if d = gcd(a, a_1 a_2...a_n) and d l gcd(d,a_1) gcd(d,a_2)...gcd(d,a_n) then why is it that d l gcd(a,a_1) gcd(a,a_2)...gcd(a,a_n)?
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Hi. Do you want to show that $\displaystyle d|gcd(a,a_1)\cdot{gcd(a,a_2)}...\cdot{gcd(a,a_n)}$? Best regards.
yes. i know that d l a and dl a_1...a_n but how do i know that d l gcd( a, a_1) ...gcd(a, a_n)
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