Question

Let a, b, c and n be natural numbers and LCM(a, b) = m. Prove that

(a) if a divides n and b divides n, then m less than or equal to n.
(b) LCM(a, b) ≤ ab.
(c) if c divides a and c divides 6, then LCM(a/c, b/c) = m/c (d) It can be shown that GCD(a, b) can be written in the form ax + by, for some integers x and y. Use this to prove Euclid's Lemma: If a divides be and GCD(a, b) = 1, then a divides c.
(e) if GCD(a, b) = 1, then LCM(a, b) = ab.
(f) if GCD(a, b) = d, then GCD(a/d, b/d) = 1
(g) LCM(a, b) GCD(a, b) = ab.

Answer 1

The student's question relates to several properties of the Least Common Multiple (LCM) and Greatest Common Divisor (GCD) in number theory, which involve proving inequalities and identities related to divisibility and common factors of natural numbers.

Step-by-step explanation:

The questions provided embrace a variety of fundamental theorems in number theory concerning the Least Common Multiple (LCM) and Greatest Common Divisor (GCD). Let's address some of the proofs mentioned:

• a ≤ n: If a divides n and b divides n, then m ≤ n because m is the smallest multiple that both a and b divide.
• LCM ≤ ab: The LCM of two numbers a and b will always be less than or equal to their product since the LCM takes only the highest powers of common factors.
• If c divides a and divides b, then LCM(a/c, b/c) = m/c follows because division by c is essentially 'scaling down' by a common factor present in both a and b.
• Euclid's Lemma: If a divides bc and GCD(a, b) = 1, a must divide c. This stems from the fact that a and b have no common factors, and thus any factor of a cannot be 'canceled out' by b.
• If GCD(a, b) = 1, then a and b are co-prime, so their LCM is their product: LCM(a, b) = ab.
• If GCD(a, b) = d, after dividing both a and b by d, no common factors remain, thus GCD(a/d, b/d) = 1.
• Finally, LCM(a, b) × GCD(a, b) = ab is a fundamental theorem which states that the product of the LCM and GCD of two numbers equals the product of those numbers.