※ 引述《PttFund (批踢踢基金只進不出)》之銘言：
: d(x,y)
: Let d be a metric. Show that d'(x,y) = ------------ is also a metric.
: 1 + d(x,y)
證明: 1. d(x,x) = 0 => d'(x,x) = 0
2. x≠y => d(x,y) > 0 => d'(x,y) > 0
3. d(x,y) = d(y,x) => d'(x,y) = d'(y,x)
d(x,y)
4. d'(x,y) = ------------
1 + d(x,y)
1
= 1 - ------------
1 + d(x,y)
1
≦ 1 - ---------------------
1 + d(x,z) + d(z,y)
d(x,z) + d(z,y)
= ---------------------
1 + d(x,z) + d(z,y)
d(x,z) d(z,y)
= --------------------- + ---------------------
1 + d(x,z) + d(z,y) 1 + d(x,z) + d(z,y)
d(x,z) d(z,y)
≦ ------------ + ------------
1 + d(x,z) 1 + d(z,y)
= d'(x,z) + d'(z,y)
d(x,y)
所以 d'(x,y) = ------------ is also a metric.
1 + d(x,y)
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