Real Analysis beginner proof

[thedarjeeling]

Ouch, the problem goes like this

"Prove that if x is irrational then there exists a y > 0 such that the open set (x - y, x + y) does not contain any rational of the form m / n, with m, n no factors in common and n is less than equal to 100."

I'm super stuck on this one. The only tools we have our disposal are stuff like we proved sqrt(2) is irrational, lower bounds, upper bounds, infimum/supremum etc.

I can sort of visualize what it's saying; it's saying pick an irrational number, and prove there's a small neighborhood around this irrational number such that there aren't any rational numbers with a certain precision that can penetrate into this zone, but I have no idea how to really prove it. I mean you'd sort of think that you should set n to 100 to get as close to another number as possible, but I don't how you'd prove that since who knows, maybe when n is 7 you would be closer to an irrational number.

 

[SlipEternal]

Consider the set \(\displaystyle A_n = \left\{\frac{m}{n}: m \in \mathbb{Z}, |x(n-1)| \le |m| \le |x(n+1)|\right\}\).
Note: \(\displaystyle A_n\) is a finite set.

Let \(\displaystyle U_n = \bigcup_{i=1}^n{A_i}\).
Note: \(\displaystyle U_n\) is a finite union of finite sets, and therefore finite.

Let \(\displaystyle D_n = \left\{|x - u|: u \in U_n\right\}\)
Note: \(\displaystyle D_n\) is in one-to-one correspondence with \(\displaystyle U_n\), and so is also finite (the function is \(\displaystyle x-u\)).

\(\displaystyle D_n\) must have a minimum element for all \(\displaystyle n \in \mathbb{N}\).
Let \(\displaystyle 0< y < \min\{D_{100}\}\).

The proof should be obvious from here.

 

[Deveno]

note the smaller n is, the farther apart rational numbers of the form m/n with gcd(m,n) = 1 are. for example, 1/3 and 2/3 are 1/3 apart, while 1/11 and 2/11 are only 1/11 apart.

so x is irrational, and we want to choose y so that (x-y,x+y) doesn't contain any "100-ths" or smaller fractions. so obviously choosing y > 1/50 is a bad idea, since that guarantees we'll have a/100 in (x-y,x+y) , and if gcd(a,100) isn't 1, then the denominator will get smaller (bad for us).

well we must have a/100 < x < (a+1)/100 for some integer a (we have the strict inequalites because both of the bounds are rational).

so our first thought might be: let b = min{|(a+1)/100 - x|, |a/100 - x|}.

but maybe some other fraction (like a 7th, maybe) is closer to x that any 100-th is.

but all is not lost: we can define a sequence of such b's:

\(\displaystyle b_k = min\{|\frac{a_k}{k} - x|, |\frac{a_k + 1}{k} - x|\}\) where \(\displaystyle a_k < kx < a_k + 1\) for k = 1,2,...,100.

this is a finite set of b's, so let \(\displaystyle y = min\{b_k\}\)

this is actually similar to what SlipEternal has done, but i have tried to make the thought-process behind how one comes up with y a bit more transparent.