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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?

3 Answers3

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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.

mathcounterexamples.net
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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.

Asinomás
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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.)

Akiva Weinberger
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  • (On the other hand, does this prove that $\sqrt2$ is irrational?) – Akiva Weinberger Aug 11 '15 at 14:47
  • Interesting suggestion, I have not seen such a proof. Have not been able to make it work, maybe more coffee will do it. – André Nicolas Aug 18 '15 at 14:06
  • @AndréNicolas Consider $(\sqrt2-1)^{1000}$; it's very small, right? – Akiva Weinberger Aug 18 '15 at 14:33
  • Indeed the expression is negligibly small. But that does not imply irrationality. You may have in mind using the fact that $(\sqrt{2}+1)^n+(\sqrt{2}-1)^n$ is an integer. But that does not seem to yield an immediate contradiction. – André Nicolas Aug 18 '15 at 14:38
  • @AndréNicolas Remember that $(\sqrt2-1)^{1000}$ can also be written as $m+n\sqrt2$ for some integers $m$ and $n$. If $\sqrt2$ were rational, what would you be able to say about the smallest numbers of the form $m+n\sqrt2$? – Akiva Weinberger Aug 18 '15 at 14:47