- #1
iamalexalright
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Homework Statement
Let a function [tex]\rho:\Re^{2}x\Re^{2}\rightarrow \Re_{+}[/tex] be defined by:
[tex]\rho((x_{1},y_{2}),(x_{2},y_{2})) = |x_{1} - x_{2}| + |y_{1} - y_{2}|[/tex]
Prove that [tex]\rho[/tex] is a metric on [tex]\Re^{2}[/tex]
Homework Equations
To be a metric it must satisfy:
1. d(x, y) ≥ 0 (non-negativity)
2. d(x, y) = 0 if and only if x = y (identity of indiscernibles)
3. d(x, y) = d(y, x) (symmetry)
4. d(x, z) ≤ d(x, y) + d(y, z)
The Attempt at a Solution
I'm not going to give a full proof of each - I just want to see if my basic ideas are correct. I will exclaim that I am not very great at writing proofs just yet (so be critical but polite please :D )
1. It's obvious (considering the absolute values) that this will be greater than zero (but that isn't a valid statement in a proof...).
Should I do it case by case, ie:
x1 > x2 > 0 implies |x1 - x2| > 0
x2 > x1 > 0 implies x1 - x2 < 0 implies |x1 - x2| = -(x1 - x2) > 0
so on and so on until I have all cases and then can assume that is correct?
2. I'm kind of at a loss for this one... maybe(by contradiction)
(x1,y1), (x2,y2) are real numbers and distinct and |x1 - x2| + |y1 - y2| = 0
|x1 - x2| = d > 0
|y1 - y2| = e > 0
if this were to equal zero then
d = -e which is a contradiction (one of these would be less than zero)
3. This one would just be algebra
4. This one is also just some algebra
Obviously I need to formally state everything but otherwise does it seem correct?