Proving the Metric Property of a Function on Real Numbers

[iamalexalright]

Homework Statement

Let a function \rho:\Re^{2}x\Re^{2}\rightarrow \Re_{+} be defined by:
\rho((x_{1},y_{2}),(x_{2},y_{2})) = |x_{1} - x_{2}| + |y_{1} - y_{2}|

Prove that \rho is a metric on \Re^{2}

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?

 

[Office_Shredder]

For 1 you don't need to do a case by case proof. |x_1-x_2| \geq 0 and |y_1-y_2| \geq 0 so |x_1-x_2|+|y_1-y_2|\geq 0.

You can use this to prove part 2 by noting that the inequalities |x_1-x_2| \geq 0 and |y_1-y_2| \geq 0 only give you zeros if x_1 = x_2 and y_1=y_2

 

[iamalexalright]

Alright... that's what I thought at first but for some reason that seemed too simple...

could my methods be considered correct (albeit longer and not as simple)?