Show that there exists $m, n \in \mathbb{Z}$ such that $$|m+n \sqrt{2}|<\frac{1}{10}$$.

Try assuming that there are no $m$ and $n$ that satisfy the inequality

Then, $|m+n \sqrt{2}|\geq \frac{1}{10} \forall m,n \in \mathbb{Z}$

Then $\frac{1}{10}$ is a lower bound of {$|m+n \sqrt{2}|$ for $m,n \in \mathbb{Z}$}

but here I'm stuck and I don't know what to do next!

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Kevin Duran
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    $5\sqrt 2 = \sqrt {50} \approx \sqrt {49}=7$, so try taking $m=-7, n=5$. – player3236 Dec 17 '20 at 05:18
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    What about the trivial solution? – Shubham Johri Dec 17 '20 at 05:23
  • @ShubhamJohri So I just can say what are the solutions? I was trying to do a proof – Kevin Duran Dec 17 '20 at 05:58
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    The linked thread has many solutions. Basically you can prove the same result for any positive bound in place of $1/10$. One idea is to observe that the powers $(\sqrt2-1)^k$ all have the form $m+n\sqrt2$ with $m,n$ depending on $k$, of course. Because $|\sqrt2-1|<1/2$ the powers converge to zero as you increase $k$. – Jyrki Lahtonen Dec 17 '20 at 06:03
  • Another would be to consider the fractional (=decimal) parts of $n\sqrt2$ for $n=1,2,\ldots,11$. That's eleven numbers in the interval $(0,11)$. By the pigeon-hole principle some two will be within $1/10$ of each others. Say $r\sqrt2$ and $s\sqrt2$ have decimal parts close to each other. This means that $(r-s)\sqrt2$ is within $1/10$ of an integer, and the claim follows. – Jyrki Lahtonen Dec 17 '20 at 06:07
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    In other words, this is not a boring calculus or real analysis question. Rather number theory and/or combinatorics :-) – Jyrki Lahtonen Dec 17 '20 at 06:08
  • Also you can consider rational numbers $r, q$ such that $r<\sqrt{2} – Ali Dursun Dec 17 '20 at 06:10
  • @AliDursun You need to be careful there. Say if we try to use $$\frac{141}{100}<\sqrt2<\frac{142}{100}.$$ It does not follow that $100\sqrt2-141$ would be small. In fact, it is about $0.42$. The trouble being that multiplying out the common denominator of $r$ and $q$ you also multiply the approximation error. – Jyrki Lahtonen Dec 17 '20 at 06:36
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    Or, seen differently. If instead of $\sqrt2$ we had a rational number $\alpha=7/5$. It would be impossible to find integers $m,n$ such that $0<|m+n\alpha|<1/5$. When $\alpha$ is rational, we can make $m+n\alpha$ exactly zero, but we cannot make it arbitrary close to but different from zero. Try it! – Jyrki Lahtonen Dec 17 '20 at 06:56
  • @JyrkiLahtonen Yes, I should be more specific about what I am saying... I mean, we should find the $m$ and $n$ satisfying (assuming $n > 0$) $-1/10 < m + nr < m + n\sqrt{2} < m + nq < 1/10$ and we get $1/10 > -m - nr$, $1/10>m+nq$. Combining these we find $n<1/(5(q-r))$ and there is such $n$ since $q-r$ could be arbitrarily small. we can find $m$, choosing such $n$ and using the inequalities above. – Ali Dursun Dec 17 '20 at 07:57
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    It is an application of the [Dirichlet's approximation theorem](https://en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem). – rtybase Dec 17 '20 at 08:45
  • @JyrkiLahtonen: I like your interpretation in last comment. That's how one can interpret a crucial difference between rationals and irrationals. – Paramanand Singh Dec 17 '20 at 09:09
  • @AliDursun I could not follow your reasoning. You seem to want to choose all of $q,r,n$ independently from each other, but $n$ appears in the chain of inequalities forming the list of requirements for $r$ and $q$. – Jyrki Lahtonen Dec 17 '20 at 11:08

1 Answers1


There are already some great answers given in the comments of the question that answer the question as stated. But just for fun, let's do a proper analysis proof. Also, $\frac1{10}$ is totally arbitrary; we're going to show that $|m+n\sqrt2|$ can always get below any arbitrary positive real $r$.

