# 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.

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.

• What are $d_2$ and $d_1$? – Parcly Taxel Aug 14 at 11:49
• I have edited the question – Epsilon Delta Aug 14 at 11:54

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 $$\frac{p+q}2$$.
• how there is only one point $\frac{p+q}{2}$ – Epsilon Delta Aug 15 at 2:56
• 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. – José Carlos Santos Aug 15 at 3:11
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 $$2$$.