$\frac{1}{ab}=\frac{s}{a}+\frac{r}{b} \overset{?}{\iff}\gcd(a,b)=1$

Question

$$\frac{1}{ab}=\frac{s}{a}+\frac{r}{b} \overset{?}{\iff} \gcd(a,b)=1$$

This seems almost painfully obvious because it is just $ar+bs=1$ in another form. This second form is the definition of coprimality, so what else is my professor looking for?

Answer 1

Hi Im not really sure what your question is I have no idea whats going on and I'm certainly no mathematician so I apologize if I'm am being really ignorant, but it looks like it could be Bezout's Identity

Answer 2

If $\gcd(a,b)=1$ then since the greatest common divisor is the smallest positive integer that can be represented as a linear combination of a and b then we have that there are integers r and s such that

$1=ra+sb$

By dividing by ab we have that

$\frac{1}{ab}=\frac{s}{a}+\frac{r}{b}$.

Now if we suppose that $\frac{1}{ab}=\frac{s}{a}+\frac{r}{b}$ then by multiplying by ab we have

$1=ra+sb$. Since $\gcd(a,b)$ divides both a and b then it divides 1 and since the greatest common divisor is non-negative then $\gcd(a,b)=1$.