87
Prove that $\sqrt{2}$ is an irrational number.
Show SolutionHide Solution↓
Let $\sqrt{2}$ be a rational number.
$\therefore \sqrt{2} = \frac{p}{q}$, where $q\neq0$ and let $p \& q$ be co-primes.
$2q^2 = p^2 \Rightarrow p^2$ is divisible by $2 \Rightarrow p$ is divisible by $2$
$\Rightarrow p = 2a$, where 'a' is some integer ----- (i)
$4a^2 = 2q^2 \Rightarrow q^2 = 2a^2 \Rightarrow q^2$ is divisible by $2 \Rightarrow q$ is divisible by $2$
$\Rightarrow q = 2b$, where 'b' is some integer ----- (ii)
(i) and (ii) leads to contradiction as 'p' and 'q' are co-primes.
$\therefore \sqrt{2}$ is an irrational number.
$\therefore \sqrt{2} = \frac{p}{q}$, where $q\neq0$ and let $p \& q$ be co-primes.
$2q^2 = p^2 \Rightarrow p^2$ is divisible by $2 \Rightarrow p$ is divisible by $2$
$\Rightarrow p = 2a$, where 'a' is some integer ----- (i)
$4a^2 = 2q^2 \Rightarrow q^2 = 2a^2 \Rightarrow q^2$ is divisible by $2 \Rightarrow q$ is divisible by $2$
$\Rightarrow q = 2b$, where 'b' is some integer ----- (ii)
(i) and (ii) leads to contradiction as 'p' and 'q' are co-primes.
$\therefore \sqrt{2}$ is an irrational number.