If $a,b,c\in\mathbb Z$, then show that $c\cdot\gcd(a,b)\le\gcd(ac,bc)$.
My try: we know that $\gcd(ca,cb)=c\cdot \gcd(a,b)$, but here I don't know what to do.
2026-04-01 14:59:35.1775055575
If $a,b,c\in\mathbb Z$ then show that $c\cdot\gcd(a,b)\leq\gcd(ac,bc)$
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You're wrong when you say $gcd(ac,bc)=c\cdot gcd(a,b)$. The correct equation is $gcd(ac,bc)=|c|\cdot gcd(a,b)$. Do you know how to prove this equation (hint: prime factorization)? From this equation the theorem follows immediately because $c\leq |c|$ holds for all $c\in\mathbb{Z}$