So I had proved that given $\gcd(a, b) = g$, if $ap + bq = g$, then $\gcd(p, q) = 1$
meaning bezout's coefficients are coprime.
I was then asked to show that if you wrote $$a = a'g, \ \ \ \ \ \ b = b'g$$ then $\gcd(a',b') = 1$ I'm not really sure how to go about this.
I currently have written out that $a = a'(gcd(a,b))$ and $b = b'(gcd(a,b))$. This means that $a'$ is a different factor of $a$ and $b'$ is a different factor of $b$. Any idea where I could go from here?
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Let $a$, $b$ be two positive integers. If $\gcd(a,b)=g$, then $g$ is the greatest positive integer to divide into both $a$ and $b$. Hence we may write $a=a'g$, $b=b'g$, for $a',\, b'\in\mathbb{Z}_{\geq1}$.
Now show that $\gcd(a',b')=1$. To this end assume some $f\in\mathbb{Z}_{>1}$ divides both $a'$ and $b'$. Then it would surely divide $a$ and $b$ and hence divide $g$, since $\gcd(a,b)=g$. Hence you have a contradiction and $f$ cannot exist, and so $\gcd(a',b')=1$.
substituting $a = a'g$, $b = b'g$ into $ap + bq = g$ you will get $a'p + b'q = 1$ which is necessary and sufficient condition for $\gcd(a',b')=1$.