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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.

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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? –  Babak S. 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 Answers 3

up vote 4 down vote accepted

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 $\mathbb{R}$.

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If $x$ taken arbitrarily? –  Git Gud Jan 10 '13 at 10:26
    
@GitGud yes, any $x\in(0,\epsilon]$ will be ok –  Adam Jan 10 '13 at 10:28

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.
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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.

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