Question:

Given: We have to show that $\sqrt{5}$ cannot be expressed as a rational number.

To Prove: $\sqrt{5}$ is irrational.

Step 1 – Assume the contrary
Assume that $\sqrt{5}$ is rational. Then it can be written as a fraction of two integers having no common factor other than 1 (i.e., in lowest terms):
$$\sqrt{5}=\frac{p}{q},\qquad p,q\in\mathbb{Z},\; q
eq0,\; \gcd(p,q)=1.$$

Step 2 – Square both sides
Squaring gives
$$5 = \frac{p^{2}}{q^{2}} \;\Rightarrow\; p^{2}=5q^{2}.$$
Thus $p^{2}$ is a multiple of 5, which implies that $p$ itself must be a multiple of 5 (if a prime divides a square, it divides the base). Let $p=5k$ for some integer $k$.

Step 3 – Substitute back
Substituting $p=5k$ in $p^{2}=5q^{2}$:
$$ (5k)^{2}=5q^{2}\;\Rightarrow\;25k^{2}=5q^{2}\;\Rightarrow\;5k^{2}=q^{2}.$$
Hence $q^{2}$ is also a multiple of 5, so $q$ must be a multiple of 5.

Step 4 – Contradiction
Both $p$ and $q$ are multiples of 5, which means they have a common factor 5. This contradicts the assumption that $p$ and $q$ are coprime (i.e., the fraction is in lowest terms).

Conclusion
The assumption that $\sqrt{5}$ is rational leads to a contradiction. Therefore, $\sqrt{5}$ is irrational.