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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 $d(x,y)=(xy)^2$ a distance? I have to prove it or not. My problem is the triangle inequality, I don't find any tool to use! $$d(x,y) \le d(x,z) + d(z,y) \longrightarrow (xy)^2 \le (xz)^2 +(zy)^2$$. Can you give me a hint? 
November 10th, 2017, 03:36 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,101 Thanks: 850 
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 $d(x,y) =(xy)^2$ is a distance in the Euclidean space.

November 10th, 2017, 10:17 AM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 1,856 Thanks: 964  Quote:
$d(\vec{x},\vec{x})\geq 0,~\text{ with equality iff }\vec{x}=\vec{0}$ $d(\vec{x},\vec{y}) = d(\vec{y},\vec{x})$ $d(a \vec{x},\vec{y}) = a\cdot d(\vec{x},\vec{y})$ $d(\vec{x}+\vec{z},\vec{y}) = d(\vec{x},\vec{y})+d(\vec{z},\vec{y})$ does $d(x,y) = (xy)^2$ 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: 6,495 Thanks: 578 
If x=y and z=0, the triangle inequality is violated.

November 11th, 2017, 05:08 AM  #6 
Newbie Joined: Oct 2017 From: Italy Posts: 15 Thanks: 0  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  
Senior Member Joined: Sep 2015 From: USA Posts: 1,856 Thanks: 964  Quote:
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) $d(x,z) \leq d(x,y) + d(y,z)$ $d(x,y) = (xy)^2$ i) is clearly satisfied ii) is clearly satisfied $d(x,z) = (xz)^2 = (x+y y  z)^2 = ((xy)+(yz))^2 = d(x,y)+d(y,z) + 2(xy)(yz)$ $z< y< x \Rightarrow 2(xy)(yz)>0\Rightarrow d(x,z) > d(x,y)+d(y,z)$ 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 
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