Thread: Show that function is a pre-metric

1. Show that function is a pre-metric

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.

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.
I for one, connot read that. Can you post the question using LaTeX?

3. ..

Originally Posted by Plato
I for one, connot read that. Can you post the question using LaTeX?
$\displaystyle \sqrt{\frac{\1}{16}(x_{2}-x_{1})^2+\frac{\1}{9}(Y_{2}-Y_{1})^2}$

4. 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 $\displaystyle \#(x,x)=0$??

5. 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

6. Originally Posted by nikie1o2
$\displaystyle \sqrt{\frac{\1}{16}(x_{2}-x_{1})^2+\frac{\1}{9}(Y_{2}-Y_{1})^2}$
$\displaystyle \#\left((x_1,x_2),(y_1,y_2)\right)=\sqrt{\frac{(x_ 2-x_1)^2}{16}+\frac{(y_2-y_1)^2}{9}}$