It is a question in functional analysis by writer Erwin Kryzic
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1Let's say I want to travel from the point $0$ to the point $5$. If I go straight there, it will cost me $d(0,5)=25$ units. On the other hand, if I go one bit at a time, it will only cost me $d(0,1)+d(1,2)+\dotsb+d(4,5)=5$ units. – Akiva Weinberger Jan 22 '17 at 08:41
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In order for it to be a metric it must follow these properties by definition,
$$d(x,y) \geq 0$$
$$d(x,y)=0 \iff x=y$$
$$d(x,y)=d(y,x)$$
$$d(x,z) \leq d(x,y)+d(y,z)$$
Does it?
Ahmed S. Attaalla
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1Suppose $x\lt y\lt z$; does $d(x,z)\le d(x,y)+d(y,z)$ *ever* hold for that funny definition of $d(x,y)?$ – bof Jan 22 '17 at 06:34
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Ok. The problem is the triangle inequality, then:
Let $x,y,z \in \mathbb{R}$, so $$d(x,z) = (x+y-y+z)^2$$ $$d(x,z) = ((x-y)+(y-z))^2$$ $$\qquad d(x,z) = d(x,y)+2(x-y)(y-z)+d(y,z)$$ $$d(x,z) \geq d(x,y)+d(y,z)\quad iff\quad (x-y)(y-z)\geq0$$
Then, suppose what $x=3$, $z=1$, $y=2$, that which does not meet the triangular inequality, therefore, $d$ is not a metric
