Question
Prove that $\sqrt{5}$ is an irrational number.
✓
Answer
$\sqrt{5}=\frac{\text{p}}{\text{q}}, p q $are coprimes $q \neq 0$
$5q^2 = p^2$
$\Rightarrow 5$ divides $p^2$
$\Rightarrow 5$ divides $p$ also
Let $p = 5a,$ for some integer $a$
$5q^2 = 25a^2$
$\Rightarrow q^2 = 5a^2$
$\Rightarrow 5$ divides $q^2$
$\Rightarrow 5$ divides $q$ also
$\therefore 5$ is a common factor of $p, q,$ which is not possible as $1\ p, q$ are coprimes.
Hence assumption is wrong $\sqrt{5}$ is irrational no.
$5q^2 = p^2$
$\Rightarrow 5$ divides $p^2$
$\Rightarrow 5$ divides $p$ also
Let $p = 5a,$ for some integer $a$
$5q^2 = 25a^2$
$\Rightarrow q^2 = 5a^2$
$\Rightarrow 5$ divides $q^2$
$\Rightarrow 5$ divides $q$ also
$\therefore 5$ is a common factor of $p, q,$ which is not possible as $1\ p, q$ are coprimes.
Hence assumption is wrong $\sqrt{5}$ is irrational no.
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