Question

Prove that √3 is an irrational number.Hence show that 2√3 - 8 is also an irrational number.

Answer 1

HEYA!!

HERE'S YOUR ANSWER.....!

$let \: \sqrt{3} \: be \: a \: rational \: number \\ \sqrt{3} = \frac{p}{q} (where \: q \: is \: not \: equal \: to \: zero) \\ then \: \sqrt{3} q = p - - - - (1) \\ squaring \: both \: sides \: we \: get \\ 3 {q}^{2} = {p}^{2} \\ here \: 3 \: divides \: {p}^{2} \\ 3 \: divides \: p - - - - (2) \\ now \: assuming \: p = 3k \: and \: substituting \: in \: {eq}^{n} (1) \: we \: get \\ \sqrt{3}q = 3k \\ now \: squaring \: both \: sides \: we \: get \\ 3 {q}^{2} = 9 {k}^{2} \\ {q}^{2} = 3 {k}^{2} \\ here \: 3 \: divides \: {q}^{2} \\ 3 \: divdes \: q - - - - - - (2) \\ from \: {eq}^{n} \: 1 \: and \: 2 \: we \: get \\ 3 \: divides \: both \: p \: and \: q \: which \: contradicts \: our \: supposition \\ hence \: \sqrt{3} \: is \: an \: irrational \: number$

now according this you can do the 2nd equation

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