Question

prove √5 is irrational....

Answer 1

Hey!! HERE IS UR ANS...
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TO PROVE √5 IRRATIONAL WE USE "CONTRADICTION" METHOD..

In which, If possible we suppose that √5 is rational number..
Therefore,
$\sqrt{5} = \frac{a}{b}$
(where a and b are co-prime integers and b is not equals to 0)
$( \sqrt{5} )^{2} = \frac{ {a}^{2} }{ {b}^{2} }$
(squaring both the side)

$5 = \frac{ {a}^{2} }{ {b}^{2} }$
${b}^{2} = \frac{ {a}^{2} }{5} - (i) \: \: \: \: (changing \: side)$

therefore,
$\: \: \: \: \: \: \: \: \: {a}^{2} \: is \: divisible \: by \: 5 \\ = > \: a \: is \: also \: divisible \: by \: 5$
$= > \: \frac{a}{5} = k \: (where \: k \: is \: some \: integer) \\ \: \: \: \: \: \: \: \: \: \: a = 5k$
FROM (i)
${b}^{2} = \frac{ {(5k)}^{2} }{5} \\ {b}^{2} = \frac{ {25k}^{2} }{5} \\ {b}^{2} = {5k}^{2} \\ \frac{ {b}^{2} }{5} = {k}^{2} \\ = > \: {b }^{2} \: is \: divisible \: by \: 5\\ = > \: b \: is \: also \: divisible \: by \: 5$
HERE, we may conclude that 5 is common factor of both a and b. This is contradiction because of our wrong assumption that √5 is a rational number
Therefore, √5 must irrational number.

#RS

HOPE THIS HELPS...☺☺

Answer 2

Hope it helps u....☺☺☺