nphw4 - divides c as required. Here is an alternative...

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Numbers and Polynomials Homework 4 Let a, b, c N with a and b coprime. Show that if a | c and b | c then ab | c . Solution As a, b are coprime, there exist m, n Z such that 1 = ma + nb and therefore c = mac + nbc . As a and b divide c , there exist x and y in N such that c = xa and c = yb , whence subtitution yields c = mayb + nbxa = ( my + nx ) ab. As Z is closed under addition and multiplication, this means that ab
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Unformatted text preview: divides c as required. Here is an alternative (briefer and neater) argument. As above, c = yb for y ∈ N . Now, a divides c = yb and is coprime to b , so a | y ; thus, y = n a for n ∈ N . It follows that c = yb = n ab so that c is a multiple of ab as required. 1...
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This note was uploaded on 11/12/2011 for the course MAS 3300 taught by Professor Staff during the Fall '08 term at University of Florida.

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