1. **Problem statement:** Prove that for any positive integer $n$, the fraction $\frac{21n + 4}{14n + 3}$ is irreducible, meaning it cannot be simplified further. 2. **Key idea:** A fraction $\frac{a}{b}$ is irreducible if and only if the greatest common divisor (gcd) of $a$ and $b$ is 1, i.e., $\gcd(a,b) = 1$. 3. **Apply to our fraction:** We want to show that $\gcd(21n + 4, 14n + 3) = 1$ for all positive integers $n$. 4. **Use the Euclidean algorithm:** $$\gcd(21n + 4, 14n + 3) = \gcd(14n + 3, (21n + 4) - (14n + 3))$$ Calculate the difference: $$(21n + 4) - (14n + 3) = 7n + 1$$ So, $$\gcd(21n + 4, 14n + 3) = \gcd(14n + 3, 7n + 1)$$ 5. **Repeat Euclidean algorithm:** $$\gcd(14n + 3, 7n + 1) = \gcd(7n + 1, (14n + 3) - 2(7n + 1))$$ Calculate the subtraction: $$(14n + 3) - 2(7n + 1) = 14n + 3 - 14n - 2 = 1$$ So, $$\gcd(14n + 3, 7n + 1) = \gcd(7n + 1, 1)$$ 6. **Since $\gcd(x,1) = 1$ for any integer $x$, we have:** $$\gcd(7n + 1, 1) = 1$$ 7. **Conclusion:** Since the gcd reduces to 1, the original fraction $\frac{21n + 4}{14n + 3}$ is irreducible for all positive integers $n$. **Final answer:** The fraction $\frac{21n + 4}{14n + 3}$ cannot be simplified for any positive integer $n$ because $\gcd(21n + 4, 14n + 3) = 1$.