# Thread: Question on proving distance formula is metric

1. ## Question on proving distance formula is metric

I'm having trouble to show the following inequality:

a, b, c, d, e, f are all some real numbers

sqrt((a-c)^2+(b-d)^2) <= sqrt((a-e)^2+(b-f)^2) + sqrt((c-e)^2+(d-f)^2)

2. Originally Posted by leojei
I'm having trouble to show the following inequality:

a, b, c, d, e, f are all some real numbers

sqrt((a-c)^2+(b-d)^2) <= sqrt((a-e)^2+(b-f)^2) + sqrt((c-e)^2+(d-f)^2)

isn't this just the triangle inequality? the left side is the distance between (a,b) and (c,d) while the right is the distance between (a,b) and (e,f) plus the distance between (e,f) and (c,d). must you prove the triangle inequality?

the easiest way i know is to use vectors (and their magnitudes):

Let the vector beginning at (a,b) and terminating at (c,d) be $\displaystyle a$
Let the vector beginning at (c,d) and terminating at (e,f) be $\displaystyle b$
Then, the vector beginning at (a,b) and terminating at (e,f) is $\displaystyle a + b$

thus we need to prove: $\displaystyle |a + b| \le |a| + |b|$

we can do this by proving: $\displaystyle |a + b|^2 \le (|a|+|b|)^2$

can you continue?

(draw a diagram to make sure my vectors match up with your points, if it doesn't, rename the vectors)

3. ## Easiest way indeed

This is the easiest way indeed.

FYI, the original question is this:
Show d[a+bi, c+di]=sqrt((a-c)^2+(b-d)^2) is metric.

Really thanks alot for the idea of using vector. I've been doing Chaos and other Dynamical Systems questions and this question really throw off on me. Thanks~