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It is hard for me to show that the set $\{\sqrt{m}-\sqrt{n}; m,n\in \Bbb N\}$ is dense in $\Bbb R$. Please help me.

Aliakbar
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    Have you tried of establishing a sequence of rational numbers approaching an element of above set in $\mathbb R$? What do you think? – Mikasa Jan 10 '13 at 10:12
  • For $x\in \Bbb R$, We should establishing a sequence of the numbers in the above set such that converges to $x$. – Aliakbar Jan 10 '13 at 10:21

3 Answers3

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Let $A=\{\sqrt{m}-\sqrt{n}; m,n\in \Bbb N\}$. We need to show that for every open interval $(a,b)$ we have $(a,b)\cap A \neq \emptyset$. To prove it we'll use two facts:

  1. For any $\epsilon \ge 0$ we can find $x=\sqrt{m}-\sqrt{n}$ for some $n,m\in\mathbb{N}$ such that $0< x\le \epsilon$. To see it just consider $$\sqrt{m+1}-\sqrt{m}=\frac{1}{\sqrt{m+1}+\sqrt{m}}$$

  2. If $x\in A$ and $k\in\mathbb{Z}$, then $kx\in A$.

Then consider open interval $(a,b)\subset\mathbb{R}$, take $\epsilon=\frac{b-a}{2}$ and by (1) there is $x$ such that $0<x\le\epsilon<b-a$. Now, as $x$ is smaller than length of $(a,b)$, we can find $k\in \mathbb{Z}$ such that $kx\in (a,b)$, and also $kx\in A$ by (2), which shows that $(a,b)\cap A$ is non-empty, and therefore $A$ is dense in $\mathbb{R}$.

Adam
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  • If $x$ taken arbitrarily? – Git Gud Jan 10 '13 at 10:26
  • @Adam, Can you please show $1.$ more rigorously? Of course intuitively it is true, but did you use the Archimedean principle or something? If so, the Archimedean principle says $0<\frac{1}{n}<\epsilon$. But generally you can't immediately conclude that $\exists k \in \mathbb{Z}$ such that$ 0<\frac{1}{k}<\epsilon \implies 0<\sqrt{m+1}-\sqrt{m} < \epsilon $ – makansij Feb 02 '19 at 03:58
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    @guimption, $\frac{1}{\sqrt{m+1} + \sqrt{m}}$ goes to $0$ as $m \to \infty$, so it can get arbitrarily close to $0$. However if you want to base it on Achimedean principle, then take $n$ such that $\frac{1}{n} < \epsilon$ and set $m = n^2$. Then $\frac{1}{\sqrt{m+1}+\sqrt{m}} < \frac{1}{\sqrt{m}} = \frac{1}{n} < \epsilon$. – Adam Feb 02 '19 at 07:23
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Hint: $\sqrt{k^2n+k^2}-\sqrt{k^2n}=\frac{k}{\sqrt{n}+\sqrt{n+1}}=\frac{k}{2\sqrt{n}+\sqrt{n+1}-\sqrt{n}}=\frac{k}{2\sqrt{n}+\frac{1}{\sqrt{n}+\sqrt{n+1}}}$

$$\sqrt{k^2n+k^2}-\sqrt{k^2n}=\frac{k}{2\sqrt{n}+\frac{1}{\sqrt{n}+\sqrt{n+1}}}$$

To show that the set is dense in $\mathbb{R}$, it suffices to estimate all rationals. To estimate the rational $\frac{p}{q}$:

  1. Get a large natural number $m$. Now we consider the rational number $\frac{2pm}{2qm}$
  2. Set $k=2pm$ and $n=(qm)^2$. It follows that $2\sqrt{n}=2qm$. Since $m$ is large, therefore $n$ would be large. Thus, the fraction $\frac{1}{\sqrt{n}+\sqrt{n+1}}$ that appears in the denominator would be negligible.
Amr
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  • Didn't you let $m = n+1$ in the hint but then later let $n=(qm)^2$? How can it be safe to assume these two constraints are valid? – Crystal Oct 14 '19 at 06:55
  • Hi. I never said. In which line of my answer do I say that I let m=n+1 ? – Amr Oct 15 '19 at 09:22
  • Hi Amr! Thanks for your reply, it looks to me like in the first line, in the hint, you require $m$ to be $n+1$, is that not right? – Crystal Oct 15 '19 at 12:33
  • No I don't require and never said that I require. The hint is about any positive integers k and n (no mention of m) – Amr Oct 16 '19 at 13:40
  • Sorry, I was mistaken. I had one more question, when we factor the $k^2$ from the square root, shouldn't we obtain the absolute value of $k$? In which case, this does not allow us to estimate the negative rational numbers. – Crystal Oct 17 '19 at 19:54
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    You are right. But the argument will still hold if you assume without loss of generality that p is nonnegative. Hence k is nonnegative – Amr Oct 17 '19 at 21:04
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Let $S$ be the set you're trying to show is dense in $\mathbb{R}$. Here are some ideas: (not necessarily the most elegant solution probably, but it is what came to my mind)

  1. It suffices to show that a subset of $S$ is a dense in $[0,1]$, since you can then do some scaling of $m,n$ to get the whole real line.

  2. You can show that elements from $S$ can approximate any element in $[0,1]$ by the following argument: notice that $\sqrt{1} - \sqrt{1}, \sqrt{2} - \sqrt{1}, \sqrt{3} - \sqrt{1}, \sqrt{4} - 1$ are elements of $S$ that subdivide $[0,1]$ using four points. Now observe that $\sqrt{4} - \sqrt{4}, \ldots, \sqrt{9} - \sqrt{4}$ subdivides $[0,1]$ using six points. Using this intuition you can show that you can always subdivide the unit interval as many times as you like, and you prove an upper bound on the width of each partition. This will prove density on the unit interval.

Christopher A. Wong
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