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Math Help - Proof Metric Spaces

  1. #1
    Junior Member
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    Proof Metric Spaces

    Let p be a fixed prime number, and de fine d : Z Z → R by d(m;m) = 0 for all
    m ∈ Z and d(m; n) = 1/r
    r when m != n and p^(r-1) is the highest power of p which
    divides m − n. Show that d is a metric on Z.

    The 1st condition holds since d(m,m)=0

    What I started to do is try and write r in terms m & n and then substitute that into d(m,n) =1/r... This seems wrong to me.. Im seriously stumped with this question.
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  2. #2
    MHF Contributor

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    Okay, you clearly have d(x,x)= 0 and that d(x,y)> 0 if x\ne y.

    Also clear is that d(x,y)= d(y,x).

    The only hard part is the "triangle inequality": d(x,y)\le d(x,z)+ d(z,y)

    If x= y or z= y or x= z, that is trivial so you can assume they are all different.

    I think the crucial part is to note that if p^n divides both x- z and z- y then it also divides x- y= (x- z)+ (z- y).
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