Proving Countability of {m+n, m,n \inZ} Using a NxN Scheme

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Homework Statement


Prove that {m+n, m,n \inZ} is countable


Homework Equations





The Attempt at a Solution

I Can prove it if I make a nxn scheme and put 1,-1,2,-2 along each side. This generates a table which when counted a long first,second etc. Diagonal hits all the numsers in the given set. But is this the formal Way to prove these kinds of things?
 
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Isn't that set just equal to Z again? Maybe I'm just misunderstanding notation...
 
aaaa202 said:

Homework Statement


Prove that {m+n, m,n \inZ} is countable


Homework Equations





The Attempt at a Solution

I Can prove it if I make a nxn scheme and put 1,-1,2,-2 along each side. This generates a table which when counted a long first,second etc. Diagonal hits all the numsers in the given set. But is this the formal Way to prove these kinds of things?
The set could also be described as {p | p = m + n, where m, n ##\in## Z}. All you need to do is to establish a one-one pairing with the integers. The things in the set are just numbers, not ordered pairs, so based on the notation you've used, your table is way more complicated than what is needed.



johnqwertyful said:
Isn't that set just equal to Z again? Maybe I'm just misunderstanding notation...
That's how I read it as well.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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