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(a) Prove that $\sqrt{2}$ is an irrational number.
OR
(b) Find which among the following numbers $a, b$ and $c$ is/are composite numbers.
$a = 7 \times 11 \times 13 + 13$
$b = 6 \times 5 \times 4 + 4$
$c = 7 \times 13 + 6$
OR
(b) Find which among the following numbers $a, b$ and $c$ is/are composite numbers.
$a = 7 \times 11 \times 13 + 13$
$b = 6 \times 5 \times 4 + 4$
$c = 7 \times 13 + 6$
Show SolutionHide Solution↓
(a) Let $\sqrt{2}$ be a rational number such that $\sqrt{2} = \frac{p}{q}$ [$\frac{1}{2}$ mark]
($p$ and $q$ are co-prime numbers, $q \neq 0$)
$\sqrt{2}q = p \Rightarrow 2q^2 = p^2$
$2$ divides $p^2 \Rightarrow 2$ divides $p$ as well [$1$ mark]
$p = 2m$ (for some integer $m$)
$2q^2 = 4m^2 \Rightarrow q^2 = 2m^2$
$2$ divides $q^2 \Rightarrow 2$ divides $q$ as well
$p$ and $q$ have a common factor $2$ which is a contradiction as $p$ and $q$ are co-prime. [$1$ mark]
$\therefore$ our assumption is wrong
Hence, $\sqrt{2}$ is an irrational number [$\frac{1}{2}$ mark]
OR
(b) $a$ and $b$ are only composite numbers. [$3$ marks]
($p$ and $q$ are co-prime numbers, $q \neq 0$)
$\sqrt{2}q = p \Rightarrow 2q^2 = p^2$
$2$ divides $p^2 \Rightarrow 2$ divides $p$ as well [$1$ mark]
$p = 2m$ (for some integer $m$)
$2q^2 = 4m^2 \Rightarrow q^2 = 2m^2$
$2$ divides $q^2 \Rightarrow 2$ divides $q$ as well
$p$ and $q$ have a common factor $2$ which is a contradiction as $p$ and $q$ are co-prime. [$1$ mark]
$\therefore$ our assumption is wrong
Hence, $\sqrt{2}$ is an irrational number [$\frac{1}{2}$ mark]
OR
(b) $a$ and $b$ are only composite numbers. [$3$ marks]