hello.

can any one proof this problem for me please?

if a|b then phi(a) | phi(b). (b is natural number).

tanx.

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- November 26th 2011, 09:58 AMmostafashaerihelp to proof : if a|b then phi(a) | phi(b)
hello.

can any one proof this problem for me please?

if a|b then phi(a) | phi(b). (b is natural number).

tanx. - November 26th 2011, 10:19 AMchisigmaRe: help to proof : if a|b then phi(a) | phi(b)
- November 26th 2011, 11:17 AMmostafashaeriRe: help to proof : if a|b then phi(a) | phi(b)
thank.

So?

i don't understand why b|a?

the problem say that a|b - November 26th 2011, 11:54 AMmostafashaeriRe: help to proof : if a|b then phi(a) | phi(b)
i don't understand why b|a . in the problem a|b but you said b|a.

please explain how to continue this proof

thanks - November 26th 2011, 05:59 PMPaulRSRe: help to proof : if a|b then phi(a) | phi(b)
I think he switched and . As for the rest, note that any prime (from that product in the formula for ) actually divides .

Another proof (back to the original ) :

Let where is the greatest integer dividing such that .

Adding to this that , we have that and must share the exact same prime divisors (and too) . Thus in fact, any number coprime to is coprime to and viceversa. It follows then that (*).

Finally since and are coprime, we have: , and since we are done.

In each set for we have exactly numbers coprime to . These sets form a partition of so indeed .