#<b>Theorem:</b> The cube root of 3 is irrational.<br><b>Proof:</b> Suppose the cube root of 3 is rational. This means the cube root of 3 is $\frac{m}{n}$ for some integers $m,n\ne 0$ with no proper common factor. We need to derive a contradiction. From the assumption we have that $3=\frac{m^3}{n^3}$. Therefore $3n^3=m^3$, so $3|m^3$. Since $3$ is prime, it follows that $3|m$ This means $m=3k$ for some integer $k$. Now $3n^3=(3k)^3$, so $n^3=3(3k^3)$. Therefore $3|n^3$, and again since 3 is prime, $3|n$. This contradicts the assumption that $m,n$ have no proper common factor. Therefore the cube root of 3 is irrational.