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

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.

• How do u define it if m/n is not an integer ? – user379195 May 9 '17 at 21:22
• true when $m/n$ is an integer AND $\gcd(n, m/n) = 1$ – Will Jagy May 9 '17 at 21:25
• In general, no. This is $\sigma(a)\sigma(b) = \sigma(ab)$. True if $a,b$ are relatively prime, but not true in general. – GEdgar May 9 '17 at 21:25
• For a counterexample, take $m=4,n=2$. – GEdgar May 9 '17 at 21:26
• Thanks for your comments. Specially to Sir Will Jagy. – Jr Antalan May 9 '17 at 21:46

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