Question

Prove that 7√5 is a rational no

Answer 1

here is your answer in the photo given above
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Answer 2

Hello friend here is your answer....

We will prove it with the help of contradiction..
$so \: let \: 7 \sqrt{5} \: is \: a \: rational \: number \\ i.e. \: it \: could \: be \: written \: as \: \frac{x}{y} where \: y \: not \: equal \: to \: zero \: \\so \\ 7 \sqrt{5} = \frac{x}{y} \\ \\ \sqrt{5} = \frac{x}{7y} \\ \\ here \sqrt{5} is \: a \: irrationl \:but \: number \: \frac{x}{7y} is \: a \: rational \: number \: \\ \\ which \: is \: contradicting \: our \: assumption \: that \: 7 \sqrt{5} is \: a \: rational \: no.$
So, **7root5 is a irrational no. **

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