Does $\sigma(\frac{m}{n})=\frac{\sigma(m)}{\sigma(n)}$ where $\sigma$ is the sum of divisor function?

Question

My question is the same as with the title. Kindly help me prove the statement below if it is true.

What I tried: For specific value of $m$ and $n$ the equality seems to hold.

Does $\sigma(\frac{m}{n})=\frac{\sigma(m)}{\sigma(n)}$ where $\sigma$ is the sum of divisor function?

If not what are the conditions for $m$ and $n$ such that the above condition is true.

Thanks a lot.

Answer 1

let $N(k)=k$ for all $k$ so that $N$ is a completely multiplicative function. by a well-known result in the elementary theory of arithmetic functions, this implies that $$ \sigma(n)=\sum_{d|n}N(d) $$ is also a multiplicative function. so, as Will Jagy pointed out, if $n|m$ and $\gcd(n,\frac{m}{n}=1)$ we have $$ \sigma(m)=\sigma(n\frac{m}{n})=\sigma(n)\sigma(\frac{m}{n}) $$ from which the result stated follows