Theorem says that if a | b or a | c, then a | bc. Is the converse true; that is, if a | bc, then a | b or a | c? Justify your conclusion.
Theorem
Let a, b, and c be integers.
(a) If a | b and a | c, then a | (b + c).
(b) If a | b and a | c, where b > c, then a | (b − c).
(c) If a | b or a | c, then a | bc.
(d) If a | b and b | c, then a | c.
Proof
(a) If a | b and a | c, then b = k1a and c = k2a for integers k1 and k2. So b + c = (k1 + k2)a and a | (b + c).
(b) This can be proved in exactly the same way as (a).
(c) As in (a), we have b = k1a or c = k2a. Then either bc = k1ac or bc = k2ab, so in either case bc is a multiple of a and a | bc.
(d) If a | b and b | c, we have b = k1a and c = k2b, so c = k2b = k2(k1a) = (k2k1)a and hence a | c.
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