Rational Numbers - Class 8 Maths - Extra Questions

There are no rational number between $$0$$ and $$1$$ such that $$q<p$$.

Let $$\dfrac{p}{q}$$ where $$q<p$$

$$p=5,q=3$$

Therefore $$\dfrac{5}{3}=1.666>1.$$

$${\textbf{Step  - 1: Defining rational numbers}}$$

                    $${\text{Rational numbers are those which can be written in the form of }}\dfrac{{\text{p}}}{{\text{q}}}{\text{, where q}} \ne {\text{ 0}}$$

$${\textbf{Step  - 2: Writing rational number which does not have reciprocal}}$$

                    $${\text{The rational number which does not has reciprocal is 0,}}$$

                    $${\text{as 0 can be written as }}\dfrac{{\text{0}}}{{\text{1}}}{\text{, so its reciprocal will be }}\dfrac{{\text{1}}}{{\text{0}}}{\text{ which is not defined}}{\text{.}}$$

$${\textbf{Thus, the rational number which does not has reciprocal is 0}}$$

Draw 10/3 on a number line.

 As the number 10/3 a positive number so it will be on right of zero. So the rational number lies between 3 and  4.

Since the given rational number is greater than one. So, it will always be represented on the right side of one on the number line. So, first of all we need to divide the number line between zero and two into 4 equal parts and the third part of the four parts will be the representation of $$\dfrac{3}{2}$$  on number line. It can be represented as shown in the number line.

Since the given rational number is greater than zero. So, it will always be represented on the right side of zero on the number line. So, first of all we need to divide the number line between zero and one into 4 equal parts and the third part of the four parts will be representation of $$\dfrac{3}{4}$$  on number line. It can be represented as.

$$ - 5 - \dfrac{1}{{ - 5}}$$

$$ =  - 5 + \dfrac{1}{5}$$

$$ = \dfrac{{1 - 25}}{5}$$

$$ = \dfrac{{ - 24}}{5}$$

So , multiplicative inverse of $$- 5 - \dfrac{1}{{ - 5}}$$ $$ = \dfrac{{ - 5}}{{24}}$$

$$L.H.S = \left( {\dfrac{5}{7} \times \dfrac{{12}}{{13}}} \right) \times \dfrac{7}{{18}}$$

$$ = \dfrac{{60}}{{11}} \times \dfrac{7}{{18}}$$

$$ = \dfrac{{420}}{{1638}}$$

$$R.H.S = \dfrac{5}{7}\left( {\dfrac{{12}}{{13}} \times \dfrac{7}{{18}}} \right) = \dfrac{5}{7}\left( {\dfrac{{84}}{{234}}} \right) = \dfrac{{420}}{{1638}}$$

$$Since\,L.H.S = R.H.S\,proved$$

Consider the following number.

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

To mark $$\dfrac{-5}{6}$$; move five parts on the left-side of zero.

Hence, this is the correct answer.

As we know that exactly halfway between two rational numbers is $$\dfrac{a+b}{2}$$

Rational numbers are $$\dfrac{1}{15}$$and $$\dfrac{1}{12}$$

Here, $$a=\dfrac{1}{15}$$and $$b=\dfrac{1}{12}$$

$$\therefore\dfrac{a+b}{2}=\dfrac{\dfrac{1}{15}+\dfrac{1}{12}}{2}$$

$$=\dfrac{\dfrac{4+5}{60}}{2}$$

$$=\dfrac{9}{60\times2}$$

$$=\dfrac{3}{20\times2}$$

$$=\dfrac{3}{40}$$

Therefore, the exactly halfway between $$\dfrac{1}{15}$$ and $$\dfrac{1}{12}$$ is $$\dfrac{3}{40}$$.

We know that the multiplicative inverse or reciprocal for a number $$x$$, denoted by $$\dfrac{1}{x}$$ where $$x\ne 0$$, is a number which when multiplied by $$x$$ yields the multiplicative identity $$1.$$

Therefore the multiplicative inverse of $$1$$ is $$\dfrac{1}{1}=1.$$ Hence, multiplicative inverse of $$1$$ is $$1$$ itself.

