NCERT Solution for class 8 chapter 1 RATIONAL NUMBER

Exercise 1.1

Exercise 1.2

EXERCISE 1.1

 

Question 1

Using appropriate properties find .

(i)  -\dfrac{2}{3}\times\dfrac{3}{5}+\dfrac{5}{2}-\dfrac{3}{5}\times\dfrac{1}{6}

sol :

\\=-\dfrac{2}{3}\times\dfrac{3}{5}+\dfrac{5}{2}-\dfrac{3}{5}\times\dfrac{1}{6}

=-\dfrac{2}{3}\times\dfrac{3}{5}-\dfrac{3}{5}\times\dfrac{1}{6}+\dfrac{5}{2} \text{[Using commutativity of rational number]}

=\Bigg(-\dfrac{3}{5}\Bigg)\times\Bigg(\dfrac{2}{3}+\dfrac{1}{6}\Bigg)+\dfrac{5}{2} \text{[Distributivity]}

=\Bigg(-\dfrac{3}{5}\Bigg)\times\Bigg(\dfrac{2\times{2}+1}{6}\Bigg)+\dfrac{5}{2}

=\Bigg(-\dfrac{3}{5}\Bigg)\times\Bigg(\dfrac{5}{6}\Bigg)+\dfrac{5}{2}

=\Bigg(-\dfrac{3}{6}\Bigg)+\dfrac{5}{2}

=\Bigg(\dfrac{-3+5\times3}{6}\Bigg)

=\dfrac{12}{6}

= 2

 

(ii)  \dfrac{2}{5}\times\Bigg(-\dfrac{3}{7}\Bigg)-\dfrac{1}{6}\times\dfrac{3}{2}+\dfrac{1}{14}\times\dfrac{2}{5}

sol :

=\dfrac{2}{5}\times\Bigg(-\dfrac{3}{7}\Bigg)+\dfrac{1}{14}\times\dfrac{2}{5}-\dfrac{1}{6}\times\dfrac{3}{2} \text{[By commutativity of rational number]}

=\dfrac{2}{5}\times\Bigg(-\dfrac{3}{7}+\dfrac{1}{14}\Bigg)-\dfrac{1}{4} \text{[by~distributivity]}

=\dfrac{2}{5}\times\Bigg(\dfrac{-3\times2+1}{14}\Bigg)-\dfrac{1}{4}

=\dfrac{2}{5}\times\Bigg(\dfrac{-5}{14}\Bigg)-\dfrac{1}{4}

=-\dfrac{1}{7}-\dfrac{1}{4}

=\dfrac{-4-7}{28}

=-\dfrac{11}{28}

 

Question 2 :

Write the additive inverse of each of the following :

(i)  \dfrac{2}{8}

sol :

Additive inverse= -\dfrac{2}{8}

(ii)  -\dfrac{5}{9}

sol :

Additive inverse=   \dfrac{5}{9}

(iii)

  \dfrac{-6}{-5}

sol :

  \dfrac{-6}{-5}=\dfrac{6}{5}

Additive inverse=  -\dfrac{6}{5}

(iv)

\dfrac{~~2}{-9}

sol :

\dfrac{~~2}{-9}=\dfrac{-2}{~~~9}

Additive inverse=\dfrac{2}{9}

(v)

\dfrac{~~19}{-6}

sol :

\dfrac{~~2}{-9}=\dfrac{-19}{~~~6}

Additive inverse =\dfrac{19}{6}

 

Question 3 :

Verify that -(-x)=x for

(i) x=\dfrac{11}{15}

sol :

The additive inverse of  x=\dfrac{11}{15} is  -x=-\dfrac{11}{15} as  \dfrac{11}{15}+\Bigg(-\dfrac{11}{15}\Bigg)=0

This equality  \dfrac{11}{15}+\Bigg(-\dfrac{11}{15}\Bigg)=0 represents that the additive inverse of  is  x=\dfrac{11}{15} or it can be said that  -\bigg(-\dfrac{11}{15}\bigg)=\dfrac{11}{15} i.e., -(-x)=x

 

(ii)  x=-\dfrac{13}{17}

sol :

The additive inverse of  x=-\dfrac{13}{17}​  is -x=\dfrac{13}{17} as -\dfrac{13}{17}+\Bigg(\dfrac{13}{17}\Bigg)=0

This equality  -\dfrac{13}{17}+\Bigg(\dfrac{13}{17}\Bigg)=0 ​ represents that the additive inverse of   x=-\dfrac{13}{17} ​ is   -x=\dfrac{13}{17}   or it can be said that  -\bigg(-\dfrac{13}{17}\bigg)=\dfrac{13}{17} i.e., -(-x)=x

 

Question 4 :

Find the multiplicative inverse of the following.

