Thread: Do metric distances have to be nonnegative?

1. Do metric distances have to be nonnegative?

I'm wondering about this, and if so, what an example would look like. They could easily satisfy the first two properties -- $d(x, y) = d(y, x), d(x, y) = 0 \iff x = y$ -- but I can't think of any instance, at least not using the real numbers, that would satisfy the triangle inequality.

2. Re: Do metric distances have to be nonnegative?

Originally Posted by phys251
I'm wondering about this, and if so, what an example would look like. They could easily satisfy the first two properties -- $d(x, y) = d(y, x), d(x, y) = 0 \iff x = y$ -- but I can't think of any instance, at least not using the real numbers, that would satisfy the triangle inequality.
metrics must be non-negative.

3. Re: Do metric distances have to be nonnegative?

Originally Posted by romsek
metrics must be non-negative.
Thanks, that's what I thought. How complicated is the proof of this? I know that any metric function has to be even, which already knocks out a lot of counterexamples...

4. Re: Do metric distances have to be nonnegative?

Originally Posted by phys251
but I can't think of any instance, at least not using the real numbers, that would satisfy the triangle inequality.
What does that statement say? I cannot understand what meaning. As was stated in reply #2 by definition a metric is a non-negative function. Therefore, you must review the definition of metric.

Here is the standard problem to teach metrics:
If $d$ is a metric on $X\times X$, then $e(x,y)=\dfrac{d(x,y)}{1+d(x,y)}$ is a metric on $X\times X$.