Question

Prove that
$\sqrt{3 + \sqrt{5} }$
Irrational​

Answer 1

Step-by-step explanation:

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$let \: \sqrt{3} + \sqrt{5} \: be \: a$

➡️where a is rational.

therefore,

➡️$\sqrt{3} + \sqrt{5} = a$

➡️${ \sqrt{3} }^{2} = {(a - \sqrt{5)} }^{2}$

$3 = {a}^{2} + 5 - 2a \sqrt{5} \\ 2a \sqrt{5} = {a}^{2} + 5 - 3 \\ 2a \sqrt{5} = {a}^{2} + 2 \\ \sqrt{5} = \frac{ {a}^{2} + 2 }{2a}$

➡️which is contradiction

as √5 is irrational and

$\frac{ {a + 2}^{2} }{2a} is \: rational$

hence proved. that √3 + √5 is irrational

$\huge{\underline{\boxed{\sf{\purple{Hope-it-helps}}}}}$