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Dirichlet's theorem on Diophantine approximation:

For any real number $x$, for integer $N>0$, there exists integers $a$ and $b>0$ with $(a,b)=1$ such that $b\leq N$ and $$|x-\frac a b|<\frac{1}{b(N+1)}.$$

Let $c$ be a fixed positive integer. Is there any version of Dirichlet's theorem to make the following restriction?

  1. $c|b$

  2. $(b,c)=1$

version 1) $c|b$

version 2) $(b,c)=1$

Dirichlet's theorem on Diophantine approximation:

For any real number $x$, for integer $N>0$, there exists integers $a$ and $b>0$ with $(a,b)=1$ such that $b\leq N$ and $$|x-\frac a b|<\frac{1}{b(N+1)}.$$

Let $c$ be a fixed positive integer. Is there any version of Dirichlet's theorem to make the following restriction?

  1. $c|b$

  2. $(b,c)=1$

Dirichlet's theorem on Diophantine approximation:

For any real number $x$, for integer $N>0$, there exists integers $a$ and $b>0$ with $(a,b)=1$ such that $b\leq N$ and $$|x-\frac a b|<\frac{1}{b(N+1)}.$$

Let $c$ be a fixed positive integer. Is there any version of Dirichlet's theorem to make the following restriction?

version 1) $c|b$

version 2) $(b,c)=1$

Source Link
7-adic
  • 3.8k
  • 24
  • 40

Variations of Dirichlet's theorem on Diophantine approximation

Dirichlet's theorem on Diophantine approximation:

For any real number $x$, for integer $N>0$, there exists integers $a$ and $b>0$ with $(a,b)=1$ such that $b\leq N$ and $$|x-\frac a b|<\frac{1}{b(N+1)}.$$

Let $c$ be a fixed positive integer. Is there any version of Dirichlet's theorem to make the following restriction?

  1. $c|b$

  2. $(b,c)=1$