A rational number between two numbers $$ a $$ and $$ b = \cfrac{(a + b)}{2} $$
So, a rational number between $$ - 1 $$ and $$ -2 = \cfrac {-1 -2}{2} = \cfrac {-3}{2} $$ 
Now, another rational number between $$ -1 $$ and $$ \cfrac {-3}{2} =\cfrac {-1 + \tfrac {-3}{2}}{2} = \cfrac {-5}{4} $$
And another rational numbers between $$\cfrac {-3}{2} $$ and $$ -2 =\cfrac { \tfrac{-3}{2} - 2 }{2} = \cfrac {-7}{4} $$

$${\textbf{Step -1: State mid point method to find rational numbers}}{\text{.}}$$

                  $${\text{Rational numbers always being in form of }}\dfrac{p}{q}.{\text{ where q}} \ne {\text{0}}{\text{.}}$$

                  $${\text{Let }}x{\text{ and }}y{\text{ be two numbers, such that }}x{\text{  <  }}y{\text{.Then in between there is one rational point}}{\text{.}}$$

                  $${\text{That is }}\dfrac{{x + y}}{2}.$$

$${\textbf{Step -2: Find rational numbers between 3and4}}{\text{.}}$$

                  $${\text{So between 3 and 4 the rational number be , }}\dfrac{{3 + 4}}{2} = \dfrac{7}{2}.$$

$${\textbf{Step -3: Find rational numbers between 3 and }}\mathbf{\dfrac{7}{2}.}$$

                  $${\text{So between 3 and }}\dfrac{7}{2}{\text{the rational number will be , }}\dfrac{{3 + \dfrac{7}{2}}}{2} = \dfrac{{\dfrac{{6 + 7}}{2}}}{2} = \dfrac{{13}}{4}.$$

$${\textbf{Step -4: Find rational numbers between 3 and }}\mathbf{\dfrac{{13}}{4}.}$$

                  $${\text{Between 3 and }}\dfrac{{13}}{4}{\text{ the rational number will be , }}\dfrac{{3 + \dfrac{{13}}{4}}}{2} = \dfrac{{25}}{8}.$$

$${\textbf{Step -5: Find rational numbers between }}\mathbf{\dfrac{7}{2}}{\textbf{ and 4}}{\textbf{.}}$$

                  $${\text{Between }}\dfrac{{\text{7}}}{{\text{2}}}{\text{ and 4 the rational  number will be , }}\dfrac{{4 + \dfrac{7}{2}}}{2} = \dfrac{{\dfrac{{7 + 8}}{2}}}{2} = \dfrac{{15}}{4}.$$

$${\textbf{Step -6: Find rational numbers between }}\mathbf{\dfrac{{15}}{4}}{\textbf{ and 4}}{\textbf{.}}$$

                  $${\text{Between }}\dfrac{{{\text{15}}}}{{\text{4}}}{\text{ and 4 , the rational number will be , }}\dfrac{{\dfrac{{15}}{4} + 4}}{2} = \dfrac{{31}}{8}.$$

$${\textbf{Hence , rational numbers between 3 and 4 are }}\mathbf{\dfrac{7}{2},\dfrac{{13}}{4},\dfrac{{15}}{4},\dfrac{{25}}{8},\dfrac{{31}}{{8}}.}$$

• $$(16+18)+20=(18+16)+20$$
• It is associative property of addition

Making equal denominators of both the dfractions.

$$\dfrac{2}{3} = \dfrac{2}{3} \times \dfrac{5}{5}$$

$$ = \dfrac{{10}}{{15}}$$

And,

$$\dfrac{4}{5} = \dfrac{4}{5} \times \dfrac{3}{3}$$

$$ = \dfrac{{12}}{{15}}$$

Now, the dfraction between $$\dfrac{{10}}{{15}}$$ and $$\dfrac{{12}}{{15}}$$ is,

$$\dfrac{{\dfrac{{10}}{{15}} + \dfrac{{12}}{{15}}}}{2} = \dfrac{{22}}{{30}}$$

So, $$\dfrac{{10}}{{15}} < \dfrac{{22}}{{30}} < \dfrac{{12}}{{15}}$$

The dfraction between $$\dfrac{{10}}{{15}}$$ and $$\dfrac{{22}}{{30}}$$ is,

$$\dfrac{{\dfrac{{10}}{{15}} + \dfrac{{22}}{{30}}}}{2} = \dfrac{{42}}{{30 \times 2}}$$

$$ = \dfrac{{21}}{{30}}$$

So, $$\dfrac{{10}}{{15}} < \dfrac{{21}}{{30}} < \dfrac{{22}}{{30}} < \dfrac{{12}}{{15}}$$

