Question

prove that 3+√5 is an irrational number​

Answer 1

If possible, then let 3+√5

3+√5 = p/q

Sqaring both sides,

(3+√5)^2 = p^2/q^2

9+5+2*3*√5 = p^2/q^2

14+6√5 = p^2/q^2

14 is a rational number, 6√5 is an irrational number, and p^2/q^2 is a rational number

Then,

6√5 = p^2/q^2 - 14

*rational - rational = rational*

Irrational number is not equal to rational number

Then, our assumption that 3+√5 is a rational number is wrong

Hence, 3+√5 is an irrational number.