Question

Prove that 2 minus 3 root 5 is an irrational number sit

Answer 1

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$let \: us \: assume \: that \: 2 - 3 \sqrt{5} </strong><strong>\:</strong><strong> </strong><strong>is \: an \: rational \: number \\ if \: 2 - 3 \sqrt{5} \: is \: rational \: number \: then \\ 2 - 3 \sqrt{5} \: = \frac{a}{b} \: \: {where \: a \: and</strong><strong> </strong><strong>\</strong><strong>:</strong><strong> </strong><strong>b \: are \: co \: prime \: number \: and \: b \: is \: not \: equal \: to \: zero} \\ - 3 \sqrt{5} = \frac{a}{b} - 2 \\ - 3 \sqrt{5} = \frac{a \: - 2b}{b} \\ \sqrt{5} = \frac{a \: - 2b}{ - 3b} \: \\ since \: a \: and \: b \: are \: coprime \: integer \: and \: b \: is \: not \: equal \: to \: zero \\ so \: \frac{a \: - 2b}{ - 3b} \: is \: a \: rational \: number \\ \: hence \: \sqrt{5} \: is \: also \: rational \\ but \: the \: contradiction \: arise \: that \: \sqrt{5} \: is \: a \: irrational \: number \\ our \: assumption \: was \: wrong \\ 2 - therefore \: 2 - 3 \sqrt{5} \: is \: a \: irrational \: number$

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Answer 2

Step-by-step explanation:

see the attachment ...

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