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This section has $6$ Short Answer (SA) type questions carrying $3$ marks each.
Prove that $\sqrt{5}$ is an irrational number.
Prove that $\sqrt{5}$ is an irrational number.
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Let $\sqrt{5}$ be a rational number.
$\therefore \sqrt{5} = \frac{p}{q}$, where $q \neq 0$ and let p & q are co-primes.
$5q^2 = p^2 \Rightarrow p^2$ is divisible by $5$
$\Rightarrow p$ is divisible by $5$----- (i)
$\Rightarrow$let $p = 5a$, where 'a' is some integer
$25a^2 = 5q^2 \Rightarrow q^2 = 5a^2 \Rightarrow q^2$ is divisible by $5$.
$\Rightarrow q$ is divisible by $5$. ----- (ii)
(i) and (ii) leads to contradiction as p and q are coprimes.
$\therefore \sqrt{5}$ is an irrational number
$\therefore \sqrt{5} = \frac{p}{q}$, where $q \neq 0$ and let p & q are co-primes.
$5q^2 = p^2 \Rightarrow p^2$ is divisible by $5$
$\Rightarrow p$ is divisible by $5$----- (i)
$\Rightarrow$let $p = 5a$, where 'a' is some integer
$25a^2 = 5q^2 \Rightarrow q^2 = 5a^2 \Rightarrow q^2$ is divisible by $5$.
$\Rightarrow q$ is divisible by $5$. ----- (ii)
(i) and (ii) leads to contradiction as p and q are coprimes.
$\therefore \sqrt{5}$ is an irrational number