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.

Re: Do metric distances have to be nonnegative?

Quote:

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.

Re: Do metric distances have to be nonnegative?

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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...

Re: Do metric distances have to be nonnegative?

Quote:

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 $X\times X$.

Doing that problem will help you understand the metric concept.