"rational points are dense on the unit ciricle". What is meant by the phrase "rational points are dense on the unit ciricle".
I know the definition of a dense set:
In topology and related areas of mathematics, a subset A of a topological space X is called dense (in X) if any point x in X belongs to A or is a limit point of A.
I don't really know any other definitions of dense let alone how to explain dense on a the unit circle.
 A: First, note what is not a rational point on the unit circle:  A point whose coordinates cannot be expressed as both rational numbers, eg., $(\sqrt{2}/2, \sqrt{2}/2)$.
All the phrase "rational points are dense on the unit circle" means is that for any point $(x,y)$ on the circle (possibly not rational), and any arbitrarily small distance $0<\delta$, one can find a rational point also on the circle within distance $\delta$ to $(x,y)$.
A: This means that given any point $(x,y)$ with $x^2 + y^2 = 1$, there exists a sequence of points $(x_n,y_n)$ such that $x_n^2 + y_n^2 = 1$ and $x_n, \, y_n \in \mathbb{Q}$ for all $n$ and such that $x_n \to x, \, y_n \to y$ as $n \to \infty$.
To construct such points, consider Pythagorean triples, i.e. integers $a,b,c$ such that $a^2 + b^2 = c^2$, and turn these into rational points $(a/c, b/c)$ on the unit circle.
The statement about density of rational points becomes massively false if you look at points satisfying $x^n + y^n = 1$ with $n > 2$, by Fermat's Last Theorem.
A: In order to understand the geometry of the rational points on $S^1=\{(x,y)\in\Bbb R^2\,|\,x^2+y^2=1\}$ it is important to remark that these can be parametrized effectively.
Consider the lines through the point $P=(1,0)\in S^1$. With the usual exception of the vertical line $x=1$ which is in fact tangent to $s^1$ at $P$, these lines have cartesian equation
$$
y=m(x-1)\qquad\qquad (\ast)
$$
where $m\in\Bbb R$ is the slope.
Each of these lines intersects $S^1$ at a second point $Q$ whose $x$-coordinate can be computed by solving the degree $2$ equation obtained plugging the equation of the line into the equation of the circle, namely
$$
x^2+m^2(x-1)^2=1. \qquad\qquad (\ast\ast)
$$
Recall that in a quadratic equation $x^2+ax+b=0$ we have $a=-\lambda_1-\lambda_2$ where $\lambda_1$ and $\lambda_2$ are the two solutions.
When $m\in\Bbb Q$ the coefficients of the equation $(\ast\ast)$ are in $\Bbb Q$ and since $\lambda_1=1$ by construction we have $\lambda_2\in\Bbb Q$ as well.
The $y$-coordinate of $Q$ can be computed plugging its $x$-coordinate into $(\ast)$ and so also the $y$-coordinate is in $\Bbb Q$.
Therefore the rational points in $S^1\setminus\{P\}$ are in one-to-one correspondence with the lines through $P$ with rational slope, i.e. with $\Bbb Q$. By continuity this shows that the rational points are dense in $S^1$.

By the way, this construction shows that the primitive Pythagorean triples can be parametrized by $\Bbb Q$ as well (see Hans Engler's answer)
