Exercise 1.1

Question 1

Insert a rational number between $\dfrac{2}{9}$ and $\dfrac{3}{8}$ arrange in descending order.

Answer

The L.C.M of 9 and 8 is 72.

$\dfrac{2}{9} = \dfrac{2 \times 8}{9 \times 8} = \dfrac{16}{72} \\[0.5em] \dfrac{3}{8} = \dfrac{3 \times 9}{8 \times 9} = \dfrac{27}{72} \\[0.5em] \text{Since } 16 \lt 27, \dfrac{2}{9} \lt \dfrac{3}{8}$

A rational number between $\dfrac{2}{9}$ and $\dfrac{3}{8}$

$= \dfrac{\dfrac{2}{9} + \dfrac{3}{8}}{2} \\[0.5em] = \dfrac{\dfrac{16 + 27}{72}}{2} \\[0.5em] = \bold{\dfrac{43}{144}} \\[0.5em]$

Numbers in descending order are:

$\bold{\dfrac{3}{8}}, \bold{\dfrac{43}{144}}, \bold{\dfrac{2}{9}}$

Question 2

Insert two rational numbers between $\dfrac{1}{3}$ and $\dfrac{1}{4}$ and arrange in ascending order.

Answer

The L.C.M of 3 and 4 is 12.

$\dfrac{1}{3} = \dfrac{1 \times 4}{3 \times 4} = \dfrac{4}{12} \\[0.5em] \dfrac{1}{4} = \dfrac{1 \times 3}{4 \times 3} = \dfrac{3}{12} \\[0.5em] \text{Since } 3 \lt 4, \dfrac{1}{4} \lt \dfrac{1}{3}$

A rational number between $\dfrac{1}{4}$ and $\dfrac{1}{3}$

$= \dfrac{\dfrac{1}{4} + \dfrac{1}{3}}{2} \\[0.5em] = \dfrac{\dfrac{4 + 3}{12}}{2} \\[0.5em] = \bold{\dfrac{7}{24}} \\[0.5em]$

A rational number between $\dfrac{1}{4}$ and $\dfrac{7}{24}$

$= \dfrac{\dfrac{1}{4} + \dfrac{7}{24}}{2} \\[0.5em] = \dfrac{\dfrac{6 + 7}{24}}{2} \\[0.5em] = \bold{\dfrac{13}{48}} \\[0.5em]$

Numbers in ascending order are:

$\bold{\dfrac{1}{4}}, \bold{\dfrac{13}{48}}, \bold{\dfrac{7}{24}}, \bold{\dfrac{1}{3}}$

Question 3

Insert two rational numbers between $-\dfrac{1}{3}$ and $-\dfrac{1}{2}$ and arrange in ascending order.

Answer

The L.C.M of 2 and 3 is 6.

$-\dfrac{1}{3} = -\dfrac{1 \times 2}{3 \times 2} = -\dfrac{2}{6} \\[0.5em] -\dfrac{1}{2} = -\dfrac{1 \times 3}{2 \times 3} = -\dfrac{3}{6} \\[0.5em] \text{Since } -3 \lt -2, -\dfrac{1}{2} \lt -\dfrac{1}{3}$

A rational number between -$\dfrac{1}{2}$ and -$\dfrac{1}{3}$

$= \dfrac{-\dfrac{1}{2} + \Big(-\dfrac{1}{3}\Big)}{2} \\[0.5em] = \dfrac{\dfrac{-3 + (-2)}{6}}{2} \\[0.5em] = \dfrac{\dfrac{-3 -2}{6}}{2} \\[0.5em] = \bold{-\dfrac{5}{12}} \\[0.5em]$

A rational number between -$\dfrac{1}{2}$ and -$\dfrac{5}{12}$

$= \dfrac{-\dfrac{1}{2} + \Big(-\dfrac{5}{12}\Big)}{2} \\[0.5em] = \dfrac{\dfrac{-6 + (-5)}{12}}{2} \\[0.5em] = \dfrac{\dfrac{-6 -5}{12}}{2} \\[0.5em] = \bold{-\dfrac{11}{24}} \\[0.5em]$

Numbers in ascending order are:

$\bold{-\dfrac{1}{2}}, \bold{-\dfrac{11}{24}}, \bold{-\dfrac{5}{12}}, \bold{-\dfrac{1}{3}}$

Question 4

Insert three rational numbers between $\dfrac{1}{3}$ and $\dfrac{4}{5}$, and arrange in descending order.

Answer

The L.C.M of 3 and 5 is 15.

$\dfrac{1}{3} = \dfrac{1 \times 5}{3 \times 5} = \dfrac{5}{15} \\[0.5em] \dfrac{4}{5} = \dfrac{4 \times 3}{5 \times 3} = \dfrac{12}{15} \\[0.5em] \text{Since } 5 \lt 12, \dfrac{1}{3} \lt \dfrac{4}{5}$

A rational number between $\dfrac{1}{3}$ and $\dfrac{4}{5}$

$= \dfrac{\dfrac{1}{3} + \dfrac{4}{5}}{2} \\[0.5em] = \dfrac{\dfrac{5 + 12}{15}}{2} \\[0.5em] = \bold{\dfrac{17}{30}} \\[0.5em]$

A rational number between $\dfrac{1}{3}$ and $\dfrac{17}{30}$

$= \dfrac{\dfrac{1}{3} + \dfrac{17}{30}}{2} \\[0.5em] = \dfrac{\dfrac{10 + 17}{30}}{2} \\[0.5em] = \bold{\dfrac{27}{60}} \\[0.5em]$

A rational number between $\dfrac{17}{30}$ and $\dfrac{4}{5}$

$= \dfrac{\dfrac{17}{30} + \dfrac{4}{5}}{2} \\[0.5em] = \dfrac{\dfrac{17 + 24}{30}}{2} \\[0.5em] = \bold{\dfrac{41}{60}} \\[0.5em]$

Numbers in descending order are:

$\bold{\dfrac{4}{5}}, \bold{\dfrac{41}{60}}, \bold{\dfrac{17}{30}}, \bold{\dfrac{27}{60}}, \bold{\dfrac{1}{3}}$

Question 5

Insert three rational numbers between 4 and 4.5.

Answer

A rational number between 4 and 4.5

$= \dfrac{4 + 4.5}{2} \\[0.5em] = \dfrac{8.5}{2} \\[0.5em] = \bold{4.25} \\[0.5em]$

A rational number between 4 and 4.25

$= \dfrac{4 + 4.25}{2} \\[0.5em] = \dfrac{8.25}{2} \\[0.5em] = \bold{4.125} \\[0.5em]$

A rational number between 4 and 4.125

$= \dfrac{4 + 4.125}{2} \\[0.5em] = \dfrac{8.125}{2} \\[0.5em] = \bold{4.0625} \\[0.5em]$

As, 4 < 4.0625 < 4.125 < 4.25 < 4.5, therefore three rational numbers between 4 and 4.5 are 4.0625, 4.125, 4.25.

Question 6

Find six rational numbers between 3 and 4.

Answer

The numbers 3 and 4 can be written as $\dfrac{3}{1}$ and $\dfrac{4}{1}$.

Since we want to find six rational numbers between the given numbers, multiplying the numerator and denominator of the above numbers by 6 + 1 i.e. by 7, we get $\dfrac{21}{7}$ and $\dfrac{28}{7}$, which are equivalent to the given numbers.

$\text{As } 21 \lt 22 \lt 23 \lt 24 \lt 25 \lt 26 \lt 27 \lt 28, \\[0.7em] \dfrac{21}{7} \lt \dfrac{22}{7} \lt \dfrac{23}{7} \lt \dfrac{24}{7} \lt \dfrac{25}{7} \lt \dfrac{26}{7} \lt \dfrac{27}{7} \lt \dfrac{28}{7}$

Therefore, six rational numbers between 3 and 4 are:

$\bold{\dfrac{22}{7}}, \bold{\dfrac{23}{7}}, \bold{\dfrac{24}{7}}, \bold{\dfrac{25}{7}}, \bold{\dfrac{26}{7}}, \bold{\dfrac{27}{7}}$

Question 7

Find five rational numbers between $\dfrac{3}{5}$ and $\dfrac{4}{5}$.

Answer

Since we want to find five rational numbers between the given numbers, multiplying the numerator and denominator of the above numbers by 5 + 1 i.e. by 6, we get $\dfrac{18}{30}$ and $\dfrac{24}{30}$, which are equivalent to the given numbers.

$\text{As } 18 \lt 19 \lt 20 \lt 21 \lt 22 \lt 23 \lt 24, \\[0.7em] \dfrac{18}{30} \lt \dfrac{19}{30} \lt \dfrac{20}{30} \lt \dfrac{21}{30} \lt \dfrac{22}{30} \lt \dfrac{23}{30} \lt \dfrac{24}{30} \\[0.7em] \Rightarrow \dfrac{3}{5} \lt \dfrac{19}{30} \lt \dfrac{2}{3} \lt \dfrac{7}{10} \lt \dfrac{11}{15} \lt \dfrac{23}{30} \lt \dfrac{4}{5}$

Therefore, six rational numbers between $\dfrac{3}{5}$ and $\dfrac{4}{5}$ are:

$\bold{\dfrac{19}{30}}, \bold{\dfrac{2}{3}}, \bold{\dfrac{7}{10}}, \bold{\dfrac{11}{15}}, \bold{\dfrac{23}{30}}$

Question 8

Find ten rational numbers between $-\dfrac{2}{5}$ and $\dfrac{1}{7}$.

Answer

Writing the given numbers with same denominator 35 (L.C.M. of 5 and 7), we get:

$-\dfrac{2}{5} = -\dfrac{14}{35} \\[0.5em] \dfrac{1}{7} = \dfrac{5}{35}$

$\text{As } -14 \lt -13 \lt -12 \lt -11 \lt -10 \lt -9 \lt -8 \lt -7 \lt 0 \lt 1 \lt 2 \lt 5, \\[0.7em] -\dfrac{14}{35} \lt -\dfrac{13}{35} \lt -\dfrac{12}{35} \lt -\dfrac{11}{35} \lt -\dfrac{10}{35} \lt -\dfrac{9}{35} \lt -\dfrac{8}{35} \lt -\dfrac{7}{35} \lt 0 \lt \dfrac{1}{35} \lt \dfrac{2}{35} \lt \dfrac{5}{35} \\[0.7em] \Rightarrow -\dfrac{2}{5} \lt -\dfrac{13}{35} \lt -\dfrac{12}{35} \lt -\dfrac{11}{35} \lt -\dfrac{2}{7} \lt -\dfrac{9}{35} \lt -\dfrac{8}{35} \lt -\dfrac{1}{5} \lt 0 \lt \dfrac{1}{35} \lt \dfrac{2}{35} \lt \dfrac{5}{35}$

Therefore, ten rational numbers between $-\dfrac{2}{5}$ and $\dfrac{1}{7}$ are:

$\bold{-\dfrac{13}{35}}, \bold{-\dfrac{12}{35}}, \bold{-\dfrac{11}{35}}, \bold{-\dfrac{2}{7}}, \bold{-\dfrac{9}{35}}, \\[0.5em] \bold{-\dfrac{8}{35}}, \bold{-\dfrac{1}{5}}, 0, \bold{\dfrac{1}{35}}, \bold{\dfrac{2}{35}}$

Question 9

Find six rational numbers between $\dfrac{1}{2}$ and $\dfrac{2}{3}$.

Answer

Writing the given numbers with same denominator 6 (L.C.M. of 2 and 3), we get:

$\dfrac{1}{2} = \dfrac{3}{6} \\[0.5em] \dfrac{2}{3} = \dfrac{4}{6}$

Since we want to find six rational numbers between the given numbers, multiplying the numerator and denominator of the above numbers by 6 + 1 i.e. by 7, we get $\dfrac{21}{42}$ and $\dfrac{28}{42}$, which are equivalent to the given numbers.

$\text{As } 21 \lt 22 \lt 23 \lt 24 \lt 25 \lt 26 \lt 27 \lt 28, \\[0.7em] \dfrac{21}{42} \lt \dfrac{22}{42} \lt \dfrac{23}{42} \lt \dfrac{24}{42} \lt \dfrac{25}{42} \lt \dfrac{26}{42} \lt \dfrac{27}{42} \lt \dfrac{28}{42} \\[0.7em] \Rightarrow \dfrac{1}{2} \lt \dfrac{11}{21} \lt \dfrac{23}{42} \lt \dfrac{4}{7} \lt \dfrac{25}{42} \lt \dfrac{13}{21} \lt \dfrac{9}{14} \lt \dfrac{2}{3}$

Therefore, six rational numbers between $\dfrac{1}{2}$ and $\dfrac{2}{3}$ are:

$\bold{\dfrac{11}{21}}, \bold{\dfrac{23}{42}}, \bold{\dfrac{4}{7}}, \bold{\dfrac{25}{42}}, \bold{\dfrac{13}{21}}, \bold{\dfrac{9}{14}}$

Exercise 1.2

Question 1

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.

Question 2

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.

Question 3

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.

Question 4

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.

Question 5

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.

Question 6

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.

Question 7

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)

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

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.

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.

Question 8

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.

Exercise 1.3

Question 1

Locate $\sqrt{10}$ and $\sqrt{17}$ on the number line.

Answer

Locating $\sqrt{10}$:

Representing 10 as the sum of squares of two natural numbers:

10 = 9 + 1 = 3² + 1²

Let l be the number line. If point O represents number 0 and point A represents number 3, then draw a line segment OA = 3 units.

At A, draw AC ⟂ OA. From AC, cut off AB = 1 unit.

We observe that OAB is a right angled triangle at A. By Pythagoras theorem, we get:

$OB^2 = OA^2 + AB^2 \\[0.5em] \Rightarrow OB^2 = 3^2 + 1^2 \\[0.5em] \Rightarrow OB^2 = 9 + 1 \\[0.5em] \Rightarrow OB^2 = 10 \\[0.5em] \Rightarrow OB = \sqrt{10} \text{units} \\[0.5em]$

With O as centre and radius = OB, we draw an arc of a circle to meet the number line l at point P.

As OP = OB = $\sqrt{10}$ units, the point P will represent the number $\sqrt{10}$ on the number line as shown in the figure above.

Locating $\sqrt{17}$:

Representing 17 as the sum of squares of two natural numbers:

17 = 16 + 1 = 4² + 1²

Let l be the number line. If point O represents number 0 and point A represents number 4, then draw a line segment OA = 4 units.

At A, draw AC ⟂ OA. From AC, cut off AB = 1 unit.

We observe that OAB is a right angled triangle at A. By Pythagoras theorem, we get:

$OB^2 = OA^2 + AB^2 \\[0.5em] \Rightarrow OB^2 = 4^2 + 1^2 \\[0.5em] \Rightarrow OB^2 = 16 + 1 \\[0.5em] \Rightarrow OB^2 = 17 \\[0.5em] \Rightarrow OB = \sqrt{17} \text{units} \\[0.5em]$

With O as centre and radius = OB, we draw an arc of a circle to meet the number line l at point P.

As OP = OB = $\sqrt{17}$ units, the point P will represent the number $\sqrt{17}$ on the number line as shown in the figure above.

Question 2

Write the decimal expansion of each of the following numbers and say what kind of decimal expansion each has:

$\begin{matrix} \text{(i)} & \dfrac{36}{100} \\[1.5em] \text{(ii)} & 4\dfrac{1}{8} \\[1.5em] \text{(iii)} & \dfrac{2}{9} \\[1.5em] \text{(iv)} & \dfrac{2}{11} \\[1.5em] \text{(v)} & \dfrac{3}{13} \\[1.5em] \text{(vi)} & \dfrac{329}{400} \\[1.5em] \end{matrix}$

Answer

$\text{(i) } \dfrac{36}{100}$

$\therefore \dfrac{36}{100} = 0.36$

Remainder becomes zero.

Decimal expansion of $\bold{\dfrac{36}{100}}$ is terminating.

$\text{(ii) } 4\dfrac{1}{8}$

$\therefore 4\dfrac{1}{8} = 4.125$

Remainder becomes zero.

Decimal expansion of $\bold{4\dfrac{1}{8}}$ is terminating.

$\text{(iii) } \dfrac{2}{9}$

$\therefore \dfrac{2}{9} = 0.2222..... = 0.\overline{2}$

Remainder is repeating.

Decimal expansion of $\bold{\dfrac{2}{9}}$ is non-terminating repeating.

$\text{(iv) } \dfrac{2}{11}$

$\therefore \dfrac{2}{11} = 0.1818..... = 0.\overline{18}$

Remainder is repeating.

Decimal expansion of $\bold{\dfrac{2}{11}}$ is non-terminating repeating.

$\text{(v) } \dfrac{3}{13}$

$\therefore \dfrac{3}{13} = 0.2307692307..... = 0.\overline{230769}$

Remainder is repeating.

Decimal expansion of $\bold{\dfrac{3}{13}}$ is non-terminating repeating.

$\text{(vi) } \dfrac{329}{400}$

$\therefore \dfrac{329}{400} = 0.8225$

Remainder becomes zero.

Decimal expansion of $\bold{\dfrac{329}{400}}$ is terminating.

Question 3

Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

$\begin{matrix} \text{(i)} & \dfrac{13}{3125} \\[1.5em] \text{(ii)} & \dfrac{17}{8} \\[1.5em] \text{(iii)} & \dfrac{23}{75} \\[1.5em] \text{(iv)} & \dfrac{6}{15} \\[1.5em] \text{(v)} & \dfrac{1258}{625} \\[1.5em] \text{(vi)} & \dfrac{77}{210} \\[1.5em] \end{matrix}$

Answer

$\text{(i) } \dfrac{13}{3125}$

The given number $\dfrac{13}{3125}$ is in its lowest form.

Prime factorization of denominator 3125:

$\begin{array}{l|l} 5 & 3125 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

3125 = 5 x 5 x 5 x 5 x 5 x 1
= 5⁵ x 1
= 1 x 5⁵
= 2⁰ x 5⁵    [∵ 2⁰ = 1]

Denominator is of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $\dfrac{13}{3125}$ has a terminating decimal expansion.

$\text{(ii) } \dfrac{17}{8}$

The given number $\dfrac{17}{8}$ is in its lowest form.

Prime factorization of denominator 8:

$\begin{array}{l|l} 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

8 = 2 x 2 x 2 x 1
= 2³ x 1
= 2³ x 5⁰    [∵ 5⁰ = 1]

Denominator is of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $\dfrac{17}{8}$ has a terminating decimal expansion.

$\text{(iii) } \dfrac{23}{75}$

The given number $\dfrac{23}{75}$ is in its lowest form.

Prime factorization of denominator 75:

$\begin{array}{l|l} 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

75 = 3 x 5 x 5 x 1
= 3 x 5² x 1
= 3 x 5² x 2⁰    [∵ 2⁰ = 1]

Denominator has a prime factor 3 other than 2 or 5.

∴ The given number $\dfrac{23}{75}$ has a non-terminating repeating decimal expansion.

$\text{(iv) } \dfrac{6}{15}$

Both numerator and denominator contain common factor 3. Reducing the number to its lowest form:

$\dfrac{6}{15} = \dfrac{\cancel{3} \times 2}{\cancel{3} \times 5} \\[0.5em] = \dfrac{2}{5}$

The Denominator 5 = 2⁰ x 5¹

Denominator is of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $\dfrac{6}{15}$ has a terminating decimal expansion.

$\text{(v) } \dfrac{1258}{625}$

The given number $\dfrac{1258}{625}$ is in its lowest form.

Prime factorization of denominator 625:

$\begin{array}{l|l} 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

625 = 5 x 5 x 5 x 5 x 1
= 5⁴ x 1
= 1 x 5⁴
= 2⁰ x 5⁴    [∵ 2⁰ = 1]

Denominator is of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $\dfrac{1258}{625}$ has a terminating decimal expansion.

$\text{(vi) } \dfrac{77}{210}$

Both numerator and denominator contain common factor 7. Reducing the number to its lowest form:

$\dfrac{77}{210} = \dfrac{\cancel{7} \times 11}{\cancel{7} \times 30} \\[0.5em] = \dfrac{11}{30}$

Prime factorization of denominator 30:

$\begin{array}{l|l} 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

30 = 2 x 3 x 5

Denominator has a prime factor 3 other than 2 or 5.

∴ The given number $\dfrac{77}{210}$ has a non-terminating repeating decimal expansion.

Question 4

Without actually performing the long division, find if $\dfrac{987}{10500}$ will have terminating or non-terminating repeating decimal expansion. Give reasons for your answer.

Answer

GCD of numerator and denominator is 21. Reducing the number to its lowest form:

$\dfrac{987}{10500} = \dfrac{\cancel{21} \times 47}{\cancel{21} \times 500} \\[0.5em] = \dfrac{47}{500}$

Prime factorization of denominator 500:

$\begin{array}{l|l} 2 & 500 \\ \hline 2 & 250 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

500 = 2 x 2 x 5 x 5 x 5 = 2² x 5³

Denominator is of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $\dfrac{987}{10500}$ has a terminating decimal expansion.

Question 5

Write the decimal expansions of the following numbers which have terminating decimal expansions:

$\begin{matrix} \text{(i)} & \dfrac{17}{8} \\[1.5em] \text{(ii)} & \dfrac{13}{3125} \\[1.5em] \text{(iii)} & \dfrac{7}{80} \\[1.5em] \text{(iv)} & \dfrac{6}{15} \\[1.5em] \text{(v)} & \dfrac{2^2 \times 7}{5^4} \\[1.5em] \text{(vi)} & \dfrac{237}{1500} \\[1.5em] \end{matrix}$

Answer

$\text{(i) } \dfrac{17}{8}$

The given number $\dfrac{17}{8}$ is in its lowest form.

Prime factorization of denominator 8:

$\begin{array}{l|l} 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

8 = 2 x 2 x 2 x 1
= 2³ x 5⁰    [∵ 5⁰ = 1]

$\dfrac{17}{8} = \dfrac{17}{2^3} \\[0.5em] = \dfrac{17 \times 5^3}{2^3 \times 5^3} \\[0.5em] = \dfrac{17 \times 125}{(2 \times 5)^3} \\[0.5em] = \dfrac{2125}{10^3} \\[0.5em] = 2.125 \\[0.5em] \bold{\therefore \dfrac{17}{8} = 2.125}$

$\text{(ii) } \dfrac{13}{3125}$

The given number $\dfrac{13}{3125}$ is in its lowest form.

Prime factorization of denominator 3125:

$\begin{array}{l|l} 5 & 3125 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

3125 = 5 x 5 x 5 x 5 x 5 x 1
= 5⁵ x 1
= 1 x 5⁵
= 2⁰ x 5⁵    [∵ 2⁰ = 1]

$\dfrac{13}{3125} = \dfrac{13}{5^5} \\[0.5em] = \dfrac{13 \times 2^5}{2^5 \times 5^5} \\[0.5em] = \dfrac{13 \times 32}{(2 \times 5)^5} \\[0.5em] = \dfrac{416}{10^5} \\[0.5em] = 0.00416 \\[0.5em] \bold{\therefore \dfrac{13}{3125} = 0.00416}$

$\text{(iii) } \dfrac{7}{80}$

The given number $\dfrac{7}{80}$ is in its lowest form.

Prime factorization of denominator 80:

$\begin{array}{l|l} 2 & 80 \\ \hline 2 & 40 \\ \hline 2 & 20 \\ \hline 2 & 10 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

80 = 2 x 2 x 2 x 2 x 5
= 2⁴ x 5
= 2⁴ x 5¹

$\dfrac{7}{80} = \dfrac{7}{2^4 \times 5^1} \\[0.5em] = \dfrac{7 \times 5^3}{2^4 \times 5^1 \times 5^3} \\[0.5em] = \dfrac{7 \times 125}{2^4 \times 5^4} \\[0.5em] = \dfrac{7 \times 125}{(2 \times 5)^4} \\[0.5em] = \dfrac{875}{10^4} \\[0.5em] = 0.0875 \\[0.5em] \bold{\therefore \dfrac{7}{80} = 0.0875}$

$\text{(iv) } \dfrac{6}{15}$

GCD of numerator and denominator is 3. Reducing the number to its lowest form:

$\dfrac{6}{15} = \dfrac{\cancel{3} \times 2}{\cancel{3} \times 5} \\[0.5em] = \dfrac{2}{5}$

$\dfrac{6}{15} = \dfrac{2}{5} \\[0.5em] = \dfrac{2 \times 2}{2 \times 5} \\[0.5em] = \dfrac{4}{10} \\[0.5em] = 0.4 \\[0.5em] \bold{\therefore \dfrac{6}{15} = 0.4}$

$\text{(v) } \dfrac{2^2 \times 7}{5^4}$

$\dfrac{2^2 \times 7}{5^4} = \dfrac{2^2 \times 7 \times 2^4}{5^4 \times 2^4} \\[0.5em] = \dfrac{4 \times 7 \times 16}{(2 \times 5)^4} \\[0.5em] = \dfrac{448}{10^4} \\[0.5em] = 0.0448 \\[0.5em] \bold{\therefore \dfrac{2^2 \times 7}{5^4} = 0.0448}$

$\text{(vi) } \dfrac{237}{1500}$

GCD of numerator and denominator is 3. Reducing the number to its lowest form:

$\dfrac{237}{1500} = \dfrac{\cancel{3} \times 79}{\cancel{3} \times 500} \\[0.5em] = \dfrac{79}{500}$

$\dfrac{237}{1500} = \dfrac{79}{500} \\[0.5em] = \dfrac{79 \times 2}{500 \times 2} \\[0.5em] = \dfrac{158}{10^3} \\[0.5em] = 0.158 \\[0.5em] \bold{\therefore \dfrac{237}{1500} = 0.158}$

Question 6

Write the denominator of the rational number $\dfrac{257}{5000}$ in the form 2^m × 5ⁿ where m, n are non-negative integers. Hence, write its decimal expansion without actual division.

Answer

The given number $\dfrac{257}{5000}$ is in its lowest form.

Prime factorization of denominator 5000:

$\begin{array}{l|l} 2 & 5000 \\ \hline 2 & 2500 \\ \hline 2 & 1250 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

5000 = 2 x 2 x 2 x 5 x 5 x 5 x 5
= 2³ x 5⁴

Hence, denominator of rational number $dfrac{257}{5000}$ in the form 2^m × 5ⁿ is 2³ x 5⁴ where m = 3 and n = 4.

$\dfrac{257}{5000} = \dfrac{257}{2^3 \times 5^4} \\[0.5em] = \dfrac{257 \times 2}{2^3 \times 5^4 \times 2} \\[0.5em] = \dfrac{514}{2^4 \times 5^4} \\[0.5em] = \dfrac{514}{(2 \times 5)^4} \\[0.5em] = \dfrac{514}{10^4} \\[0.5em] = 0.0514 \\[0.5em] \bold{\therefore \dfrac{257}{5000} = 0.0514}$

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Exercise 1.4

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Exercise 1.5

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