Question:

Given: We have to show that $\sqrt{3}$ cannot be expressed as a rational number.

To Prove: $\sqrt{3}$ is irrational.

Step 1 – Assume the contrary
Assume that $\sqrt{3}$ is rational. Then there exist two integers $p$ and $q$ (with $q
eq 0$) such that
$$\sqrt{3}=\frac{p}{q}$$
where the fraction $\frac{p}{q}$ is in its lowest terms, i.e., $\gcd(p,q)=1$.

Step 2 – Square both sides
Squaring the equality gives
$$3 = \frac{p^{2}}{q^{2}} \quad \Rightarrow \quad p^{2}=3q^{2}.$$
Thus $p^{2}$ is a multiple of $3$.

Step 3 – Use the property of prime divisibility
If a prime divides a square, it must divide the number itself. Hence, from $3\mid p^{2}$ we conclude $3\mid p$.
Let $p = 3k$ for some integer $k$.

Step 4 – Substitute back
Substituting $p = 3k$ in $p^{2}=3q^{2}$ gives
$$ (3k)^{2}=3q^{2} \quad \Rightarrow \quad 9k^{2}=3q^{2} \quad \Rightarrow \quad 3k^{2}=q^{2}.$$
Thus $q^{2}$ is also a multiple of $3$, which implies $3\mid q$.

Step 5 – Arrive at a contradiction
We have shown that both $p$ and $q$ are divisible by $3$. This contradicts the initial assumption that $p$ and $q$ have no common factor (i.e., $\gcd(p,q)=1$).

Conclusion
The assumption that $\sqrt{3}$ is rational leads to a contradiction. Hence the assumption is false and $\sqrt{3}$ is irrational.

(If the question literally asks to prove that the number 3 itself is irrational, the answer is that 3 is a rational number because it can be expressed as $\frac{3}{1}$. The intended statement in the textbook is to prove that $\sqrt{3}$ is irrational.)