NCERT Solutions for Class 8 Maths Chapter 1 - Rational Numbers

Exercise 1.1

1. Using appropriate properties find:

i. \[\text{-}\dfrac{\text{2}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{5}}\text{+}\dfrac{\text{5}}{\text{2}}\text{-}\dfrac{\text{3}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{6}}\]

Ans: Given 

\[\text{-}\dfrac{\text{2}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{5}}\text{+}\dfrac{\text{5}}{\text{2}}\text{-}\dfrac{\text{3}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{6}}\]  

By using commutative property of rational numbers i.e $\text{a+b=b+a}$

\[\text{=-}\dfrac{\text{2}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{5}}\text{-}\dfrac{\text{3}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{6}}\text{+}\dfrac{\text{5}}{\text{2}}\]                     

Now, by using distributive property of rational numbers i.e $\text{a }\!\!\times\!\!\text{ b+b }\!\!\times\!\!\text{ c=b }\!\!\times\!\!\text{ }\left( \text{a+c} \right)$

\[\text{=}\left( \text{-}\dfrac{\text{3}}{\text{5}} \right)\text{ }\!\!\times\!\!\text{ }\left( \dfrac{\text{2}}{\text{3}}\text{+}\dfrac{\text{1}}{\text{6}} \right)\text{+}\dfrac{\text{5}}{\text{2}}\]                  

\[\text{=}\left( \text{-}\dfrac{\text{3}}{\text{5}} \right)\text{ }\!\!\times\!\!\text{ }\left( \dfrac{\text{2 }\!\!\times\!\!\text{ 2+1}}{\text{6}} \right)\text{+}\dfrac{\text{5}}{\text{2}}\]

\[\text{=}\left( \text{-}\dfrac{\text{3}}{\text{5}} \right)\text{ }\!\!\times\!\!\text{ }\left( \dfrac{\text{5}}{\text{6}} \right)\text{+}\dfrac{\text{5}}{\text{2}}\]

\[\text{=}\left( \text{-}\dfrac{\text{3}}{\text{6}} \right)\text{+}\dfrac{\text{5}}{\text{2}}\]

\[\text{=}\left( \dfrac{\text{-3+5 }\!\!\times\!\!\text{ 3}}{\text{6}} \right)\]

\[\text{=}\left( \dfrac{\text{-3+15}}{\text{6}} \right)\]

\[\text{=}\dfrac{\text{12}}{\text{6}}\]

\[\text{=2}\]

ii. \[\dfrac{\text{2}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\left( \text{-}\dfrac{\text{3}}{\text{7}} \right)\text{-}\dfrac{\text{1}}{\text{6}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{2}}\text{+}\dfrac{\text{1}}{\text{14}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{2}}{\text{5}}\]

Ans: Given 

\[\dfrac{\text{2}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\left( \text{-}\dfrac{\text{3}}{\text{7}} \right)\text{-}\dfrac{\text{1}}{\text{6}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{2}}\text{+}\dfrac{\text{1}}{\text{14}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{2}}{\text{5}}\]

By using commutative property of rational numbers i.e $\text{a+b=b+a}$

\[\text{=}\dfrac{\text{2}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\left( \text{-}\dfrac{\text{3}}{\text{7}} \right)\text{+}\dfrac{\text{1}}{\text{14}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{2}}{\text{5}}\text{-}\dfrac{\text{1}}{\text{6}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{2}}\]

Now, by using distributive property of rational numbers i.e $\text{a }\!\!\times\!\!\text{ b+b }\!\!\times\!\!\text{ c=b }\!\!\times\!\!\text{ }\left( \text{a+c} \right)$

\[\text{=}\dfrac{\text{2}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\left( \text{-}\dfrac{\text{3}}{\text{7}}\text{+}\dfrac{\text{1}}{\text{14}} \right)\text{-}\dfrac{\text{1}}{\text{4}}\]                          

\[\text{=}\dfrac{\text{2}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\left( \dfrac{\text{-3 }\!\!\times\!\!\text{ 2+1}}{\text{14}} \right)\text{-}\dfrac{\text{1}}{\text{4}}\]

\[\text{=}\dfrac{\text{2}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\left( \dfrac{\text{-5}}{\text{14}} \right)\text{-}\dfrac{\text{1}}{\text{4}}\]

\[\text{=-}\dfrac{\text{1}}{\text{7}}\text{-}\dfrac{\text{1}}{\text{4}}\]

\[\text{=}\dfrac{\text{-4-7}}{\text{28}}\]

  \[\text{=}\dfrac{\text{-11}}{\text{28}}\]

2. Write the additive inverse of each of the following:

i. \[\dfrac{\text{2}}{\text{8}}\]

Ans: Since, additive inverse of $\dfrac{\text{a}}{\text{b}}\text{=}\dfrac{\text{-a}}{\text{b}}$

Therefore, additive inverse of  \[\dfrac{\text{2}}{\text{8}}\]is \[\text{-}\dfrac{\text{2}}{\text{8}}\].

ii. \[\text{-}\dfrac{\text{5}}{\text{9}}\]

Ans: Since, additive inverse of $\text{-}\dfrac{\text{a}}{\text{b}}\text{=}\dfrac{\text{a}}{\text{b}}$

Therefore, additive inverse of  \[\text{-}\dfrac{\text{5}}{\text{9}}\]is \[\dfrac{\text{5}}{\text{9}}\].

iii. \[\dfrac{\text{-6}}{\text{-5}}\]

Ans: Since, additive inverse of $\dfrac{\text{-a}}{\text{-b}}\text{=-}\dfrac{\text{a}}{\text{b}}$

Therefore, additive inverse of \[\dfrac{\text{-6}}{\text{-5}}\]is \[\text{-}\dfrac{\text{6}}{\text{5}}\].

iv. \[\dfrac{\text{2}}{\text{-9}}\]

Ans: Since, additive inverse of $\dfrac{\text{a}}{\text{-b}}\text{=}\dfrac{\text{a}}{\text{b}}$

Therefore, additive inverse of  \[\dfrac{\text{2}}{\text{-9}}\] is  \[\dfrac{\text{2}}{\text{9}}\].

v. \[\dfrac{\text{19}}{\text{-6}}\]

Ans: Since, additive inverse of $\dfrac{\text{a}}{\text{-b}}\text{=}\dfrac{\text{a}}{\text{b}}$

Therefore, additive inverse of  \[\dfrac{\text{19}}{\text{-6}}\] is \[\dfrac{\text{19}}{\text{6}}\].

3. Verify that \[\text{-(-x)=x}\] for.

i. \[\text{x=}\dfrac{\text{11}}{\text{15}}\]

Ans: The additive inverse of  \[\text{x=}\dfrac{\text{11}}{\text{15}}\] is \[\text{-x=-}\dfrac{\text{11}}{\text{15}}\] 

Now,  \[\dfrac{\text{11}}{\text{15}}\text{+}\left( \text{-}\dfrac{\text{11}}{\text{15}} \right)\text{=0}\]

Thus , it represents that the additive inverse of \[\text{-}\dfrac{\text{11}}{\text{15}}\] is \[\dfrac{\text{11}}{\text{15}}\] i.e., \[\text{-}\left( \text{-}\dfrac{\text{11}}{\text{15}} \right)\text{=}\dfrac{\text{11}}{\text{15}}\] .

Hence, \[\text{-(-x)=x}\] holds for \[\text{x=}\dfrac{\text{11}}{\text{15}}\].

ii. \[\text{x=-}\dfrac{\text{13}}{\text{17}}\]

Ans: The additive inverse of \[\operatorname{x}=-\dfrac{13}{17}\] is \[\text{-x=}\dfrac{\text{13}}{\text{17}}\]

Now,  \[\text{-}\dfrac{\text{13}}{\text{17}}\text{+}\dfrac{\text{13}}{\text{17}}\text{=0}\]

Thus , it represents that the additive inverse of \[\dfrac{\text{13}}{\text{17}}\] is \[\text{-}\dfrac{\text{13}}{\text{17}}\] i.e., \[\text{-}\left( \text{-}\dfrac{\text{13}}{\text{17}} \right)\text{=}\dfrac{\text{13}}{\text{17}}\] .

Hence, \[\text{-(-x)=x}\] holds for \[\text{x=-}\dfrac{\text{13}}{\text{17}}\].

4. Find the multiplicative inverse of the following.

i. \[\text{-13}\]

Ans: Since, multiplicative inverse of $\text{-a=-}\dfrac{\text{1}}{\text{a}}$.

Therefore, the multiplicative inverse of  \[\text{-13}\] is \[\text{-}\dfrac{\text{1}}{\text{13}}\].

ii. \[\text{-}\dfrac{\text{13}}{\text{19}}\]

Ans: Since, multiplicative inverse of $\text{-}\dfrac{\text{a}}{\text{b}}\text{=-}\dfrac{\text{b}}{\text{a}}$.

Therefore, multiplicative inverse of  \[\text{-}\dfrac{\text{13}}{\text{19}}\] is \[\text{-}\dfrac{\text{19}}{\text{13}}\].

iii. \[\dfrac{\text{1}}{\text{5}}\]

Ans: Since, multiplicative inverse of $\dfrac{\text{a}}{\text{b}}\text{=}\dfrac{\text{b}}{\text{a}}$.

Therefore, the multiplicative inverse of  \[\dfrac{\text{1}}{\text{5}}\] is \[\text{5}\].

iv. \[\text{-}\dfrac{\text{5}}{\text{8}}\text{ }\!\!\times\!\!\text{ -}\dfrac{\text{3}}{\text{7}}\]

Ans: It can be written as \[\text{-}\dfrac{\text{5}}{\text{8}}\text{ }\!\!\times\!\!\text{ -}\dfrac{\text{3}}{\text{7}}\text{=}\dfrac{\text{15}}{\text{56}}\].

Since, multiplicative inverse of $\dfrac{\text{a}}{\text{b}}\text{=}\dfrac{\text{b}}{\text{a}}$.

Therefore, multiplicative inverse of  \[\text{-}\dfrac{\text{5}}{\text{8}}\text{ }\!\!\times\!\!\text{ -}\dfrac{\text{3}}{\text{7}}\] is \[\dfrac{\text{56}}{\text{15}}\].

v. \[\text{-1 }\!\!\times\!\!\text{ -}\dfrac{\text{2}}{\text{5}}\]

Ans: It can be written as\[\text{-1 }\!\!\times\!\!\text{ -}\dfrac{\text{2}}{\text{5}}\text{=}\dfrac{\text{2}}{\text{5}}\].

Since, multiplicative inverse of $\dfrac{\text{a}}{\text{b}}\text{=}\dfrac{\text{b}}{\text{a}}$.

Therefore, multiplicative inverse of  \[\text{-1 }\!\!\times\!\!\text{ -}\dfrac{\text{2}}{\text{5}}\] is \[\dfrac{\text{5}}{\text{2}}\].

vi. \[\text{-1}\]

Ans: Since, multiplicative inverse of $\text{-}\dfrac{\text{a}}{\text{b}}\text{=-}\dfrac{\text{b}}{\text{a}}$.

Therefore, the multiplicative inverse of  \[\text{-1}\] is \[\text{-1}\].

5. Name the property under multiplication used in each of the following:

i. \[\dfrac{\text{-4}}{\text{5}}\text{ }\!\!\times\!\!\text{ 1=1 }\!\!\times\!\!\text{ }\dfrac{\text{-4}}{\text{5}}\text{=}\dfrac{\text{-4}}{\text{5}}\]

Ans: Since, after multiplying by \[\text{1}\] , we are getting the same number.

Therefore, \[\text{1}\] is the multiplicative identity.

ii. \[\text{-}\dfrac{\text{13}}{\text{17}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{-2}}{\text{7}}\text{=}\dfrac{\text{-2}}{\text{7}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{-13}}{\text{17}}\]

Ans: Since, $\text{a }\!\!\times\!\!\text{ b=b }\!\!\times\!\!\text{ a}$.

Therefore, its Commutative property.

iii. \[\text{-}\dfrac{\text{19}}{\text{29}}\text{ }\!\!\times\!\!\text{ -}\dfrac{\text{29}}{\text{19}}\text{=1}\]

Ans: Since, \[\text{-a }\!\!\times\!\!{\dfrac{1}{-a}}\text{=1}\].

Therefore, the property is Multiplicative inverse.

6. Multiply \[\dfrac{\text{6}}{\text{13}}\] by the reciprocal of \[\text{-}\dfrac{\text{7}}{\text{16}}\]

Ans: Reciprocal of \[\text{-}\dfrac{\text{7}}{\text{16}}\]  is \[\text{-}\dfrac{\text{16}}{\text{7}}\] .

Thus, \[\dfrac{\text{6}}{\text{13}}\text{ }\!\!\times\!\!\text{ -}\dfrac{\text{16}}{\text{7}}\]

 \[\text{=-}\dfrac{\text{96}}{\text{91}}\] .

7. Tell what property allows you to compute \[\dfrac{\text{1}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\left( \text{6 }\!\!\times\!\!\text{ }\dfrac{\text{4}}{\text{3}} \right)\] as \[\left( \dfrac{\text{1}}{\text{3}}\text{ }\!\!\times\!\!\text{ 6} \right)\text{ }\!\!\times\!\!\text{ }\dfrac{\text{4}}{\text{3}}\].

Ans: Since, $\text{a }\!\!\times\!\!\text{ }\left( \text{b }\!\!\times\!\!\text{ c} \right)\text{=}\left( \text{a }\!\!\times\!\!\text{ b} \right)\text{ }\!\!\times\!\!\text{ c}$ .

Therefore, its associative property.

8. Is \[\dfrac{\text{8}}{\text{9}}\] the multiplicative inverse of \[\text{-1}\dfrac{\text{1}}{\text{8}}\]? Why or why not?

Ans: We know, If it is the multiplicative inverse, then the product should be \[\text{1}\].

Now,

\[\dfrac{\text{8}}{\text{9}}\text{ }\!\!\times\!\!\text{ }\left( \text{-1}\dfrac{\text{1}}{\text{8}} \right)\text{=}\dfrac{\text{8}}{\text{9}}\text{ }\!\!\times\!\!\text{ }\left( \text{-}\dfrac{\text{9}}{\text{8}} \right)\]

\[\text{=-1}\] 

Since, the product is not \[\text{1}\] .

Therefore, \[\dfrac{\text{8}}{\text{9}}\] is not the multiplicative inverse of \[\text{-1}\dfrac{\text{1}}{\text{8}}\] .

9. Is \[\text{0}\text{.3}\] the multiplicative inverse of \[\text{3}\dfrac{\text{1}}{\text{3}}\]? Why or why not?

Ans: We know, If it is the multiplicative inverse, then the product should be \[\text{1}\].

Now,

\[\text{0}\text{.3 }\!\!\times\!\!\text{ 3}\dfrac{\text{1}}{\text{3}}\text{=0}\text{.3 }\!\!\times\!\!\text{ }\dfrac{\text{10}}{\text{3}}\]

\[\text{=}\dfrac{\text{3}}{\text{10}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{10}}{\text{3}}\]

\[\text{=1}\]

Since, the product is \[\text{1}\] .

Therefore, \[\text{0}\text{.3}\] is the multiplicative inverse of \[\text{3}\dfrac{\text{1}}{\text{3}}\].

10. Write:

i. The rational number that does not have a reciprocal.

Ans: \[\text{0}\] is a rational number but its reciprocal is not defined.

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

Ans: \[\text{1}\] and \[\text{-1}\] are the rational numbers that are equal to their reciprocals.

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

Ans: \[\text{0}\] is the rational number that is equal to its negative.

11. Fill in the blanks.

i. Zero has __________ reciprocal.

Ans: No

ii. The numbers __________ and __________ are their own reciprocals.

Ans: \[\text{1,-1}\]

iii. The reciprocal of \[\text{-5}\] is __________.

Ans: \[\text{-}\dfrac{\text{1}}{\text{5}}\]

iv. Reciprocal of \[\dfrac{\text{1}}{\text{x}}\], where \[\text{x}\ne \text{0}\] is __________.

Ans: \[\text{x}\]

v. The product of two rational numbers is always a __________.

Ans: Rational number

vi. The reciprocal of a positive rational number is __________.

Ans: Positive rational number

Refer to page 9 - 14 for exercise 1.1 in the PDF

Exercise 1.2

1. Represent these numbers on the number line.

i. $\dfrac{7}{4}$

Ans: Since, $\dfrac{7}{4}$ is approximately equal to $1.75$.

Therefore, $\dfrac{7}{4}$ can be represented on the number line as follows:

ii. $-\dfrac{5}{6}$

Ans: Since, $-\dfrac{5}{6}$ is approximately equal to $-0.833$.

Thus, $-\dfrac{5}{6}$ can be represented on the number line as follows:

2. Represent \[\text{-}\dfrac{\text{2}}{\text{11}}\text{,-}\dfrac{\text{5}}{\text{11}}\text{,-}\dfrac{\text{9}}{\text{11}}\] on the number line.

Ans: We can divide interval between $\text{0}$ and $\text{-1}$ in $\text{11}$ parts to get \[\text{-}\dfrac{\text{2}}{\text{11}}\text{,-}\dfrac{\text{5}}{\text{11}}\text{,-}\dfrac{\text{9}}{\text{11}}\] on number line.

Thus, \[\text{-}\dfrac{\text{2}}{\text{11}}\text{,-}\dfrac{\text{5}}{\text{11}}\text{,-}\dfrac{\text{9}}{\text{11}}\] can be represented on the number line as follows:

3. Write five rational numbers which are smaller than \[\text{2}\].

Ans: Since, we have to write five rational numbers which are less than \[\text{2}\] .

Therefore, we can multiply and divide \[\text{2}\] by \[7\].

Now, \[\text{2}\] becomes \[\dfrac{\text{14}}{\text{7}}\].

Thus, five rational numbers which are smaller than \[\text{2}\] are given as - 

\[\dfrac{\text{13}}{\text{7}}\text{,}\dfrac{\text{12}}{\text{7}}\text{,}\dfrac{\text{11}}{\text{7}}\text{,}\dfrac{\text{10}}{\text{7}}\text{,}\dfrac{\text{9}}{\text{7}}\].

4. Find ten rational numbers between \[\dfrac{\text{-2}}{\text{5}}\] and \[\dfrac{\text{1}}{\text{2}}\].

Ans: We can make denominator of  \[\dfrac{\text{-2}}{\text{5}}\] and \[\dfrac{\text{1}}{\text{2}}\] same.

Therefore, multiplying and dividing \[\dfrac{\text{-2}}{\text{5}}\] by $\text{4}$ and \[\dfrac{\text{1}}{\text{2}}\] by $\text{10}$.

Thus, now \[\dfrac{\text{-2}}{\text{5}}\] becomes \[\dfrac{\text{-8}}{\text{20}}\]  and  \[\dfrac{\text{1}}{\text{2}}\] becomes \[\dfrac{\text{10}}{\text{20}}\] .

Hence, ten rational numbers between \[\dfrac{\text{-2}}{\text{5}}\] and \[\dfrac{\text{1}}{\text{2}}\] are-

\[\text{-}\dfrac{\text{7}}{\text{20}}\text{,-}\dfrac{\text{6}}{\text{20}}\text{,-}\dfrac{\text{5}}{\text{20}}\text{,-}\dfrac{\text{4}}{\text{20}}\text{,-}\dfrac{\text{3}}{\text{20}}\text{,-}\dfrac{\text{2}}{\text{20}}\text{,-}\dfrac{\text{1}}{\text{20}}\text{,0,}\dfrac{\text{1}}{\text{20}}\text{,}\dfrac{\text{2}}{\text{20}}\].

5. Find five rational numbers between-

i. \[\dfrac{\text{2}}{\text{3}}\] and \[\dfrac{\text{4}}{\text{5}}\]

Ans: We can make denominator of  \[\dfrac{\text{2}}{\text{3}}\] and \[\dfrac{\text{4}}{\text{5}}\] same.

Therefore, multiplying and dividing \[\dfrac{\text{2}}{\text{3}}\] by $\text{15}$ and \[\dfrac{\text{4}}{\text{5}}\] by $9$.

Thus, now \[\dfrac{\text{2}}{\text{3}}\] becomes \[\dfrac{\text{30}}{\text{45}}\]  and  \[\dfrac{\text{4}}{\text{5}}\] becomes \[\dfrac{\text{36}}{\text{45}}\] .

Hence, ten rational numbers between \[\dfrac{\text{2}}{\text{3}}\] and \[\dfrac{\text{4}}{\text{5}}\] are-

\[\dfrac{\text{31}}{\text{45}}\text{,}\dfrac{\text{32}}{\text{45}}\text{,}\dfrac{\text{33}}{\text{45}}\text{,}\dfrac{\text{34}}{\text{45}}\text{,}\dfrac{\text{35}}{\text{45}}\].

ii. \[\dfrac{\text{-3}}{\text{2}}\] and \[\dfrac{\text{5}}{\text{3}}\].

Ans: We can make denominator of  \[\text{-}\dfrac{\text{3}}{\text{2}}\] and \[\dfrac{\text{5}}{\text{3}}\] same.

Therefore, multiplying and dividing \[\text{-}\dfrac{\text{3}}{\text{2}}\] by $3$ and \[\dfrac{\text{5}}{\text{3}}\] by $2$.

Thus, now \[\text{-}\dfrac{\text{3}}{\text{2}}\] becomes \[\text{-}\dfrac{\text{9}}{\text{6}}\]  and  \[\dfrac{\text{5}}{\text{3}}\] becomes \[\dfrac{\text{10}}{\text{6}}\] .

Hence, ten rational numbers between \[\text{-}\dfrac{\text{3}}{\text{2}}\] and \[\dfrac{\text{5}}{\text{3}}\] are-

\[\text{-}\dfrac{\text{8}}{\text{6}}\text{,-}\dfrac{\text{7}}{\text{6}}\text{,-1,-}\dfrac{\text{5}}{\text{6}}\text{,-}\dfrac{\text{4}}{\text{6}}\].

iii. \[\dfrac{\text{1}}{\text{4}}\] and \[\dfrac{\text{1}}{\text{2}}\]

Ans: We can make denominator of  \[\dfrac{\text{1}}{\text{4}}\] and \[\dfrac{\text{1}}{\text{2}}\] same.

Therefore, multiplying and dividing \[\dfrac{\text{1}}{\text{4}}\] by $8$ and \[\dfrac{\text{1}}{\text{2}}\] by $16$.

Thus, now \[\dfrac{\text{1}}{\text{4}}\] becomes \[\dfrac{\text{8}}{\text{32}}\]  and  \[\dfrac{\text{1}}{\text{2}}\] becomes \[\dfrac{\text{16}}{\text{32}}\] .

Hence, ten rational numbers between \[\dfrac{\text{1}}{\text{4}}\] and \[\dfrac{\text{1}}{\text{2}}\] are-

\[\dfrac{\text{9}}{\text{32}}\text{,}\dfrac{\text{10}}{\text{32}}\text{,}\dfrac{\text{11}}{\text{32}}\text{,}\dfrac{\text{12}}{\text{32}}\text{,}\dfrac{\text{13}}{\text{32}}\].

6. Write five rational numbers greater than \[\text{-2}\].

Ans: Since, we have to write five rational numbers which are greater than \[\text{-2}\].

Therefore, we can multiply and divide \[\text{-2}\] by \[7\].

Now, \[\text{-2}\] becomes \[\text{-}\dfrac{\text{14}}{\text{7}}\].

Thus, five rational numbers greater than \[\text{-2}\] are given as-

\[\text{-}\dfrac{\text{13}}{\text{7}}\text{,-}\dfrac{\text{12}}{\text{7}}\text{,-}\dfrac{\text{11}}{\text{7}}\text{,-}\dfrac{\text{10}}{\text{7}}\text{,-}\dfrac{\text{9}}{\text{7}}\].

7. Find ten rational numbers between \[\dfrac{\text{3}}{\text{5}}\] and \[\dfrac{\text{3}}{\text{4}}\].

Ans: We can make denominator of  \[\dfrac{\text{3}}{\text{5}}\] and \[\dfrac{\text{3}}{\text{4}}\] same.

Therefore, multiplying and dividing \[\dfrac{\text{3}}{\text{5}}\] by $16$ and \[\dfrac{\text{3}}{\text{4}}\] by $20$.

Thus, now \[\dfrac{\text{3}}{\text{5}}\] becomes \[\dfrac{\text{48}}{\text{80}}\]  and  \[\dfrac{\text{3}}{\text{4}}\] becomes \[\dfrac{\text{60}}{\text{80}}\] .

Hence, ten rational numbers between \[\dfrac{\text{3}}{\text{5}}\] and \[\dfrac{\text{3}}{\text{4}}\] are-

\[\dfrac{\text{49}}{\text{80}}\text{,}\dfrac{\text{50}}{\text{80}}\text{,}\dfrac{\text{51}}{\text{80}}\text{,}\dfrac{\text{52}}{\text{80}}\text{,}\dfrac{\text{53}}{\text{80}}\text{,}\dfrac{\text{54}}{\text{80}}\text{,}\dfrac{\text{55}}{\text{80}}\text{,}\dfrac{\text{56}}{\text{80}}\text{,}\dfrac{\text{57}}{\text{80}}\text{,}\dfrac{\text{58}}{\text{80}}\].

NCERT Solutions For Class 8 Maths Chapter 1 Rational Numbers - PDF Download

Points to Remember to Solve Chapter 1 of Class 8 NCERT

Rational Number: Any number that can be expressed in the form of p/q, where p and q are integers and q ≠ 0, is known as a rational number. The collection or group of rational numbers is denoted by Q.

Properties of a Rational Number

• Example: Let p and q be any two rational numbers. Then their sum, difference and product will also be a rational number. This is known as the Closure property.

• Commutativity: Rational numbers will be commutative under addition and multiplication. 

Let p and q be any two rational numbers, then

Commutative law under addition is p + q = q + p.

Commutative law under multiplication is p x q = q x p.

(Note: Rational numbers, integers and whole numbers are commutative under addition and multiplication. Also, they are non-commutative under subtraction and division.)

• Associativity: Rational numbers will be associative under addition and multiplication. 

Let p, q and r be the rational numbers, then

Associative property under addition is: p + (q + r) = (p + q) + r

Associative property under multiplication is: p(qr) = (pq)r

• Role of Zero and One: 0 will be the additive identity for rational numbers. 1 will be the multiplicative identity for the rational numbers.

• Multiplicative Inverse: When the product of two rational numbers is 1, then they are called as the multiplicative inverse of each other.

We Cover All The Exercises in Chapter Given Below:

• Exercise 1.1 - 11 Questions with Solutions.

• Exercise 1.2 - 7 Questions with Solutions.
  
  

NCERT Solutions for Class 8 Maths - Chapterwise Solutions

• Chapter 2 - Linear Equations in One Variable

• Chapter 3 - Understanding Quadrilaterals

• Chapter 4 - Practical Geometry

• Chapter 5 - Data Handling

• Chapter 6 - Squares and Square Roots

• Chapter 7 - Cubes and Cube Roots

• Chapter 8 - Comparing Quantities

• Chapter 9 - Algebraic Expressions and Identities

• Chapter 10 - Visualising Solid Shapes

• Chapter 11 - Mensuration

• Chapter 12 - Exponents and Powers

• Chapter 13 - Direct and Inverse Proportions

• Chapter 14 - Factorisation

• Chapter 15 - Introduction to Graphs

• Chapter 16 - Playing with Numbers

Along with this, students can also download additional study materials provided by Vedantu, for Chapter 1 of CBSE Class 8 Maths Solutions –

• Chapter 1: Important Questions

• Chapter 1: Important Formulas

• Chapter 1: Revision Notes

• Chapter 1: NCERT Exemplar Solutions

• Chapter 1: RD Sharma Solutions

Benefits of NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers

Vedantu’s NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers are curated and prepared by our best teachers to cater to students’ need for concise and accurate maths solutions to all the questions given in the NCERT textbook for this chapter. Students going through these solutions will have a higher chance of scoring well in their Class 8 Maths exams.

Quick Revision

The following are the key points to remember while studying the chapter on Rational Numbers.

• Rational numbers are enclosed within addition, subtraction, and multiplication operations.

• The operations related to addition and multiplication are:

1. Commutative property of rational numbers

2. Associative property of rational numbers

• The rational number 0 is the additive identity for all rational numbers.

• The rational number 1 is the multiplicative identity for all rational numbers.

• For all rational numbers a, b, and c, a (b + c) = ab + ac and a (b – c) = ab – ac.

• A number line can be used to represent rational numbers.

• There can be countless rational numbers between any two given rational numbers. Mean aids in determining rational numbers between two rational numbers.

List of Formulas

The following is a list of important formulas that students need to keep in mind while studying the chapter on Rational Numbers. 

Did You Know?

Pythagoras was confident that using whole numbers of a small enough unit would measure everything. That means any two measurements or lengths will have a rational ratio. Through Pythagoras Theorem, they discovered that a rational number could not represent a unit square’s diagonal measurement. They had to come up with new solutions that can avoid showing their false assumptions and complied with facts. This historical event is a first of its kind which gives strength to the fact that mathematicians always look for evidence. You will never be able to understand this without learning Rational Numbers!

Conclusion:

NCERT Solutions for Class 8 Maths Rational Numbers offer comprehensive guidance for students to master this important mathematical concept. With clear explanations, practice exercises, and problem-solving techniques, these solutions facilitate a deeper understanding of rational numbers. By utilizing these solutions, students can build a solid foundation in mathematics and excel in their academic endeavors.