(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) Let prime factorisation of $a = p_1 p_2 \dots p_k$. $a^2 = p_1^2 p_2^2 \dots p_k^2$. By uniqueness of FTA, $p$ must be one of $p_1, \dots, p_k$.<br>(ii) Contradiction proof for $\sqrt{3}$.
Stepwise Solution:
(i) Let prime factorisation of $a = p_1 p_2 \dots p_k$, where $p_1, \dots, p_k$ are primes. Then $a^2 = (p_1 p_2 \dots p_k)^2 = p_1^2 p_2^2 \dots p_k^2$. We are given that prime $p$ divides $a^2$. By Fundamental Theorem of Arithmetic, the prime factors of $a^2$ are unique and are $p_1, p_2, \dots, p_k$. So $p$ must be one of $p_1, p_2, \dots, p_k$. Since $a = p_1 p_2 \dots p_k$, $p$ divides $a$. Proved! [3.0 Marks]
(ii) Suppose $\sqrt{3} = \dfrac{x}{y}$ where $x, y \in \mathbb{Z}, y
eq 0$ and $\text{gcd}(x,y) = 1$. Then $x^2 = 3y^2 \Rightarrow 3 \mid x^2 \Rightarrow 3 \mid x$ (by part i). Let $x = 3k$. Then $9k^2 = 3y^2 \Rightarrow y^2 = 3k^2 \Rightarrow 3 \mid y^2 \Rightarrow 3 \mid y$. Thus $3$ is a common factor of $x$ and $y$, contradicting $\text{gcd}(x,y) = 1$. Hence $\sqrt{3}$ is irrational. Proved! [2.0 Marks]
Marking Scheme:
• Part (i) Prime factorisation of $a$ and $a^2$: 1.5 Marks
• Part (i) Uniqueness by FTA to conclude $p \mid a$: 1.5 Marks
• Part (ii) Proof by contradiction for $\sqrt{3}$ using part (i): 2.0 Marks
Correct Answer: