Diophantine approximation

 

Thought

$x^{2} - Dy^{2} = 1$ for some $x, y \in \mathbb{N}$

$\Rightarrow$  $(x+y\sqrt{D})(x-y\sqrt{D}) = 1$

$\Rightarrow$  $x-y\sqrt{D} = \frac{1}{x+y\sqrt{D}} \leq \frac{1}{y}$

$\Rightarrow$  $\left|\frac{x}{y} - \sqrt{D} \right| \leq \frac{1}{y^{2}}$

 

 

 

정리1 Dirichlet's Diophantine approximation theorem

Let $\alpha$ be an irrational number.

Then there exists infinitely many intergers $x$  and  $y$ $(y>0)$  s.t.

$\left| x - y \alpha \right| < \frac{1}{y}$    $( \Leftrightarrow \left|\frac{x}{y} - \alpha \right| < \frac{1}{y^{2}})$

 

pf)

It suffices to show that for give positive interger $N$, there exists intergers $x$  and  $y$  s.t.  

$\left| x - y \alpha \right| < \frac{1}{N}$  and $0< y \leq N$  

 

Notation

For a real number $A$,  $\left \lfloor A \right \rfloor = $ largest interger $ \leq A$  $\left\{A \right\} = A - \left \lfloor A \right \rfloor$     (i.e.  $0 \leq \left\{A \right\} < 1$)

 

For $k = 0, 1, 2, \cdots , N$, let

$n_{k} = \left \lfloor k \alpha \right \rfloor$

$f_{k} = \left\{k\alpha \right\} = k\alpha - n_{k}$

 

There are $N+1$ $f_{k}'s$  and  $N$ subintervals at the same length on $[0,1)$.

(i.e.  $[0, \frac{1}{N}), \; [\frac{1}{N}, \frac{2}{N}), \; \cdots, \; [\frac{N-1}{N}, 1)$)

 

By pigeon hole principle, there exists $f_{i}, f_{j} \in [ \frac{k}{N}, \frac{k+1}{N})$ for some $i, j$  $(i>j)$  and  $k$  s.t.  

$\left|f_{j} - f_{i} \right| < \frac{1}{N}$

 

So  $\left|(j\alpha - n_{j}) - (i\alpha -n_{i}) \right| < \frac{1}{N}$

$\Rightarrow$  $\left|(n_{i} - n_{j}) - (i-j)\alpha \right| < \frac{1}{N}$

 

Take $x = n_{i} - n_{j} \in \mathbb{Z}$  and  $y = i - j \in \mathbb{N}$.

Since $0 \leq j < i  \leq N$, so $0 < y \leq N$.

 

$\therefore$  For give positive interger $N$, there exists intergers $x$  and  $y$  s.t.  $\left| x - y \alpha \right| < \frac{1}{N}$  and $0< y \leq N$