- #1

Math100

- 676

- 180

- Homework Statement:
- Prove that if gcd(a, b)=1, then gcd(a+b, ab)=1.

- Relevant Equations:
- None.

Proof: Suppose for the sake of contradiction that gcd(a, b) \neq 1.

Then there exists a prime number k that divides both a+b and ab.

Note that k divides either a or b.

Since k divides a+b,

it follows that k divides b.

Thus, this is a contradiction because a and b are relatively prime.

Therefore, if gcd(a, b)=1, then gcd(a+b, ab)=1.

Then there exists a prime number k that divides both a+b and ab.

Note that k divides either a or b.

Since k divides a+b,

it follows that k divides b.

Thus, this is a contradiction because a and b are relatively prime.

Therefore, if gcd(a, b)=1, then gcd(a+b, ab)=1.