a question on dense subsets of the (0,1) interval in number theory A well-known fact:
If $\alpha$ is irrational, then $\{n\alpha\bmod{1}|n\in\mathbb{N}\}$ is dense in $(0,1)$.
In particular, $\{n\sqrt{2}\bmod{1}|n\in\mathbb{N}\}$ is dense in $(0,1)$.
What if we replace 1 in "mod 1" with something else, e.g. is $\{n\sqrt{2}\bmod{\sqrt{3}}|n\in\mathbb{N}\}$ dense in $(0,\sqrt{3})$ ?
Is $\{n\alpha\bmod{\beta}|n\in\mathbb{N}\}$ dense in $(0,\beta)$ in case $\frac{\alpha}{\beta}$ irrational ?
I did not found results modulo something different than 1, I'd be grateful for references.
 A: To your specific question:  you are asking to show that the set $$S=\{m\sqrt 2 +m\sqrt 3\; |\;  m,n\in \mathbb Z\}$$ is dense in $\mathbb R$.  We remark that $S$ is closed under internal addition and subtraction and under multiplication by integers. 
Note that if $S$ had an accumulation point then $0$ would be an accumulation point (since $S$ is closed under subtraction).  If $0$ is an accumulation point then $S$ is dense, clearly.  Thus, in order to get a contradiction, let us assume that $S$ has no accumulation points.
In that case, there exists a least positive element $\alpha \in S$.  But then it is easy to see that every element in $S$ must be an integer multiple of $\alpha$ (Pf:  were it otherwise then we'd have some $s\in S$ and $n\in \mathbb Z$ for which $n\alpha < s < (n+1)\alpha$ in which case $s-n\alpha $ would be a smaller positive element of $S$).
But that is impossible:  were it so we could write $$r\alpha = \sqrt 2 \quad s\alpha = \sqrt 3$$ for $r,s\in \mathbb Z$ which would imply that $\frac {\sqrt 2}{\sqrt 3}\in \mathbb Q$ which is the desired contradiction.
A: $(\{n\alpha\} \mathrm{mod} 1)$ is equidistributed iff $\alpha$ is irrational. Fix a real $q>0$. So $$(\left\{q\{n\alpha\}\right\} \mathrm{mod}\, q\,)=(\{nq\alpha\} \mathrm{mod}\, q\,)$$ is equidistributed iff $q\alpha$ is irrational. In your case, $q\alpha=\sqrt{6}$ is irrational, hence yes.
