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How do I go about showing that the cardinality of the set of natural numbers and the cardinality of the cartesian product of integers is the same?:

|N|=|Z x Z|

Directly |N| = Aleph-null and I can separate the right side like this: |Z|*|Z|, and because the cardinality of the set of integers is Aleph-null, Aleph-null times Aleph-null is still Aleph-null and thus they are equal. However, I need to show an example how a bijection can be used here? How do I construct a map to let me see a bijection is being used/provide that it is surjective/injective?

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I know that, but how do I do it? –  Shawn S. Rana Apr 24 '14 at 22:45
Do you know any bijections between $\mathbb Z$ and $\mathbb N$? How about between $\mathbb N$ and $\mathbb N\times\mathbb N$? –  Andres Caicedo Apr 24 '14 at 22:48
I know that they both have bijections but I do not how to prove it/find it –  Shawn S. Rana Apr 24 '14 at 22:52
I can set up a table such as the top row resembles integers: 0,1,-1,2,-2,3,-3,4,-4... and the bottom row resembles natural numbers: 1,2,3,4,5,6,7,8... There is a bijection –  Shawn S. Rana Apr 24 '14 at 22:58
Don't get distracted by that. Your task is to find one. Never mind that there are many, we just need one. A common example, that has a nice intuitive picture attached to it, is Cantor's pairing, see here. One that is easier to verify uses that each positive integer is the product of an odd number and a power of two, $n=2^a(2b+1)$. This gives you a bijection $n\mapsto(a,b)$ between the positive integers and $\mathbb N\times \mathbb N$. –  Andres Caicedo Apr 24 '14 at 23:37

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