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0

$begingroup$

Let $d:mathbbZ$ x $mathbbZ rightarrow mathbbR_≥0$

$d(x,y)= $

*$0$ if $x=y$

*$2^-m$ if $x≠y$ with $2^m$ divides $(y-x)$ and $2^m+1$ does not divide $(y-x)$

Now $d(x,y)=0$ iff $x=y$ is clear from the condition of the metric,

$d(x,y)=d(y,x)=2^-m$
and then triangle inequality looks good to me.

But this second condition confuse me a little bit, I think there has to be more work behind it, but I'm not sure what I have to consider. Any help please?

linear-algebra general-topology metric-spaces

share|cite|improve this question

asked Mar 13 at 18:51

DadaDada

9010

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You should check me. But I believe this is called the $2$-adic metric (you can do it with any prime $p$).
  $endgroup$
  – Yanko
  Mar 13 at 18:53
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  mathworld.wolfram.com/p-adicNorm.html (this website defines it for any rational number but I guess it can be useful)
  $endgroup$
  – Yanko
  Mar 13 at 18:55
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $d:mathbbZ$ x $mathbbZ rightarrow mathbbR_≥0$

$d(x,y)= $

*$0$ if $x=y$

*$2^-m$ if $x≠y$ with $2^m$ divides $(y-x)$ and $2^m+1$ does not divide $(y-x)$

Now $d(x,y)=0$ iff $x=y$ is clear from the condition of the metric,

$d(x,y)=d(y,x)=2^-m$
and then triangle inequality looks good to me.

But this second condition confuse me a little bit, I think there has to be more work behind it, but I'm not sure what I have to consider. Any help please?

linear-algebra general-topology metric-spaces

share|cite|improve this question

asked Mar 13 at 18:51

DadaDada

9010

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You should check me. But I believe this is called the $2$-adic metric (you can do it with any prime $p$).
  $endgroup$
  – Yanko
  Mar 13 at 18:53
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  mathworld.wolfram.com/p-adicNorm.html (this website defines it for any rational number but I guess it can be useful)
  $endgroup$
  – Yanko
  Mar 13 at 18:55
  

  
  
  
  

add a comment | 

$begingroup$

Let $d:mathbbZ$ x $mathbbZ rightarrow mathbbR_≥0$

$d(x,y)= $

*$0$ if $x=y$

*$2^-m$ if $x≠y$ with $2^m$ divides $(y-x)$ and $2^m+1$ does not divide $(y-x)$

Now $d(x,y)=0$ iff $x=y$ is clear from the condition of the metric,

$d(x,y)=d(y,x)=2^-m$
and then triangle inequality looks good to me.

But this second condition confuse me a little bit, I think there has to be more work behind it, but I'm not sure what I have to consider. Any help please?

linear-algebra general-topology metric-spaces

share|cite|improve this question

asked Mar 13 at 18:51

DadaDada

9010

$endgroup$

share|cite|improve this question

asked Mar 13 at 18:51

DadaDada

9010

share|cite|improve this question

asked Mar 13 at 18:51

DadaDada

9010

asked Mar 13 at 18:51

DadaDada

9010

asked Mar 13 at 18:51

DadaDada

9010