Therefore, 1√2 is irrational number.
(ii) 7√5
Let us assume 7√5 is rational.
So, we can write this number as
7√5=ab ---- (1)
Here, a and b are two co-prime numbers and b is not equal to zero.
Simplify the equation (1) divide by 7 both sides, we get
√5=a7b
Here, a and b are integers, so a7b is a rational
number, so √5 should be a rational number.
But √5 is a irrational number, so it is contradictory.
Therefore, 7√5 is irrational number.
(iii) 6+√2
Let us assume 6+√2 is rational.
So we can write this number as
6+√2=ab ---- (1)
Here, a and b are two co-prime number and b is not equal to zero.
Simplify the equation (1) subtract 6 on both sides, we get
√2=ab−6
√2=a−6bb
Here, a and b are integers so, a−6bb is a rational
number, so √2 should be a rational number.
But √2 is a irrational number, so it is contradictory.
Therefore, 6+√2 is irrational number.