How to show that gcd(|a|,|b|) = gcd(a,b) ? I tried many ways but not convinced that im doing it right. i tried to look at different scenarios where :
a,b>0
a,b<0
a>0,b<0
but I know I am doing it totally wrong.
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2Hinr. $a$ and $|a|$ have exactly the same divisors. – Ethan Bolker Oct 01 '18 at 00:14
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Same as here in the linked dupe, i.e. since $,d\mid |x|\iff d\mid x,$ we infer that $|a|,|b|$ and $,a,b,$ have the same set of common divisors so they have the same greatest common divisor. – Bill Dubuque Nov 25 '23 at 18:12
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Show that every common divisor of $a$ and $b$ is also a common divisor of $|a|$ and $|b|$. Then vice verse. Since the set of common divisors of $a$ and $b$ is the same as the set of common divisors of $|a|$ and $|b|$, the largest element in each set has to be the same.
Another way is to invoke Bezout. If $d=\gcd(a,b)$ then there exist integers $x$ and $y$ so that $d=ax+by$. By adjusting the signs on $x$ and $y$ you can get a similar expression for $\gcd(|a|,|b|).$
B. Goddard
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