What about property $2$:
$d(x,y)=0$ iff $x=y$
$|x^3-y^3|=0$
$x^3-y^3=0$
$(x-y)(x^2+xy+y^2)=0$
If $x-y=0$ then $x=y$. But if $x^2+xy+y^2=0$ then $x$ does not equal to $y$.
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1$\begingroup$ Are there any real solutions to $x^2+xy+y^2=0$ other than $(0,0)$? $\endgroup$ – lulu Nov 22 '19 at 16:48
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$\begingroup$ Also take a look at math.stackexchange.com/questions/3234082/dx-y-fx-fy-on-mathbbr $\endgroup$ – Tom Nov 22 '19 at 16:50
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$\begingroup$ A year too late but $x^2 +xy + y^2 = 0\iff x =\frac {y \pm\sqrt{y^2 -4y^2}}2=\frac {y\pm\sqrt{-3y^2}}2$. If we are restricting to reals then $-3y^2 \le 0$ so this is only possible if $y = 0$ and therefore $x=0$. ... and If we are not restriction to reals, then $|x^3-y^3|$ is not a metric on $\mathbb C$. $\endgroup$ – fleablood Oct 18 '20 at 19:42
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You don’t need to factorize that polynomial to see that there aren’t any nontrivial solutions to the equation $x^3 - y^3 = 0$ - just ask yourself, “Are there any pairs of distinct numbers $x, y$ such that $x^3 = y^3$? The answer is no, because cubing is an injective function. Therefore the property is satisfied
answered Nov 22 '19 at 17:05
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Note that $$x^2+xy+y^2= (x+\frac {y}{2} )^2 + \frac {3}{4}y^2$$
Thus $$x^2+xy+y^2=0 \iff x=y=0$$
Therefore there is no problem with your metric.
answered Nov 22 '19 at 17:09
