Question

prove that ✓5 is irrational no.

Let us assumed
$\: \: let \: us \: assumed \: \sqrt{5} \: \: \: \: \:is \: rational \: \: \:$
number
$\sqrt{5} = \frac{a}{b \: }$
where a and b are co prime number
and b not 0

$\sqrt{5} = \frac{a}{b}$
squaring both side,
$\\ ( \sqrt{5)} {}^{2} = ( \frac{a}{b} ) {}^{2}$
$b {}^{2} = \frac{a {}^{2} }{5}$
$a {}^{2} \: \: is \: divisible \: by \: 5 \: \: so \: \: a \: is \: also$
divisible by 5
let a = 5m
$\sqrt{5} = ( \frac{5m}{b} )$
squaring both side
$5 = \frac{25m {}^{2} }{b {}^{2} } \\ \\ b { }^{2} = \frac{25 \: m {}^{2} }{5} \\ \\ \frac{5m {}^{2} }{1} \\ \\ m {}^{2} = \frac{b {}^{2} }{5} \\ \\ 5 \: is \: common \: factor \: of \: a \: and \: \: b \\ \\ so \: our \: assumptionis \: \: is \: wrong \: hence \: \\ \\ \sqrt{5} \: is \: irrational \: \: numbers$




​

Answer 1

are Bhai kehna kya chahte ho