Rational and Irrational Numbers — Chapter Test

Without actual division, find whether the following rational numbers are terminating decimals or recurring decimals :

$\begin{matrix} \text{(i)} & \dfrac{13}{45} \\[1.5em] \text{(ii)} & -\dfrac{5}{56} \\[1.5em] \text{(iii)} & \dfrac{7}{125} \\[1.5em] \text{(iv)} & -\dfrac{23}{80} \\[1.5em] \text{(v)} & -\dfrac{15}{66} \\[1.5em] \end{matrix}$

In case of terminating decimals, write their decimal expansions.

Answer

$\text{(i) } \dfrac{13}{45}$

The given number $\dfrac{13}{45}$ is in its lowest form.

Prime factorization of denominator 45:

$\begin{array}{l|l} 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

45 = 3 x 3 x 5 x 1
= 3² x 5 x 1

Denominator is not of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $\dfrac{13}{45}$ is recurring decimal.

$\text{(ii) } -\dfrac{5}{56}$

The given number $-\dfrac{5}{56}$ is in its lowest form.

Prime factorization of denominator 56:

$\begin{array}{l|l} 2 & 56 \\ \hline 2 & 28 \\ \hline 2 & 14 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

56 = 2 x 2 x 2 x 7 x 1
= 2³ x 7 x 1

Denominator is not of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $-\dfrac{5}{56}$ is recurring decimal.

$\text{(iii) } \dfrac{7}{125}$

The given number $\dfrac{7}{125}$ is in its lowest form.

Prime factorization of denominator 125:

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

125= 5 x 5 x 5 x 1
= 5³ x 1
= 5³ x 2⁰

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

$\dfrac{7}{125} = \dfrac{7}{2^0 × 5^3} \\[1.5em] = \dfrac{7 × 2^3}{2^3 × 5^3} \\[1.5em] = \dfrac{56}{(2 × 5)^3} \\[1.5em] = \dfrac{56}{(10)^3} \\[1.5em] = \dfrac{56}{1000} \\[1.5em] = 0.056$

∴ The given number $\dfrac{7}{125}$ is a terminating decimal and its decimal expansion is 0.056.

$\text{(iv) } \dfrac{-23}{80}$

The given number $-\dfrac{23}{80}$ 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 x 1
= 2⁴ x 5¹

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

$-\dfrac{23}{80} = -\dfrac{23}{2^4 × 5^1} \\[1.5em] = -\dfrac{23 × 5^3} {2^4 × 5^4} \\[1.5em] = -\dfrac{23 × 125}{(2 × 5)^4} \\[1.5em] = -\dfrac{2875}{(10)^4} \\[1.5em] = -\dfrac{2875}{10000} = -0.2875$

∴ The given number $-\dfrac{23}{80}$ is a terminating decimal and its decimal expansion is -0.2875.

$\text{(v) } -\dfrac{15}{66}$

The given number $-\dfrac{15}{66}$ is in its lowest form.

Prime factorization of denominator 66:

$\begin{array}{l|l} 2 & 66 \\ \hline 3 & 33 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

66 = 2 x 3 x 11 x 1
= 2 x 3 x 11

Denominator is not of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $-\dfrac{15}{66}$ is recurring decimal.

Express the following recurring decimals as vulgar fractions :

(i) $1.3\overline{45}$

(ii) $2.\overline{357}$

Answer

(i) Let x = $1.3\overline{45}$ = 1.3454545 ... $\qquad \text{....(i)}$

So multiplying both sides of (i) by 10

we get,

10x = 13.4545...$\qquad \text{....(ii)}$

Again multiply by 100 on both sides ,

1000x =1345.4545.....$\qquad \text{....(iii)}$

Subtracting (ii) from (iii), we get

1000x - 10x = 1345.4545... - 13.4545...

990x = 1332

x = $\dfrac{1332}{990}$ = $\bold{\dfrac{74}{55}}$

which is in the form of $\dfrac{p}{q}$, q ≠ 0

(ii) Let x = $2.\overline{357}$ = 2.357357... $\qquad \text{....(i)}$

So multiplying both sides of (i) by 1000,

we get,

1000x = 2357.357357...$\qquad \text{....(ii)}$

Subtracting (i) from (ii), we get

1000x - x = 2357.357357... - 2.357357...

999x = 2355

x = $\bold{\dfrac{2355}{999}}$

which is in the form of $\dfrac{p}{q}$, q ≠ 0.

Insert a rational number between $\dfrac{5}{9}$ and $\dfrac{7}{13}$, and arrange in ascending order.

Answer

The L.C.M. of 9 and 13 is 117.

$\dfrac{5}{9} = \dfrac{5 \times 13}{9 \times 13} = \dfrac{65}{117} \\[1.5em] \dfrac{7}{13} = \dfrac{7 \times 9}{13 \times 9} = \dfrac{63}{117} \\[1.5em] \text{Since } 9 \lt 13, \dfrac{1}{13} \lt \dfrac{1}{9}$

A rational number between $\dfrac{5}{9}$ and $\dfrac{7}{13}$

$= \dfrac{\dfrac{5}{9} + \dfrac{7}{13}}{2} \\[1.5em] = \dfrac{\dfrac{65 + 63}{117}}{2} \\[1.5em] = \dfrac{128}{117 × 2} \\[1.5em] = \bold{\dfrac{64}{117}} \\[1.5em]$

$\therefore\dfrac{7}{13} \lt \dfrac{64}{117} \lt \dfrac{5}{9}$

Hence, numbers in ascending order are:

$\bold{\dfrac{7}{13}}, \bold{\dfrac{64}{117}}, \bold{\dfrac{5}{9}}$

Insert four rational numbers between $\dfrac{4}{5}$ and $\dfrac{5}{6}$.

Answer

The L.C.M of 5 and 6 is 30.

$\dfrac{4}{5} = \dfrac{4 \times 6}{5 \times 6} = \dfrac{24}{30} \\[1.5em] \dfrac{5}{6} = \dfrac{5 \times 5}{6 \times 5} = \dfrac{25}{30} \\[1.5em] \text{Since } 24 \lt 25, \\[1.5em] \therefore \dfrac{4}{5} \lt \dfrac{5}{6}$

To find four rational number between $\dfrac{4}{5}$ and $\dfrac{5}{6}$ multiply the numerator and denominator of $\dfrac{24}{30}$ and $\dfrac{25}{30}$ by 4 + 1 i.e. by 5 we get, $\dfrac{120}{150}$ and $\dfrac{125}{150}$

$\text{Since}, 120 \lt 121 \lt 122 \lt 123 \lt 124 \lt 125 \\[1.5em] \Rightarrow\dfrac{120}{150} \lt \dfrac{121}{150} \lt \dfrac{122}{150} \lt \dfrac{123}{150} \lt \dfrac{124}{150} \lt \dfrac{125}{150} \\[1.5em] \Rightarrow\dfrac{4}{5} \lt \dfrac{121}{150} \lt \dfrac{61}{75} \lt \dfrac{123}{150} \lt \dfrac{62}{75} \lt \dfrac{5}{6} \\[1.5em]$

Four rational number between $\dfrac{4}{5}$ and $\dfrac{5}{6}$ are:

$\bold{\dfrac{121}{150}}, \bold{\dfrac{61}{75}}, \bold{\dfrac{123}{150}}, \bold{\dfrac{62}{75}}$

Prove that the reciprocal of an irrational number is irrational.

Answer

Let us consider, x as an irrational number.

Reciprocal of x is $\dfrac{1}{x}$.

Let us consider $\dfrac{1}{x}$ to be a non-zero rational number.

Then, $x × \dfrac{1}{x}$ will also be an irrational number as product of non-zero rational number and an irrational number is also an irrational number.

But $x × \dfrac{1}{x}$ = 1 is a rational number.

Hence, our supposition is wrong. So, $\dfrac{1}{x}$ is an irrational number.

Prove that the following numbers are irrational:

(i) $\sqrt{8}$

(ii) $\sqrt{14}$

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

Answer

(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.

Prove that $\sqrt{3}$ is an irrational number. Hence show that 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)

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

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

$(3m)^2 = 3q^2 \\[1.5em] \Rightarrow 9m^2 = 3q^2 \\[1.5em] \Rightarrow 3m^2 = q^2 \\[1.5em]$

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

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

$\Rightarrow - \sqrt{3} = r - 5 \\[1.5em] \Rightarrow \sqrt{3} = 5 - r \\[1.5em]$

As r is rational and r ≠ 0, so 5 - r is rational

$\Rightarrow \sqrt{3}$ is rational

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

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

Prove that the following numbers are irrational:

(i) $3 + \sqrt{5}$

(ii) $15 - 2\sqrt{7}$

(iii) $\dfrac{1}{3 - \sqrt5}$

Answer

$\text{(i) } 3 + \sqrt{5}$

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

Then,

$3 + \sqrt{5} = r \\[1.5em] \Rightarrow \sqrt{5} = r - 3$

As r is rational, r - 3 is rational

$\Rightarrow \sqrt{5}$ is rational

But this contradicts the fact that $\sqrt{5}$ is irrational.

Hence, our assumption is wrong.

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

$\text{(ii) } 15 - 2\sqrt{7}$

Let us assume that $15 - 2\sqrt{7}$ is a rational number, say r.

Then,

$15 - 2\sqrt{7} = r \\[1.5em] \Rightarrow 2\sqrt{7} = 15 - r \\[1.5em] \Rightarrow \sqrt{7} = \dfrac{15 - r}{2} \\[1.5em]$

As r is rational, 15 - r is rational

$\Rightarrow \dfrac{15 - r}{2}$ is rational

$\Rightarrow \sqrt{7}$ is rational

But this contradicts the fact that $\sqrt{7}$ is irrational.

Hence, our assumption is wrong.

∴ $\bold{15 - 2\sqrt{7}}$ is an irrational number.

$\text{(iii) }\dfrac{1}{3 - \sqrt5}$

Let us rationalise the denominator

$\dfrac{1}{3 - \sqrt5} = \dfrac{1}{3 - \sqrt5} × \dfrac{3 + \sqrt5}{3 + \sqrt5} \\[1.5em] = \dfrac{3 + \sqrt5}{(3)^2 - (\sqrt5)^2} \\[1.5em] = \dfrac{3 + \sqrt5}{9 - 5} \\[1.5em] = \dfrac{3 + \sqrt5}{4} \\[1.5em]$

Let us assume that $\dfrac{3 + \sqrt5}{4}$ is a rational number, say r.

Then,

$\dfrac{3 + \sqrt5}{4} = r \\[1.5em] \Rightarrow 3 +\sqrt{5} = 4r \\[1.5em] \Rightarrow \sqrt{5} = 4r - 3 \\[1.5em]$

As r is rational, 4r is rational

$\Rightarrow$ 4r - 3 is also rational

$\Rightarrow \sqrt{5}$ is rational

But this contradicts the fact that $\sqrt{5}$ is irrational.

Hence, our assumption is wrong.

$\Rightarrow \dfrac{3 + \sqrt5}{4}$ is an irrational number.

∴ $\bold{\dfrac{1}{3 - \sqrt5}}$ is an irrational number.

Rationalise the denominator of the following :

$\begin{matrix} \text{(i)} & \dfrac{10}{2\sqrt{2} + \sqrt{3}} \\[1.5em] \text{(ii)} & \dfrac{7\sqrt{3} - 5\sqrt{2}}{\sqrt{48} + \sqrt{18}} \\[1.5em] \text{(iii)} & \dfrac{1}{\sqrt{3} - \sqrt{2} + 1} \\[1.5em] \end{matrix}$

