**Step 1 — Assume the contrary (method of contradiction).**
Let us assume that $\sqrt5$ is **rational**.

Then $\sqrt5 = \dfrac{a}{b}$, where $a$ and $b$ are co-prime integers ($\text{HCF}(a,b)=1$) and $b \neq 0$.

**Step 2 — Rearrange and square.**
$$\sqrt5 \, b = a \ \Rightarrow\ 5b^2 = a^2 \qquad \dots(i)$$

**Step 3 — Conclude that 5 divides $a$.**
From $(i)$, $5$ divides $a^2$.
$\Rightarrow$ $5$ divides $a$ (since 5 is prime).

So let $a = 5c$ for some integer $c$.

**Step 4 — Substitute back into (i).**
$$5b^2 = (5c)^2 = 25c^2 \ \Rightarrow\ b^2 = 5c^2$$

**Step 5 — Conclude that 5 divides $b$.**
$5$ divides $b^2$ $\Rightarrow$ $5$ divides $b$.

**Step 6 — Reach the contradiction.**
Now $5$ is a common factor of both $a$ and $b$.
This contradicts our assumption that $a$ and $b$ are co-prime.

**Step 7 — Conclusion.**
Our assumption is wrong.

$$\therefore\ \sqrt5 \text{ is an irrational number. } \textbf{Hence proved.}$$