Show that √5 is irrational. Why does the same argument fail for √9?

Step 1. Suppose. $\sqrt{5} = \dfrac{p}{q}$, with $p$ and $q$ sharing no factor. Then $p^2 = 5q^2$.

Step 2. $p$. $5$ divides $p^2$, and $5$ is prime, so $5$ divides $p$. Write $p = 5k$.

Step 3. $q$. Then $25k^2 = 5q^2$, so $q^2 = 5k^2$, and $5$ divides $q$ as well.

Step 4. Contradiction. $p$ and $q$ share the factor $5$, against what we assumed. So $\sqrt{5}$ is irrational.

Step 5. Why not $\sqrt{9}$? $p^2 = 9q^2$ only tells us $3$ divides $p$; there is no contradiction, since $\sqrt{9} = \dfrac{3}{1}$ really is a fraction.