1
$\begingroup$

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...

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.