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Subject: "Properties of GCD and LCM: Lemma proof"     Previous Topic | Next Topic
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John
guest
Sep-25-10, 00:08 AM (EST)
 
"Properties of GCD and LCM: Lemma proof"
 
   I'm confused by the proof of the Lemma for the Chinese Remainder Theorem on the page titled "Properties of GCD and LCM." The Lemma is (or the part that's proven)
"lcm(gcd(N1, M), gcd(N2, M), ..., gcd(Nk, M)) = gcd(lcm(N1, ..., Nk), M)."

I know I'm missing something, but I don't see how it was actaully proven. The proof given forces me to conclude only this: that if a prime p dvides the left hand side, then it divides the right hand side--and vice versa. But this is not the same as the left hand side equals the right hand side, which is what needs to be proven. Could someone explain why the proof works?


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alexbadmin
Charter Member
2620 posts
Sep-26-10, 00:34 AM (EST)
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1. "RE: Properties of GCD and LCM: Lemma proof"
In response to message #0
 
   John, you are right here and the proof needs a fixing. Will do that tomorrow.

When p is composite, it could be factored (not uniquely) into divisors of N_i (and, naturally, M). For every prime divisor q of p there is at least one i such that N_i is divisible by the highest power of q in lcm. So, for every i, N_i is associated with the product of maximum powers of q's.

Something's along these lines.


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alexbadmin
Charter Member
2620 posts
Sep-26-10, 10:08 AM (EST)
Click to EMail alexb Click to send private message to alexb Click to view user profileClick to add this user to your buddy list  
2. "RE: Properties of GCD and LCM: Lemma proof"
In response to message #1
 
   I have modified https://www.cut-the-knot.org/arithmetic/GcdLcmProperties.shtml#lemma

In fact the only change that was needed was replacing prime p with a power pα.

LCM of several numbers is divisible by pα iff one of the numbers is. One needs only to pick the number with the largest α.

Thank you for pointing out the mistake.


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