Hi,

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

Thanks!

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- January 27th 2010, 06:11 PMrhemoDense subset
Hi,

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

Thanks! - January 27th 2010, 06:14 PMDrexel28
- January 28th 2010, 04:27 AMrhemo
I can’t explain in English very well, but I will try:

A={m +n}

I make ( or do?):

(1) >0 A (0, ) is true, if x A (0, ) then, sequence {x, 2x,3x...}is arithmetic progression d**=**. r R, k Z; kx (r,r+ ).

(1) u and v in A, |u-v|< . A is closed for addition. Then, w in A, |w|< and -w in A.

Ufff.... - January 31st 2010, 04:09 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 . But i got stuck when i found p such that .

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 ? How would it be different? What if in the set definition you could use ?

Oh, yes, and i assumed that . - February 2nd 2010, 06:27 PMxalk
- February 3rd 2010, 06:07 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 with respect to addition. Then one and only one of the following is valid:**

- such that
- G is dense in

**Proof.**Let denote . Clearly is non-empty and bounded below. Let . Then either*a>0*or*a=0*.

Suppose, first, that*a>0*. Then , because otherwise there would be*g*,*h*in such that*a<g<h<a+(a/2)*, and that would imply , a contradiction.

Furthermore, let . 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 and q is an integer, then . But if*r>0*then . Therefore , 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 , there is such that . Then there should be such that . In fact, let . The well ordering principle assures that there's . It's obvious that . Let's show that .

It is sufficient to show that . The first inequality is valid due to the definition of A. Now suppose . Then , but that contradicts the definition of n.

Now let x be any negative real number. We have shown that, since -x>0, . Equivalently, there's g in G such that . But since G is a group, -h is also in G and:

Therefore (since 0 is in G), G is dense in R if .

QED.

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

Ok. Now the problem gets easier: first, we show that is an additive subgroup of . Then it is easy to get a contradiction by supposing that . That is equivalent, by Lemma 2, to asserting that is dense in . We then use this fact to prove that is dense in . These proofs are all easy and I won't write them, since i've already written the dirty part. - February 3rd 2010, 03:57 PMxalk
- February 3rd 2010, 04:10 PMDefunkt
- February 3rd 2010, 04:41 PMxalk
- February 8th 2010, 07:51 PMJoachimAgrell
If you agree that my previous post is correct, then there are, as i said, only a few basic things to do. Let .

Then you can show that 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.