Mathematics Solutions Solutions for Class 8 Maths Chapter 1 Rational And Irrational Numbers are provided here with simple step-by-step explanations. These solutions for Rational And Irrational Numbers are extremely popular among Class 8 students for Maths Rational And Irrational Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 8 Maths Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 8 Maths are prepared by experts and are 100% accurate.

#### Page No 2:

#### Question 1:

Show the following numbers on a number line. Draw a separate number line for each example.

(1) $\frac{3}{2},\frac{5}{2},-\frac{3}{2}$

(2) $\frac{7}{5},\frac{-2}{5},\frac{-4}{5}$

(3) $\frac{-5}{8},\frac{11}{8}$

(4) $\frac{13}{10},\frac{-17}{10}$

#### Answer:

$\left(1\right)\frac{3}{2},\frac{5}{2},-\frac{3}{2}$ can be represented on the number line as follows.

$\left(2\right)\frac{7}{5},-\frac{2}{5},-\frac{4}{5}$ can be represented on the number line as follows.

$\left(3\right)-\frac{5}{8},\frac{11}{8}$ can be represented on the number line as follows.

$\left(4\right)\frac{13}{10},-\frac{17}{10}$ can be represented on the number line as follows.

#### Page No 2:

#### Question 2:

Observe the number line and answer the questions.

(1) Which number is indicated by point B?

(2) Which point indicates the number $1\frac{3}{4}$?

(3) State whether the statement, 'the point D denotes the number $\frac{5}{2}$, is true of false.

#### Answer:

(1) We observe that each unit on the number line is divided into 4 equal parts.

Now, B is the tenth point on the left of 0.

So, B indicates $-\frac{10}{4}$ on the number line .

$\left(2\right)1\frac{3}{4}=\frac{7}{4}=7\times \frac{1}{4}$

So, the seventh point on the right of 0 is C that indicates $1\frac{3}{4}$ on the number line.

(3) The point D is the tenth point on the right of 0. So, D indicates $\frac{10}{4}$ on the number line.

Now, $\frac{10}{4}=\frac{5}{2}$

So, D denotes $\frac{5}{2}$ on the number line. Hence, the given statement is true.

#### Page No 3:

#### Question 1:

Compare the following numbers.

(1) −7, −2

(2) 0, $\frac{-9}{5}$

(3) $\frac{8}{7}$, 0

(4) $\frac{-5}{4}$, $\frac{1}{4}$

(5) $\frac{40}{29}$, $\frac{141}{29}$

(6) −$\frac{17}{20}$, $\frac{-13}{20}$

(7) $\frac{15}{12}$, $\frac{7}{16}$

(8) $\frac{-25}{8}$, $\frac{-9}{4}$

(9) $\frac{12}{15}$, $\frac{3}{5}$

(10) $\frac{-7}{11}$, $\frac{-3}{4}$

#### Answer:

(1) We know that, 7 > 2

∴ −7 < −2.

(2) We know that, a negative number is always less than 0.

∴ $0>-\frac{9}{5}$.

(3) We know that, a positive number is always greater than 0.

∴ $\frac{8}{7}>0$

(4) We know that, −5 < 1.

∴ $-\frac{5}{4}<\frac{1}{4}$.

(5) We know that, 40 < 141.

∴ $\frac{40}{29}<\frac{141}{29}$.

(6) We know that, −17 < −13.

∴$-\frac{17}{20}<-\frac{13}{20}$.

$\left(7\right)\frac{15}{12}=\frac{15\times 4}{12\times 4}=\frac{60}{48};\frac{7}{16}=\frac{7\times 3}{16\times 3}=\frac{21}{48}$

Now, $\frac{60}{48}>\frac{21}{48}$

∴ $\frac{15}{12}>\frac{7}{16}$.

(8) Let us first compare $\frac{25}{8}$ and $\frac{9}{4}$.

$\frac{25}{8}=\frac{25\times 1}{8\times 1}=\frac{25}{8};\frac{9}{4}=\frac{9\times 2}{4\times 2}=\frac{18}{8}$

Now, $\frac{25}{8}>\frac{18}{8}$

∴ $\frac{25}{8}>\frac{9}{4}$

∴ $-\frac{25}{8}<-\frac{9}{4}$.

$\left(9\right)\frac{12}{15}=\frac{12\times 1}{15\times 1}=\frac{12}{15};\frac{3}{5}=\frac{3\times 3}{5\times 3}=\frac{9}{15}$

Now, $\frac{12}{15}>\frac{9}{15}$

∴ $\frac{12}{15}>\frac{3}{5}$.

(10) Let us first compare $\frac{7}{11}$ and $\frac{3}{4}$.

$\frac{7}{11}=\frac{7\times 4}{11\times 4}=\frac{28}{44};\frac{3}{4}=\frac{3\times 11}{4\times 11}=\frac{33}{44}$

Now, $\frac{28}{44}<\frac{33}{44}$

∴ $\frac{7}{11}<\frac{3}{4}$

∴ $-\frac{7}{11}>-\frac{3}{4}$.

#### Page No 4:

#### Question 1:

Write the following rational numbers in decimal form.

(1) $\frac{9}{37}$

(2) $\frac{18}{42}$

(3) $\frac{9}{14}$

(4) $\frac{-103}{5}$

(5) −$\frac{11}{13}$

#### Answer:

(1) The given number is $\frac{9}{37}$.

