Kenneth Rosen Edition 7 Exercise 6.2 Question 45 (Page No. 407)

Question

Let $x$ be an irrational number. Show that for some positive integer $j$ not exceeding the positive integer $n,$ the absolute value of the difference between $j x$ and the nearest integer to $j x$ is less than $1/n.$

Answer 1

Not yet the complete answer.

For ease of notation, let's call $$\min{(j|x|-\lfloor j|x|\rfloor,\lceil j|x|\rceil-j|x|)}=f(j); \left\lfloor\frac{1}{f(j)}\right\rfloor=g(j).$$

  We want to show that \(\forall n\in |N,~\exists j,0<j\leq n,~f(j)<\frac{1}{n}\) or, in other words, \(g(j)>n\). Suppose \(\left\lfloor\frac{1}{f(1)}\right\rfloor=m.\) That is, \(m\) is the largest positive integer such that the absolute difference between \(x\) and its nearest integer is less than \(\frac{1}{m}\). Obviously, for \(n\leq m\), the desired \(j\) is 1. Let's consider \(n>m\).
  
  First note that \(f(j)\) cannot be 0 or \(\frac{1}{n}\) for any \(n\in |N\). Being a positive fraction that is less than its complement, \(f(j)\in(0,0.5)\) always.

  Secondly, \(j\neq k\Rightarrow f(j)\neq f(k)\), because \(x\) is irrational.
  
  Thus,

1. \(f(j)\) must be a sequence of positive irrational fractions within the open interval \((0,0.5)\) in which no member is repeated ever.
2. Most importantly, this sequence \(f(j)\) is completely determined by \(f(1)=p\) Thus, \(f_x(j)=f_p(j)\) when \(p,x\) are related as above.
3. Each \(f(j)\) is in the open subinterval \(\frac{k}{n},\frac{k+1}{n}\) where \(k\equiv\left\lfloor\frac{jp}{\max{g(i):i\leq n}}\right\rfloor\mod {\max{g(i):i\leq n}\).
4. The number of such open subintervals in \((0,0.5)\) is \(<n\).
5. \(p\) being a fraction, \(k\) will become 0 for some \(j\).

Now what remains to be shown is that for any \(n\), \(g(j)\) has to go above \(n\) for some \(j\leq n\). It seems that if \(g(1)=\left\lfloor\frac{1}{p}\right\rfloor<n\) then there will be a point \(j<n\) at which the \(g(j)\) sequence, diminishing before to 2 and possibly staying constant for some duation, shoots up to a high value. It seems elusively easy but is not.