Respuesta :
According to Euclidean's algorithm:
GCD(a, b) = GCD(a-b, b), a > b
GCD(a, b) = GCD(a, b-a), b > a
GCD(2n+7, n+3) = GCD(2n+7-n-3, n+3) = GCD(n+4, n+3) =
= GCD(n+4-n-3, n+3) = GCD(1, n+3) = 1
⇒ The fraction is irreductible.
According to Euclidean's algorithm:
GCD(a, b) = GCD(a-b, b), a > b
GCD(a, b) = GCD(a, b-a), b > a
GCD(2n+7, n+3) = GCD(2n+7-n-3, n+3) = GCD(n+4, n+3) =
= GCD(n+4-n-3, n+3) = GCD(1, n+3) = 1
⇒ The fraction is irreductible.