**Solution: **

**(i) 1/**√**2**

Let us assume 1/√2 is rational.

Then we can find co-prime x and y (y ≠ 0) such that 1/√2 = x/y

Rearranging, we get,

√2 = y/x

Since x and y are integers, thus, √2 is a rational number, which contradicts the fact that √2 is irrational.

Therefore, we can conclude that 1/√2 is irrational.

Hence proved.

**(ii) 7**√**5**

Let us assume 7√5 is a rational number.

Then we can find co-prime a and b (b ≠ 0) such that 7√5 = x/y

Rearranging, we get,

√5 = x/7y

Since x and y are integers, thus, √5 is a rational number, which contradicts the fact that √5 is irrational.

Therefore, we can conclude that 7√5 is irrational.

Hence proved.

**(iii) 6 +**√**2**

Let us assume 6 +√2 is a rational number.

Then we can find co-primes x and y (y ≠ 0) such that 6 +√2 = x/y⋅

Rearranging, we get,

√2 = (x/y) – 6

Since, x and y are integers, thus (x/y) – 6 is a rational number and therefore, √2 is rational. This contradicts the fact that √2 is an irrational number.

Therefore, we can conclude that 6 +√2 is irrational.

Hence proved.