Question

Let a and b be positive integers. Then ab = gcd(a, b)∙lcm(a, b). prove it and show the intermediate steps . [Hint: Use the prime factorizations of a and b and the formulae for gcd(a, b) and lcm(a, b) in terms of these factorizations.

Answer 1

To prove ab = gcd(a, b) * lcm(a, b) for positive integers a and b, we examine their prime factorizations. The gcd contains prime factors at their minimum power, while lcm contains them at maximum power. Multiplying gcd and lcm together, each prime factor regains its original power in a and b, proving the original statement.

Step-by-step explanation:

To prove the statement ab = gcd(a, b) * lcm(a, b), we will use the prime factorizations of a and b along with the formulas for gcd(a, b) and lcm(a, b).

Let's start by expressing a and b in terms of their prime factorizations:

a = p1a1 * p2a2 * ... * pnan

b = p1b1 * p2b2 * ... * pnbn

Where pi are distinct prime numbers and ai and bi are positive integers.

The greatest common divisor, gcd(a, b), can be obtained by taking the minimum of the exponents for each prime factor:

gcd(a, b) = p1min(a1, b1) * p2min(a2, b2) * ... * pnmin(an, bn)

The least common multiple, lcm(a, b), can be obtained by taking the maximum of the exponents for each prime factor:

lcm(a, b) = p1max(a1, b1) * p2max(a2, b2) * ... * pnmax(an, bn)

Multiplying gcd(a, b) and lcm(a, b) together:

gcd(a, b) * lcm(a, b) = (p1min(a1, b1) * p2min(a2, b2) * ... * pnmin(an, bn)) * (p1max(a1, b1) * p2max(a2, b2) * ... * pnmax(an, bn))

Using the rules of exponents, we can rewrite the expression as:

gcd(a, b) * lcm(a, b) = p1min(a1, b1) + max(a1, b1) * p2min(a2, b2) + max(a2, b2) * ... * pnmin(an, bn) + max(an, bn)

Since min(x, y) + max(x, y) = x + y, we can simplify further:

gcd(a, b) * lcm(a, b) = p1a1 + b1 * p2a2 + b2 * ... * pnan + bn

Which is equal to ab. Therefore, we have proven that ab = gcd(a, b) * lcm(a, b).

Answer 2

To prove ab = gcd(a, b) * lcm(a, b) for positive integers a and b, we examine their prime factorizations. The gcd contains prime factors at their minimum power, while lcm contains them at maximum power. Multiplying gcd and lcm together, each prime factor regains its original power in a and b, proving the original statement.

Step-by-step explanation:

To prove that for any two positive integers a and b, the product ab is equal to the product of their greatest common divisor (gcd(a, b)) and their least common multiple (lcm(a, b)), we can use their prime factorizations. Let's say that the prime factorization of a is p1^x * p2^y * ... and the prime factorization of b is p1^s * p2^t * ... where p1, p2, ... are prime factors and x, y, s, t, ... are their respective powers.

By definition, gcd(a, b) is the product of common prime factors raised to the minimum powers found in both a and b. So it would have the prime factorization p1^min(x,s) * p2^min(y,t) * .... On the other hand, lcm(a, b) is the product of the prime factors raised to their maximum powers in either a or b, thus having the prime factorization p1^max(x,s) * p2^max(y,t) * .... When we multiply the gcd and lcm, each prime factor p1, p2, ... is raised to the power of min(x,s) + max(x,s), which simplifies to x + s, and the same applies for all other prime factors.

Observing the multiplication, we see that it restores each prime factor to its respective power as it appears in the factorization of both a and b. Therefore, the multiplication of gcd(a, b) and lcm(a, b) results in the original numbers a and b multiplied together, leading to the conclusion that ab = gcd(a, b) * lcm(a, b).