triangle inequality

created by mattbw
(idea) by mattbw (3 mon) (print)   (I like it!) Mon Jul 31 2000 at 17:58:20
A fundamental part of mathematical analysis. Given vectors a and b with length defined as ||a|| and ||b||, then

||a+b|| <= ||a|| + ||b||
(idea) by srkorn (2.6 y) (print)   (I like it!) Tue Aug 10 2004 at 18:53:36
In the magic world of complex numbers, the proof of the Triangle Inequality is somewhat more elegant, and its geometrical interpretation is more straightforward. Note that for a complex number z=x+iy, we denote its distance from the origin by the handy formula sqrt(x2+y2), and denote this value by |z|.
Theorem: For any w and z in C, |z+w| ≤ |z| + |w|.
Helpful ASCII diagram:


 w   | w+z (in accordance with the Parallelogram Law)
  \  |  /\ 
   \ | /  \ <-- vector w emanating from z instead of origin 
    \|/    \
-----+======--------
            z 
So, the idea is just that any leg of a triangle cannot be longer than the sum of the other sides.
Proof: Letting z = x+iy and w = s+it, we have

|z+w|2 = (x+s)2+(y+t)2
       = x2+s2+y2+t2+2(xs+yt)
       = (x2+y2)+(s2+t2)+2*Re(zw*)
       = |z|2+|w|2+2*Re(zw*)
       ≤ |z|2+|w|2+2|zw*|
       = |z|2+|w|2+2|z||w|
       = (|z| + |w|)2.
We finish off the proof by taking the square root of both sides.
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