(i) 1/√2 is a rational number

we find two integers b such as 1/√2 = a/b

where a and b co-prime integers (b ≠ 0)

Square both sides

(1/√2)² = (a/b)² 

⇒ 1/2 = a²/b²

b² = 2a²

So, b² divides by 2.

.’. So, b also divides by 2.

Now let b = 2c, where c is any integer

b² = (2c)²

⇒ b² = 4c²

⇒ 2a² = 4c² (∴ b² = 2a²)

⇒ a² = 2c²

Hence, a², divides by 2.

a also divides by 2.

Hence, 2 is a common  factor of a and b.

But, this contradicts the fact that a and b have no common factor other than 1.

This contradiction arises by assuming that 1/√2 is rational,

Hence 1/√2 is a irrational number.

(ii) Let 6 + √2 is a rational number

we can find two integers a and b (b ≠ 0)

Such as

6 + √2 = a/b

√2 = a/b - 6

√2 = (a - 6b)/b

a, b and 6 all are integers.

(a - 6b)/b is a rational numbers

√2 is also a rational number

But this contradicts the fact that √2 is an irrational number.

Our hypothesis is wrong.

Hence, 6 + √2 is an irrational number.

(iii) Let 3√2 is a rational number we find two integers a and b such as

3√2 = a/b (where a and b co-prime integers)

⇒ √2 = a/3b

⇒ a, b and 3 are integers.

a/3b is a rational number

√2 is a rational number

3√2 will be also a rational numbers.

But this contradicts the fact that √2 is an irrational number.

So, our hypothesis is wrong.

So, 3√2 is an irrational number.