Mathematics Solutions Solutions for Class 8 Math Chapter 1 - Rational And Irrational Numbers

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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:

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

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

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

$(4) \frac{13}{10}, - \frac{17}{10}$ 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 .
$(2) 1 \frac{3}{4} = \frac{7}{4} = 7 \times \frac{1}{4}$
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.

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

$(7) \frac{15}{12} = \frac{15 \times 4}{12 \times 4} = \frac{60}{48}; \frac{7}{16} = \frac{7 \times 3}{16 \times 3} = \frac{21}{48}$
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, $\frac{25}{8} > \frac{18}{8}$
∴ $\frac{25}{8} > \frac{9}{4}$
∴ $- \frac{25}{8} < - \frac{9}{4}$ .

$(9) \frac{12}{15} = \frac{12 \times 1}{15 \times 1} = \frac{12}{15}; \frac{3}{5} = \frac{3 \times 3}{5 \times 3} = \frac{9}{15}$
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.... = $0.243$
The decimal form of $\frac{9}{37}$ is $0.243$ .

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

∴ $\frac{18}{42}$ = 0.428571428571.... = $0.428571$
The decimal form of $\frac{18}{42}$ is $0.428571$ .

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

∴ $\frac{9}{14}$ = 0.6428571428571.... = $0.6428571$
The decimal form of $\frac{9}{14}$ is $0.6428571$ .

(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.... = $0.846153$
The decimal form of $- \frac{11}{13}$ is $- 0.846153$ .

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:

âˆ™ The point Q on the number line shows the number ........
âˆ™ A line perpendicular to the number line is drawn through the point Q. Point R is at unit distance from Q on the line.
âˆ™ Right angled âˆ† ORQ is obtained by drawing seg OR.
âˆ™ l(OQ) = $\sqrt{2}$ , l(QR) = 1

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

0

² +

0

²=

0

0

0

∴ l(OR) =

0

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}

Answer:

âˆ™ The point Q on the number line shows the number $\sqrt{2}$ .
âˆ™ A line perpendicular to the number line is drawn through the point Q. Point R is at unit distance from Q on the line.
âˆ™ Right angled âˆ† ORQ is obtained by drawing seg OR.
âˆ™ l(OQ) = $\sqrt{2}$ , l(QR) = 1
∴ by Pythagoras theorem,
[l(OR)]² = [l(OQ)]² + [l(QR)]²
= ${\sqrt{2}}^{2} + 1^{2} = 2 + 1$
= 3
∴ l(OR) = $\sqrt{3}$
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, âˆ†OQR is a right angled triangle.
By Pythagoras theorem, we have
OR² = OQ² + QR²
= (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, âˆ†OBC is a right angled triangle.
In âˆ†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, âˆ†ODE is a right angled triangle.
In âˆ†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, âˆ†OFG is a right angled triangle.
In âˆ†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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