Here is another way.
Now take the square root.
Hi I am working on my homework for Discrete Math and I don't 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.
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.