Discrete Math Proof: |x + y| <= |x| + |y| for all real x, y

[dude15129]

Hi I am working on my homework for Discrete Math and I dont' know how to do this proof. It says, "Use proof by cases that |x+y|? |x| + |y| for all real numbers x and y." I know what proof by cases is but I don't know where to start.

 

[stapel]

Start by considering the cases. Is x + y > 0, so |x + y| = x + y, or is x + y < 0, so |x + y| = -(x + y)? Is x > 0, so |x| = x, or is x < 0, so |x| = -x? And so forth.

Eliz.

 

[dude15129]

Ok....I have started doing that and this is what I have so far:

Case 1: x + y>0
|x+y|>0 ? |x+y| = x + y.

x + y cannot be > |x| + |y|.
? x + y ? |x| + |y|


Case 2: x + y = 0
|x + y| = 0 ? x=x & y= -x

? x + (-x) ? |x| + |y|

Case 3: x + y < 0
|x + y| < 0 ? x + y = -(x + y).

I have gotten this far, but I don’t know where to go from here. I know it makes sense in my head, but I don’t know how to put it into words.