Let $p_d(x)$ be a polynomial of degree $d$ in one variable $x$, where the polynomial coefficients are in $\mathbb{Q}$. Let $R_d$ be the set of roots of all $p_d(x)=0$.
Q. Is it the case that the roots $R_d$ of all those polynomials, $d \ge 0$, even though their coeffcients are rational $\in \mathbb{Q}$, nevertheless are dense in $\mathbb{R}$?
The rational numbers $\mathbb{Q}$ are a dense subset of $\mathbb{R}$. We can show that $\mathbb{Q}\subset R_d$, and then it's trivial to prove that $R_d$ is dense in $\mathbb{R}$. Let $P = \{p(x) = x^d - q^d : q\in \mathbb{Q}\}$. The roots of all polynomials in $P$ are exactly $\mathbb{Q}$.
Certainly. Just look at the polynomials $x^d - a$, where $a$ is an $d$th power of a rational number. Since every rational number thus satisfies a degree $d$ polynomial and the rationals are dense in the reals, the answer is yes.
Sure. And if you’re hoping that $p_d$ was $\Bbb Q$-irreducible, as is not the case for Vik78’s answer, just take any real irrationality $\alpha$ of degree $d$ over $\Bbb Q$, and look at all the rational multiples of $\alpha$. Every one of these is root of an irreducible polynomial of degree $d$, and certainly they make up a dense subset of $\Bbb R$.
