Is d' = d/(d+1) a Valid Metric?

[Daveyboy]

d is a metric show d'= d/d+1 is a metric

I know d(x,z)\leqd(x,y)+d(y,z). And have been trying to make it fall out of this.

I have been fooling around with the terms but it has not provided to be useful. Any direction would be helpful.

 

[Office_Shredder]

Which parts of the definition of metric have you been able to demonstrate?

 

[Daveyboy]

d'(x,x)=0
d'(x,y)>0
d'(x,y)=d'(y,x)

They follow immediately from d being a metric. But the triangle inequality is providing to be more difficult.

 

[snipez90]

I actually saw this problem a bit ago as well. If we let a = d(x,z), b = d(x,y) and c = d(y,z), then we know a\leq b + c and it comes down to proving \frac{a}{1+a} \leq \frac{b}{1+b} + \frac{c}{1+c}, which is easy if you just multiply out, I think. Anyways there is probably a nicer way to do this, but I think the above works (I didn't really double check) and suits me.

*EDIT* By the way, I think awhile ago you were trying to prove all of axioms of a metric spaces using only 2 of them. At first I said you had to have a rearrangement of the triangle inequality, but later I realized you could do it using the triangle inequality by switching around a few variables. If you're still interested, I could show you.

 

[Daveyboy]

I figured that one out... eventually. But thanks I'm impressed that you remember.
Ya, that does work, I guess I missed the most obvious approach.