# Shouldn't this definition of a metric include a square root?

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Mathworld.Wolfram says that there is a metric on an inner product space (with inner product <.,.>) defined by <v-w,v-w>. Shouldn't that be the square root of <v-w,v-w>?
In https://mathworld.wolfram.com/InnerProduct.html, it states
"Every inner product space is a metric space. The metric is given by
g(v,w)= <v-w,v-w>."
In https://en.wikipedia.org/wiki/Inner_product_space , on the other hand,
"As for every normed vector space, an inner product space is a metric space, for the distance defined by
d(x,y) = ||y-x||"
after it had defined ||x|| as sqrt(<x,x>)

(I am assuming that a metric is the same as a distance function.)

The Mathworld definition appears to be the square of the Wikipedia definition. Whereas one can have more than one metric on a vector space, the former definition would seem to violate the triangle inequality
[Example: take the inner product on the one-dimensional vector space (the number line) as the dot product (which reduces to multiplication of the coordinates of the representative vectors from the origin), and check the triangle inequality for the a=0, b=2, c=5: 4+9 < 25. ]

Did Wolfram make a mistake, or am I missing some elementary and obvious point?

Did Wolfram make a mistake?
It looks like it.

• I haven't checked the triangle identity, but isn't the square of the Euclidean metric a metric, too?

I haven't checked the triangle identity, but isn't the square of the Euclidean metric a metric, too?
It doesn't obey the triangle inequality.

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