Let m; n; z 2 N such that m and n divide z and gcd(m; n) = 1. Prove that m � n divides z.
To prove that m * n divides z when m and n are natural numbers that divide z and when gcd(m, n) = 1, we can use the following approach:
1. Let's assume that m and n are natural numbers that divide z, which means that z is divisible by both m and n. Therefore, we can express z as z = m * x and z = n * y, where x and y are integers.
2. We are given that gcd(m, n) = 1, which means that m and n have no common factors other than 1. In other words, their greatest common divisor is 1.
3. Since m and n have no common factors other than 1, it implies that m and n are coprime or relatively prime.
4. Now, let's consider the product of m * n. Since m and n are coprime, their product m * n will also have no common factors other than 1.
5. Now, let's consider the number z = m * x = n * y. Since z is equal to both m * x and n * y, we can equate these two expressions: m * x = n * y.
6. Since m and n are coprime, it means that m does not divide n and n does not divide m. Therefore, the only way for the equation m * x = n * y to hold true is if m divides y and n divides x.
7. Since m divides y and n divides x, it implies that y can be written as y = m * p and x can be written as x = n * q, where p and q are integers.
8. Substituting the values of y and x in the equation m * x = n * y, we get m * (n * q) = n * (m * p).
9. By simplifying the equation, we find that m * n * q = m * n * p.
10. Canceling out m * n from both sides of the equation, we are left with q = p.
11. Since q = p, it means that both x and y are divisible by both m and n.
12. Therefore, we can conclude that m * n divides z, as z can be expressed as z = m * x = m * (n * q) = (m * n) * q, where q is an integer.
Hence, we have proved that if m and n are natural numbers that divide z and gcd(m, n) = 1, then m * n divides z.