Which of the following is a metric on S? d^2 or d^(1/2)

[yxgao]

(For every set S and every metric d on S)

The answer is d^(1/2)

How do you prove this mathematically?

[NateTG]

For d to be a metric you need to show that:
d(a,b) = 0 \iff a=b
(which is easy in this case)
d(a,b) \geq 0
(also easy)
d(a,c) \leq d(a,b)+d(b,c)
Which is the only one that really needs any looking into in this case.

The does not necessarily hold for d'=d^2 since if you have d(a,b)=1 and d(b,c)=1 and d(a,b)=2, then the triangle inequality does not hold for d'.

To prove that the triangle inequality holds for d'=d^{\frac{1}{2}}, start with the triangle inequality for d, complete the square on the RHS, and take the square root of both sides.

[yxgao]

that makes perfect sense.
thanks