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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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 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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 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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 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  Tags distance, metric, proving, space Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Vaki Linear Algebra 3 June 13th, 2017 08:07 AM giokakocha Real Analysis 4 March 3rd, 2017 09:27 AM alejandrofigueroa Real Analysis 5 September 29th, 2013 10:03 AM jugger3 Linear Algebra 2 August 29th, 2013 05:24 PM semeevics Real Analysis 1 September 17th, 2009 07:22 PM

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