Show that the set of all algebraic numbers is countable.
Note: Part (a) asks you to prove the very fun and interesting fact that $\sqrt{2}+\sqrt{3}$ is an algebraic number. More generally, the algebraic numbers are closed under addition and, in fact, form a field.
We did part $\texttt{(a)}$ in Abstract Algebra. I show it here for everyone. Note for $\sqrt{2}$, the polynomial $p(x)=x^2-2$ works and for $\sqrt[3]{2}$, the polynomial $p(x)=x^3-2$ works. The interesting one is $\sqrt{2}+\sqrt{3}$. Let $x=\sqrt{3} + \sqrt{2}$. We do some algebra and get $x^2=2+2\sqrt{6}+3\implies x^2-5=2\sqrt{6}$. Squaring both sides again we get $x^4-10x^2+25=24\implies x^4-10x^2+1=0$. Therefore our polynomial is $p(x)=x^4-10x^2+1$. You can plug $\sqrt{2}+\sqrt{3}$ in for $x$ and see it works.
Very nice!!!!!!!!!!!!!
I'm not really going to prove this, but I can give an intuitive idea.
Def: A number $x\in\mathbb{R}$ is algebraic if there exists $a_0,a_1,\dots ,a_n\in \mathbb{Z}$ such that $$f(x)=a_nx^n+a_{n-1}x^{n-1}+\dots +a_1x+a_0=0$$
We want to prove the set of algebraic numbers is countable. Note that Exercise 1.4.10 asks us to do basically the same thing, the only difference in this case is we look at something closer to tuples than sets because values can be repeated for $a_i$ and order matters. Also, we do it over the integers instead of the natural numbers which makes the orderings a little bit harder. However, I think it is quite clear that this set requires a very similar proof to Exercise 1.4.10 because the polynomials are finite.