An injective function going from N to the set of algebraic numbers

hangainlover
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Homework Statement


Prove that the set of algebraic numbers is countably infinite.


Homework Equations


If there exists a bijective map between N and a set A, N and A have the same cardinality


The Attempt at a Solution


Rather than coming up with a bijective map between S =the set of algebraic numers and N =natural numbers, I proved S is countable but i also have to prove that S is infinite.
So, I wanted to design an injective function f:N to S.
Can anyone come up with sucn an injective function?
 
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How about the map that sends n to n...?
 
Ahh...
$f(x,n)=x-n$\\
$k(n) ={a:f(a,n)=0}$\\

Then, $k(1)=1, k(2)=2,k(3)=3, ... k(n) = n$.\\
this should be an injective map from N to the set of algebraic numbers...
 
thank you so much!
 
No problem.

But really you shouldn't be thinking of maps and such: just try to think about why there are infinitely many algebraic numbers. Every rational number is algebraic. But there's more. Every number of the form ##\sqrt[n]{r}## with r rational is also algebraic. And there's more still..
 
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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