 9 '12 at 6:25
    
Oh, okay... I haven't had much linear algebra in my studies, sorry :) –  K.. Oct 9 '12 at 6:29

1 Answer 1

up vote 2 down vote accepted

It is technically not a vector space, since $\mathbb{Z}$ is not a field.

There are many ways to partition. For example, throw $(x,y,z)$ and $(u,v,w)$ into the same bin if $x+y+z$ and $u+v+w$ have the same remainder on division by $n$.

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So simple... thank you :) and if I wanted this independent from the z value, I would throw (x,y,z) and (u,v,w) into the same bin when x+y and u+v have the same remainder? –  K.. Oct 9 '12 at 6:33
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Sure, there are any number of variant ideas. And I chose remainders because I am number-theoretically minded, but something quite different may be more suitable for the application you have in mind. If that were more closely specified (preferably in another question) I am sure you would be given more directly relevant ideas. –  André Nicolas Oct 9 '12 at 6:37
    
I will do this, thank you :) –  K.. Oct 9 '12 at 6:39

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