(i) Let assume that \(\frac{1}{\sqrt2}\) is rational

Therefore it can be expressed in the form of \(\frac{p}{q}\), where p and q are integers and q ≠ 0

Therefore we can write \(\frac{1}{\sqrt2}\) = \(\frac{p}{q}\)

√2 = \(\frac{q}{p}\)

\(\frac{q}{p}\) is a rational number as p and q are integers.

Therefore √2 is rational which contradicts the fact that √2 is irrational.

Hence, our assumption is false and we can say that \(\frac{1}{\sqrt2}\) is irrational.

(ii) Let assume that 7√5 is rational therefore it can be written in the form of \(\frac{p}{q}\)

where p and q are integers and q ≠ 0.

Moreover, let p and q have no common divisor > 1.

7√5 = \(\frac{p}{q}\) for some integers p and q

Therefore √5 = \(\frac{p}{7q}\)

\(\frac{p}{7q}\)is rational as p and q are integers, therefore √5 should be rational.

This contradicts the fact that √5 is irrational.

Therefore our assumption that 7√5 is rational is false.

Hence 7√5 is irrational.

(iii) Let assume that 6 + √2 is rational therefore it can be written in the form of \(\frac{p}{q}\)

where p and q are integers and q ≠ 0.

Moreover, let p and q have no common divisor > 1.

6 + √2 = \(\frac{p}{q}\) for some integers p and q

Therefore √2 = \(\frac{p}{q}\) - 6

Since p and q are integers therefore \(\frac{p}{q}\) - 6 is rational, hence √2 should be rational.

This contradicts the fact that √2 is irrational.

Therefore our assumption is false.

Hence, 6 + √2 is irrational.

(iv) Let us assume that 3 - √5 is rational

Therefore 3 - √5 can be written in the form of \(\frac{p}{q}\) where p and q are integers and q ≠ 0

3 - √5 = \(\frac{p}{q}\) ⇒ \(\frac{p}{q}\) - 3 = √5

⇒ \(\frac{p-3q}{q}\) = √5

Since p, q and 3 are integers therefore \(\frac{p-3q}{q}\) is rational number

But we know √5 is irrational number,

Therefore it is a contradiction.

Hence 3 - √5 is irrational