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(a) Prove that $\sqrt{5}$ is an irrational number.
OR
(b) State the "Fundamental Theorem of Arithmetic" and use it to find LCM of $36$ and $54$.
OR
(b) State the "Fundamental Theorem of Arithmetic" and use it to find LCM of $36$ and $54$.
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Solution: (a) Let $\sqrt{5}$ be a rational number such that $\sqrt{5} = \frac{p}{q}$ ($p$ and $q$ are co-prime numbers, $q \neq 0$) [1/2 mark]
$\sqrt{5}q = p \Rightarrow 5q^2 = p^2$
$5$ divides $p^2 \Rightarrow 5$ divides $p$ as well [1 mark]
$p = 5m$ (for some integer $m$)
$5q^2 = 25m^2 \Rightarrow q^2 = 5m^2$
$5$ divides $q^2 \Rightarrow 5$ divides $q$ as well [1 mark]
$p$ and $q$ have a common factor $5$ which is a contradiction as $p$ and $q$ are co-prime.
$\therefore \text{our assumption is wrong}$
Hence, $\sqrt{5}$ is an irrational number [1/2 mark]
OR
(b) Statement: "Every composite number can be factorized as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur." [1 mark]
$36 = 2^2 \times 3^2$ [1/2 mark]
$54 = 2 \times 3^3$ [1/2 mark]
$\text{LCM}(36, 54) = 2^2 \times 3^3$ or $108$ [1 mark]
$\sqrt{5}q = p \Rightarrow 5q^2 = p^2$
$5$ divides $p^2 \Rightarrow 5$ divides $p$ as well [1 mark]
$p = 5m$ (for some integer $m$)
$5q^2 = 25m^2 \Rightarrow q^2 = 5m^2$
$5$ divides $q^2 \Rightarrow 5$ divides $q$ as well [1 mark]
$p$ and $q$ have a common factor $5$ which is a contradiction as $p$ and $q$ are co-prime.
$\therefore \text{our assumption is wrong}$
Hence, $\sqrt{5}$ is an irrational number [1/2 mark]
OR
(b) Statement: "Every composite number can be factorized as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur." [1 mark]
$36 = 2^2 \times 3^2$ [1/2 mark]
$54 = 2 \times 3^3$ [1/2 mark]
$\text{LCM}(36, 54) = 2^2 \times 3^3$ or $108$ [1 mark]