Question

prove that is an irrational nos
$4 + \sqrt{3}$
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Answer 1

Answer:

$let \: us \: suppose \: that \: 4 + \sqrt{3} \\ is \: rational \: number \: which \: \: \\ means \: it \: can \: be \: written \\ \: in \: the \: form \: of \: 4 + \sqrt{3} = \frac{a}{b} \\ where a \: and \: b \: are \: co \: prime \: and \\ b \: is \: not \: equal \: to \: 0 \\ \\ so \\ \\ 4 + \sqrt{3} = \frac{a}{b} \\ \\ 4 - \frac{a}{b} = - \sqrt{3} \\ \\ \frac{4b - a}{b} = - \sqrt{3} \\ \\ since \: a \: and \: b \: integers \: we \\ \\ get \: \frac{4b - a}{b \:} \: is \: rational \: number \\ \\ and \: - \sqrt{3} \: \: is \: rational .\\ \\ \: but \\ \\ this \: contradict \: the \: fact \: \\ \\ that \: - \sqrt{3} \: is \: irrational \: . \\ \\ so \: we \: conclude \: that \: 4 + \sqrt{3} \\ \\ \: is \: irrational$