(i) -13

sol :

Multiplicative inverse =-\dfrac{1}{13}

(ii) -\dfrac{13}{19}

sol :

Multiplicative inverse = -\dfrac{19}{13}

(iii) \dfrac{1}{5}

 

sol :

Multiplicative inverse =5

(iv) \dfrac{-5}{~~~8}\times\dfrac{-3}{~~~7}

sol :

\dfrac{-5}{~~~8}\times\dfrac{-3}{~~~7}=\dfrac{15}{56}

Multiplicative inverse =\dfrac{56}{15}

(v) -1\times\dfrac{-2}{~~~5}

sol :

-1\times\dfrac{-2}{~~~5}=\dfrac{2}{5}

Multiplicative inverse =\dfrac{5}{2}

(vi) -1

sol :

Multiplicative inverse =-1

 

Question 5 :

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

(i) \dfrac{-4}{5}\times{1}=1\times\dfrac{-4}{5}=-\dfrac{4}{5}

sol :  1 is a multiplicative identity.

 

(ii) -\dfrac{13}{17}\times\dfrac{-2}{7}=\dfrac{-2}{7}\times\dfrac{-13}{17}

sol : Commutativity

 

(iii) \dfrac{-19}{29}\times\dfrac{~~~29}{-19}=1

sol : Multiplicative inverse

 

Question 6 :

Multiply \dfrac{6}{13}​ by the reciprocal of -\dfrac{7}{16}​.

sol :

*** QuickLaTeX cannot compile formula:


\\=\dfrac{6}{13}\times\Bigg(Reciprocal~of~-\dfrac{7}{16}\Bigg)\\=\dfrac{6}{13}\times-\dfrac{16}{7}\\=-\dfrac{96}{91}



*** Error message:
Missing $ inserted.

 

Question 7 :

Tell what property allow you to compute

\dfrac{1}{3}\times\Bigg(6\times\dfrac{4}{3}\Bigg) as \Bigg(\dfrac{1}{3}\times6\Bigg)\times\dfrac{4}{3}

sol :   Associativity

 

Question 8 :

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

sol : If it is the multiplicative inverse , then the product should be 1. However, here the product is not 1 as

\dfrac{8}{9}\times\Bigg(-1\dfrac{1}{8}\Bigg)=\dfrac{8}{9}\times\Bigg(-\dfrac{9}{8}\Bigg)=-1\not=1

 

Question 9 :

Is 0.3 the multiplicative inverse of  3\dfrac{1}{3} ? Why or Why not ?

sol : 

*** QuickLaTeX cannot compile formula:


\\=3\dfrac{1}{3}=\dfrac{10}{3}\\=0.3\times3\dfrac{1}{3}=0.3\times\dfrac{10}{3}\\=\dfrac{10}{3}\times\dfrac{3}{10}\\=1



*** Error message:
Missing $ inserted.

Here is the product is 1 . Hence , 0.3 is the multiplicative inverse of 3\dfrac{1}{3}

 

Question 10 :

Write :

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

sol : 0 is a rational number but its reciprocal is not defined.

  

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

sol : 1 and -1 are the rational numbers that are equal to their reciprocals .

 

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

sol : 0 is the rational number that is equal to its negative.

 

Question 11 :

Fill in the blanks.

(i) Zero has No reciprocal.

(ii) The numbers 1 and -1 are their own reciprocals.

(iii) The reciprocal of -5 is -\dfrac{1}{5}

(iv) Reciprocal of \dfrac{1}{x} , where x\not=0~is~x

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

 

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

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