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$Z$ is the set of non-negative integers including $0$. Show that $Z \times Z \times Z$ is countable by constructing the actual bijection $f: Z\times Z\times Z \to \mathbb{N}$ ($\mathbb{N}$ is the set of all natural numbers). There is no need to prove that it is a bijection.

After searching for clues on how to solve this, I found $(x+y-1)(x+y-z)/z+y$ but that is only two dimensional and does not include $0$. Any help on how to solve this?

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    If you have too bijections $g: Z\times Z\to Z$ and $id:Z\to Z$, can you find a way to combine them into a bijection $h: Z\times Z\times Z\to Z\times Z$? If you have such bijections, what can you say about the composition $g\circ h$?2012-04-03
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    There's nothing which says that it needs to be a polynomial. The question could be rephrased by saying "Let $A,B,C$ be countable sets: show that $A\times B\times C$ is countable." Have you seen the proof that $\mathbb{N}\times\mathbb{N}$ is countable?2012-04-03
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    Do you mean $(x+y-1)(x+y-2)/2+y$?2012-04-03
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    Note that the question sentence is somewhat non-sensical. A bijection is not countable, a set is countable. (Yes, in some technical sense, a bijection is a set, but nowhere here are you referring to such a set.) What you want is a bijection that shows that the set $Z\times Z\times Z$ is countable.2012-04-03
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    Please fix the title along the lines of what Thomas suggests so that it makes sense! :)2012-04-03
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    @user25800 : It's been more than six hours since the senselessness of your title was pointed out. I've replaced it.2012-04-04

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