Question

assume that |a+b| <= |a| + |b|

using mathematical induction, show that

|x_1 + x_2 + x_3 + ... + x_n| <= |x_1| + |x_2| + ... + |x_n|

Answer 1

Last one, then I'm hitting the hay. 5:00 is a little late for me :)
Assume it's true for some integer n. Then we seek to show that this implies it holds for n+1.

\[|x_1 + x_2 + x_3 + ... + x_n + x_{n+1} | \]
let
\[a = x_1 + x_2 + x_3 + ... + x_n\]
and
\[b = x_{n+1} \]
via our assumption, since |a+b| <= |a| + |b|, then
\[|x_1 + x_2 + x_3 + ... + x_n + x_{n+1} | \leq | x_1 + x_2 + x_3 + ... + x_n| + |x_{n+1}|\]

Answer 2

great,

wish I was as good as you in math; since the AP calculus test I've forgotten everything... especially since I didn't get to college last fall :(

Answer 3

But since we assumed that the statement holds for n.... I'm lazy and sleepy, the rest should be fairly obvious. :) It can slip away alarmingly fast without practice, but I'm sure it'll all come back soon enough!

Answer 4

right; maybe after a hard weekend of working through the book again, it will come back to me.

Answer 5

I'll just say that \[|x_1 + x_2 + ... + x_n| + |x_{n+1}| \leq |x_1| + |x_2| + ... + |x_n| + |x_{n+1}|\]

Answer 6

which is just adding the |x_n+1| part to both sides of the inequality, right?

Answer 7

Yes. And you can do that, because we assumed at the beginning that it holds for n, which means that
\[ | x_1 + x_2 + x_3 + ... + x_n| \leq |x_1| + |x_2| + |x_3| + ... + |x_n| \]

Answer 8

So you've proven that, if that inequality holds true for some integer n, it also holds true for n+1. All that remains is to show that it holds for some n (in this case, n=2) but that's just the triangle inequality which we take for granted here, so you're done. It's true for 2, which implies its true for 3, which implies its true for 4, etc etc etc.

Answer 9

right