Answer

$\text{(i)}$

$\dfrac{10}{2\sqrt{2} + \sqrt{3}}$

Let us rationalise the denominator,

Then,

$\dfrac{10}{2\sqrt{2} + \sqrt{3}} = \dfrac{10}{2\sqrt{2} + \sqrt{3}} × \dfrac{2\sqrt{2} - \sqrt{3}}{2\sqrt{2} - \sqrt{3}} \\[1.5em] = \dfrac{10({2\sqrt{2} - \sqrt{3}})}{(2\sqrt{2})^2 - (\sqrt{3})^2} \\[1.5em] = \dfrac{10 × 2\sqrt{2} - 10 × \sqrt{3}}{(2\sqrt{2})^2 - (\sqrt{3})^2} \\[1.5em] = \dfrac{10(2\sqrt{2} - \sqrt{3})}{8 - 3} \\[1.5em] = \dfrac{10(2\sqrt{2} - \sqrt{3})}{5} \\[1.5em] \bold{= 2(2\sqrt{2} - \sqrt{3}) } \\[1.5em]$

$\text{(ii)}$ Since, it is given that

$\dfrac{7\sqrt{3} - 5\sqrt{2}}{\sqrt{48} + \sqrt{18}}$

Let us rationalise the denominator,

$\dfrac{7\sqrt{3} - 5\sqrt{2}}{\sqrt{48} + \sqrt{18}} × \dfrac{\sqrt{48} - \sqrt{18}}{\sqrt{48} - \sqrt{18}} \\[1.5em] = \dfrac{7\sqrt{3} × \sqrt{48} - 7\sqrt{3} × \sqrt{18} - 5\sqrt{2} × \sqrt{48} + 5\sqrt{2} × \sqrt{18} }{(\sqrt{48})^2 -(\sqrt{18})^2} \\[1.5em] = \dfrac{7\sqrt{144} - 7\sqrt{54} - 5\sqrt{96} +5\sqrt{36}}{48 - 18} \\[1.5em] = \dfrac{7 × 12 - 7\sqrt{2 × 3 × 3 × 3 } - 5\sqrt{2 × 2 × 2 × 2 × 2 × 3} +5\sqrt{2 × 2 × 3 × 3}}{30} \\[1.5em] = \dfrac{84 - 21\sqrt{6} - 20\sqrt{6} + 30}{30} \\[1.5em] = \dfrac{84 - 21\sqrt{6} - 20\sqrt{6} + 30}{30} \\[1.5em] = \dfrac{114 - 41{\sqrt6}}{30} \\[1.5em] = \dfrac{114}{30} - \dfrac{41\sqrt{6}}{30} \\[1.5em] = \bold{\dfrac{57}{15} - \dfrac{41\sqrt{6}}{30}} \\[1.5em]$

$\text{(iii)}$

$\dfrac{1}{\sqrt{3} - \sqrt{2} + 1}$

Let us rationalise the denominator,

Then,

$\dfrac{1}{\sqrt{3} - \sqrt{2} + 1} = \dfrac{1}{\sqrt{3} - (\sqrt{2} - 1)} × \dfrac{\sqrt{3} + (\sqrt{2} - 1)}{\sqrt{3} + (\sqrt{2} - 1)} \\[1.5em] =\dfrac{{\sqrt{3} + \sqrt{2}} - 1}{{(\sqrt{3})^2 }- (\sqrt{2} - 1)^2} \\[1.5em] =\dfrac{\sqrt{3} + \sqrt{2} - 1}{(\sqrt{3})^2 - ( (\sqrt{2})^2 - 2 × \sqrt{2} + 1)} \\[1.5em] =\dfrac{\sqrt{3} + \sqrt{2} - 1}{3 - (2 - 2\sqrt{2} + 1)} \\[1.5em] =\dfrac{\sqrt{3} + \sqrt{2} - 1}{3 - 2 + 2\sqrt{2} -1} \\[1.5em] =\dfrac{\sqrt{3} + \sqrt{2} - 1}{2\sqrt{2}} \\[1.5em] =\dfrac{\sqrt{3} + \sqrt{2} - 1}{2\sqrt{2}} × \dfrac{\sqrt{2}}{\sqrt{2}} \\[1.5em] =\dfrac{\sqrt{2}(\sqrt{3} + \sqrt{2} - 1)}{2\sqrt{2} × \sqrt{2}} \\[1.5em] =\dfrac{\sqrt{2} × \sqrt{3} + \sqrt{2} × \sqrt{2} - \sqrt{2}}{2\sqrt{2} × \sqrt{2}} \\[1.5em] = \dfrac{\sqrt{6} + 2 -\sqrt{2}}{4} \\[1.5em] \bold{{=} \dfrac{2 + \sqrt{6} -\sqrt{2}}{4}} \\[1.5em]$

If p, q are rational numbers and $p - \sqrt{15}q = \dfrac{2\sqrt{3} - \sqrt{5}}{4\sqrt{3} - 3\sqrt{5}}$, find the values of p and q.

Answer

Since, it is given that

$\dfrac{2\sqrt{3} - \sqrt{5}}{4\sqrt{3} - 3\sqrt{5}} = p - \sqrt{15} q$

On solving,

$\dfrac{2\sqrt{3} - \sqrt{5}}{4\sqrt{3} - 3\sqrt{5}} × \dfrac{4\sqrt{3} + 3\sqrt{5}}{4\sqrt{3} + 3\sqrt{5}} \\[1.5em] = \dfrac{2\sqrt{3} × 4\sqrt{3}+ 2\sqrt{3} × 3\sqrt{5} - \sqrt{5} × 4\sqrt{3} - \sqrt{5} × 3\sqrt{5} }{(4\sqrt{3})^2 -(3\sqrt{5})^2} \\[1.5em] = \dfrac{24 + 6\sqrt{15} - 4\sqrt{15} - 15 }{(4\sqrt{3})^2 -(3\sqrt{5})^2} \\[1.5em] = \dfrac{24 - 15 + 6\sqrt{15} - 4\sqrt{15} }{48 - 45} \\[1.5em] = \dfrac{9 + 2{\sqrt{15}}}{3} \\[1.5em] = \dfrac{9}{3} + \dfrac{2\sqrt{15}}{3} \\[1.5em] = 3 + \dfrac{2}{3}{\sqrt{15}} \\[1.5em] \therefore 3 - \Big(-\dfrac{2}{3}\Big){\sqrt{15}} = p - q\sqrt{15}$

Hence, p = 3 and q = $-\dfrac{2}{3}$.

If x = $\dfrac{1}{3 + 2\sqrt2}$ , then find the value of $x - \dfrac{1}{x}$.

Answer

Given,

$x =\dfrac{1}{3 + 2\sqrt{2}} \qquad \text{....(i)} \\[1.5em]$

Let us rationalise the denominator,

$x = \dfrac{1}{3 + 2\sqrt{2}} ×\dfrac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} \\[1.5em] = \dfrac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2} \\[1.5em] = \dfrac{3 - 2\sqrt{2}}{9 - 8} \\[1.5em] = (3 - 2\sqrt{2}) \\[1.5em] \therefore x = (3 - 2\sqrt{2})$

From (i) we get,

$\dfrac{1}{x} = 3 + 2\sqrt{2} \\[1.5em] \therefore x - \dfrac{1}{x} = (3 - 2\sqrt{2}) - (3 + 2\sqrt{2}) = 3 - 2\sqrt{2} - 3 - 2\sqrt{2}\\[1.5em] \Rightarrow\bold{ x - \dfrac{1}{x} = -4\sqrt{2}} \\[1.5em]$

(i) If x = $\dfrac{7 + 3\sqrt{5}}{7- 3\sqrt{5}}$ , find the value of $x^2 + \dfrac{1}{x^2}$

(ii) If x = $\dfrac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}$ and y = $\dfrac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}$ , find the value of x² + xy + y²

(iii) If x = $\dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$ and y = $\dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}$ , find the value of x³ + y³.

Answer

(i) Given x = $\dfrac{7 + 3\sqrt{5}}{7- 3\sqrt{5}}$

Rationalising the denominator,

$\dfrac{7 + 3\sqrt{5}}{7- 3\sqrt{5}} = \dfrac{7 + 3\sqrt{5}}{7- 3\sqrt{5}} × \dfrac{7 + 3\sqrt{5}}{7 + 3\sqrt{5}} \\[1.5em] = \dfrac{({7 + 3\sqrt{5}})^2}{(7)^2 - (3\sqrt{5})^2} \\[1.5em] = \dfrac{7^2 + 2 × 7 × 3\sqrt{5} + (3\sqrt{5})^2}{49 - 45} = \dfrac{49 + 42\sqrt{5} + 45}{4} = \dfrac{94 + 42\sqrt{5}}{4} \\[1.5em] = \dfrac{47 + 21\sqrt{5}}{2} \\[1.5em] \therefore x = \dfrac{47 + 21\sqrt{5}}{2} \\[1.5em] \Rightarrow \dfrac{1}{x} = \dfrac{2}{47 + 21\sqrt{5}} \\[1.5em]$

Rationalising denominator of $\dfrac{1}{x}$,

$\dfrac{1}{x} = \dfrac{2}{47 + 21\sqrt{5}} × \dfrac{47 - 21\sqrt{5}}{47 - 21\sqrt{5}} \\[1.5em] = \dfrac{{2}(47 - 21\sqrt{5})} {(47)^2 - (21\sqrt{5})^2} \\[1.5em] = \dfrac{{2}(47 - 21\sqrt{5})} {2209 - 2205} \\[1.5em] = \dfrac{{2}(47 - 21\sqrt{5})} {4} \\[1.5em] = \dfrac{(47 - 21\sqrt{5})} {2} \\[1.5em]$

Now,

${\Big(x + \dfrac{1}{x}\Big)} = \dfrac{47 + 21\sqrt{5}}{2} + \dfrac{(47 - 21\sqrt{5})}{2} \\[1.5em] = \dfrac{47 + 21\sqrt{5} + 47 -21\sqrt{5}}{2} \\[1.5em] = \dfrac{94}{2} = 47 \\[1.5em] \therefore {\Big(x + \dfrac{1}{x}\Big)} = 47 \qquad \text{....(i)}$

We know that ${\Big(x + \dfrac{1}{x}\Big)}^2 = x^2 + \dfrac{1}{x^2} + 2$

$\Rightarrow x^2 + \dfrac{1}{x^2} = {\Big(x + \dfrac{1}{x}\Big)}^2 -2 \\[1.5em] \Rightarrow x^2 + \dfrac{1}{x^2} = (47)^2 - 2 \qquad ..... \text{using(i)} \\[1.5em] \Rightarrow x^2 + \dfrac{1}{x^2} = 2209 - 2 = 2207 \\[1.5em] \bold{x^2 + \dfrac{1}{x^2} = 2207} \\[1.5em]$

$\text{(ii) } x = \dfrac{\sqrt{5} - \sqrt{2} }{\sqrt{5} + \sqrt{2}} \text{ and y} = \dfrac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} \\[1.5em] \therefore x+y = \dfrac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}} +\dfrac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} \\[1.5em] \Rightarrow\dfrac{(\sqrt{5} - \sqrt{2})^2 + (\sqrt{5} + \sqrt{2})^2}{(\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2})} \\[1.5em] \Rightarrow\dfrac{(\sqrt{5})^2 - 2 × \sqrt{2} × \sqrt{5} + (\sqrt{2})^2+ (\sqrt{5})^2 + 2 × \sqrt{2} × \sqrt{5} + (\sqrt{2})^2 }{(\sqrt{5})^2 - (\sqrt{2})^2 } \\[1.5em] \Rightarrow\dfrac{5 - 2\sqrt{10} + 2 + 5 + 2\sqrt{10} + 2}{5 - 2} = \dfrac{14}{3} \\[1.5em] \Rightarrow x + y = \dfrac{14}{3} \qquad \text{....(i)} \\[1.5em] \text{Also } xy = \dfrac{\sqrt{5} - \sqrt{2} }{\sqrt{5} + \sqrt{2}} ×\dfrac{\sqrt{5} + \sqrt{2} }{\sqrt{5} - \sqrt{2}} = 1 \qquad \text{....(ii)} \\[1.5em]$

We need to find the value of ${x^2 + xy + y^2}$ ${x^2 + xy + y^2} = x^2 + y^2 + 2xy - xy \\[1.5em] \Rightarrow {x^2 + xy + y^2} = (x+y)^2 - xy \qquad \text{....(iii)} \\[1.5em]$

Substituting the values from (i) and (ii) in (iii),

${x^2 + xy + y^2} = \Big(\dfrac{14}{3}\Big)^2 - 1 = \dfrac{196}{9} - 1 = \dfrac{196 - 9}{9} = \dfrac{187}{9} \\[1.5em] \therefore \bold{x^2 + xy + y^2} = \bold{\dfrac{187}{9}} \\[1.5em]$

$\text{(iii) } x = \dfrac{\sqrt{3} - \sqrt{2} }{\sqrt{3} + \sqrt{2}} \text{ and y} = \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \\[1.5em] \therefore x+y = \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} +\dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \\[1.5em] = \dfrac{(\sqrt{3} - \sqrt{2})^2 + (\sqrt{3} + \sqrt{2})^2}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})} \\[1.5em] = \dfrac{(\sqrt{3})^2 - 2 × \sqrt{2} × \sqrt{3} + (\sqrt{2})^2+ (\sqrt{3})^2 + 2 × \sqrt{2} × \sqrt{3} + (\sqrt{2})^2 }{(\sqrt{3})^2 - (\sqrt{2})^2 } \\[1.5em] = \dfrac{3 - 2\sqrt{6} + 2 + 3 + 2\sqrt{6} + 2}{3 - 2} = \dfrac{10}{1} = 10 \\[1.5em] \therefore x + y = 10 \qquad \text{....(i)} \\[1.5em] \text{Also } xy = \dfrac{\sqrt{3} - \sqrt{2} }{\sqrt{3} + \sqrt{2}} ×\dfrac{\sqrt{3} + \sqrt{2} }{\sqrt{3} - \sqrt{2}} = 1 \qquad \text{....(ii)} \\[1.5em]$

We need to find the value of ${x^3 + y^3}$ ${x^3 + y^3} = (x+y)^3 - 3xy(x + y) \qquad \text{....(iii)} \\[1.5em]$

Substituting the values from (i) and (ii) in (iii),

${x^3+ y^3} = (10)^3 - 3 × 1 × 10 \\[1.5em] = 1000 - 30 \\[1.5em] = 970 \\[1.5em] \therefore \bold{x^3+ y^3 = 970}$

Write the following real numbers in descending order :

$\sqrt{2}, 3.5, \sqrt{10}, -\dfrac{5}{\sqrt{2}}, {\dfrac{5}{2}}{\sqrt{3}}$

Answer

Write all the numbers as square root under one radical :

$\sqrt{2} = \sqrt{2} \\[1.5em] 3.5 = \sqrt{12.25} \\[1.5em] \sqrt{10} = \sqrt{10} \\[1.5em] {-\dfrac{5}{\sqrt{2}}} = - \sqrt{\dfrac{25}{2}} = -\sqrt{12.5} \\[1.5em] {\dfrac{5}{2}}{\sqrt{3}} = \sqrt{\dfrac{25 × 3}{4}} = \sqrt{\dfrac{75}{4}} = \sqrt{18.75} \\[1.5em] \text{Since} , 18.75 \gt 12.25 \gt 10 \gt 2 \gt - 12.5 \\[1.5em] \Rightarrow \sqrt{18.75} \gt \sqrt{12.25} \gt \sqrt{10} \gt \sqrt{2} \gt -\sqrt{12.5} \\[1.5em] \Rightarrow {\dfrac{5}{2}}{\sqrt{3}}\gt 3.5 \gt \sqrt{10} \gt \sqrt{2} \gt -\dfrac{5}{\sqrt{2}}$

Hence, the given numbers in descending order are $\bold{\dfrac{5}{2}{\sqrt{3}}} , \bold{3.5} , \bold{\sqrt{10}} ,\bold{\sqrt{2}} , \bold{-\dfrac{5}{\sqrt{2}}}$.

Find a rational number and an irrational number between $\sqrt{3}$ and $\sqrt{5}$.

Answer

Consider the squares of $\sqrt{3}$ and $\sqrt{5}$

$(\sqrt{3})^2$ = 3 and $(\sqrt{5})^2$ = 5

Here, 4 is rational number between 3 and 5

$\sqrt{4}$ = 2 is rational number between $\sqrt{3}$ and $\sqrt{5}$.

Irrational number between $\sqrt{3}$ and $\sqrt{5}$ = $\dfrac{\sqrt{3} + \sqrt{5}}{2}$

Insert three irrational numbers between $2\sqrt{3}$ and $2\sqrt{5}$ , and arrange in descending order.

Answer

Consider the squares of $2\sqrt{3}$ and $2\sqrt{5}$.

$(2\sqrt{3})^2$ = 4 × 3 = 12 and $(2\sqrt{5})^2$ = 4 × 5 = 20

As , $18 \gt 17 \gt 15$ it follows that

$\sqrt{18} \gt \sqrt{17} \gt \sqrt{15}$ , therefore

$\sqrt{18}$ , $\sqrt{17}$ , $\sqrt{15}$ lie between $\sqrt{12}$ and $\sqrt{20}$ i.e. $2\sqrt{3}$ and $2\sqrt{5}$.

Hence, three irrational number between $2\sqrt{3}$ and $2\sqrt{5}$ in descending order are $\sqrt{18}$ , $\sqrt{17}$ , $\sqrt{15}$.

Give an example each of two different irrational numbers , whose

(i) sum is an irrational number.

(ii) product is an irrational number.

Answer

Let a = $\sqrt{2}$ and b = $\sqrt{3}$ are two different irrational numbers :

(i) a + b = $\sqrt{2}$ + $\sqrt{3}$ is also an irrational number.

(ii) a × b = $\sqrt{2}$ × $\sqrt{3}$ = $\sqrt{6}$ is also an irrational number.

Give an example of two different irrational numbers , a and b where $\dfrac{a}{b}$ is a rational number.

Answer

Let a = $2\sqrt{3}$ and b = $5\sqrt{3}$ be two different irrational numbers

Here, $\dfrac{a}{b}$ = $\dfrac{2\sqrt{3}}{5\sqrt{3}}$ = $\dfrac{2}{5}$

$\therefore \bold{\dfrac{2}{5}}$ is a rational number.

If 34.0356 is expressed in the form $\dfrac{p}{q}$, where p and q are coprime integers, then what can you say about the factorisation of q ?

Answer

34.0356 can be expressed in the form $\dfrac{p}{q}$

This can be written as 34.0356 = $\dfrac{340356}{10000}$ = $\dfrac{85089}{2500}$

Here, 85089 and 2500 are coprime integers .

Since , it is terminating decimal

It is Rational number and the prime factors of its denominator q will be 2 or 5 or both .

In each case, state whether the following numbers are rational or irrational. If they are rational and expressed in the form $\dfrac{p}{q}$, where p and q are coprime integers, then what can you say about the prime factors of q?

$\begin{matrix} \text{(i)} & 279.034 \\[1.5em] \text{(ii)} & 76.\overline{17893} \\[1.5em] \text{(iii)} & 3.010010001... \\[1.5em] \text{(iv)} & 39.546782 \\[1.5em] \text{(v)} & 2.3476817681... \\[1.5em] \text{(vi)} & 59.120120012000... \\[1.5em] \end{matrix}$

Answer

(i) 279.034

This can be written as 279.034 = $\dfrac{279034}{1000}$

Since, it is terminating decimal

It is Rational number and the prime factors of its denominator q will be 2 or 5 or both .

(ii) $76.\overline{17893}$

Since it is non-terminating recurring decimal,

$76.\overline{17893}$ = 76.1789317893...

It is a rational number which is non-terminating and repeating. Its denominator q will have prime factors other than 2 or 5.

(iii) 3.010010001...

Since, it is non-terminating non-repeating decimal number

∴ It is an Irrational number.

(iv) 39.546782

This can be written as 39.546782 = $\dfrac{39546782}{1000000}$

Since , it is terminating decimal

It is Rational number and the prime factors of its denominator q will be 2 or 5 or both .

(v) 2.3476817681... = $2.34\overline{7681}$

Since, it is a non-terminating repeating decimal number,

∴ It is a Rational number and its denominator q will have prime factors other than 2 or 5.

(vi) 59.120120012000...

Since, it is non-terminating non-repeating decimal number

∴ It is an Irrational number.