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I am aware that this has been addressed plenty of times here on this site and elsewhere. However, I want to offer a different proof for your examination. To iterate the problem is:

A complex number $z$, is algebraic if there are integers $a_{0}$,$a_{1}$,$…,a_{n}$ not all zero, such that $a_{0}z^{n}+ a_{1}z^{n-1}+ a_{2}z^{n-2}+…+ a_{n}=0$

Prove that the set of such algebraic numbers is countable.

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My proof follows the following lines:

  1. Let $S$ be the collection of all such coefficients $a_{0},a_{1},…,a_{n}$, each representing an equation of the above form.
  2. By diagonalisation, we can show $S$ is countably infinite.
  3. Let $X$ be the collection of sets of complex numbers which are algebraic
  4. Individual sets of the collection, $x \in X$ such that $x$ is the set of roots of the above equation.
  5. The above equation is a 1 to 1 relation from $X$ to $S$ such that every element of $S$ maps to an element in $X$ uniquely, where both elements are sets in themselves : element in set $S$ represent coefficients and element in $X$ represent roots
  6. Since $S$ is countable then it implies $X$ is countable as there is an injective relationship between the two.

Edits:

  1. Changed $X$ to collection
  2. Expanded point 5
  3. Corrected a typo in point 6
Soham
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    It is not clear to me how you establish a 1-to-1 relation between $X$ and $S$, since distinct polynomials can have common roots. – Martin R Oct 21 '22 at 09:04
  • Moreover, you did not define a map from $S$ to $X$ but (vaguely) from $S$ to the power set of $X.$ – Anne Bauval Oct 21 '22 at 09:08
  • @MartinR I miswrote . Apologies. $X$ is the collection of sets where each element $x$ is a set of roots (all algebraic by definition) ${z_{0},z_{1}, …, z_{n}}$. The elements of $S$ map on a 1 to 1 basis to the sets in $X$. – Soham Oct 21 '22 at 16:01
  • @AnneBauval Now that you mention it, you are right. I was assuming the equation itself is the relation which will maps the two sets together. – Soham Oct 21 '22 at 16:13

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