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