Rational and Irrational Numbers — Exercise 1.3

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 below:

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 below:

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.

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.

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.

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:

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

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

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.

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

Write the decimal expansion of $\dfrac{1}{7}$. Hence, write the decimal expansions of $\dfrac{2}{7}$, $\dfrac{3}{7}$, $\dfrac{4}{7}$, $\dfrac{5}{7}$ and $\dfrac{6}{7}$.

Answer

The fraction : $\dfrac{1}{7}$

Decimal expansion of $\dfrac{1}{7}$ = $\bold{0.\overline{142857}}$

Since, it is recurring

$\dfrac{2}{7}=2\times\dfrac{1}{7}=2\times0.\overline{142857}=\bold{0.\overline{285714}}$

$\dfrac{3}{7}=3\times\dfrac{1}{7}=3\times0.\overline{142857}=\bold{0.\overline{428571}}$

$\dfrac{4}{7}=4\times\dfrac{1}{7}=4\times0.\overline{142857}=\bold{0.\overline{571428}}$

$\dfrac{5}{7}=5\times\dfrac{1}{7}=5\times0.\overline{142857}=\bold{0.\overline{714285}}$

$\dfrac{6}{7}=6\times\dfrac{1}{7}=6\times0.\overline{142857}=\bold{0.\overline{857142}}$

Express the following numbers in the form $\dfrac{p}{q}$, where p and q are both integers and q ≠ 0.

$\begin{matrix} \text{(i)} & 0.\overline{3} \\[1.5em] \text{(ii)} & 5.\overline{2} \\[1.5em] \text{(iii)} & 0.404040... \\[1.5em] \text{(iv)} & 0.4\overline{7} \\[1.5em] \text{(v)} & 0.1\overline{34} \\[1.5em] \text{(vi)} & 0.\overline{001} \\[1.5em] \end{matrix}$

Answer

(i) Let x = $0.\overline{3}$ = 0.333333... $\qquad \text{....(i)}$

As there is one repeating digit after decimal point,

So multiplying both sides of (i) by 10

we get,

10x = 3.3333...$\qquad \text{....(ii)}$

Subtracting (i) from (ii), we get

9x = 3

x = $\dfrac{3}{9}$ = $\bold{\dfrac{1}{3}}$,

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

(ii) Let x = $5.\overline{2}$ = 5.2222... $\qquad \text{....(i)}$

As there is one repeating digit after decimal point,

So multiplying both sides of (i) by 10

we get,

10x = 52.2222...$\qquad \text{....(ii)}$

Subtracting (i) from (ii), we get

9x = 47

x = $\bold{\dfrac{47}{9}}$,

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

(iii) Let x = $0.\overline{40}$ = 0.4040... $\qquad \text{....(i)}$

As there are two repeating digit after decimal point,

So multiplying both sides of (i) by 100

we get,

100x = 40.4040...$\qquad \text{....(ii)}$

Subtracting (i) from (ii), we get

99x = 40

x = $\bold{\dfrac{40}{99}}$,

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

(iv) Let x = $0.4\overline{7}$ = 0.477777... $\qquad \text{....(i)}$

As there is one repeating digit after decimal point ,

So multiplying both sides of (i) by 10

we get,

10x=4.7777...$\qquad \text{....(ii)}$

Multiply by 100 on both sides

100x=47.77777.....$\qquad \text{....(iii)}$

subtracting (ii) from (iii), we get

100x-10x=47.7777.. -4.777...

90x= 43

x = $\bold{\dfrac{43}{90}}$

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

(v) Let x = $0.1\overline{34}$ = 0.13434 ... $\qquad \text{....(i)}$

So multiplying both sides of (i) by 10

we get,

10x=1.343434...$\qquad \text{....(ii)}$

Again multiply by 100 on both sides ,

1000x =134.3434.....$\qquad \text{....(iii)}$

Subtracting (ii) from (iii), we get

1000x - 10x = 134.3434... - 1.3434...

990x = 133

x = $\bold{\dfrac{133}{990}}$

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

(vi) Let x = $0.\overline{001}$ = 0.001001001... $\qquad \text{....(i)}$

So multiplying both sides of (i) by 1000,

we get,

1000x = 1.001001...$\qquad \text{....(ii)}$

Subtracting (i) from (ii), we get

1000x - x = 1.001001... - 0.001001...

999x = 1

x = $\bold{\dfrac{1}{999}}$

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

Classify the following numbers as rational or irrational:

$\begin{matrix} \text{(i)} & \sqrt{23} \\[1.5em] \text{(ii)} & \sqrt{225} \\[1.5em] \text{(iii)} & 0.3796 \\[1.5em] \text{(iv)} & 7.478478... \\[1.5em] \text{(v)} & 1.101001000100001... \\[1.5em] \text{(vi)} & 345.0\overline{456} \\[1.5em] \end{matrix}$

Answer

Rational numbers are in the form c, q ≠ 0, and p and q are integers.

(i) $\bold{\sqrt{23}}$ is an irrational number , as it is not a perfect square so it cannot be written in the form $\dfrac{p}{q}$ , q ≠ 0.

(ii) $\sqrt{225}$ = $\sqrt{15×15}$ = 15 = $\dfrac{15}{1}$,

As it can be written in the form $\dfrac{p}{q}$ , q ≠ 0.

∴ $\bold{\sqrt{225}}$ is a rational number .

(iii) 0.3796 = $\dfrac{3796}{1000}$

Since, the decimal expansion is terminating decimal.

∴ 0.3796 is a rational number .

(iv) 7.478478

Let x = 7.478478 $\text{....(i)}$

Since there is three repeating digit after decimal point,

Multiplying both sides by 1000, we get

1000x = 7478.478478... $\text{....(ii)}$

Subtracting (i) from (ii) we get,

999x = 7471

x = $\bold{\dfrac{7471}{999}}$

∴ It is non terminating , repeating rational number .

(v) 1.101001000100001...

Since number of 0's are increasing between two consecutive terms as we move further , So it is non terminating, non repeating decimal.

∴ 1.101001000100001... is an irrational number.

(vi) $345.0\overline{456}$

$345.0\overline{456}$ = 345.0456456...

Let x = 345.0456456...

Multiply both sides by 10, we get

10x = 3450.456456.. $\text{....(i)}$

Since, after decimal there are three repeating digit:

Multiply both sides by 1000, we get

10000x = 3450456.456456... $\text{....(ii)}$

Subtracting (i) from (ii) ,

9990x = 3447006

x = $\bold{\dfrac{3447006}{9990}}$

since, it is non-terminating , repeating decimal.

∴ $\bold{345.0\overline{456}}$ is a Rational number .

The following real numbers have decimal expansions as given below. In each case, state whether they are rational or not. If they are rational and expressed in the form $\dfrac{p}{q}$, where p, q are integers, q ≠ 0 and p, q are co-prime, then what can you say about the prime factors of q ?

$\begin{matrix} \text{(i)} & 37.09158 \\[1.5em] \text{(ii)} & 423.\overline{04567}\\[1.5em] \text{(iii)} & 8.9010010001... \\[1.5em] \text{(iv)} & 2.3476817681... \\[1.5em] \end{matrix}$

Answer

(i) 37.07158

This can be written as 37.09158 = $\dfrac{3709158}{100000}$

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) $423.\overline{04567}$

since it has non-terminating recurring decimal,

$423.\overline{04567}$ = 423.0456704567...

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

(iii) 8.9010010001...

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

∴ It is not a Rational number.

(iv) 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.

Insert an irrational number between the following :

(i) $\dfrac{1}{3}$ and $\dfrac{1}{2}$

(ii) $-\dfrac{2}{5}$ and $\dfrac{1}{2}$

(iii) 0 and 0.1

Answer

(i) One irrational number between $\dfrac{1}{3}$ and $\dfrac{1}{2}$

$\dfrac{1}{3}$ = 0.333...

$\dfrac{1}{2}$ = 0.5

So there are infinite irrational number between $\dfrac{1}{3}$ and $\dfrac{1}{2}$

One irrational number among them can be $\bold{0.4040040004...}$

(ii) One irrational number between $-\dfrac{2}{5}$ and $\dfrac{1}{2}$

$-\dfrac{2}{5}$ = -0.4

$\dfrac{1}{2}$ = 0.5

So, there are infinite irrational numbers between $-\dfrac{2}{5}$ and $\dfrac{1}{2}$

One irrational number among them can be $\bold{0.2020020002...}$

(iii) One irrational number among 0 and 0.1 , can be $\bold{0.050050005...}$

Insert two irrational number between 2 and 3.

Answer

Consider, the squares $(2)^2$ = 4 and $(3)^2$ = 9

Two irrational numbers can be te squares root of any natural number between 4 and 9.

As, As, $4 \lt 5 \lt 6\lt 9$ , it follows that

$\sqrt{4} \lt \sqrt{5} \lt \sqrt{6} \lt \sqrt{9}$

therefore , $\sqrt{5}$ and $\sqrt{6}$ lie between 2 and 3

$2 \lt \sqrt{5} \lt \sqrt{6} \lt \\ 3$

Hence, two irrational number between 2 and 3 or $\sqrt{4}$ and $\sqrt{9}$ are $\sqrt{5}$ , $\sqrt{6}$ .

Write two irrational numbers between $\dfrac{4}{9}$ and $\dfrac{7}{11}$.

Answer

$\dfrac{4}{9}$ is expressed as 0.4444...

$\dfrac{7}{11}$ is expressed as 0.636363..

So, two irrational number between $\dfrac{4}{9}$ and $\dfrac{7}{11}$ are 0.5050050005... and 0.6060060006...

Find a rational number between $\sqrt{2}$ and $\sqrt{3}$.

Answer

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

${(\sqrt2)^2}$ = 2 and ${(\sqrt3)^2}$ = 3

Take any rational number between 2 and 3 which is a perfect squares of a rational number,

One such number is 2.25 and

2.25 = $(1.5)^2$

$\sqrt{2.25}$ = 1.5

As, $2 \lt 2.25 \lt 3$ , it follows that

$\sqrt{2} \lt \sqrt{2.25} \lt \sqrt{3}$

$\sqrt{2} \lt 1.5 \lt \sqrt{3}$

Hence , one rational number between $\sqrt{2}$ and $\sqrt{3}$ is 1.5 .

Find two rational numbers between $2\sqrt{3}$ and $\sqrt{15}$.

Answer

$2\sqrt{3}$ = $\sqrt{4 × 3}$ = $\sqrt{12}$

So, we need to find two irrational number between $\sqrt{12}$ and $\sqrt{15}$

$\text{Since}, 12 \lt 12.25 \lt 12.96 \lt 15 \\[0.5em] \Rightarrow \sqrt{12} \lt \sqrt{12.25} \lt \sqrt{12.96} \lt \sqrt{15}$

Hence, two rational number between $\sqrt{12}$ and $\sqrt{15}$ are $\bold{\sqrt{12.25}}$ and $\bold{\sqrt{12.96}}$ .

Insert an irrational number between $\sqrt{5}$ and $\sqrt{7}$.

Answer

consider the squares of $\sqrt{5}$ and $\sqrt{7}$

${(\sqrt5)^2}$ = 5 and ${(\sqrt7)^2}$ = 7

As, $5 \lt 6\lt 7$ , it follows that

$\sqrt{5} \lt \sqrt{6} \lt \sqrt{7}$ therefore , $\sqrt{6}$ lie between $\sqrt{5}$ and $\sqrt{7}$

Hence, irrational number between $\sqrt{5}$ and $\sqrt{7}$ is $\bold{\sqrt{6}}$ .

Insert two irrational numbers between $\sqrt{3}$ and $\sqrt{7}$.

Answer

consider the squares of $\sqrt{3}$ and $\sqrt{7}$

${(\sqrt3)^2}$ = 3 and ${(\sqrt7)^2}$ = 7

As, $3 \lt 5 \lt 6\lt 7$ , it follows that

$\sqrt{3} \lt \sqrt{5} \lt \sqrt{6} \lt \sqrt{7}$ therefore , $\sqrt{5}$ and $\sqrt{6}$ lie between $\sqrt{3}$ and $\sqrt{7}$

Hence, two irrational number between $\sqrt{3}$ and $\sqrt{7}$ is $\bold{\sqrt{5}}$ and $\bold{\sqrt{6}}$ .