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anonymous
 4 years ago
"suppose \(c(a+b)\) where \(a,b,c\in\mathbb{Z}\). then \(c(pa+qb)\) where \(p,q\in\mathbb{Z}\)."
can you really make such assumption!?
anonymous
 4 years ago
"suppose \(c(a+b)\) where \(a,b,c\in\mathbb{Z}\). then \(c(pa+qb)\) where \(p,q\in\mathbb{Z}\)." can you really make such assumption!?

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0"for SOME integers \(p\) and \(q\)" i meant to add

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i don't seem to get why x.x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No we can't make such assumption.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0like a=2, b=4, c=3 c(a+b) > 36 which is true but c(2a+3b) > 316 is false :(

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.3\[ c  (a+b) \implies c  (1\cdot a + 1\cdot b) \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(3(4+5)\) but 3 does not divide \((4\times 3+5 \times 5) \)

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.3so if the question is there exist at least one pair p,q such your statement is true, then yes.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But this question is asking for for all \(p,q \in \mathbb{Z}\), hence incorrect.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.3ffm, prealgebra says immediately below "for SOME p and q ..."

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hey, EDIT feature is a must!!!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have a tendency not to read any comment/answer before trying it on my own.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0"for SOME integers p and q" that's true.
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