Check my proof : gcd(a,b)=1=gcd(x,y) => (xa,yb)=gcd(x,b) gcd(y,a)

Question

Note: (x,y) means gcd(x,y)

I managed to prove the next

Proposition: Let $(a,b)=1=(x,y)$. Then $(x a,y b)=(x,b)(y,a)$.

It can be easily be generalized for the case that $(a,b)\neq1$ and or $(x,y)\neq1$ using: $(x a,y b) =(x,y)(a,b)(\frac{x}{(x,y)}\frac{a}{(a,b)},\frac{y}{(x,y)}\frac{b}{(a,b)})$

and applying the proposition to the last factor we get this

Corolary: $(x a,y b) =(x,y)(a,b)(\frac{x}{(x,y)},\frac{b}{(a,b)})(\frac{y}{(x,y)},\frac{a}{(a,b)})$

This is my proof of the proposition:

Let $(a,b)=1=(x,y)$. Then $(x a,y b)=(x,b)(y,a)(\frac{x}{(x,b)}\frac{a}{(y,a)},\frac{y}{(y,a)}\frac{b}{(x,b)})$ being $(\frac{x}{(x,b)}\frac{a}{(y,a)},\frac{y}{(y,a)}\frac{b}{(x,b)})=1$ because $(\frac{x}{(x,b)} ,\frac{y}{(y,a)})=(\frac{x}{(x,b)},\frac{b}{(x,b)})=(\frac{a}{(y,a)},\frac{y}{(y,a)})=(\frac{a}{(y,a)},\frac{b}{(x,b)})=1$

Is correct the proof?
Is easy to understand?
Can you think an alternate proof?

cheers...

Answer 1

Yours is a good proof as any and of course it is correct. You can equate the gcd to any number instead of one.