Question

HELP !
http://prntscr.com/4fwk7o

Answer 1

@iambatman

Answer 2

I need to show :
\[\large f(p_1*p_2) = f(p_1)*f(p_2)\]

Answer 3

yeah yeah

Answer 4

not p
any two values

Answer 5

but as u said gcd need to be 1

Answer 6

say \(\large n = ab\) such that \(\large \gcd(a,b) = 1\)

Answer 7

need to show
\[\large f(a*b) = f(a)*f(b)\]

fine ?

Answer 8

\[f(p_1*p_2) = (p_1*p_2)^k=etc...\]

Answer 9

so
\(f(p_1 p_2)=( p_1p_2)^k=p_1^k p_2^k = f(p_1)f(p_2)\)

Answer 10

BSwan <3

Answer 11

<\3

Answer 12

:P buttt

Answer 13

WOW :(

Answer 14

http://puu.sh/aPIQS.jpg

Answer 15

i couldnt forget the laugh when somebody here said cardiod is butt shaped :D

Answer 16

Haha more like a heart no? :P

Answer 17

lol buttriod :P

Answer 18

>_>

Answer 19

look at first result in search
https://www.google.co.in/search?q=site%3Aopenstudy.com+ganeshie8%2Bbutt&oq=site%3Aopenstudy.com+ganeshie8%2Bbutt&aqs=chrome..69i57j69i58.17607j0j4&sourceid=chrome&es_sm=0&ie=UTF-8

Answer 20

lool

Answer 21

hehe , i remember that

Answer 22

organic math

Answer 23

It exposes your true identity! Haha.

Answer 24

haha :P
who cares

Answer 25

getting back to the actual question,
so that proves the multiplicative property is it ?
but here gcd need not be 1 i guess it works always : (ab)^n = a^n*b^n regardless of gcd right ?

Answer 26

it always work , exp properties

Answer 27

^

Answer 28

i like this movie songs >.<

Answer 29

which movie songs ?

next question looks similar but it maybe tough
http://prntscr.com/4fwmnb

Answer 30

jab we met

Answer 31

haha hahaa haaha ahhhaaaha haha

thats my fav song too ^^

Answer 32

here is actual song
https://www.youtube.com/watch?v=xdUv-CwmZPs

Answer 33

im watching themovie , actully its the 100 time im watching it hehe
im just inlove with it lol , songs amazing
still dint reach that part i think within 30 mnts it will start in movie

Answer 34

ok the question hehe

Answer 35

what does `identically zero` mean ?

Answer 36

not identically zero means

\(\large g(p^k) g^{-1} (p^k)\neq 0\)
or
\(\large g( p^k p^{-k} )\neq 0 \)


the song started

Answer 37

ok enjoy :) we can work this afterwards, im going for lunch

Answer 38

shure

Answer 39

its 1-1 function

let
\(\large n= p_1^{k_1} p_2^{k_2} p_3^{k_3}... p_r^{k_r} \)
\(\large f(n)=f( p_1^{k_1} p_2^{k_2} p_3^{k_3}... p_r^{k_r}) \\\large = f(p_1^{k_1} )f(p_2^{k_2}) f(p_3^{k_3})... f(p_r^{k_r})\\\large = g(p_1^{k_1} )g(p_2^{k_2}) g(p_3^{k_3})... g(p_r^{k_r}) \\\large = g( p_1^{k_1} p_2^{k_2} p_3^{k_3}... p_r^{k_r}) = g(n) \)

Answer 40

Hence :-
\(\forall n \in N , f(n)=g(n) \iff f=g \)

is it correct ?

Answer 41

looks perfect :)
all it means is that if you show that two functions give same value for all prime powers, then the functions are equal.

we don't need to prove them for each and every integer

Answer 42

:)

Answer 43

*valid only for multiplicative functions

Answer 44

if the functions are NOT multiplicative, then f(p^k) = g(p^k) need not imply f=g

Answer 45

what if for all p and k ?

Answer 46

i cant imagin any two continues function that
f(x)=g(x) for all x in domain
need not imply f=g >.<

Answer 47

what about f(6^k)

Answer 48

if `f` is not multiplicative, you canNOT write f(6^k) = f(2^k)*f*3^k)

so clearly f(2^k) = g(2^k) and f(3^k) = g(3^k) NEED not imply f(6^k) = g(6^k)

Answer 49

hmm , its not my point :o

Answer 50

then ?

Answer 51

f=g , iff f(x)=g(x) for all x in domain , right ?

Answer 52

you said something like :
f(p^k) = g(p^k) means f=g even if the functions are NOT multiplicative right ?

Answer 53

no , this is what i mean
http://prntscr.com/4fxako
like this
f=g , iff f(x)=g(x) for all x in domain

Answer 54

yes
\[\large f(x) = g(x)\iff f=g\]

Answer 55

but below is true only for multiplicative functions :

\[\large f(p^k) = g(p^k) \iff f=g\]

Answer 56

so since , f(n) and g(n) multiplicative , and we already given f(p^k)=g(p^k)
and any number n can be represented as product of primes , hence we could show
f(n)=g(n) for any n in domain , thus f=g

Answer 57

that makes sense