4. (god and Icm) Prove the following statements. ON a For a, b, d E Z+, if da, db, and ab, then 212 a b) For a, b, k E Z+, if akb, then k. god(a, b) c) For all a, b E Zt, we have Icm(a, b) = kb, where k = min {j EZ+ | aljb} . d) For a, b E Z+, we have god(a, b) . Icm(a, b) = ab.a) For a, b, d E Z+, ifd a, d b, and a b, then RIP Proof: 1. Since d | a and d |b, there exist integers m and n such that a = md and b = nd. 2. Given a | b, then b = ka for some integer k. 3. Substituting a from step 1 into the equation b = ka, we get: nd = k(md) Simplifying, we find: n = km 4. Hence, = 1 nd = n and = md = m. 5. From n = km, it follows that n is a multiple of m, thus m | n or a | 3.Statement (b) For a, b, k E Z+, if a | kb, then gca god (a, b) | k. Proof: We start by establishing a relationship between a, b, and & using the property of divisibility and the greatest common divisor (god). Step 1: Express a and b in terms of their god . Let g = gcd(a, b). . Then, we can express a and b in terms of g such that: a = ga', b =gb' where a' and b' are integers such that god(a', b') = 1. This is because any common factor of a' and b' multiplied by g would contradict the definition of g as the god of a and b. Step 2: Simplify the given divisibility condition . Given that a | kb, it implies: ga' | k(gb') . Simplifying, we get: a | kb'Step 3: Apply properties of coprime numbers Since god(a', b') = 1 (i.e., a' and b' are coprime), the property of coprimality ensures that a' dividing kb' implies a' | k. This follows from a key property of coprime numbers where if a my for coprime x and y, then x | m. Step 4: Relate the conclusion back to the original numbers . From a' | k and since a' = & or a' = a ged(a,b) . we conclude that: a god(a, b) I k Conclusion: Thus, god(a,b) , which simplifies a by removing the common factors it shares with b, indeed divides k. This relationship reflects how the factors of a that are independent of b (given by a ) must be factors of k, emphasizing the role of god in simplifying the divisibility conditions between numbers. This proof leverages the fundamental theorem of arithmetic and properties of god and Icm to establish a clear and rigorous conclusion.Statement (c) For all a, b E Z+, we have Icm(a, b) = kb, where k = min{je Z+laj | b} Proof: To prove this statement, we need to establish the relationship between k, a, and b and demonstrate how k is related to the least common multiple (LCM). Step 1: Definition and setup Define k as k = min{j E Z+laj | b}. This means k is the smallest positive integer for which aj is a multiple of b. Step 2: Properties of k and divisibility By definition, ak | b. Therefore, b = m(ak) for some integer m. . Since b = m(ak), it's evident that ak is a common multiple of a and b. Step 3: Establishing ak as the least common multiple To show that ak is the least common multiple of a and b, we must prove that any other common multiple of a and b is a multiple of ak. . Let c be another common multiple of a and b. Then, c = xa and c = yb for some integers x, y Since ak divides b (from step 2), and b | c (since c = yb), it follows that ak divides c.Step 4: Minimal property of &k e The minimality of k implies no smaller j exists such that aj | b, making ak the smallest (least) common multiple satisfying this condition. If there were a smaller common multiple, say aj, with 7