Question

PROVE that 3+√5 is irrational

Thanks

Answer 1

Let take that 6 + √2 is a rational number.
So we can write this number as
6 + √2 = a/b
Here a and b are two co prime number and b is not equal to 0
Subtract 6 both side we get
√2 = a/b – 6
√2 = (a-6b)/b
Here a and b are integer so (a-6b)/b is a rational number so √2 should be a rational number But √2 is a irrational number so it is contradict
Hence result is 6 + √2 is a irrational number


similarly u can prove ur ques.. try it...!!

Answer 2

Let 3+√5 be rational. So it can be written in the form p/q where p and q are rational
Now,
3+√5=p/q
√5 =(p/q)-3
Therefore √5 will be rational but we know that it is irrational. This contradiction ha arisen because of our wrong assumption of 3+√5 being rational.
Therefore 3+√5 is irrational.