Divisor Sum Function is Multiplicative/Proof 2

Proof
Because $a$ and $b$ have no common divisor, the divisors of $a b$ are integers of the form $a_i b_j$, where $a_i$ is a divisor of $a$ and $b_j$ is a divisor of $v$.

That is, any divisor $d$ of $a b$ is in the form:
 * $d = a_i b_j$

in a unique way, where $a_i \mathrel \backslash a$ and $b_j \mathrel \backslash b$.

We can express $a$ and $b$ as:
 * $\displaystyle a = \prod_{i \mathop = 1}^r a_i$


 * $\displaystyle b = \prod_{j \mathop = 1}^s b_j$

and so their $\sigma$ function values are:


 * $\displaystyle \sigma \left({a}\right) = \sum_{i \mathop = 1}^r a_i$


 * $\displaystyle \sigma \left({b}\right) = \sum_{j \mathop = 1}^s b_j$

Consider all divisors of $a b$ with the same $a_i$.

Their sum is $\displaystyle a_i \sigma \left({b}\right) = \sum_{j \mathop = 1}^s b_j$

Summing for all $a_i$: