# NCERT solution for class 8 maths rational numbers ( Chapter 1)

#### Solution for Exercise 1.1

$$\frac{-2}{3}$$*$$\frac{3}{5}$$+$$\frac{5}{2}$$-$$\frac{3}{5}$$*$$\frac{1}{6}$$

$$\frac{-2}{3}$$*$$\frac{3}{5}$$+$$\frac{5}{2}$$-$$\frac{3}{5}$$*$$\frac{1}{6}$$
$$\frac{-2}{3}$$*$$\frac{3}{5}$$-$$\frac{3}{5}$$*$$\frac{1}{6}$$+$$\frac{5}{2}$$
$$\frac{3}{5}$$*($$\frac{-2}{3}$$-$$\frac{1}{6}$$)+$$\frac{5}{2}$$
$$\frac{3}{5}$$*($$\frac{-2*2}{3*2}$$-$$\frac{1*6}{6*1}$$)+$$\frac{5}{2}$$
$$\frac{3}{5}$$*($$\frac{1}{2}$$

1.Using appropriate properties find:
(i)$$-\frac{2}3×\frac{3}5+\frac{5}2-\frac{3}5×\frac{1}6$$

(ii)$$\frac{2}5×(-\frac{3}7)-\frac{1}6×\frac{3}2+\frac{1}{14}×\frac{2}5$$

(i) We have: $$-\frac{2}3×\frac{3}5+\frac{5}2-\frac{3}5×\frac{1}6$$

$$= -\frac{2}3×\frac{3}5-\frac{3}5×\frac{1}6+\frac{5}2 (By \quad regrouping)$$

$$= \frac{3}5×(-\frac{2}3-\frac{1}6)+\frac{5}2 (by\quad distributive\quad property)$$

$$= \frac{3}5×(\frac{-2×2}{3×2}-\frac{1×1}{6×1})+\frac{5}2$$

$$= \frac{3}5×(\frac{-4}{6}-\frac{1}{6})+\frac{5}2= \frac{3}5×(\frac{-4-1}{6})+\frac{5}2$$

$$= \frac{3}5×(\frac{-5}{6})+\frac{5}2$$

$$= -\frac{3×5}{5×6}+\frac{5}2$$

$$= -\frac{1}2 +\frac{5}2=(\frac{-1+5}2)=\frac{4}2 =2$$

so, the value obtained is =2.

(ii) we have:$$\frac{2}5×(-\frac{3}7)-\frac{1}6×\frac{3}2+\frac{1}{14}×\frac{2}5$$

$$= \frac{2}5×(-\frac{3}7)+\frac{1}{14}×\frac{2}5 -\frac{1}6×\frac{3}2 (by \quad regrouping)$$

$$= \frac{2}5×(\frac{-3}7+\frac{1}{14})-\frac{1}6×\frac{3}2 (using \quad distributive \quad property)$$

$$= \frac{2}5×(\frac{-3×2}{7×2}+\frac{1×1}{14×1})-\frac{1}6×\frac{3}2$$

$$= \frac{2}5×(\frac{-6}{14}+\frac{1}{14})-\frac{1}6×\frac{3}2$$

$$= \frac{2}5×(\frac{-6+1}{14})-\frac{3}{12}$$

$$= \frac{2}5×(\frac{-5}{14})-\frac{1}{4}$$

$$= -\frac{1}7-\frac{1}{4}= \frac{-1×4}{7×4}-\frac{1×7}{4×7}$$

$$= \frac{-4}{28}-\frac{7}{28}=\frac{-4-7}{28}=\frac{-11}{28}$$

Thus, the value obtained is $$\frac{-11}{28}$$.

2.Write the additive inverse of each of the following:
(i)$$\frac{2}8$$

(ii)$$\frac{-5}9$$

(iii)$$\frac{-6}{-5}$$

(iv)$$\frac{2}{-9}$$

(v)$$\frac{19}{-6}$$

We have additive inverse of a as a=-a .So, we have:

(i) Additive inverse of $$\frac{2}8 = \frac{-2}8$$

(ii)Additive inverse of $$\frac{-5}9 = -(\frac{-5}9)= \frac{5}9$$

(iii) we have $$\frac{-6}{-5}= \frac{6}5$$

∴ the additive inverse of $$\frac{6}5 =\frac{-6}5$$

(iv)We have $$\frac{2}{-9}=\frac{-2}9$$

∴ Additive inverse of $$\frac{-2}{9} = -(\frac{-2}9)= \frac{2}9$$

(v)We have $$\frac{19}{-6}=\frac{-19}6$$

∴ Additive inverse of $$\frac{-19}{6} = -(\frac{-19}6)= \frac{19}6$$

3.verify that -(-x)=x for
(i)x=$$\frac{11}{15}$$

(ii)x=$$\frac{-13}{17}$$

(i) We have x=$$\frac{11}5$$

∴ -x=$$\frac{-11}{15}$$

-(-x)=$$-(\frac{-11}{15}) = \frac{11}{15}=x \quad$$[∵ (-)×(-)=(+)]

Hence verified

(ii)We have x=$$\frac{-13}{17}$$

∴ -x=$$-(\frac{-13}{17})=\frac{13}{17}$$[∵ (-)×(-)=(+)]

-(-x)=$$-(\frac{13}{17})=x (Verified)$$

4.Find the multiplicative inverse of the following :
(i)-13

(ii)$$\frac{-13}{19}$$

(iii)$$\frac{1}5$$

(iv)$$\frac{-5}8×\frac{-3}{7}$$

(v)$$-1×\frac{-2}5$$

(vi)-1

We know that multiplicative inverse of a is $$\frac{1}a \qquad$$[∵ $$a ×\frac{1}a=1$$]

(i) Multiplicative inverse of $$-13=\frac{-1}{13} \qquad$$[∵ $$-13 ×\frac{-1}{13}=1$$]

(ii) Multiplicative inverse of $$\frac{-13}{19}=\frac{-19}{13} \qquad$$[∵ $$\frac{-13}{19} ×\frac{-19}{13}=1$$]

(iii) Multiplicative inverse of $$\frac{1}{5}=5 \qquad$$[∵ $$\frac{1}{5} ×5=1$$]

(iv) Multiplicative inverse of $$\frac{-5}8×\frac{-3}{7}=\frac{-8}5×\frac{-7}{3}$$

we have $$\frac{-5}8×\frac{-3}{7}=\frac{15}{56}$$

and we have the multiplicative inverse as $$\frac{-8}5×\frac{-7}{3}=\frac{56}{15}$$

∴ $$\frac{15}{56} ×\frac{56}{15}=1$$

(v) We have $$-1\frac{-2}5=\frac{2}5$$

∴Multiplicative inverse of $$\frac{2}5=\frac{5}2 \qquad$$[∵ $$\frac{2}{5} ×\frac{5}{2}=1$$]

(vi)Multiplicative inverse of $$-1=\frac{1}{-1}=-1 \qquad$$[∵ $$-1 ×\frac{1}{-1}=1$$]

5.Name the property under multiplication used in each of the following:
(i)$$\frac{-4}5×1=1×\frac{-4}5=\frac{-4}5$$

(ii)$$\frac{-13}{17}×\frac{-2}7=\frac{-2}7×\frac{-13}{17}$$

(iii)$$\frac{-19}{29}×\frac{29}{-19}=1$$

(i)Commutative property of multiplication.

(ii)Commutative property of multiplication.

(iii)Multiplicative inverse property.

6.Multiply $$\frac{6}{13}$$ by the reciprocal of $$\frac{-7}{16}$$

We have reciprocal of $$\frac{-7}{16}=\frac{16}{-7}=\frac{-16}7$$

∴ $$\frac{6}{13}×\frac{-16}{7}=\frac{6×(-16)}{13×7}=-\frac{96}{91}$$

7.Tell what property allows you to compute
$$\frac{1}3×(6×\frac{4}3)as (\frac{1}3×6)×\frac{4}3$$

We know that a × (b× c) = (a × b) × c shows the associative property of multiplications.

∴$$\frac{1}3× (6× \frac{4}3)=(\frac{1}3× 6)× \frac{4}3$$ shows the associative property of multiplication.

8.Is $$\frac{8}9$$ the multiplicative inverse of $$-1\frac{1}8$$? Why or why not?

We have $$-1\frac{1}8=\frac{-9}{8}$$

Since multiplication inverse of $$\frac{8}9 is \frac{9}8$$ but not $$\frac{-9}8$$

Because, $$\frac { 8 }{ 9 } \times -1\frac { 1 }{ 8 } =\frac { 8 }{ 9 } \times \frac { -9 }{ 8 } =-1\neq 1$$

So, $$\frac{8}9$$ is not the multiplicative inverse of $$-1\frac{1}8$$.

9. Is 0.3 the multiplicative inverse of $$3\frac { 1 }{ 3 }$$? Why or why not?

Yes, 0.3 is multiplicative inverse of $$3\frac { 1 }{ 3 }$$.
Because $$0.3\times 3\frac { 1 }{ 3 } =\frac { 3 }{ 10 } \times \frac { 10 }{ 3 } =1$$.

10.Write:
(i) The rational number that does not have a reciprocal.

(ii) The rational numbers that are equal to their reciprocals.

(iii) The rational number that is equal to its negative.

(i) The rational number 0 has no reciprocal because there is no rational number which when multiplied with 0, gives 1.

(ii) We know that the reciprocal of 1 is 1 and the reciprocal of -1 is -1. Thus,1 and -1 are the only rational numbers which are their own reciprocals.

(iii) 0 is the rational number which equal to its negative.

11.Fill in the blanks :

(i) Zero has ……………. reciprocal.

(ii) The numbers ……………. and ………. are their own reciprocals.

(iii) The reciprocal of – 5 is …………….

(iv) Reciprocal of $$\frac { 1 }{ x }$$ , where $$x\neq 0$$ is ………

(v) The product of two rational numbers is always a ………….

(vi) The reciprocal of a positive rational number is ………..

(i) No
(ii) 1, -1

(iii)$$\frac { -1 }{ 5 }$$

(iv) x

(v) rational number

(vi) positive

#### Solution for Exercise 1.2

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1.Represent these numbers on the number line.
(i) $$\frac { 7 }{ 4 }$$

(ii)$$\frac { -5 }{ 6 }$$

(i)For $$\frac { 7 }{ 4 }$$ ,as we have the positive number, we make 7 markings of distance each on the right of zero and starting from 0. The seven markings are

Here, point P represents $$\frac { 7 }{ 4 }$$on the number line.
(ii) For $$\frac { -5 }{ 6 }$$ , as the number is negative,we make 5 markings of distance $$\frac { 1 }{ 6 }$$ each on the left and starting from 0. The fifth marking $$\frac { -5 }{ 6 }$$.

The point P represents $$\frac { -5 }{ 6 }$$ on the number line.

2.Represent $$\frac { -2 }{ 11 } ,\frac { -5 }{ 11 } ,\frac { -9 }{ 11 }$$ on the number line.

As the given numbers are negative so mark a point 0 and mark eleven points till the point -1 on the left of 0 taking each division of $$\frac{1}{11}$$ distance.The points marked are as follows.

Hence, point A represents $$\frac { -2 }{ 11 }$$, point B represents $$\frac { -5 }{ 11 }$$, point C represents $$\frac { -9 }{ 11 }$$

3.Write five rational numbers which are smaller than 2.

There can be many rational numbers as example. Few of them are:1, $$\frac{1}2$$,0,$$\frac{1}3$$,$$\frac{1}6$$.

4.Find ten rational numbers between $$\frac { -2 }{ 5 }$$ and $$\frac { 1 }{ 2 }$$.

We have rational numbers as $$\frac{-2}{5}$$ and $$\frac{1}{2}$$
Making same denominator:

⇒ $$\frac{-2}5 =\frac{-2×4}{5×4}=\frac{-8}{20}$$

and $$\frac{1}2 =\frac{1×10}{2×10}=\frac{10}{20}$$

Now, we have to find ten rational numbers between $$\frac{-8}{20}$$ and$$\frac{10}{20}$$

So,these rational numbers can be $$\frac{-7}{20},\frac{-6}{20},\frac{-5}{20},\frac{-4}{20},\frac{-3}{20},\frac{-2}{20},\frac{-1}{20},0,\frac{1}{20},\frac{2}{20}$$

5.Find five rational numbers between
(i)$$\frac { 2 }{ 3 }$$ and $$\frac { 4 }{ 5 }$$

(ii) $$\frac { -3 }{ 2 }$$and $$\frac { 5 }{ 3 }$$

(iii) $$\frac { 1 }{ 4 }$$ and $$\frac { 1 }{ 2 }$$

(i) We have rational numbers as: $$\frac{2}3$$ and$$\frac{4}5$$

Making the denominator same for both the rational numbers:

$$\frac{2}3=\frac{2}3×\frac{20}{20}=\frac{40}{60}$$ and $$\frac{4}5=\frac{4×12}{5×12}=\frac{48}{60}$$

So now we have to see five rational number between $$\frac{40}{60}$$ and $$\frac{48}{60}$$

These can be, $$\frac{41}{60},\frac{42}{60},\frac{43}{60},\frac{44}{60},\frac{46}{60}$$

(ii) We have rational numbers as: $$\frac{-3}2$$ and$$\frac{5}3$$

Making the denominator same for both the rational numbers:

$$\frac{-3}2=\frac{-3×3}{2×3}=\frac{-9}{6}$$ and $$\frac{5}3=\frac{5×2}{3×2}=\frac{10}{6}$$

So now we have to see five rational number between $$\frac{-9}{6}$$ and $$\frac{10}{6}$$

These can be, $$\frac{-8}{6},\frac{-7}{6},\frac{-6}{6},\frac{3}{6},\frac{4}{6}$$

(iii) We have rational numbers as: $$\frac{1}4$$ and$$\frac{1}2$$

Making the denominator same for both the rational numbers:

$$\frac{1}4=\frac{1×8}{2×8}=\frac{8}{32}$$ and $$\frac{1}2=\frac{1×16}{2×16}=\frac{16}{32}$$

So now we have to see five rational number between $$\frac{8}{32}$$ and $$\frac{16}{32}$$

These can be, $$\frac{9}{32},\frac{10}{32},\frac{11}{32},\frac{12}{32},\frac{13}{32}$$

6.Write five rational numbers greater than -2.

Rational numbers greater than -2 are many and few of them are: -1, 0, 1, $$\frac{1}4$$,$$\frac{3}4$$.

7.Find ten rational numbers between $$\frac { 3 }{ 5 }$$ and $$\frac { 3 }{ 4 }$$.

We have rational numbers as: $$\frac{3}5$$ and$$\frac{3}4$$
$$\frac{3}5=\frac{3×20}{5×20}=\frac{60}{100}$$ and $$\frac{3}4=\frac{3×25}{4×25}=\frac{75}{100}$$
So now we have to see ten rational number between $$\frac{60}{100}$$ and $$\frac{75}{100}$$
So, these can be, $$\frac{61}{100},\frac{62}{100},\frac{63}{100},\frac{64}{100},\frac{65}{100},\frac{66}{100},\frac{67}{100},\frac{68}{100},\frac{69}{100},\frac{70}{100}$$