Rational and Irrational Numbers — Exercise 1.2

Prove that $\sqrt{5}$ is an irrational number.

Answer

Let $\sqrt{5}$ be a rational number, then

$\sqrt{5} = \dfrac{p}{q},$

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

$\Rightarrow 5 = \dfrac{p^2}{q^2} \\[0.5em] \Rightarrow p^2 = 5q^2 \qquad \text{....(i)}$

As 5 divides 5q², so 5 divides p² but 5 is prime

$\Rightarrow 5 \text{ divides } p \qquad \text{(Theorem 1)}$

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

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

$(5m)^2 = 5q^2 \\[0.5em] \Rightarrow 25m^2 = 5q^2 \\[0.5em] \Rightarrow 5m^2 = q^2 \\[0.5em]$

As 5 divides 5m², so 5 divides q² but 5 is prime

$\Rightarrow 5 \text{ divides } q \qquad \text{(Theorem 1)}$

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

Hence, $\sqrt{5}$ is not a rational number. So, we conclude that $\sqrt{5}$ is an irrational number.

Prove that $\sqrt{7}$ is an irrational number.

Answer

Let $\sqrt{7}$ be a rational number, then

$\sqrt{7} = \dfrac{p}{q},$

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

$\Rightarrow 7 = \dfrac{p^2}{q^2} \\[0.5em] \Rightarrow p^2 = 7q^2 \qquad \text{....(i)}$

As 7 divides 7q², so 7 divides p² but 7 is prime

$\Rightarrow 7 \text{ divides } p \qquad \text{(Theorem 1)}$

Let p = 7m, where m is an integer.

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

$(7m)^2 = 7q^2 \\[0.5em] \Rightarrow 49m^2 = 7q^2 \\[0.5em] \Rightarrow 7m^2 = q^2 \\[0.5em]$

As 7 divides 7m², so 7 divides q² but 7 is prime

$\Rightarrow 7 \text{ divides } q \qquad \text{(Theorem 1)}$

Thus, p and q have a common factor 7. This contradicts that p and q have no common factors (except 1).

Hence, $\sqrt{7}$ is not a rational number. So, we conclude that $\sqrt{7}$ is an irrational number.

Prove that $\sqrt{6}$ is an irrational number.

Answer

Suppose that $\sqrt{6}$ is a rational number, then

$\sqrt{6} = \dfrac{p}{q},$

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

$\Rightarrow 6 = \dfrac{p^2}{q^2} \\[0.5em] \Rightarrow p^2 = 6q^2 \qquad \text{....(i)}$

As 2 divides 6q², 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

$(2k)^2 = 6q^2 \\[0.5em] \Rightarrow 4k^2 = 6q^2 \\[0.5em] \Rightarrow 2k^2 = 3q^2 \\[0.5em]$

As 2 divides 2k², so 2 divides 3q²

$\Rightarrow$ 2 divides 3 or 2 divides q²

But 2 does not divide 3, 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, $\sqrt{6}$ is not a rational number. So, we conclude that $\sqrt{6}$ is an irrational number.

Prove that $\dfrac{1}{\sqrt{11}}$ is an irrational number.

Answer

Let $\dfrac{1}{\sqrt{11}}$ be a rational number, then

$\dfrac{1}{\sqrt{11}} = \dfrac{p}{q},$

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

$\Rightarrow \dfrac{1}{11} = \dfrac{p^2}{q^2} \\[0.5em] \Rightarrow q^2 = 11p^2 \qquad \text{....(i)}$

As 11 divides 11p², so 11 divides q² but 11 is prime

$\Rightarrow 11 \text{ divides } q \qquad \text{(Theorem 1)}$

Let q = 11m, where m is an integer.

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

$(11m)^2 = 11p^2 \\[0.5em] \Rightarrow 121m^2 = 11p^2 \\[0.5em] \Rightarrow 11m^2 = p^2 \\[0.5em]$

As 11 divides 11m², so 11 divides p² but 11 is prime

$\Rightarrow 11 \text{ divides } p \qquad \text{(Theorem 1)}$

Thus, p and q have a common factor 11. This contradicts that p and q have no common factors (except 1).

Hence, $\dfrac{1}{\sqrt{11}}$ is not a rational number. So, we conclude that $\dfrac{1}{\sqrt{11}}$ is an irrational number.

Prove that $\sqrt{2}$ is an irrational number. Hence, show that $3 - \sqrt{2}$ is an irrational number.

Answer

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)

$\Rightarrow 2 = \dfrac{p^2}{q^2} \\[0.5em] \Rightarrow p^2 = 2q^2 \qquad \text{....(i)}$

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

$(2m)^2 = 2q^2 \\[0.5em] \Rightarrow 4m^2 = 2q^2 \\[0.5em] \Rightarrow 2m^2 = q^2 \\[0.5em]$

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.

Suppose that $3 - \sqrt{2}$ is a rational number, say r.

Then, $3 - \sqrt{2}$ = r (note that r ≠ 0)

$\Rightarrow - \sqrt{2} = r - 3 \\[0.5em] \Rightarrow \sqrt{2} = 3 - r \\[0.5em]$

As r is rational and r ≠ 0, so 3 - r is rational

$\Rightarrow \sqrt{2}$ is rational

But this contradicts that $\sqrt{2}$ is irrational. Hence, our supposition is wrong.

∴ $3 - \sqrt{2}$ is an irrational number.

Prove that $\sqrt{3}$ is an irrational number. Hence, show that $\dfrac{2}{5}\sqrt{3}$ is an irrational number.

Answer

Let $\sqrt{3}$ be a rational number, then

$\sqrt{3} = \dfrac{p}{q},$

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

$\Rightarrow 3 = \dfrac{p^2}{q^2} \\[0.5em] \Rightarrow p^2 = 3q^2 \qquad \text{....(i)}$

As 3 divides 3q², so 3 divides p² but 3 is prime

$\Rightarrow 3 \text{ divides } p \qquad \text{(Theorem 1)}$

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

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

$(3m)^2 = 3q^2 \\[0.5em] \Rightarrow 9m^2 = 3q^2 \\[0.5em] \Rightarrow 3m^2 = q^2 \\[0.5em]$

As 3 divides 3m², so 3 divides q² but 3 is prime

$\Rightarrow 3 \text{ divides } q \qquad \text{(Theorem 1)}$

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

Hence, $\sqrt{3}$ is not a rational number. So, we conclude that $\sqrt{3}$ is an irrational number.

Suppose that $\dfrac{2}{5}\sqrt{3}$ is a rational number, say r.

Then, $\dfrac{2}{5}\sqrt{3}$ = r (note that r ≠ 0)

$\Rightarrow \sqrt{3} = \dfrac{5}{2}r \\[0.5em]$

As r is rational and r ≠ 0, so $\dfrac{5}{2}r$ is rational
[∵ System of rational numbers is closed under all four fundamental arithmetic operations (except division by zero)]

$\Rightarrow \sqrt{3}$ is rational

But this contradicts that $\sqrt{3}$ is irrational. Hence, our supposition is wrong.

∴ $\dfrac{2}{5}\sqrt{3}$ is an irrational number.

Prove that $\sqrt{5}$ is an irrational number. Hence, show that $-3 + 2\sqrt{5}$ is an irrational number.

Answer

Let $\sqrt{5}$ be a rational number, then

$\sqrt{5} = \dfrac{p}{q},$

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

$\Rightarrow 5 = \dfrac{p^2}{q^2} \\[0.5em] \Rightarrow p^2 = 5q^2$

As 5 divides 5q², so 5 divides p² but 5 is prime

$\Rightarrow 5 \text{ divides } p \qquad \text{(Theorem 1)}$

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

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

$(5m)^2 = 5q^2 \\[0.5em] \Rightarrow 25m^2 = 5q^2 \\[0.5em] \Rightarrow 5m^2 = q^2 \\[0.5em]$

As 5 divides 5m², so 5 divides q² but 5 is prime

$\Rightarrow 5 \text{ divides } q \qquad \text{(Theorem 1)}$

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

Hence, $\sqrt{5}$ is not a rational number. So, we conclude that $\sqrt{5}$ is an irrational number.

Suppose that $-3 + 2\sqrt{5}$ is a rational number, say r.

Then, $-3 + 2\sqrt{5}$ = r (note that r ≠ 0)

$\Rightarrow 2\sqrt{5} = r + 3 \\[0.5em] \Rightarrow \sqrt{5} = \dfrac{r + 3}{2} \\[0.5em]$

As r is rational and r ≠ 0, so $\dfrac{r + 3}{2}$ is rational

$\Rightarrow \sqrt{5}$ is rational

But this contradicts that $\sqrt{5}$ is irrational. Hence, our supposition is wrong.

∴ $-3 + 2\sqrt{5}$ is an irrational number.

Prove that the following numbers are irrational:

$\begin{matrix} \text{(i)} & 5 + \sqrt{2} \\[0.5em] \text{(ii)} & 3 - 5\sqrt{3} \\[0.5em] \text{(iii)} & 2\sqrt{3} - 7 \\[0.5em] \text{(iv)} & \sqrt{2} + \sqrt{5} \end{matrix}$

Answer

$\text{(i) } 5 + \sqrt{2}$

Let us assume that $5 + \sqrt{2}$ is a rational number, say r.

Then,

$5 + \sqrt{2} = r \\[0.5em] \Rightarrow \sqrt{2} = r - 5$

As r is rational, r - 5 is rational

$\Rightarrow \sqrt{2}$ is rational

But this contradicts the fact that $\sqrt{2}$ is irrational.

Hence, our assumption is wrong.

∴ $5 + \sqrt{2}$ is an irrational number.

$\text{(ii) } 3 - 5\sqrt{3}$

Let us assume that $3 - 5\sqrt{3}$ is a rational number, say r.

Then,

$3 - 5\sqrt{3} = r \\[0.5em] \Rightarrow 5\sqrt{3} = 3 - r \\[0.5em] \Rightarrow \sqrt{3} = \dfrac{3 - r}{5} \\[0.5em]$

As r is rational, 3 - r is rational

$\Rightarrow \dfrac{3 - r}{5}$ is rational

$\Rightarrow \sqrt{3}$ is rational

But this contradicts the fact that $\sqrt{3}$ is irrational.

Hence, our assumption is wrong.

∴ $3 - 5\sqrt{3}$ is an irrational number.

$\text{(iii) } 2\sqrt{3} - 7$

Let us assume that $2\sqrt{3} - 7$ is a rational number, say r.

Then,

$2\sqrt{3} - 7 = r \\[0.5em] \Rightarrow 2\sqrt{3} = r + 7 \\[0.5em] \Rightarrow \sqrt{3} = \dfrac{r + 7}{2} \\[0.5em]$

As r is rational, r + 7 is rational

$\Rightarrow \dfrac{r + 7}{2}$ is rational

$\Rightarrow \sqrt{3}$ is rational

But this contradicts the fact that $\sqrt{3}$ is irrational.

Hence, our assumption is wrong.

∴ $2\sqrt{3} - 7$ is an irrational number.

$\text{(iv) } \sqrt{2} + \sqrt{5}$

Let us assume that $\sqrt{2} + \sqrt{5}$ is a rational number, say r.

Then,

$\sqrt{2} + \sqrt{5} = r \\[0.5em] \Rightarrow \sqrt{5} = r - \sqrt{2} \\[0.5em] \Rightarrow (\sqrt{5})^2 = (r - \sqrt{2})^2 \\[0.5em] \Rightarrow 5 = r^2 + 2 - 2\sqrt{2}r \\[0.5em] \Rightarrow 2\sqrt{2}r = r^2 + 2 - 5 \\[0.5em] \Rightarrow 2\sqrt{2}r = r^2 - 3 \\[0.5em] \Rightarrow \sqrt{2} = \dfrac{r^2 - 3}{2r} \\[0.5em]$

As r is rational,

$\Rightarrow \dfrac{r^2 - 3}{2r}$ is rational

$\Rightarrow \sqrt{2}$ is rational

But this contradicts the fact that $\sqrt{2}$ is irrational.

Hence, our assumption is wrong.

∴ $\sqrt{2} + \sqrt{5}$ is an irrational number.