Prove that $\frac{21n+4}{14n+3}$ is in lowest terms for any natural $n$.

Question

Prove that the fraction $\dfrac{21n+4}{14n+3}$ is in lowest terms for any natural value of $n$.

Answer 1

For starters such as yourself, you can begin by assuming otherwise. That is the fraction is reducible. So $\exists k \in \mathbb{N}, k > 1$ such that $k \mid 21n + 4, k \mid 14n+3\implies 21n+4 = ak, 14n+3 = bk$ for some natural numbers $a,b$. Thus: $42n+8 = 2ak, 42n+9 = 3bk\implies 3bk - 2ak = 1\implies k(3b-2a) = 1\implies k = 1$, contradicting the assumption that $k > 1$. This means $\text{gcd}(21n+4,14n+3) = 1$ or $\dfrac{21n+4}{14n+3}$ is irreducible.

Answer 2

Writing $(n,m)$ for the GCD of $m$ and $n$, immediately one has $(n,m)=(n,m-kn)$ for any $k\in\mathbb{Z}$. Then $$ (21n+4,14n+3)=(7n+1,14n+3)=(7n+1,1)=1. $$

Thus $21n+4$ and $14n+3$ are coprime, so their ratio is in lowest terms.