Question
Prove that $\sqrt{5}$ is an irrational number.
✓
Answer
Let $\sqrt{5}$ be a rational number.
$\therefore \sqrt{5}=\frac{p}{q}$, where $q \neq 0$ and $p \& q$ are coprime.
$5 q^2=p^2 \Rightarrow p^2$ is divisible by 5
⟹ p is divisible by 5----- (i)
⟹ p = 3a, where ‘a’ is a postive integer
$25 a^2=5 q^2 \Rightarrow q^2=5 a^2 \Rightarrow q^2$ is divisible by 5
⟹ q is divisible by 5 ----- (ii)
(i) and (ii) leads to contradiction as ‘p’ and ‘q’ are coprime.
$\therefore \sqrt{5}$ is an irrational number.
$\therefore \sqrt{5}=\frac{p}{q}$, where $q \neq 0$ and $p \& q$ are coprime.
$5 q^2=p^2 \Rightarrow p^2$ is divisible by 5
⟹ p is divisible by 5----- (i)
⟹ p = 3a, where ‘a’ is a postive integer
$25 a^2=5 q^2 \Rightarrow q^2=5 a^2 \Rightarrow q^2$ is divisible by 5
⟹ q is divisible by 5 ----- (ii)
(i) and (ii) leads to contradiction as ‘p’ and ‘q’ are coprime.
$\therefore \sqrt{5}$ is an irrational number.
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