∴ $\frac{9}{37}$ = 0.243243.... = $0.\overline{)243}$

The decimal form of $\frac{9}{37}$ is $0.\overline{)243}$.

(2) The given number is $\frac{18}{42}$.

∴ $\frac{18}{42}$ = 0.428571428571.... = $0.\overline{)428571}$

The decimal form of $\frac{18}{42}$ is $0.\overline{)428571}$.

(3) The given number is $\frac{9}{14}$.

∴ $\frac{9}{14}$ = 0.6428571428571.... = $0.6\overline{)428571}$

The decimal form of $\frac{9}{14}$ is $0.6\overline{)428571}$.

(4) The given number is $-\frac{103}{5}$.

∴ $\frac{103}{5}$ = 20.6

The decimal form of $-\frac{103}{5}$ is −20.6.

(5) The given number is $-\frac{11}{13}$.

∴ $\frac{11}{13}$ = 0.846153846153.... = $0.\overline{)846153}$

The decimal form of $-\frac{11}{13}$ is $-0.\overline{)846153}$.

#### Page No 5:

#### Question 1:

The number $\sqrt{2}$ is shown on a number line. Steps are given to show $\sqrt{3}$ on the number line using $\sqrt{2}$. Fill in the boxes properly and complete the activity.

Activity:

∙ The point Q on the number line shows the number ........

∙ A line perpendicular to the number line is drawn through the point Q. Point R is at unit distance from Q on the line.

∙ Right angled ∆ ORQ is obtained by drawing seg OR.

∙ *l*(OQ) = $\sqrt{2}$, *l*(QR) = 1

[

*l*(OR)]

^{2}= [

*l*(OQ)]

^{2}+ [

*l*(QR)]

^{2}

^{2}+ $\overline{){0}}$

^{2 }= $\overline{){0}}$ + $\overline{){0}}$

∴

*l*(OR) = $\overline{){0}}$

#### Answer:

∙ The point Q on the number line shows the number $\overline{)\sqrt{2}}$.

∙ A line perpendicular to the number line is drawn through the point Q. Point R is at unit distance from Q on the line.

∙ Right angled ∆ ORQ is obtained by drawing seg OR.

∙ *l*(OQ) = $\sqrt{2}$, *l*(QR) = 1

∴ by Pythagoras theorem,

[*l*(OR)]^{2} = [*l*(OQ)]^{2} + [*l*(QR)]^{2}

= ${\overline{)\sqrt{2}}}^{2}+{\overline{)1}}^{2}=\overline{)2}+\overline{)1}$

= 3

∴ *l*(OR) = $\overline{)\sqrt{3}}$

Draw an arc with centre O and radius OR. Mark the point of intersection of the line and the arc as C. The point C shows the number line $\sqrt{3}$.

#### Page No 6:

#### Question 2:

Show the number $\sqrt{5}$ on the number line.

#### Answer:

Draw a number line as shown in the figure. Let the point O represent 0 and point Q represent 2. Draw a perpendicular QR at Q on the number line such that QR = 1 unit. Join OR. Now, ∆OQR is a right angled triangle.

By Pythagoras theorem, we have

OR^{2} = OQ^{2} + QR^{2}

= (2)^{2 }+ (1)^{2}

= 4 + 1

= 5

∴ OR = $\sqrt{5}$

Taking O as the centre and radius OR = $\sqrt{5}$, draw an arc cutting the number line at C.

Clearly, OC = OR = $\sqrt{5}$.

Hence, C represents $\sqrt{5}$ on the number line.

#### Page No 6:

#### Question 3:

Show the number $\sqrt{7}$ on the number line.

#### Answer:

Draw a number line as shown in the figure and mark the points O, A and B on it such that OA = AB = 1 unit. The point O represents 0 and B represents 2. At B, draw CB perpendicular on the number line such that BC = 1 unit. Join OC. Now, ∆OBC is a right angled triangle.

In ∆OBC, by Pythagoras theorem

(OC)^{2} = (OB)^{2} + (BC)^{2}

= (2)^{2} + (1)^{2}

= 4 + 1

= 5

∴ OC = $\sqrt{5}$

Taking O as centre and radius OC = $\sqrt{5}$, draw an arc cutting the number line at D.

Clearly, OC = OD = $\sqrt{5}$

At D, draw ED perpendicular on the number line such that ED = 1 unit. Join OE. Now, ∆ODE is a right angled triangle.

In ∆ODE, by Pythagoras theorem

(OE)^{2} = (OD)^{2} + (DE)^{2}

= ($\sqrt{5}$)^{2} + (1)^{2}

= 5 + 1

= 6

∴ OE = $\sqrt{6}$

Taking O as centre and radius OE = $\sqrt{6}$, draw an arc cutting the number line at F.

Clearly, OE = OF = $\sqrt{6}$

At F, draw GF perpendicular on the number line such that GF = 1 unit. Join OG. Now, ∆OFG is a right angled triangle.

In ∆OFG, by Pythagoras theorem

(OG)^{2} = (OF)^{2} + (FG)^{2}

= ($\sqrt{6}$)^{2} + (1)^{2}

= 6 + 1

= 7

∴ OG = $\sqrt{7}$

Taking O as centre and radius OG = $\sqrt{7}$, draw an arc cutting the number line at H.

Clearly, OG = OH = $\sqrt{7}$

Hence, H represents $\sqrt{7}$ on the number line.

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