91Ó°ÊÓ

For all integers \(a, b,\) and \(c,\) if \(a \mid c\) and \(b \mid c,\) then \((ab) \mid c\). We can prove this proposition false by providing a counterexample. Let's consider the integers \(a = 2, b = 3, c = 6\). \(a \mid c\) implies \(2 \mid 6\) which is true, as \(\frac{6}{2} = 3\). \(b \mid c\) implies \(3 \mid 6\) which is also true, as \(\frac{6}{3} = 2\). Now let's verify if \((a b) \mid c,\) which means if \((2 \cdot 3) \mid 6\) or \(6 \mid 6\). While this is technically true since \(\frac{6}{6} = 1\), we can find another counterexample to show that this proposition is not always valid. Let's consider \(a = 2, b = 4, c = 8\). \(a \mid c\) implies \(2 \mid 8\) which is true, as \(\frac{8}{2} = 4\). \(b \mid c\) implies \(4 \mid 8\) which is also true, as \(\frac{8}{4} = 2\). Now let's verify if \((a b) \mid c,\) which means if \((2 \cdot 4) \mid 8\) or \(8 \mid 16\). This is false since \(\frac{16}{8} = 2\), which means that 8 does not divide 16 evenly. Thus, the proposition (a) is false.

For all integers \(a, b,\) and \(c,\) if \(a \mid c, b \mid c\) and \(\operatorname{gcd}(a, b) = 1,\) then \((ab) \mid c\). To determine the truthfulness of this statement, let's first analyze the given conditions. The condition \(\operatorname{gcd}(a, b) = 1\) means that \(a\) and \(b\) are coprime, that is, their greatest common divisor is 1. This implies that they do not share any common factors other than 1. Now, let's prove the proposition (b). Since \(a \mid c,\) we can write \(c= ak\) for some integer \(k\); and since \(b \mid c,\) we can also write \(c = bl\) for some integer \(l\). We are given that \(\operatorname{gcd}(a,b) = 1\). Using Bézout's identity, we can write it as the linear combination of a and b: \(ax + by = 1\), where x and y are some integers. Now, let's multiply both sides of this equation by \(c\): \(a(cx) + b(cy) = c\) Our goal is to prove that \((ab) \mid c\), so let's rewrite the equation above as: \(acx + bcy = c\) Now, notice that \(ab \mid acx\) and \(ab \mid bcy\) because a multiple of \(a\) or \(b\) multiplied by any other integer yields a multiple of \(ab\). Therefore, since both terms on the left side of the equation are divisible by \(ab\), their sum must also be divisible by \(ab\); so we can conclude that \(ab \mid c\). Thus, the proposition (b) is true.