Question

14. Prove that
$\sqrt{3}$
is an irrational number.​

Answer 1

Step-by-step explanation:

$let \: us \: assume \: that \: \sqrt{3} \: is \: not \: irrational \: \\ therefore \: \sqrt{3} \: is \: rational \: . \\ \sqrt{3} = \frac{a}{b} \: \: \: b \: not \: equal \: to \: 0 \: and \: a \: b \: are \: co \: primes \: \\ squaring \: on \: both \: sides \: \\ ({ \sqrt{3} })^{2} = | \frac{a}{b} | {}^{2} \\ 3 = \frac{a {}^{2} }{b {}^{2} } \\ {3b}^{2} = {a}^{2} \\ {a }^{2} = 3 {b}^{2} \\ {a}^{2} \: divde \: 3 \\ a \: divide \: 3 \\ now \: \\ let \: a \: = 3c \: . \\ substitute \: a = 3c \: in \: {a}^{2} = 3 {b}^{2} \\ (3 {c}^{2} ) = 3 {b}^{2} \\ 9 {c}^{2} = 3 {b}^{2} \\ 3 {c}^{2} = {b}^{2} \\ {b}^{2} = 3 {c}^{2} \: \\ {b}^{2} \: divide \: 3 \\ b \: divide \: 3 \\ from \: equation \: \: 1 \: and \: 2 \: a \: \: and \: b \: are \: common \: factor \: of \: 3 \\ this \: contradiction \: arisen \: because \: of \: our \: wrong \: assumption \: that \: \sqrt{3} \: is \: rational \: \\ there fore \: \sqrt{3} \: is \: irrational \: \\ hence \: proved$