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

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Homework Help Overview

The discussion revolves around proving that the set of algebraic numbers is countably infinite, with a focus on constructing an injective function from the natural numbers to this set.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to design an injective function from the natural numbers to the set of algebraic numbers, seeking input on possible mappings. Some participants suggest specific mappings, while others encourage a broader consideration of the nature of algebraic numbers.

Discussion Status

Participants are exploring various mappings and discussing the infinite nature of algebraic numbers. Some guidance has been offered regarding the existence of algebraic numbers beyond rational numbers, but there is no explicit consensus on a specific injective function.

Contextual Notes

The original poster is required to prove both that the set of algebraic numbers is countable and that it is infinite, which influences the direction of the discussion.

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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..
 

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