0

$$ d(A, B) = |A-B|^2 $$

I'm told to consider the following: For the Euclidean metric, denoted by $d_2$, and defined by

$$ d_2(A,B) = \sqrt{(x_A - x_B)^2 + (y_A - y_B)^2} $$

How are they related, the later seems to be ($|A^2+B^2|$)

enter image description here

pepsi-maniac
  • 33
  • Actually your $A,B \in \mathbb{R}$. So, check if the given function satisfies the triangle inequality? As such, it has nothing to do with the metric $d_2$ because for that your $A,B \in \mathbb{R}^2$. – Anurag A Nov 20 '22 at 20:28
  • How do the vectors A and B translate to this formula $|A-B|^2$ – pepsi-maniac Nov 20 '22 at 20:30
  • As I mentioned earlier, for the $d(A,B)$ function, your $A,B$ are real numbers (as in example 1.3.2) and not vectors in $\mathbb{R}^2$. – Anurag A Nov 20 '22 at 20:31
  • 1
    How about choosing $A=0, B=1$ and $C=2$? Now check if $d(A,C) \leq d(A,B)+d(B,C)$? Note: deleting your comments is not useful as it breaks the chain for another user. – Anurag A Nov 20 '22 at 20:38
  • $A=0,B=1,C=2$ $d(a,c)=2 \le (d(a,b) = 1 + d(b, c) = 1) = 2$, It holds, its equal – pepsi-maniac Nov 20 '22 at 20:40
  • 1
    $d(A,C)=|0-2|^{2}=4$. – Anurag A Nov 20 '22 at 20:40
  • But what is C, i'm only given two points. A and B, 0, and 1. Am I just picking C arbitrarily? – pepsi-maniac Nov 20 '22 at 20:51

1 Answers1

2

For a map $d:X^2\to[0,\infty)$ to be a metric on $X$, it must satisfy the conditions for all pairs of points $(x,y)\in X^2$. So a single counterexample suffices. In this example, if $a=0$, $b=1$, $c=2$ then $$ d(a,c) = |0-2|^2 = 4 > 2 = |0-1|^2 + |1-2|^2 = d(a,b)+d(a,c). $$ It does not matter how we arrive at the counterexample, only that it is valid. Which in this case it is, in showing that the map $(x,y)\mapsto|x-y|^2$ is not a metric on $\mathbb R$, as the triangle inequality does not hold.

Math1000
  • 36,983