Mathematics Class 8th Chapter 1 - Rational And Irrational Numbers

Question 1:

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

(1) 32, 52, −32$\frac{3}{2}, \frac{5}{2}, -\frac{3}{2}$

(2) 75, −25, −45$\frac{7}{5}, \frac{-2}{5}, \frac{-4}{5}$

(3) −58, 118$\frac{-5}{8}, \frac{11}{8}$

(4) 1310, −1710$\frac{13}{10}, \frac{-17}{10}$

ANSWER:

(1) 32,52,−32$\left(1\right) \frac{3}{2},\frac{5}{2},-\frac{3}{2}$ can be represented on the number line as follows.

 
(2) 75,−25,−45$\left(2\right) \frac{7}{5},-\frac{2}{5},-\frac{4}{5}$ can be represented on the number line as follows.

 
(3) −58,118$\left(3\right) -\frac{5}{8},\frac{11}{8}$ can be represented on the number line as follows.

 
(4) 1310,−1710$\left(4\right) \frac{13}{10},-\frac{17}{10}$ can be represented on the number line as follows.
 

Queston 2

Compare the following numbers.
(1) −7, −2

(2) 0, −95$\frac{-9}{5}$

(3) 87$\frac{8}{7}$, 0

(4) −54$\frac{-5}{4}$, 14$\frac{1}{4}$

(5) 4029$\frac{40}{29}$, 14129$\frac{141}{29}$

(6)  −1720$\frac{17}{20}$, −1320$\frac{-13}{20}$

(7) 1512$\frac{15}{12}$, 716$\frac{7}{16}$

(8) −258$\frac{-25}{8}$, −94$\frac{-9}{4}$

(9) 1215$\frac{12}{15}$, 35$\frac{3}{5}$

(10) −711$\frac{-7}{11}$, −34$\frac{-3}{4}$

ANSWER:

(1) We know that, 7 > 2

∴ −7 < −2.

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

∴ 0>−95$0>-\frac{9}{5}$.

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

∴ 87>0$\frac{8}{7}>0$

(4) We know that, −5 < 1.

∴ −54<14$-\frac{5}{4}<\frac{1}{4}$.

(5) We know that, 40 < 141.

∴ 4029<14129$\frac{40}{29}<\frac{141}{29}$.

(6) We know that, −17 < −13.

∴−1720<−1320$-\frac{17}{20}<-\frac{13}{20}$.

(7) 1512=15×412×4=6048; 716=7×316×3=2148$\left(7\right) \frac{15}{12}=\frac{15\times 4}{12\times 4}=\frac{60}{48}; \frac{7}{16}=\frac{7\times 3}{16\times 3}=\frac{21}{48}$
Now, 6048>2148$\frac{60}{48}>\frac{21}{48}$
∴ 1512>716$\frac{15}{12}>\frac{7}{16}$.

(8) Let us first compare 258$\frac{25}{8}$ and 94$\frac{9}{4}$.
258=25×18×1=258; 94=9×24×2=188$\frac{25}{8}=\frac{25\times 1}{8\times 1}=\frac{25}{8}; \frac{9}{4}=\frac{9\times 2}{4\times 2}=\frac{18}{8}$
Now, 258>188$\frac{25}{8}>\frac{18}{8}$
∴ 258>94$\frac{25}{8}>\frac{9}{4}$
∴ −258<−94$-\frac{25}{8}<-\frac{9}{4}$.

(9) 1215=12×115×1=1215; 35=3×35×3=915$\left(9\right) \frac{12}{15}=\frac{12\times 1}{15\times 1}=\frac{12}{15}; \frac{3}{5}=\frac{3\times 3}{5\times 3}=\frac{9}{15}$
Now, 1215>915$\frac{12}{15}>\frac{9}{15}$
∴ 1215>35$\frac{12}{15}>\frac{3}{5}$.

(10) Let us first compare 711$\frac{7}{11}$ and 34$\frac{3}{4}$.
711=7×411×4=2844; 34=3×114×11=3344$\frac{7}{11}=\frac{7\times 4}{11\times 4}=\frac{28}{44}; \frac{3}{4}=\frac{3\times 11}{4\times 11}=\frac{33}{44}$
Now, 2844<3344$\frac{28}{44}<\frac{33}{44}$
∴ 711<34$\frac{7}{11}<\frac{3}{4}$
∴ −711>−34$-\frac{7}{11}>-\frac{3}{4}$.

Question 3

Write the following rational numbers in decimal form.
(1) 937$\frac{9}{37}$

(2) 1842$\frac{18}{42}$

(3) 914$\frac{9}{14}$

(4) −1035$\frac{-103}{5}$

(5) −1113$\frac{11}{13}$

ANSWER:

(1) The given number is 937$\frac{9}{37}$.

∴ 937$\frac{9}{37}$ = 0.243243.... = 0.243$0.\overline{243}$
The decimal form of 937$\frac{9}{37}$ is 0.243$0.\overline{243}$.

(2) The given number is 1842$\frac{18}{42}$.

∴ 1842$\frac{18}{42}$ = 0.428571428571.... = 0.428571$0.\overline{428571}$
The decimal form of 1842$\frac{18}{42}$ is 0.428571$0.\overline{428571}$.

(3) The given number is 914$\frac{9}{14}$.

∴ 914$\frac{9}{14}$ = 0.6428571428571.... = 0.6428571$0.6\overline{428571}$
The decimal form of 914$\frac{9}{14}$ is 0.6428571$0.6\overline{428571}$.

(4) The given number is −1035$-\frac{103}{5}$.

∴ 1035$\frac{103}{5}$ = 20.6
The decimal form of −1035$-\frac{103}{5}$ is −20.6.

(5) The given number is −1113$-\frac{11}{13}$.

∴ 1113$\frac{11}{13}$ = 0.846153846153.... = 0.846153$0.\overline{846153}$
The decimal form of −1113$-\frac{11}{13}$ is −0.846153$-0.\overline{846153}$.

Question 1:

The number 2–√$\sqrt{2}$ is shown on a number line. Steps are given to show 3–√$\sqrt{3}$ on the number line using 2–√$\sqrt{2}$. 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) = 2–√$\sqrt{2}$, l(QR) = 1

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

=   0   $\boxed{ 0 }$2 +   0   $\boxed{ 0 }$2 =   0   $\boxed{ 0 }$ +    0  $\boxed{ 0 }$
 

=    0  $\boxed{ 0 }$

∴ l(OR) =    0  $\boxed{ 0 }$

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 3–√$\sqrt{3}$.

 

ANSWER:

∙ The point Q on the number line shows the number 2–√$\underline{\sqrt{2}}$.
∙ 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) = 2–√$\sqrt{2}$, l(QR) = 1
∴ by Pythagoras theorem,
[l(OR)]2 = [l(OQ)]2 + [l(QR)]2
               = 2–√2+12=2+1${\boxed{\sqrt{2}}}^2+{\boxed{1}}^2=\boxed{2}+\boxed{1}$ 
               = 3
∴ l(OR) = 3–√$\boxed{\sqrt{3}}$
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 3–√$\sqrt{3}$.

Question 4

Show the number 5–√$\sqrt{5}$ 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
OR2 = OQ2 + QR2
= (2)2 + (1)2
= 4 + 1
= 5
∴ OR = 5–√$\sqrt{5}$
Taking O as the centre and radius OR = 5–√$\sqrt{5}$, draw an arc cutting the number line at C.
Clearly, OC = OR = 5–√$\sqrt{5}$.
Hence, C represents 5–√$\sqrt{5}$ on the number line.

Question 5

Show the number 7–√$\sqrt{7}$ 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 = 5–√$\sqrt{5}$
Taking O as centre and radius OC = 5–√$\sqrt{5}$, draw an arc cutting the number line at D.
Clearly, OC = OD = 5–√$\sqrt{5}$
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
= (5–√$\sqrt{5}$)2 + (1)2
= 5 + 1
= 6
∴ OE = 6–√$\sqrt{6}$
Taking O as centre and radius OE = 6–√$\sqrt{6}$, draw an arc cutting the number line at F.
Clearly, OE = OF = 6–√$\sqrt{6}$
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
= (6–√$\sqrt{6}$)2 + (1)2
= 6 + 1
= 7
∴ OG = 7–√$\sqrt{7}$
Taking O as centre and radius OG = 7–√$\sqrt{7}$, draw an arc cutting the number line at H.
Clearly, OG = OH = 7–√$\sqrt{7}$
Hence, H represents 7–√$\sqrt{7}$ on the number line.