Let # : PxP->R defined as #((x1,y1),(x2,y2))=SquareRoot of : (1/16(x2-x1)^2+1/9(y2-y1)^2. Show that # is a premetric but you dont have to check the triangle inequality.
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Originally Posted by nikie1o2 Let # : PxP->R defined as #((x1,y1),(x2,y2))=SquareRoot of : (1/16(x2-x1)^2+1/9(y2-y1)^2. I for one, connot read that. Can you post the question using LaTeX?
Originally Posted by Plato I for one, connot read that. Can you post the question using LaTeX? $\displaystyle \sqrt{\frac{\1}{16}(x_{2}-x_{1})^2+\frac{\1}{9}(Y_{2}-Y_{1})^2}$
Originally Posted by nikie1o2 Let # : PxP->R defined as #((x1,y1),(x2,y2))=SquareRoot of : (1/16(x2-x1)^2+1/9(y2-y1)^2. Show that # is a premetric but you dont have to check the triangle inequality. Come on is $\displaystyle \#(x,y)\geqslant 0$ and $\displaystyle \#(x,x)=0$??
Originally Posted by nikie1o2 $\displaystyle \sqrt{\frac{\1}{16}(x_{2}-x_{1})^2+\frac{\1}{9}(Y_{2}-Y_{1})^2}$ I am trying to but i keep getting an error... not to technological savvy
Originally Posted by nikie1o2 $\displaystyle \sqrt{\frac{\1}{16}(x_{2}-x_{1})^2+\frac{\1}{9}(Y_{2}-Y_{1})^2}$ $\displaystyle \#\left((x_1,x_2),(y_1,y_2)\right)=\sqrt{\frac{(x_ 2-x_1)^2}{16}+\frac{(y_2-y_1)^2}{9}}$
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