Proof:

Assume, to the contrary, that $\sqrt{5}$ is rational.

Then we can write $\sqrt{5} = a/b$, where $a$ and $b$ are coprime integers and $b \neq 0$.

Squaring both sides:

$$5 = \frac{a^2}{b^2}$$

Therefore,

$$a^2 = 5b^2$$

This means 5 divides $a^2$.

By Theorem 1.2: If a prime $p$ divides $a^2$, then $p$ divides $a$.

Since 5 divides $a^2$, it follows that 5 divides $a$.

Let $a = 5c$ for some integer $c$.

Substituting in $a^2 = 5b^2$:

$$(5c)^2 = 5b^2$$ $$25c^2 = 5b^2$$ $$5c^2 = b^2$$

This means 5 divides $b^2$.

Again by Theorem 1.2, 5 divides $b$.

Contradiction: We have shown that 5 divides both $a$ and $b$.

But this contradicts our assumption that $a$ and $b$ are coprime (have no common factor other than 1).

This contradiction has arisen because of our incorrect assumption that $\sqrt{5}$ is rational.

∴ Conclusion: $\sqrt{5}$ is irrational.