By using piegonhole principal, prove that for any positive irrational number r and positive real numbers x , y ∈ ( 0 , 1 ) , x &lt; y, there exists a positive integer n such that x ≤ n r − [ n r ] ≤ y, where [ n r ] is the integral part of n r.

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By using piegonhole principal, prove that for any positive irrational number   r and positive real numbers   x  ,  y  ∈      (          0      ,      1        )  ,   x  &amp;lt;  y, there exists a positive integer   n such that   x  ≤  n  r  −      [          n      r        ]    ≤  y, where       [          n      r        ]   is the integral part of   n  r.

Hadley Cunningham · Accepted Answer

Consider   0  ,  {  r  }  ,  {  2  r  }  ,  …  ,  {  n  r  }  ,  1 and divide   (  0  ,  1  ) in   n  +  1 equal parts. By the pigeonhole principle, at least one part will contain two numbers, say   {  p  r  } and   {  q  r  }; then      |    {    p    r    }    −    {    q    r    }    |    &amp;lt;      1          n      +      1        .  {  p  r  }  =  p  r  −      ⌊    p    r    ⌋   This is Dirichlet&#039;s approximation theorem.

By using piegonhole principal, prove that for any positive irrational number r and positive re

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