Given $p\in\mathbb{N}^+, p>1$, find irrational numbers $r$ such that the set of all numbers of the form $\{r\cdot p^n\} $, where $n\in\mathbb{N}$ and $\{x\}:=x-[x]$, is dense in $(0,1)$.
Obviously not every irrational number possesses this property. But does there exist such numbers at all? What are some special (and interesting) cases? Would $r=\sqrt n$ (for $n$ not a perfect square) do?
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First, note that we may assume $r$ is between $0$ and $1$ (since adding an integer to $r$ will change nothing). Write $r$ in base $p$ expansion: $$r = 0.a_1a_2a_3a_4\dots,$$ where each $a_i$ satisfies $0\leq a_i<p$. Upon multiplying by $p^n$, this becomes $$r\cdot p^n=a_1a_2\dots a_n.a_{n+1}a_{n+2}a_{n+3}\dots,\\\{r\cdot p^n\}=0.a_{n+1}a_{n+2}a_{n+3}\dots$$ Therefore we see that whether the set you ask about is dense in $(0,1)$ is equivalent to the question of which patterns of digits appear in its base $p$ expansion. More precisely, it's not hard to prove the following:
This is closely related to being a base $p$ normal number. There are known normal numbers, the simplest example being the (base $p$) Champernowne constant (mentioned by Hagen von Eitzen in another answer). Moreover, there are numbers known to be normal in all bases $p>1$, and the Wikipedia page on normal numbers gives you some references.
Let me also address the question of $\sqrt{n}$ - whether this is a normal number in any base is open, and indeed I am fairly sure that it is also open whether it contains arbitrary strings in any base. The thing is that, in some sense, decimal expansions (or expansions in different bases) are, from algebraic point of view, very dfifficult to work with. There are barely any examples of normal numbers apart from the ones which have been artificially constructed, even though we believe many known constants (including irrational square roots, and more generally irrational algebraic numbers) are all normal, but we are very far from being able to prove that.