Let a and b be natural numbers that are co-prime. Prove that (b-a) and b must also be co-prime

Question

If gcd(a,b)=1 then gcd(b-a,b)=1. I'm kind of lost... Any hint or advice on how to start this proof?

Thanks!

Answer 1

Suppose $d$ divides $b-a$ and also divides $b$. This means that $d$ divides their diffence $b- (b-a)=a$ as well. So $d$ divides $a$ and $b$ hence their gcd...

Answer 2

Let $d=(b-a,b)$. Then $d | b$ and $d | a$. Since the only positive divisor of $a$ and $b$ is $1$, $d=1$.

Answer 3

Let $d=\gcd(b-a,b)$. Suppose $b-a=dc_1$ and $b=dc_2$， where $\gcd(c_1,c_2)=1$. Then $$a=-(b-a)+b=-dc_1+dc_2=d(c_2-c_1).$$ So $d\mid a$ and $d\mid b$. It follows from the definition of $\gcd$ that $d\mid \gcd(a,b)=1$. It must have $d=1$, i.e., $\gcd(b-a,b)=1$.