Proving the Countability of Nx{0}

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Homework Statement


Proove that Nx{0} is countable. x stand for a product i.e. like the cartesian product NxN

Homework Equations


N is countable.

The Attempt at a Solution


This is so obvious since Nx{0} is just (1,0),(2,0) etc. But how do you write a proof formally?
 
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aaaa202 said:

Homework Statement


Proove that Nx{0} is countable. x stand for a product i.e. like the cartesian product NxN


Homework Equations


N is countable.


The Attempt at a Solution


This is so obvious since Nx{0} is just (1,0),(2,0) etc. But how do you write a proof formally?

Show there is a 1-1 correspondence between Nx{0} and N. Write an explicit bijection. Yes, it is easy.
 
So I can define the map:
f: N->Nx{0}
f(x)=(x,0)
and state that this is clearly bijective as a proof?
 
aaaa202 said:
So I can define the map:
f: N->Nx{0}
f(x)=(x,0)
and state that this is clearly bijective as a proof?
I suggest writing the extra few lines to prove injectivity and surjectivity instead of writing "clearly." The problem statement itself seems so obvious that it's tempting to write "clearly" as a one-word proof, but clearly that isn't the intent.
 
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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