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9. Set Slip [6 points] Proposition 3. Let a, b be integers and let S be a finitely-large set of integers such that 3s ES as and s2 ES b | $2 . Then ab divides the product of all the integers in S. Incorrect Proof. We can list the elements of S as S = {$1, $2, ..., Sc} , where we choose to write s1 (divided by a) first and $2 (divided by b) second. So we have s1 = ak, and $2 = 6k2 for some integers k, and k2. The product of all the integers in S is S18253 . .. Sc = (ak1) (bk2) $3 . . . Sc = (ab) . (kik253 . . . Sc) . Since K1, K2, $3, ..., Sc are all integers, we have that Kik2$3 ... S is an integer, which shows that ab divides the product of the integers in S