Showing that not all reals are algebraic using countability 
Definition: A number $z\in\mathbb{C}$ is said to be algebraic if there exist integers $a_0,a_1,\dots,a_n$ not all zero, such that
  $$
a_nz^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0=0
$$

I want to show that not every real number is algebraic. I thought in a simple argument but don't feel it's right: if every real was algebraic, for any $x\in\mathbb{R}$ we can find a sequence of integers $(a_0, a_1,\dots,a_n)\in\mathbb{Z}^{n+1}$ such that the equality above holds for $z=x$. So we can define a function $f:\cup_{n\in\mathbb{N}}\mathbb{Z}^n\to\mathbb{R}$ that for each real associate a sequence of integers as described in the definition. So $f$ would be surjective, then once $\cup_{n\in\mathbb{N}}\mathbb{Z}^n$ is countable, so would be $\mathbb{R}$, a contradiction because $\mathbb{R}$ is uncountable. So not all real numbers are algebraic.
Is this line of reasoning right? Thanks in advance.
 A: You’re pretty much there. Note that you use contradiction here, but there’s really no need:
Since $\bigcup_{n\in\mathbb{N}}\mathbb{Z}^n$ is countable, its image is countable. Moreover, the image must contain the set of algebraic numbers, and so algebraic numbers are countable. Since the reals are uncountable, the sets cannot be equal.
Fun note: this is a nice proof to see why algebraic numbers are measure zero in $\mathbb{R}$!
A: Show there are countably many polynomials with integer coefficients of degree n.  As there are at most n solutions to each polynomial, there are countably many algebraic numbers that are a solution to a n-th degree polynomial with integer coefficients.  
Collecting all those algebraic numbers for each n together shows altogether there are just countably many algebraic numbers. It is trivial to show that there are infinitely countable many algebraic numbers.  
Consequently, since the number of reals is uncountable, there are real numbers that are not algebraic.  In fact most of them are transcendent:  such an overwhelming majority that the real line is almost everywhere transdental, almost nowhere algebraic. 
A: Let $ \mathcal{A} $ be the set of all algebraic numbers. If $ \mathbb{R} \backslash \mathcal{A} $ empty, then $ \mathcal{A} = \mathbb{R} $ which is impossible, since $ \mathcal{A} $ is countable but $ \mathbb{R} $ is not. So, $ \mathbb{R} \backslash \mathcal{A} $ is not empty. Or in other word, there exist real numbers which are not algebraic.
