Show that function is a pre-metric

• Apr 1st 2010, 02:32 PM
nikie1o2
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.
• Apr 1st 2010, 02:43 PM
Plato
Quote:

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?
• Apr 1st 2010, 03:22 PM
nikie1o2
..
Quote:

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

$\sqrt{\frac{\1}{16}(x_{2}-x_{1})^2+\frac{\1}{9}(Y_{2}-Y_{1})^2}$
• Apr 1st 2010, 03:23 PM
Drexel28
Quote:

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 $\#(x,y)\geqslant 0$ and $\#(x,x)=0$??
• Apr 1st 2010, 03:25 PM
nikie1o2
Quote:

Originally Posted by nikie1o2
$\sqrt{\frac{\1}{16}(x_{2}-x_{1})^2+\frac{\1}{9}(Y_{2}-Y_{1})^2}$

I am trying to but i keep getting an error... not to technological savvy
• Apr 1st 2010, 03:27 PM
Drexel28
Quote:

Originally Posted by nikie1o2
$\sqrt{\frac{\1}{16}(x_{2}-x_{1})^2+\frac{\1}{9}(Y_{2}-Y_{1})^2}$

$\#\left((x_1,x_2),(y_1,y_2)\right)=\sqrt{\frac{(x_ 2-x_1)^2}{16}+\frac{(y_2-y_1)^2}{9}}$