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Neha claimed that there does not exist any irrational number between 1 and 2. Raunak claimed that $\sqrt{2}$ lies between 1 and 2 and $\sqrt{2}$ is an irrational number. Who do you think is correct? Justify by proving either $\sqrt{2}$ as an irrational number or otherwise.
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Solution: Raunak is correct (1/2 Mark)
Let $\sqrt{2}$ be a rational number such that $\sqrt{2} = \frac{a}{b}$, where $a$ and $b$ are coprime and $b \neq 0$ (1/2 Mark)
$\sqrt{2}b = a$
$2b^2 = a^2$
2 divides $a^2$
2 divides $a$ as well. (1/2 Mark)
Let $a = 2p$ (for some integer $p$)
$a^2 = 4p^2$
$2b^2 = 4p^2$
$b^2 = 2p^2$
2 divides $b^2$
2 divides $b$ as well (1 Mark)
$\therefore$ 2 is a common factor of $a$ and $b$ which is a contradiction as $a$ and $b$ are coprime.
$\therefore$ Our assumption is wrong. Hence, $\sqrt{2}$ is an irrational number. (1/2 Mark)
Let $\sqrt{2}$ be a rational number such that $\sqrt{2} = \frac{a}{b}$, where $a$ and $b$ are coprime and $b \neq 0$ (1/2 Mark)
$\sqrt{2}b = a$
$2b^2 = a^2$
2 divides $a^2$
2 divides $a$ as well. (1/2 Mark)
Let $a = 2p$ (for some integer $p$)
$a^2 = 4p^2$
$2b^2 = 4p^2$
$b^2 = 2p^2$
2 divides $b^2$
2 divides $b$ as well (1 Mark)
$\therefore$ 2 is a common factor of $a$ and $b$ which is a contradiction as $a$ and $b$ are coprime.
$\therefore$ Our assumption is wrong. Hence, $\sqrt{2}$ is an irrational number. (1/2 Mark)