Mathematics Solutions Solutions for Class 8 Math 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 Math 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 Math 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 Math are prepared by experts and are 100% accurate.

#### Question 1:

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

(1)

(2)

(3)

(4)

can be represented on the number line as follows.

can be represented on the number line as follows.

can be represented on the number line as follows.

can be represented on the number line as follows.

#### 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.

(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 .

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

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

(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.

$-\frac{17}{20}<-\frac{13}{20}$.

Now, $\frac{60}{48}>\frac{21}{48}$
∴ $\frac{15}{12}>\frac{7}{16}$.

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

Now, $\frac{25}{8}>\frac{18}{8}$
∴ $\frac{25}{8}>\frac{9}{4}$
∴ $-\frac{25}{8}<-\frac{9}{4}$.

Now, $\frac{12}{15}>\frac{9}{15}$
∴ $\frac{12}{15}>\frac{3}{5}$.

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

Now, $\frac{28}{44}<\frac{33}{44}$
∴ $\frac{7}{11}<\frac{3}{4}$
∴ $-\frac{7}{11}>-\frac{3}{4}$.

#### 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) −$\frac{11}{13}$

(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 $-0.\overline{)846153}$.

#### 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

∴ by Pythagoras theorem,
[l(OR)]2 = [l(OQ)]2 + [l(QR)]2
2 + 2 = +

=

l(OR) =
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 $\sqrt{3}$.

∙ 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 $\sqrt{3}$.

#### Question 2:

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

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
OR2 = OQ2 + QR2
= (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.

#### Question 3:

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

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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