Question

if l x+y l is smaller than or equal to l x l + l y l
prove l x+y+...+xn l is smaller than or equal to l x l + l y l + ... + l xn l where n is Natural number

Answer 1

the "if" part probably means you are supposed to do this by induction right?

Answer 2

yes by PIM

Answer 3

whatever that means
in any case do you know how to do a proof by induction?

Answer 4

umm not quite

Answer 5

I know how to do it when there is equal sign and without absolute value

Answer 6

ok first step is base case ,but you are given that as an "if " part, ie. you get to assume
\[|x_1+x_2|\leq |x_1|+|x_2|\]
normally you would have to prove that first

Answer 7

squaring each side?

Answer 8

there are lots of proofs, that is one of them, but you don't need to do it
that came under the heading of IF
normally you would, but you are spared that here because of the way the question was written

Answer 9

oh ok thx
then how would you proof other parts?
I mean the text book itself says to set that x is S and prove them by x+1

Answer 10

then you get to ASSUME it is true if \(n=k\) i.e you assume
\[|x_1+x_2+...+x_k|\leq |x_1|+|x_2|+...+|x_k|\] and now show it is true for \(n=k+1\)

Answer 11

if we put \(n=k+1\) we are trying to show that \[|x_1+x_2+...+x_k+x_{k+1}|\leq |x_1|+|x_2|+...+|x_k|+|x_{k+1}|\]

Answer 12

break part in to two pieces so you can use \[|x_1+x_2+...+x_k|\leq |x_1|+|x_2|+...+|x_k|\]
just maybe put parentheses

Answer 13

put parentheses on where?

Answer 14

\[|(x_1+x_2+...+x_k)+x_{k+1}|\]

Answer 15

you get to assume (because after all \((x_1+x_2+...+x_k)\) is just some number) that \[|(x_1+x_2+...+x_k)+x_{k+1}|\leq |(x_1+x_2+...+x_k)|+|x_{k+1}|\]

Answer 16

then the "induction hypothesis" will finish it, because you get to assume that \[|x_1+x_2+...+x_k|\leq |x_1|+|x_2|+...+|x_k|\]

Answer 17

thank you it helped me alot

Answer 18

but do you need closure for this?

Answer 19

yw
with these problems it is best to write exactly what the induction hypothesis is, so you can figure out how to get back to it

Answer 20

closure?
lost me there

Answer 21

Like assuming that (x 1 +x 2 +...+x k ) to some number in the end of the proof

Answer 22

no you don't have to worry about it being a number
it is a number