Question

HELP!
http://prntscr.com/4fxiq7

Answer 1

for coprime integers a,b i need to prove :
\[\large (fg)(ab) = (fg)(a) * (fg)(b)\]

Answer 2

\(f(n_1n_2) g(n_1n_2)=f(n_1)f(n_2) g(n_1)g(n_2)=f(n_1)g(n_1) g(n_1)g(n_2)\)

Answer 3

\(=f(n_1)g(n_1) f(n_2)g(n_2)=fg(n_1) fg(n_2)\)

Answer 4

we get stuck there, it looks circular to me :

\[f(n_1)g(n_1) = fg(n_1)\]

this is precicely the thing we need to prove, so we CANNOT use it

Answer 5

hey :O noooo

Answer 6

its not fog
its only f.g

Answer 7

I think you're over thinking this rational.

Answer 8

yesh he ish

Answer 9

f(ab) = f(a)*f(b)
g(ab) = g(a)*g(b)
-------------------------
thats all we're given. we don't know anything about fg

Answer 10

Rashhhh

Answer 11

no no no no nononononononononoooooooooooooooooooooooo

Answer 12

how can you say f(a)*g(a) = fg(a) ?

Answer 13

lets start over

let R(x)=f(x) g(x)

Answer 14

as i said before its only product not after notation.

\(R(n_1 n_2)=f(n_1n_2) g(n_1n_2)= f(n_1)f(n_2)g(n_1)g(n_2)\\=f(n_1)g(n_1)f(n_1)g(n_1)= R(n_1) R(n_2)\)

Answer 15

i see it now :) but that doesn't prove anything :/

Answer 16

made typo
last line
\(f(n_1)g(n_1)f(n_2)g(n_2)\)

Answer 17

are you serouse :'(

Answer 18

Oh you mean :
\[\large (fg)(ab) = f(ab)*g(ab)= (fg)(a) * (fg)(b) \]

Answer 19

?

Answer 20

ARE YOU SEROUSE =(

R(X) is multiplicative or not :(
dint we prove that R(x) is multiplicative :(

Answer 21

these proofs are making me more dyslexic :/

Answer 22

i know hehe i remember we dint study these things together :P
i did them on my own last semister :P

Answer 23

ok ok u got this one ?

Answer 24

yes got it :)

whats going on with this proof from MIT prof :
http://prntscr.com/4fxohu

?

Answer 25

here is the actual link
http://ocw.mit.edu/courses/mathematics/18-781-theory-of-numbers-spring-2012/lecture-notes/MIT18_781S12_lec13.pdf

Answer 26

the shame shing :PPPP

Answer 27

but they used the sum notation wait ill show u why

Answer 28

:\

Answer 29

why are they not using the fact that `f*g(ab) = f(ab) * g(ab)` ?

Answer 30

well , idk :P
i never study in MIT :P
but that does not mean ,they are smarter or more perfect hehehehe

http://prntscr.com/4fxq4y

Answer 31

forget about MIT link, its about `convolution` proof, not product of two functions >.<

they're using \(\large *\) for convolution operator, look the definition above that theorem

Answer 32

:D

Answer 33

for part b :

\[\large \dfrac{f}{g}(ab) = \dfrac{f(ab)}{g(ab)} = \dfrac{f(a).f(b)}{g(a).g(b) } = \dfrac{f}{g}(a) . \dfrac{f}{g}(b)\]

Answer 34

bingo

Answer 35

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