There is, of course, the trivial solution $m=n=0$, but let's suppose for the sake of argument that you want something else.

With this in mind, let's consider the set $S=\{|m+n\sqrt2|\colon m,n\in\mathbb{Z}, (m,n)\neq(0,0)\}$. Every element in this set is bounded from below (by $0$), and so there must be some real number $r$ which serves as a greatest lower bound for $S$. Let's assume that $r>0$. Then there are two cases: either $r\in S$ or $r\not\in S$.

In the first case, there are integers $m,n$ such that $m+n\sqrt2=r$. (Notice that we can drop the absolute value, because if $m+n\sqrt2=-r$ then $-m-n\sqrt2=r$.) Now, I claim that $S=\{kr\colon k\in\mathbb{Z}\}$, in other words, $S$ is just all of the multiples of $r$. The proof of this is as follows:

Clearly every integer multiple of $r$ is an element of $S$. Suppose that there was some $s=a+b\sqrt 2\in S$ such that $\frac sr\not\in\mathbb{Z}$. Then $\frac sr=k+x$, where $k\in \mathbb{Z}$ and $x\in(0,1)$. So $s=a+b\sqrt2=r(n+x)=kr+rx$. But that means that $s-kr=(a-m)+(b-n)\sqrt2=rx$. Therefore, we have $s-kr\in S$ and $0<s-kr<r$. But this contradicts the fact that $r$ is a greatest lower bound, so such an $s$ cannot exist, and $S$ is the set of integer multiples of $r$.

But we know that $1\in S$ and $\sqrt 2\in S$, and so therefore there are integers $k_1,k_2$ suchthat $1=k_1r$ and $\sqrt2=k_2r$, so $\frac{\sqrt2}1=\sqrt2=\frac{k_1}{k_2}$. So $\sqrt2$ is the ratio of two integers, and therefore $\sqrt2$ is a rational number. This is a contradiction, so we can move on to case $2$.

In our second case, we have $r\not\in S$. But because $r$ is a greatest lower bound of $S$, for every real $\epsilon>0$ there exists some $s\in S$ such that $s\in (r,r+\epsilon)$. (Again, notice that we are justified in dropping the absolute value.) So let $\epsilon=r$, and then we have some $s_1=m_1+n_1\sqrt2\in S$ such that $s_1\in (r,2r)$; in other words, $s_1=r+\epsilon_1$, where $0<\epsilon_1<r$.

But now, we do the same trick, this time letting $\epsilon=\epsilon_1$. What that means is that there is some $s_2=m_2+n_2\sqrt2\in S$ such that $s_2\in (r,r+\epsilon_1)$. So we have $s_2=r+\epsilon_2$, where $0<\epsilon_2<\epsilon_1$. But then, $s_1-s_2=(m_1-m_2)+(n_1-n_2)\sqrt2\in S$, and $s_1-s_2=\epsilon_1-\epsilon_2$. But we know that $0<\epsilon_2<\epsilon_1<r$, and so $0<\epsilon_1-\epsilon_2<r$. In other words, we have again found an element of $S$ which is less than $r$, contradicting the "lower-boundedness" of $r$!

So clearly, such a greatest lower bound cannot exist, and therefore, for any real $r$, we can find integers $m,n$, not both zero, such that $|m+n\sqrt2|<r$. As a special case, letting $r=\frac1{10}$ solves the problem given.

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    Basically proving that a non-trivial additive subgroup of $\Bbb{R}$ is either cyclic or dense :-) – Jyrki Lahtonen Dec 17 '20 at 06:40
  • @JyrkiLahtonen I hadn't thought of it that way! Given that, I'm sure there's another argument to be made involving linear independence over $\mathbb{Q}$, but I can't seem to figure it out. I keep running into that "approximation error" issue that you brought up in the question comments. – tromben Dec 17 '20 at 06:47