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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.

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$(\{n\alpha\} \mathrm{mod} 1)$ 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.

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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.

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