# Euclidean distance vs Squared

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?

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Please pick an answer different than mine to be the correct one, as my answer deals with norms instead of metrics. – JavaMan Oct 23 '11 at 22:03

## 2 Answers

The square of the distance does not obey the triangle inequality: $1^2+1^2<(1+1)^2$

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I know it doesn't by why doesn't it obey it? – Laciel Oct 23 '11 at 20:43
To 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. – JavaMan Oct 23 '11 at 20:46
I see. Thank you. I think Scott explained the lingering question. – Laciel Oct 23 '11 at 20:54

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 $(2,0)$.

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