Let $\mathbb{R}$ be the set of all real numbers under the usual metric $d$ defined as follows: $$d(x,y) \colon= |x-y|$$ for all $x$, $y$ in $\mathbb{R}$, and let $\mathbb{Q}$ be the set of all rational numbers.
Then how to rigorously prove that, given a positive integer $n$ which is a non-perfect square, the number $\sqrt{n}$ is in the closure of $\mathbb{Q}$?
For example, how to show that $\sqrt{3}$ is in the closure of $\mathbb{Q}$?
Generalising the above, how to show that, for a given positive integer $n$ which is not the $k$th power of any positive integer, the number $\sqrt[k]{n}$ is in the closure of $\mathbb{Q}$?
How to explicitly find a sequence of rational numbers that converges to our number? And how to rigorously demonstrate this convergence?
Best Answer
$\mathbb{Q}$ is dense in $\mathbb{R}$. That is for any $x \in \mathbb{R}$ you can find a sequence of rationals converging to $x$.
To see this, let $\lfloor x \rfloor$ denote the floor function. Note that $$\lfloor x \rfloor \leq x < \lfloor x \rfloor + 1.$$
Show that $$ l_n := \frac{\lfloor 2^n x \rfloor}{2^n} \to x. $$