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(a) Prove that $\sqrt{3}$ is an irrational number.
OR
(b) The factor tree of a number $x$ is shown below :
Find the values of $x, y, a$ and $b$. Hence, write the product of the prime factors of the number $x$ so obtained.
OR
(b) The factor tree of a number $x$ is shown below :
Find the values of $x, y, a$ and $b$. Hence, write the product of the prime factors of the number $x$ so obtained.
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Solution: (a) Let $\sqrt{3}$ be a rational number such that $\sqrt{3} = \frac{p}{q}$ ($p$ and $q$ are co-prime numbers, $q \neq 0$)
$\sqrt{3}q = p \Rightarrow 3q^2 = p^2$
$3$ divides $p^2 \Rightarrow 3$ divides $p$ as well
Let, $p = 3m$ (for some integer $m$)
$3q^2 = 9m^2 \Rightarrow q^2 = 3m^2$
$3$ divides $q^2 \Rightarrow 3$ divides $q$ as well
$p$ and $q$ have a common factor $3$, which is a contradiction as $p$ and $q$ are co-prime.
$\therefore$ our assumption is wrong
Hence, $\sqrt{3}$ is an irrational number
OR
(b) $b = 7$
$a = 3$
$y = 420$
$x = 840$
$x = 840 = 2^3 \times 3 \times 5 \times 7$
$\sqrt{3}q = p \Rightarrow 3q^2 = p^2$
$3$ divides $p^2 \Rightarrow 3$ divides $p$ as well
Let, $p = 3m$ (for some integer $m$)
$3q^2 = 9m^2 \Rightarrow q^2 = 3m^2$
$3$ divides $q^2 \Rightarrow 3$ divides $q$ as well
$p$ and $q$ have a common factor $3$, which is a contradiction as $p$ and $q$ are co-prime.
$\therefore$ our assumption is wrong
Hence, $\sqrt{3}$ is an irrational number
OR
(b) $b = 7$
$a = 3$
$y = 420$
$x = 840$
$x = 840 = 2^3 \times 3 \times 5 \times 7$