Proving gcd(ab, b) 1 if gcd(a, b) 1

To prove that if gcd(a, b) 1, then gcd(ab, b) 1, we can use the properties of coprime numbers. Let's delve into the proof step by step.

Definitions

- We say that two integers a and b are coprime or relatively prime if their greatest common divisor gcd is 1, i.e., gcd(a, b) 1.

- Our goal is to show that if gcd(a, b) 1, then gcd(ab, b) 1.

Proof

Assumption: Let d gcd(ab, b). By definition, d divides both ab and b.

Divisibility:

Since d divides b, we can express b as b kd for some integer k. Since d also divides ab, we can write:

[ab md quad text{for some integer } m.]

Rearranging: From ab md, we can rearrange this to find a:

[a frac{ab}{b} frac{md}{kd} frac{m}{k}.]

This shows that a can be expressed as a multiple of d. Thus d divides a as well. Conclusion: Now we have d dividing both a and b. But we started with the assumption that gcd(a, b) 1, which means the only positive integer that divides both a and b is 1. Hence d must be 1.

Final Result: Therefore, we have proved that if gcd(a, b) 1, then gcd(ab, b) 1.

Alternative Proof

Let's consider another method to prove the same statement.

- Let gcd(a, b) d. Then we have:

[d | a quad text{and} quad d | b.]

- This implies that d is a common divisor of both a and b when gcd(a, b) 1.

- Since gcd(a, b) 1, d must be 1. Hence, we have:

[gcd(a, b) 1.]

- Now, if a common divisor x divides both ab and b, then it must also divide the difference a.

- The set of common divisors of a and b is equal to the set of common divisors of ab and b because any divisor of a and b is also a divisor of ab and any divisor of ab and b is also a divisor of a and b.

- Therefore, gcd(a, b) gcd(ab, b) 1.