Question

Prove√3+√5 irrational​

Answer 1

Answer:

let √3 is rational where p and q r co-prime and q is not equals to 0

$\sqrt{3} = \frac{p}{q}$

squaring both side

$3 = \frac{ {p}^{2} }{ {q}^{2} } \\ {q}^{2} = \frac{ {p}^{2} }{3}$

p² is divisible by 3. By Euclid division algorithm p is also divisible by 3.

so we write p=3a

${q}^{2} = \frac{ {(3a)}^{2} }{3} \\{q}^{2} = \frac{ {9a}^{2} }{3} \\ {q}^{2} = 3 {a}^{2} \\ {a}^{2} = \frac{ {q}^{2} }{3}$

q² is divisible by 3. By Euclid division algorithm q ia also devisible by 3.

so, we conclude that √3 is irrational.

as like this, we can prove √5 is irrational