show that 5-7 root 3 is a irrational number
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Answer:
let it be a rational number
5 - \frac{3}{7} \sqrt{3} = \frac{a}{b}5−
7
3
3
=
b
a
= > - \frac{3}{7} \sqrt{3 } = \frac{a}{b} - 5=>−
7
3
3
=
b
a
−5
= > \frac{3}{7} \sqrt{3} = \frac{5b - a}{b}=>
7
3
3
=
b
5b−a
\sqrt{3} = \frac{7(5b - a)}{3b}
3
=
3b
7(5b−a)
it seems that √3 is rational
so our assumption get wrong because √3 is irrational number so 5-3/7√3 is irrational
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