Prove the multiplicativity of the totient function. For His sake, don't use fields or rings T_T.

- ParthKohli

Prove the multiplicativity of the totient function. For His sake, don't use fields or rings T_T.

- jamiebookeater

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- ParthKohli

What I mean is I wanna prove this:\[\phi(ab\cdots yz) = \phi(a) \cdot \phi(b) \cdots \phi(y) \cdot \phi(z) \]

- ParthKohli

If \(\gcd(a,b\cdots,y,z) = 1\)

- ParthKohli

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## More answers

- anonymous

|dw:1361115089981:dw|

- ParthKohli

No... coprime numbers.

- anonymous

|dw:1361115307691:dw|U mean

- ParthKohli

No.\[\phi(7) = 6\]

- anonymous

|dw:1361115408680:dw|

- ParthKohli

Yeah.

- ParthKohli

LOL wait, I realized it lol

- ParthKohli

\[\phi(a) = \left(1 - \frac{1}{p_1}\right)\left(1 - \frac{1}{p_2}\right)\cdots\left(1 - \frac{1}{p_n}\right)\]And when you multiply all those phis, you get another phi.

- ParthKohli

Too lazy to type it out, you get the point.

- ParthKohli

I realized the proof, but I am pretty lazy at the moment. I'd explain to you how it works.

- ParthKohli

It basically says that if \(a\) has factors \(a_1,a_2\cdots a_n\) then phi(a) is obviously (1 - 1/a1)(1 - 1/a2)...(1/an)
Same for phi(b) and so on
Since a * b ... y * z's set of factors the union of the sets of factors of each a,b....y,z, so the factors are a1,a2...b1,b2.......x1,x2...y1,y2...y_n and the phi of that is (1 - a1)(1 - a2)...(1 - b1)....... = phi(a) * phi(b) ... phi(y) * phi(z)

- ParthKohli

Too lazy to do LaTeX.

- ParthKohli

Maybe I'd write a whole document on this someday. Not now, just.

- anonymous

|dw:1361117190858:dw|But how

- ParthKohli

I don't remember the proof but it's related to how \(\dfrac{\phi(N)}{N}\) is the probablity that a number less than or equal to \(N\) is coprime to it

- ParthKohli

Correction |dw:1361117896513:dw|

- terenzreignz

It may suffice to show that it is multiplicative for any two distinct prime numbers.
First, let's get rid of the case where the prime numbers are the same.
Let p and q be distinct prime numbers. It can be shown that T(p) = p-1, for all primes.
T(p*p) =
Well, consider all numbers less than p squared, and there are p^2 - 1 of them
p^2 - 1
But
p^2 - 1 = (p+1)(p-1) which is not equal to (p-1)(p-1) = T(p)T(p)
Then it doesn't hold for two prime numbers which are the same.
Consider T(pq)
Well, now consider all positive integers less than pq, and there are pq-1 of them.
(pq - 1)
But you have to take away all the multiples of p, up to (q-1)p, since these are not coprime with pq, obviously.
(pq - 1) - (q - 1)
You also have to take away all the multiples of q, up to (p-1)q, since these are not coprime with pq, clearly
(pq - 1) - (q - 1) - (p - 1)
And everything else will be coprime with pq, since both p and q are prime.
= pq -1 - q + 1 - p + 1 = pq - q - p + 1 = (p - 1)(q - 1) = T(p)T(q)
It holds for any two distinct prime numbers!!!

- terenzreignz

I may have been overenthusiastic with that one... my bad :)

- experimentX

that seems okay .. use induction on that.

- terenzreignz

That's where I'm lost, unfortunately :(

- ParthKohli

Is my proof OK?

- experimentX

i didn't read all your post ... i think it should be something like ... if it holds for two prime number, then rest of integers are just product of primes, i should hold for any number.

- terenzreignz

Maybe you can do more with my little result than I'm capable of :)

- experimentX

|dw:1361120997117:dw|

- experimentX

rest of the number ... say a number z is just a product of distinct prime number, of course it should hold.

- terenzreignz

I wasn't aware of the probability bit that Parth mentioned, so... poor me :(
Good thing you guys could make use of it :D

- experimentX

|dw:1361121296399:dw|

- ParthKohli

@experimentX That was my proof...

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