Let $V$ be any irreducible variety over $\mathbb{R}$, prove that if $dim_{\mathbb{R}}V(\mathbb{R})= dim(V)$, then $V(\mathbb{R})$ is dense in $V$.
Any hints to make a start?
Let $V$ be any irreducible variety over $\mathbb{R}$, prove that if $dim_{\mathbb{R}}V(\mathbb{R})= dim(V)$, then $V(\mathbb{R})$ is dense in $V$.
Any hints to make a start?
This is not a solution, but is more of an extended comment:
You might want to consider some examples before trying to prove this in general. Here are some that you can try (in some of the them the hypothesis of your problem holds, in others it doesn't):
(1) $V$ is the affine curve given by the equation $x^2 + y^2 = 0$.
(2) $V$ is the affine curve given by the equation $y^2 = x^2(x-1)$.
(3) $V$ is the affine curve given by the equation $y^2 = x^3 - x$.
In each case you should plot the real points and see what dimension they are, and then consider whether or not they are Zariski dense in $V$ (i.e. in the set of complex points).