Answer to Question #98934 in Combinatorics | Number Theory for Dorothy

Question #98934
If f is a multiplicative arithmetic function then prove that
f([m, n])f((m, n)) = f(m)f(n)
for all positive integers m and n.
Here [m, n] is the least common multiple of m and n and (m, n) is the
greatest common divisor of m and n.
1
Expert's answer
2019-11-18T12:11:46-0500

Let "m=\\prod\\limits_{i=1}^u p_i^{k_i}" and "n=\\prod\\limits_{i=1}^u p_i^{l_i}".

We have "f(m)f(n)=f(\\prod\\limits_{i=1}^u p_i^{k_i})f(\\prod\\limits_{i=1}^u p_i^{l_i})=\\prod\\limits_{i=1}^u f(p_i^{k_i})\\prod\\limits_{i=1}^u f(p_i^{l_i})="

"=\\prod\\limits_{i=1}^u f(p_i^{k_i})f(p_i^{l_i})=\\prod\\limits_{i=1}^u f(p_i^{\\min\\{k_i,l_i\\}})f(p_i^{\\max\\{k_i,l_i\\}})="

"=\\prod\\limits_{i=1}^u f(p_i^{\\min\\{k_i,l_i\\}})\\prod\\limits_{i=1}^u f(p_i^{\\max\\{k_i,l_i\\}})="

"=f(\\prod\\limits_{i=1}^u p_i^{\\min\\{k_i,l_i\\}})f(\\prod\\limits_{i=1}^u p_i^{\\max\\{k_i,l_i\\}})=f((m,n))f([m,n])"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS