Prove that the following numbers are irrational:

(i) $\sqrt{8}$

(ii) $\sqrt{14}$

(iii) $\sqrt[3]{2}$

(i) $\sqrt{8}$ can be written as $2\sqrt{2}$, now we are going to show that $\sqrt{2}$ is an irrational number.

Let $\sqrt{2}$ be a rational number, then

$\sqrt{2} = \dfrac{p}{q},$

where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)

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

As 2 divides 2q², so 2 divides p² but 2 is prime

$\Rightarrow 2 \text{ divides } p \qquad \text{(Theorem 1)}$

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

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

$(2m)^2 = 2q^2 \\[1.5em] \Rightarrow 4m^2 = 2q^2 \\[1.5em] \Rightarrow 2m^2 = q^2 \\[1.5em]$

As 2 divides 2m², so 2 divides q² but 2 is prime

$\Rightarrow 2 \text{ divides } q \qquad \text{(Theorem 1)}$

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

Hence, $\sqrt{2}$ is not a rational number. So, we conclude that $\sqrt{2}$ is an irrational number.

Since, product of non-zero rational number and an irrational number is an irrational number.

And $\sqrt{2}$ is an irrational number this implies that $2\sqrt{2}$ = $\bold{\sqrt{8}}$ is an irrational number.

(ii) Suppose that $\sqrt{14}$ is a rational number, then

$\sqrt{14} = \dfrac{p}{q},$

where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)

$\Rightarrow 14 = \dfrac{p^2}{q^2} \\[1.5em] \Rightarrow p^2 = 14q^2 \qquad \text{….(i)}$

As 2 divides 14q², so 2 divides p² but 2 is prime

$\Rightarrow 2 \text{ divides } p \qquad \text{(Theorem 1)}$

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

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

$(2k)^2 = 14q^2 \\[1.5em] \Rightarrow 4k^2 = 14q^2 \\[1.5em] \Rightarrow 2k^2 = 7q^2 \\[1.5em]$

As 2 divides 2k², so 2 divides 7q²

$\Rightarrow$ 2 divides 7 or 2 divides q²

But 2 does not divide 7, therefore, 2 divides q²

$\Rightarrow$ 2 divides q      (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. Therefore, $\sqrt{14}$ is not a rational number. So, we conclude that $\bold{\sqrt{14}}$ is an irrational number.

(iii) 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.