Let p be a fixed prime number, and define 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.