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Calculus and Beyond Homework Help
Does the GCD of a set divide the GCD of any subset?
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[QUOTE="andrewkirk, post: 5477150, member: 265790"] It's not the most elegant way, but it's quick and easy: try proof by contradiction. Let ##a=gcd(A)## so that ##\forall x\in A:\ a|x## and ##b=gcd(B)## so that ##\forall x\in B:\ b|x## Assume that ##a\nmid b##. By looking at prime power divisors of ##a## that are not divisors of ##b## you should be able to find a way of making a common divisor of ##B## that is bigger than ##b##, thereby contradicting the supposition that ##b## is a GCD of ##B##. [/QUOTE]
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Calculus and Beyond Homework Help
Does the GCD of a set divide the GCD of any subset?
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