Prove that the following numbers are irrational:

$\begin{matrix} \text{(i)} & \sqrt[3]{2} \\[1.5em] \text{(ii)} & \sqrt[3]{3} \\[1.5em] \text{(iii)} & \sqrt[4]{5} \\[1.5em] \end{matrix}$

(i) Suppose that $\sqrt[3]{2}$ = $\dfrac{p}{q}$, where p, q are integers , q ≠ 0 , p and q have no common factors (except 1)

$\Rightarrow 2 = \Big(\dfrac{p}{q}\Big)^3 \\[1.5em] \Rightarrow p^3 = 2q^3 \qquad \text{….(i)}$

As 2 divides 2q³ $\Rightarrow$ 2 divides p³
$\Rightarrow$ 2 divides p    (using generalisation of theorem 1)

Let p = 2k , where k is an integer.

Substituting this value of p in (i), we get

$\phantom{\Rightarrow}$(2k)³ = 2q³
$\Rightarrow$ 8k³ = 2q³
$\Rightarrow$ 4k³ = q³

As 2 divides 4k³ $\Rightarrow$ 2 divides q³

$\Rightarrow$ 2 divides q    (using generalisation of theorem 1)

Thus, p and q have a common factor 2. This contradicts that p and q have no common factors (except 1).

Hence, our supposition is wrong. It follows that $\sqrt[3]{2}$ cannot be expressed as $\dfrac{p}{q}$, where p, q are integers, q > 0, p and q have no common factors (except 1).

∴ $\bold{\sqrt[3]{2}}$ is an irrational number.

(ii) Suppose that $\sqrt[3]{3}$ = $\dfrac{p}{q}$, where p, q are integers, q ≠ 0, p and q have no common factors (except 1)

$\Rightarrow 3 = \Big(\dfrac{p}{q}\Big)^3 \\[1.5em] \Rightarrow p^3 = 3q^3 \qquad \text{….(i)}$

As 3 divides 3q³ $\Rightarrow$ 3 divides p³
$\Rightarrow$ 3 divides p    (using generalisation of theorem 1)

Let p = 3k, where k is an integer.

Substituting this value of p in (i), we get

$\phantom{\Rightarrow}$(3k)³ = 3q³
$\Rightarrow$ 27k³ = 3q³
$\Rightarrow$ 9k³ = q³

As 3 divides 9k³ $\Rightarrow$ 3 divides q³
$\Rightarrow$ 3 divides q    (using generalisation of theorem 1)

Thus, p and q have a common factor 3. This contradicts that p and q have no common factors (except 1).

Hence, our supposition is wrong . It follows that $\sqrt[3]{3}$ cannot be expressed as $\dfrac{p}{q}$, where p, q are integers, q > 0, p and q have no common factors (except 1).

Therefore,$\bold{\sqrt[3]{3}}$ is an irrational number.

(iii) Suppose that $\sqrt[4]{5}$ = $\dfrac{p}{q}$, where p, q are integers, q ≠ 0, p and q have no common factors (except 1)

$\Rightarrow 5 = \Big(\dfrac{p}{q}\Big)^4 \\[1.5em] \Rightarrow p^4 = 5q^4 \qquad \text{….(i)}$

As 5 divides 5q⁴ $\Rightarrow$ 5 divides p⁴
$\Rightarrow$ 5 divides p    (using generalisation of theorem 1)

Let p= 5k, where k is an integer.

Substituting this value of p in (i), we get

$\phantom{\Rightarrow}$(5k)⁴ = 5q⁴
$\Rightarrow$ 625k⁴ = 5q⁴
$\Rightarrow$ 125k⁴ = q⁴

As 5 divides 125k⁴ $\Rightarrow$ 5 divides q⁴
$\Rightarrow$ 5 divides q    (using generalisation of theorem 1)

Thus, p and q have a common factor 5. This contradicts that p and q have no common factors (except 1).

Hence, our supposition is wrong . It follows that $\sqrt[4]{5}$ cannot be expressed as $\dfrac{p}{q}$, where p, q are integers, q > 0, p and q have no common factors (except 1).

Therefore,$\bold{\sqrt[4]{5}}$ is an irrational number .