Question:

Stepwise Solution:

Let us assume, on the contrary, that $\sqrt{2}$ is rational. Then $\sqrt{2} = \dfrac{p}{q}$, where $p, q \in \mathbb{Z}, q
eq 0$, and $p, q$ are coprime (have no common factor other than $1$). [0.5 Mark]

Squaring both sides: $2 = \dfrac{p^2}{q^2} \Rightarrow p^2 = 2q^2$. This means $2$ divides $p^2$, so $2$ divides $p$ (by theorem: if prime $p$ divides $a^2$, then $p$ divides $a$). [1.0 Mark]

Let $p = 2k$ for some integer $k$. Substituting $p = 2k$ into $p^2 = 2q^2$: $(2k)^2 = 2q^2 \Rightarrow 4k^2 = 2q^2 \Rightarrow q^2 = 2k^2$. This means $2$ divides $q^2$, so $2$ divides $q$. [1.0 Mark]

Thus, $2$ is a common factor of both $p$ and $q$. But this contradicts our assumption that $p$ and $q$ are coprime. Hence, our assumption is false, so $\sqrt{2}$ is irrational. [0.5 Mark]

Marking Scheme:

• Initial contradiction assumption and coprime setup: 0.5 Mark
• Proving $p$ is divisible by 2: 1.0 Mark
• Proving $q$ is divisible by 2: 1.0 Mark
• Contradiction statement and final conclusion: 0.5 Mark