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Prove that $(\mathbb{R}^2,d_2)$ and $(\mathbb{R}^2,d_1)$, are not isometric where $d_2$ is euclidean metric and $d_1$ is absolute value metric.

I have to show that there cannot exist a map from $(\mathbb{R}^2,d_2)$ to $(\mathbb{R}^2,d_1)$ that preserves length. But I didn't get any idea.

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    $\begingroup$ What are $d_2$ and $d_1$? $\endgroup$ – Parcly Taxel Aug 14 at 11:49
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    $\begingroup$ I have edited the question $\endgroup$ – Epsilon Delta Aug 14 at 11:54
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Consider the points $(1,0)$ and $(0,1)$. Then $d_1\bigl((1,0),(0,1)\bigr)=2$ and both points $(0,0)$ and $(1,1)$ are such that their distance to both of them is equal to $1$.

However, in $(\mathbb R^2,d_2)$, whenever you have two points $p$ and $q$ such that $d_2(p,q)=2$, there is one and only one point whose distance to both of them is $1$, which is $\frac{p+q}2$.

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  • $\begingroup$ how there is only one point $\frac{p+q}{2}$ $\endgroup$ – Epsilon Delta Aug 15 at 2:56
  • $\begingroup$ I don't understand what you are asking. $\endgroup$ – José Carlos Santos Aug 15 at 3:01
  • $\begingroup$ Can you prove that there is only one point whose distance from both p and q is 1? $\endgroup$ – Epsilon Delta Aug 15 at 3:02
  • $\begingroup$ If the distance from a point $r$ to $p$ is $1$, then $r$ belongs to the circle centered at $p$ with radius $1$. And if the distance from $r$ to $q$ is also $1$, then $r$ belongs to the circle centered at $q$ with radius $1$. But the distance from $p$ to $q$ is $2$, and so those two circles have a single intersection point. $\endgroup$ – José Carlos Santos Aug 15 at 3:11
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With $d_1\bigl((x,y),(u,v)\bigr)=|x-u|+|y-v|$, we have $$ (0,0),\quad (1,0), \quad (\tfrac12,\tfrac12)$$ and $$ (0,0),\quad (0,1), \quad (\tfrac12,\tfrac12),$$ that is, two equilateral triangles with side length $1$ and common edge and the other vertices have distance $2$.

Under the Euclidean metric, we would always have distance $\sqrt 3$ instead of $2$.

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