To show:
If every Prime-Divisor of m is a prime-Divisor of n ist, then φ(mn) = mφ(n).
φ(n) is the totient-function introduced by euler.
To show:
If every Prime-Divisor of m is a prime-Divisor of n ist, then φ(mn) = mφ(n).
φ(n) is the totient-function introduced by euler.
Here is my own answer:
Let $n = p^{a_1}_1 p^{a_2}_2...p^{a_i}_i p^{a_i+1}_{i+1}...p^{a_s}_s$
Every Prime-Divisor of m is a Prime-Divisor of n. So $m = p^{b_1}_1 p^{b_2}_2...p^{b_i}$
n can have more prime divisors than m, thats why it can have the extra $p^{a_i+1}_{i+1}...p^{a_s}_s$.
We know φ(k) = k $\prod_{p|k} (1-1/p)$
so φ(mn) = mn $\prod_{p|mn} (1-1/p)$ $=mn(1-1/p_1)(1-1/p_2)...(1-1/p_i)...(1-1/p_s)$
also φ(n) $= n(1-1/p_1)(1-1/p_2)...(1-1/p_i)...(1-1/p_s)$
so it follows φ(mn) = m φ(n)
q.e.d
Correct?