Question

prove that if x1,x2,...xn are n real numbers in the closed interval [a,b], then (x1+x2+...+xn)/n is in the closed interval [a,b]

Answer 1

this is real analysis problem. any ideas?

Answer 2

i know need to use math induction... getting stuck at the end....
missing something

Answer 3

go joe!

Answer 4

I would start by noting that if all n numbers were "a", then we would have the sum:
\[a+a+a+\cdots+a = na\]This is also the smallest sum we could have.

Likewise, if all the numbers were b, we would have:
\[b+b+b+\cdots +b = nb\]which is the largest sum we could possible have.

So:
\[a+a+a+\cdots +a\leq x_1+x_2+\cdots +x_n\leq b+b+b+\cdots +b\]
\[\iff na\leq x_1+x_2+\cdots +x_n \leq nb\]
\[\iff a\leq \frac{x_1+x_2+\cdots+x_n}{n}\leq b\]

Answer 5

\[\min(\{x_1,x_2,\ldots,x_n\})\leq\bar{x}\leq\max(\{x_1,x_2,\ldots,x_n\})\]

Answer 6

joe thats cute i like it

Answer 7

lolol :) zarkons is so much better looking though (as always :P)

Answer 8

i like yours better
no offense zarkon

Answer 9

its remedial is why i like it

Answer 10

nothing wrong with what you have...I just have a tighter bound

Answer 11

really the same Idea

Answer 12

like zarkon uses doctor notation

Answer 13

he thinks he better than all of us non-doctors

Answer 14

jk

Answer 15

yeah. @myin for some reason that sounds like an insult :P lolol i know its not, just saying

Answer 16

lol

Answer 17

no its not an insult

Answer 18

"down to earth" language? lolol

Answer 19

i was just trying to say you broke it all the way down

Answer 20

totally easy to understand

Answer 21

i mean zarkon's was easy to understand too

Answer 22

just ignore me

Answer 23

haha

Answer 24

thank you, guys! awesome....
i suppose to use induction method, though...
I got p(1)- true; p(k) - assumed to be true.
then need to proof:
a<(x1+x2+...xk+x(k+1))/(k+1) <b

Answer 25

oh its required to prove by induction? eek

Answer 26

it really is the same idea...replace a in joe's proof with \[x_{(1)}\]

and b with \[x_{(n)}\]

then proceed with the rest of his proof.

Answer 27

close, but not the same

Answer 28

i got:
a<(x1+x2+...x(k+1))/(k+1)<b
a+a/k<p(k) + x(k+1)/(k+1) <b+b/k
x(k+1)= means index not multipl

Answer 29

not to worry... I got it!
THANK YOU EVERYBODY!

Answer 30

im posting my inductive proof anywhos <.< i spent some time writing it out lolol

Answer 31

Skipped the base step, just showed the inductive step

Answer 32

you posted the same thing twice...missing last page?

Answer 33

whoops! where'd it go....

Answer 34

my bad >.<

Answer 35

if you add a Basis Step then it will be perfect ;)

Answer 36

a doctor wanting to add a basic step
i don't know
sounds fishy

Answer 37

yeah, if this was my homework it would totally be there. Most times when people need help with induction, its the inductive step thats killing them, so ive gotten used to skipping to it >.< lolol

Answer 38

that means its not really basic

Answer 39

nice work Joe!

Answer 40

yes joe is awesome

Answer 41

and so is zarkon

Answer 42

thank you :) you guys are the awesome ones <.<

Answer 43

i'm going to sleep

Answer 44

sounds like a good idea...I started back to work this week

Answer 45

i missed a cool trig problem below this thread :(

Answer 46

i have class tomorrow morning >.< and some people want to meet early and get Number Theory/Problem SOlving work done >.<

Answer 47

I don't teach on Thursdays :)

Answer 48

i was too busy picking on joe and i didn't realize

Answer 49

lucky o.O

Answer 50

lol <.<

Answer 51

ok later

Answer 52

later

Answer 53

lates, im out as well.

Answer 54

good night Joe/myininaya