Does d(x;y)=(x-y)^2 define a metric on the set of real numbers?

Question

Does d(x;y)=(x-y)^2  define a metric on the set of real numbers?

anonymous · Answer

i have explaind this topic somewhere, let me try and find it for you

zugzwang · Answer

It seems to comply with the first three conditions... the fourth one, the triangle inequality, is worth some inspection...

zugzwang · Answer

Given any x, y, and z
Is it always true that
(x - z)² ≤ (x - y)² + (y - z)²
?

anonymous · Answer

i have solved those properties somewhere i am still looking for them, then i will give you the link

zugzwang · Answer

Going with that, we get
x² + z² - 2xz ≤ x² + y² - 2xy + y² + z² - 2yz

Cancelling out gives
-2xz ≤ 2y² - 2xy - 2yz

zugzwang · Answer

Bringing all the terms to the right-hand-side
0 ≤ 2y² + 2xz - 2xy - 2yz

anonymous · Answer

here we go http://openstudy.com/study#/updates/51110b6be4b0d9aa3c483dda

zugzwang · Answer

But supposedly, this is true for any x, y, and z
Then let 
x = -1
y = 0
z = 1

We get
0 ≤ 2(0)² + 2(-1)(1) - 2(-1)(0) - 2(0)(1)
0 ≤ -2

An impossibility...
Therefore, the fourth condition (the triangle inequality) doesn't hold for this particular function d(x,y)

walters · Answer

so it means is not metric space

zugzwang · Answer

Yeah.

walters · Answer

thnx