My Math Forum Proving a distance of a metric space

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 November 10th, 2017, 12:46 AM #1 Newbie   Joined: Oct 2017 From: Italy Posts: 15 Thanks: 0 Proving a distance of a metric space Let be in an Euclidean metric space ($\mathbb{R}^2$). Is the distance $d(x,y)=(x-y)^2$ a distance? I have to prove it or not. My problem is the triangle inequality, I don't find any tool to use! $$d(x,y) \le d(x,z) + d(z,y) \longrightarrow (x-y)^2 \le (x-z)^2 +(z-y)^2$$. Can you give me a hint?
 November 10th, 2017, 03:36 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 If x and y are points in $\displaystyle R^2$, what do you mean by "$\displaystyle x- y$"? And if that is not a single number (if you are subtracting coordinates separately), how are you squaring it? Last edited by Country Boy; November 10th, 2017 at 03:38 AM.
 November 10th, 2017, 04:15 AM #3 Newbie   Joined: Oct 2017 From: Italy Posts: 15 Thanks: 0 Sorry, I misread the exercise. It says: Prove or disprove that the function $d(x,y) =(x-y)^2$ is a distance in the Euclidean space.
November 10th, 2017, 10:17 AM   #4
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Quote:
 Originally Posted by Berker Sorry, I misread the exercise. It says: Prove or disprove that the function $d(x,y) =(x-y)^2$ is a distance in the Euclidean space.
There are four (the last 2 are just considered linearity) properties an inner product, or "distance", must satisfy.

$d(\vec{x},\vec{x})\geq 0,~\text{ with equality iff }\vec{x}=\vec{0}$

$d(\vec{x},\vec{y}) = d(\vec{y},\vec{x})$

$d(a \vec{x},\vec{y}) = a\cdot d(\vec{x},\vec{y})$

$d(\vec{x}+\vec{z},\vec{y}) = d(\vec{x},\vec{y})+d(\vec{z},\vec{y})$

does $d(x,y) = (x-y)^2$ satisfy all these?

It's pretty clear the first one isn't satisfied.

 November 10th, 2017, 12:07 PM #5 Global Moderator   Joined: May 2007 Posts: 6,821 Thanks: 722 If x=-y and z=0, the triangle inequality is violated. Thanks from topsquark
November 11th, 2017, 05:08 AM   #6
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Quote:
 Originally Posted by romsek There are four (the last 2 are just considered linearity) properties an inner product, or "distance", must satisfy. $d(\vec{x},\vec{x})\geq 0,~\text{ with equality iff }\vec{x}=\vec{0}$ It's pretty clear the first one isn't satisfied.
I don't agree with you about this, because I am taught that the first condition is that $d(\vec{x},\vec{y})\geq 0,~\text{ with the equality only if }\vec{x}=\vec{y}$, that in my case is satisfied.

November 11th, 2017, 11:33 AM   #7
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Quote:
 Originally Posted by Berker I don't agree with you about this, because I am taught that the first condition is that $d(\vec{x},\vec{y})\geq 0,~\text{ with the equality only if }\vec{x}=\vec{y}$, that in my case is satisfied.
you're right, I screwed that up.

the requirements for a metric are

i) $d(x,y)\geq 0, \text{ with equality iff }x=y$

ii) $d(x,y)=d(y,x)$

iii) $d(x,z) \leq d(x,y) + d(y,z)$

$d(x,y) = (x-y)^2$

i) is clearly satisfied

ii) is clearly satisfied

$d(x,z) = (x-z)^2 = (x+y -y - z)^2 = ((x-y)+(y-z))^2 = d(x,y)+d(y,z) + 2(x-y)(y-z)$

$z< y< x \Rightarrow 2(x-y)(y-z)>0\Rightarrow d(x,z) > d(x,y)+d(y,z)$

and thus (iii) can be violated.

 November 12th, 2017, 12:45 AM #8 Newbie   Joined: Oct 2017 From: Italy Posts: 15 Thanks: 0 Thanks a lot

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