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anonymous
 4 years ago
Could somebody help me understand this question?
"Show that if \(a_1, a_2, ..., a_n\) are pairwise relatively prime integers, then \([a_1, a_2, ..., a_n]=a_1a_2...a_n\)."
anonymous
 4 years ago
Could somebody help me understand this question? "Show that if \(a_1, a_2, ..., a_n\) are pairwise relatively prime integers, then \([a_1, a_2, ..., a_n]=a_1a_2...a_n\)."

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JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1what's the definition of [a_1, ..., a_n] ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That is the least common multiple of those numbers.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1right. I'm sure there's a couple of ways to do this. But I would use prime factorization. Suppose \[ x = p_1^{a_1}p_2^{a_2}...p_n^{a_n}, \ \ y = p_1^{b_1}p_2^{b_2}...p_n^{b_n} \] where the \( p_i \) are distinct prime numbers and the \[ a_1, a_2, ... a_n, b_1, ..., b_n \] are nonnegative integers. Then the l.c.m. of x and y \[ [x,y] = \Pi_i \ p_i^{\max(a_i,b_i)} \] For example, \( 12 = 2^23^1 \) and \( 50 = 2^15^2 \), then \[ [12,5] = 2^23^15^2 = 300 \] Now, use that definition for your problem and it should be straightforward.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Perfect. I'll try that. Thank you very much.
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