Michaelmoo
cbff...
- Joined
- Sep 23, 2008
- Messages
- 590
- Gender
- Male
- HSC
- 2009
Hey. Ok. I've done a proof but I'm not sure if it would be accepted in an exam. I mean it uses logic to prove a particular solution.
Show that |z1 +z2 +..................+zn| < |z1| + |z2| + ............+ |zn|
My proof:
It is known that
|z1 + z2| < |z1| + |z2| ----- (1)
|z3 + z4| < |z3| + |z4| ------(2)
.......
|zn-1 + zn| < |zn-1| + |zn| if n is even or |zn| = |zn| is n is odd ------(k)
1 + 2 + ...........+ k
|z1 + z2| +|z3 + z4| +..........+ |zn-1 + zn| < |z1| + |z2| + |z3| + |z4| +...+|zn|
now from this:
|z1 + z2 + z3 +z4| < |z1 + z2| +|z3 + z4| + < |z1| + |z2| + |z3| + |z4|
Continue the series and therefore:
|z1 +z2 +..................+zn| < |z1| + |z2| + ............+ |zn|
I understand that the best way to do this is by MI. But we havn't done MI nor sequences and series. If there is any other way you could do it, that would be great.
Thanks.
Show that |z1 +z2 +..................+zn| < |z1| + |z2| + ............+ |zn|
My proof:
It is known that
|z1 + z2| < |z1| + |z2| ----- (1)
|z3 + z4| < |z3| + |z4| ------(2)
.......
|zn-1 + zn| < |zn-1| + |zn| if n is even or |zn| = |zn| is n is odd ------(k)
1 + 2 + ...........+ k
|z1 + z2| +|z3 + z4| +..........+ |zn-1 + zn| < |z1| + |z2| + |z3| + |z4| +...+|zn|
now from this:
|z1 + z2 + z3 +z4| < |z1 + z2| +|z3 + z4| + < |z1| + |z2| + |z3| + |z4|
Continue the series and therefore:
|z1 +z2 +..................+zn| < |z1| + |z2| + ............+ |zn|
I understand that the best way to do this is by MI. But we havn't done MI nor sequences and series. If there is any other way you could do it, that would be great.
Thanks.