Prove that $f(x)=0$ for all $x\in\mathbb{R}$ The problem at which I am currently stuck is,

Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that $f(m+n\sqrt{2})=0$ for all $m,n\in\mathbb{Z}$. Prove that $f(x)=0$ for all $x\in\mathbb{R}$.

I have noted that to solve this problem what I need to show is that the set $\{m+n\sqrt{2}\mid m,n\in\mathbb{Z}\}$ is dense in $\mathbb{R}$ but I can't prove it. Can anyone help me? 
 A: Hint.
The additive subgroups of $\mathbb R$ are either dense or discrete.
Prove that if $$S=\{m+n\sqrt{2}\mid m,n\in\mathbb{Z}\}$$ is discrete then $\sqrt{2}$ would be rational which is not.
So $S$ is dense and a continuous function which vanishes on a dense subset is always vanishing.
A: It suffices to show that we can find $0<n+m\sqrt 2<a$ for every positive delta.(with this we can prove the density of the set, pretty much in the same way we prove density of $\mathbb Q$ in $\mathbb R$.
The result is true if we change $\sqrt2$ for any irrational $x$. Consider the set $\{xm\}|m\in\mathbb Z$, we have to prove it has no lower bound $a$ greater than $0$, suppose it did. This set is infinite so we can find $m,n$ such that $0<\{xm\}-\{xn\}<a$. Then $\{x(m-n)\}=\{xm\}-\{xn\}<a$. A contradiction.
A: Hint: Note that $\sqrt2-1=0.414\dots<1$. Also note that $(\sqrt2-1)^N$ is of the form $m+n\sqrt2$ for all $N$.
(This doesn't generalize well to, say, $m+n\pi$ or other irrational things, which is annoying.)
