Let he function # : PxP->R be defined as #((x1,y1),(x2,y2))=((x2-x1)^3+(y2-y1)^3)^(1/3) Show that # violates all four conditions that define a metric
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Originally Posted by nikie1o2 Let he function # : PxP->R be defined as #((x1,y1),(x2,y2))=((x2-x1)^3+(y2-y1)^3)^(1/3) Show that # violates all four conditions that define a metric You really need to learn to use LaTeX! Metrics are non-negative. Find points that give a negative value.
Originally Posted by nikie1o2 Let he function # : PxP->R be defined as #((x1,y1),(x2,y2))=((x2-x1)^3+(y2-y1)^3)^(1/3) Show that # violates all four conditions that define a metric For the second condition note that $\displaystyle (-1,0)\ne(1,0)$ but $\displaystyle \#((-1,0),(1,0))=0$
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