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    • Thread Starter
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    I have in my notes:

    a divides y, b divides y, gcd(a,b) = 1 thus ab divides y.

    Can someone explain this to me.

    Thanks
    • PS Helper
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    Suppose a = p_1^{k_1}p_2^{k_2}...p_n^{k_n} and b = q_1^{l_1}q_2^{l_2}...q_m^{l_m}. Then since \gcd(a,b) = 1, we have that each of the p_i, q_j are distinct (for all i,j). If a|y then y = \alpha p_1^{k_1}p_2^{k_2}...p_n^{k_n} and if b|y then y = \beta q_1^{l_1}q_2^{l_2}...q_m^{l_m} (for some constant integers \alpha, \beta), but because of unique prime factorisation we must have \alpha = \gamma q_1^{l_1}q_2^{l_2}...q_m^{l_m} (for some constant integer \gamma), and hence y = \gamma \beta p_1^{k_1}p_2^{k_2}...p_n^{k_n} q_1^{l_1}q_2^{l_2}...q_m^{l_m} = \lambda a b (for some constant integer \lambda, and hence ab|y.

    This is a bit messy and there's probably a neater way of showing it than this, but you get the idea.
 
 
 
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Updated: January 21, 2010
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