(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
