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**TheAbstractionist** I’d also say yes is the answer. If $\displaystyle \alpha\in\mathbb R,$ then for any real interval $\displaystyle I_\epsilon=(\alpha-\epsilon,\,\alpha+\epsilon),$ there is a rational number $\displaystyle \beta$ such that $\displaystyle \alpha<\beta<\epsilon;$ any rational number is algebraic, therefore $\displaystyle \beta\in\mathbb A\cap\mathbb R$ and $\displaystyle \beta\in I_\epsilon\setminus\{\alpha\}.$ So every open neighbourhood of any number $\displaystyle \alpha\in\mathbb R$ contains a point in $\displaystyle \mathbb A\cap\mathbb R$ other than $\displaystyle \alpha,$ i.e. $\displaystyle \mathbb A\cap\mathbb R$ is dense in $\displaystyle \mathbb R.$