(i) Prove that if a prime number $p$ divides $a^2$ (where $a$ is a positive integer), then $p$ divides $a$. [3 Marks]
(ii) Use the above theorem to prove that $\sqrt{3}$ is an irrational number. [2 Marks]
Step-by-Step Solution
Key Concept: (i) Unique prime factorisation of $a$ and $a^2$. (ii) Contradiction proof for $\sqrt{3}$.
(i) Let $a = p_1 p_2 \dots p_k$. $a^2 = p_1^2 p_2^2 \dots p_k^2$. Since prime $p \mid a^2$, by FTA uniqueness, $p$ must be one of $p_1, \dots, p_k$. Thus $p \mid a$. [3.0 Marks]
(ii) Let $\sqrt{3} = x/y$ (coprime $x,y$). $x^2 = 3y^2 \Rightarrow 3 \mid x^2 \Rightarrow 3 \mid x$. Let $x = 3k \Rightarrow 9k^2 = 3y^2 \Rightarrow y^2 = 3k^2 \Rightarrow 3 \mid y^2 \Rightarrow 3 \mid y$. Contradicts $\text{gcd}(x,y)=1$. Hence $\sqrt{3}$ is irrational. [2.0 Marks]
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🎯 Official CBSE Marking Scheme:
Part (i) Proof using FTA uniqueness: 3.0 Marks
Part (ii) Contradiction proof for $\sqrt{3}$: 2.0 Marks
Correct Answer: