Hi,

If $\displaystyle a $ is irrational, show that the set {m$\displaystyle a$+n; m $\displaystyle \in$N, n$\displaystyle \in$Z} is dense in R.

Thanks!

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- Jan 27th 2010, 05:11 PMrhemoDense subset
Hi,

If $\displaystyle a $ is irrational, show that the set {m$\displaystyle a$+n; m $\displaystyle \in$N, n$\displaystyle \in$Z} is dense in R.

Thanks! - Jan 27th 2010, 05:14 PMDrexel28
- Jan 28th 2010, 03:27 AMrhemo
I can’t explain in English very well, but I will try:

A={m$\displaystyle \alpha$+n}

I make ( or do?):

(1)$\displaystyle \forall$$\displaystyle \epsilon$ >0 A$\displaystyle \cap$(0,$\displaystyle \epsilon$) is true, if x$\displaystyle \in$A$\displaystyle \cap$(0,$\displaystyle \epsilon$) then, sequence {x, 2x,3x...}is arithmetic progression d**=**$\displaystyle \epsilon$. $\displaystyle \forall$r$\displaystyle \in$R, $\displaystyle \exists$k$\displaystyle \in$Z; kx$\displaystyle \in$(r,r+$\displaystyle \epsilon$).

(1)$\displaystyle \Rightarrow$$\displaystyle \exists$u and v in A, |u-v|<$\displaystyle \epsilon$. A is closed for addition. Then, $\displaystyle \exists$w in A, |w|<$\displaystyle \epsilon$ and $\displaystyle \Rightarrow$ -w in A.

Ufff.... - Jan 31st 2010, 03:09 PMJoachimAgrell
I tried to show it this way but i couldn't even start. Then i tried to find successive approximations for the real number x using the fact that there is a natural number k such that $\displaystyle k\alpha\leq x\leq(k+1)\alpha$. But i got stuck when i found p such that $\displaystyle k\alpha+p\leq x\leq k\alpha+(p+1)$.

What i thought was that i could make the first coefficient (the one that multiplies the irrational number) arbitrarily big if i made the second coefficient small enough, and, that way, i could build a sequence converging to x.

But, uh, it didn't work. A few things came to my mind: what if $\displaystyle \alpha\in\mathbb{Q}\backslash\mathbb{Z}$? How would it be different? What if in the set definition you could use $\displaystyle m\in\mathbb{Z}$?

Oh, yes, and i assumed that $\displaystyle \alpha>0$. - Feb 2nd 2010, 05:27 PMxalk
- Feb 3rd 2010, 05:07 AMJoachimAgrell
Ok, I'm not him, but i'll show you the progress i've made.

First, we need to prove two facts.

**Lemma 1. (modified euclidean algorithm) Let***x*,*a*be real numbers. Then there is an integer*q*and a non-negative real number*r*such that*x=a*q+r*and r<a.

**Proof.**Not hard and hasn't got much to do with the problem.

**Lemma 2. Let G be a non-trivial (i.e., not {0}) subgroup of $\displaystyle \mathbb{R}$ with respect to addition. Then one and only one of the following is valid:**

- $\displaystyle \exists g\in G$ such that $\displaystyle G = g\mathbb{Z}$
- G is dense in $\displaystyle \mathbb{R}$

**Proof.**Let $\displaystyle G^+$ denote $\displaystyle G\cap\mathbb{R}^+$. Clearly $\displaystyle G^+$ is non-empty and bounded below. Let $\displaystyle a=\inf{G^+}$. Then either*a>0*or*a=0*.

Suppose, first, that*a>0*. Then $\displaystyle a\in G$, because otherwise there would be*g*,*h*in $\displaystyle G^+$ such that*a<g<h<a+(a/2)*, and that would imply $\displaystyle \tfrac{a}{2}>g-h\in G^+$, a contradiction.

Furthermore, let $\displaystyle g\in G$. Then, by Lemma 1, there is an integer*q*and a non-negative real number*r*such that*g=aq+r*with*r<a*. Since both*a*and*g*are in $\displaystyle G$ and q is an integer, then $\displaystyle 0\leq r=g-aq\in G$. But if*r>0*then $\displaystyle r\in G^+$. Therefore $\displaystyle 0<r<a=\inf{G^+}$, a contradiction. Therefore the only possibility is*r=0*, i.e., g=aq.

Now suppose a=0. Let x be any real number. Suppose, first, that x is positive. Given $\displaystyle \epsilon>0$, there is $\displaystyle g\in G^+$ such that $\displaystyle 0<g<\epsilon$. Then there should be $\displaystyle n\in\mathbb{Z}$ such that $\displaystyle ng\in (x-\epsilon,x+\epsilon)\cap G$. In fact, let $\displaystyle A=\{t\in\mathbb{N}: tg>x-\epsilon\}$. The well ordering principle assures that there's $\displaystyle n=\min{A}$. It's obvious that $\displaystyle ng\in G$. Let's show that $\displaystyle ng\in (x-\epsilon,x+\epsilon)$.

It is sufficient to show that $\displaystyle x-\epsilon<ng<x+\epsilon$. The first inequality is valid due to the definition of A. Now suppose $\displaystyle ng\geq x+\epsilon$. Then $\displaystyle ng\geq x+\epsilon>x+g\implies (n-1)g>x$, but that contradicts the definition of n.

Now let x be any negative real number. We have shown that, since -x>0, $\displaystyle \exists h\in (-x-\epsilon,-x+\epsilon)\cap G$. Equivalently, there's g in G such that $\displaystyle -x-\epsilon < h < -x+\epsilon$. But since G is a group, -h is also in G and:

$\displaystyle -x-\epsilon < h < -x+\epsilon \Leftrightarrow x-\epsilon < (-h) < x+\epsilon\Leftrightarrow (-h)\in(x-\epsilon,x+\epsilon)$

Therefore (since 0 is in G), G is dense in R if $\displaystyle 0=\inf{G^+}$.

QED.

-------------

Ok. Now the problem gets easier: first, we show that $\displaystyle S_1 = \{m\alpha+n|m,n\in\mathbb{Z}\}$ is an additive subgroup of $\displaystyle \mathbb{R}$. Then it is easy to get a contradiction by supposing that $\displaystyle S_1 = k\mathbb{Z}$. That is equivalent, by Lemma 2, to asserting that $\displaystyle S_1$ is dense in $\displaystyle \mathbb{R}$. We then use this fact to prove that $\displaystyle S_1\supset S_2=\{m\alpha+n|m\in\mathbb{N},n\in\mathbb{Z}\}$ is dense in $\displaystyle \mathbb{R}$. These proofs are all easy and I won't write them, since i've already written the dirty part. - Feb 3rd 2010, 02:57 PMxalk
- Feb 3rd 2010, 03:10 PMDefunkt
- Feb 3rd 2010, 03:41 PMxalk
- Feb 8th 2010, 06:51 PMJoachimAgrell
If you agree that my previous post is correct, then there are, as i said, only a few basic things to do. Let $\displaystyle S=\{m\alpha+n:m\in\mathbb{Z^{+}}, n\in\mathbb{Z}\}$.

Then you can show that $\displaystyle \inf{S\cap\mathbb{R^+}}=\sup{S\cap\mathbb{R^{-}}}=0$ and use an argument similar to the one used in Lemma 2 to show that an open interval with arbitrary radius centered at an arbitrary element of**R**contains at least one element of S.