Question

prove that 7 root 3 plus root 5 is irrational number

Answer 1

Heya...!
Here is ur answer...
Lets assume to the contrary that
$7 \sqrt{3} + \sqrt{5} \: s \: rational$
Then there must be coprime a and b such that
$7 \sqrt{3} + \sqrt{5} = \frac{a}{b}$
$= \sqrt{3} + \sqrt{5} = \frac{a}{7b}$
We know that root 3 and root 5 are irrational..so,
$\sqrt{3} + \sqrt{5} \: must \: be \: irrational$
But a, b and 7 are integers and so they are rational...then how can they equalize an irrational no.
Therefore, this equation contradicts.
This contradiction has arisen due to our incorrect assumption that
$7 \sqrt{3} + \sqrt{5} \: is \: a \: rational \: number.$
Therefore, 7root3+root5 is an irrational no.
Hope it is satisfactory....
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VIDHI☺️