# example showing Minkowski distance with $p<1$ is not a metric

The Minkowski distance: $$\left(\sum_i |x_i-x_i'|^p \right)^{1/p},\ \text{where}\ p\ge1$$

is only a metric for $p\ge1$. Can someone give me a quick example why the triangle inequality doesn't hold in other cases?

The Wikipedia article has the following example:

Let $x := (0, 0)$, $y := (1, 1)$, $z := (0, 1)$.

Then:

• $d(x, y) = 2^{1/p}$, which when $p < 1$ is more than $2$.
• $d(x, z) = d(y, z) = 1$.

So $d(x, y) > d(x, z) + d(z, y)$.

• Hey, cool, I get to upvote you twice. – Sycorax says Reinstate Monica Jan 22 '16 at 18:52
• @Dougal thanks! For some reason I only found the definition without the example. – Dahlai Jan 22 '16 at 19:08