So I understand that Euclidean distance is valid for all of properties for a metric. But why doesn't the square hold the same way?
Euclidean distance vs Squared
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metric-spaces
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2Please pick an answer different than mine to be the correct one, as my answer deals with norms instead of metrics. – 2011-10-23
2 Answers
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The square of the distance does not obey the triangle inequality: $1^2+1^2<(1+1)^2$
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0I know it doesn't by why doesn't it obey it? – 2011-10-23
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3To show that the triangle inequality is obeyed means to show that it holds for _all_ possible examples. To show that the triangle does not hold, you need only _at least one_ counterexample. – 2011-10-23
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0I see. Thank you. I think Scott explained the lingering question. – 2011-10-23
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You lose the triangle inequality if you don’t take the square root: the ‘distance’ from the origin to $(2,0)$ would be $4$, which is greater than $2$, the sum of the ‘distances’ from the origin to $(1,0)$ and from $(1,0)$ to $(2,0)$.