# Do metric distances have to be nonnegative?

• Apr 26th 2014, 01:41 PM
phys251
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.
• Apr 26th 2014, 01:52 PM
romsek
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.
• Apr 26th 2014, 02:06 PM
phys251
Re: Do metric distances have to be nonnegative?
Quote:

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...
• Apr 26th 2014, 02:18 PM
Plato
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 $X\times X$.