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Prove that $\sqrt{5}$ is an irrational number.
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Sol. Let $\sqrt{5}$ be a rational number.
$\therefore \sqrt{5} = \frac{p}{q}$, where $q\neq0$ and let $p \& q$ be co-primes.
$5q^2 = p^2 \Rightarrow p^2$ is divisible by $5 \Rightarrow p$ is divisible by $5$
$\Rightarrow p = 5a$, where ‘a' is some integer
dots (i)
$25a^2 = 5q^2 \Rightarrow q^2 = 5a^2 \Rightarrow q^2$ is divisible by $5 \Rightarrow q$ is divisible by $5$
$\Rightarrow q = 5b$, where ‘b' is some integer
dots (ii)
(i) and (ii) leads to contradiction as 'p' and 'q' are co-primes.
$\therefore \sqrt{5}$ is an irrational number.
$\therefore \sqrt{5} = \frac{p}{q}$, where $q\neq0$ and let $p \& q$ be co-primes.
$5q^2 = p^2 \Rightarrow p^2$ is divisible by $5 \Rightarrow p$ is divisible by $5$
$\Rightarrow p = 5a$, where ‘a' is some integer
dots (i)
$25a^2 = 5q^2 \Rightarrow q^2 = 5a^2 \Rightarrow q^2$ is divisible by $5 \Rightarrow q$ is divisible by $5$
$\Rightarrow q = 5b$, where ‘b' is some integer
dots (ii)
(i) and (ii) leads to contradiction as 'p' and 'q' are co-primes.
$\therefore \sqrt{5}$ is an irrational number.