The dfraction between $$\dfrac{{22}}{{30}}$$ and $$\dfrac{{12}}{{15}}$$ is,

$$\dfrac{{\dfrac{{22}}{{30}} + \dfrac{{12}}{{15}}}}{2} = \dfrac{{46}}{{30 \times 2}}$$

$$ = \dfrac{{23}}{{30}}$$

$$\dfrac{{10}}{{15}} < \dfrac{{21}}{{30}} < \dfrac{{22}}{{30}} < \dfrac{{23}}{{30}} < \dfrac{{12}}{{15}}$$

The dfraction between $$\dfrac{{10}}{{15}}$$ and $$\dfrac{{21}}{{30}}$$ is,

$$\dfrac{{\dfrac{{10}}{{15}} + \dfrac{{21}}{{30}}}}{2} = \dfrac{{41}}{{30 \times 2}}$$

$$ = \dfrac{{41}}{{60}}$$

So, $$\dfrac{{10}}{{15}} < \dfrac{{41}}{{60}} < \dfrac{{21}}{{30}} < \dfrac{{22}}{{30}} < \dfrac{{23}}{{30}} < \dfrac{{12}}{{15}}$$

The dfraction between $$\dfrac{{23}}{{30}}$$ and $$\dfrac{{12}}{{15}}$$.

$$\dfrac{{\dfrac{{23}}{{30}} + \dfrac{{12}}{{15}}}}{2} = \dfrac{{47}}{{30 \times 2}}$$

$$ = \dfrac{{47}}{{60}}$$

So, $$\dfrac{{10}}{{15}} < \dfrac{{41}}{{60}} < \dfrac{{21}}{{30}} < \dfrac{{22}}{{30}} < \dfrac{{23}}{{30}} < \dfrac{{47}}{{60}} < \dfrac{{12}}{{15}}$$

Therefore, the five dfractions between the given numbers are $$\dfrac{{41}}{{60}},\dfrac{{21}}{{30}},\dfrac{{22}}{{30}},\dfrac{{23}}{{30}},\dfrac{{47}}{{60}}$$.

Consider given the given binary operation,

$$a*b=a-b$$

Commutative –

  $$ a*b=a-b $$

 $$ b*a=b-a $$

 $$ a-b\ne b-a $$

 $$ a*b\ne b*a $$

Hence, given operation is not commutative.

Associative-

  $$ a*b=a-b $$

 $$ a*\left( b*c \right)=a*\left( b-c \right) $$

 $$ =a-\left( b-c \right)=a-b+c $$

 $$ \left( a*b \right)*c=\left( a-b \right)*c=\left( a-b \right)-c $$

 $$ =a-b-c $$

Now,

  $$ a*\left( b*c \right)\ne \left( a*b \right)*c $$

 $$ \because a-b+ca\ne a-b-c $$

Hence, given operation is not associative.

$$L.H.S = \dfrac{{ - 13}}{{24}} \times \left( {\dfrac{{12}}{5} \times \dfrac{{35}}{{36}}} \right)$$

$$ = \dfrac{{ - 13}}{{24}} \times \left( {\dfrac{{ - 7}}{3}} \right)$$

$$ = \dfrac{{91}}{{72}}$$

$$R.H.S = \left( {\dfrac{{ - 13}}{{24}} \times \dfrac{{ - 12}}{5}} \right) \times \dfrac{{35}}{{36}}$$

$$ = \dfrac{{13}}{{10}} \times \dfrac{{35}}{{36}}$$

$$ = \dfrac{{91}}{{72}}$$

$$Since,\,R.H.S = L.H.S$$

Making the same denominator of the rational numbers $$\dfrac{4}{5}$$ and $$\dfrac{2}{3}$$.

$$\dfrac{4}{5} = \dfrac{{4 \times 3}}{{5 \times 3}} = \dfrac{{12}}{{15}}$$

$$\dfrac{2}{3} = \dfrac{{2 \times 5}}{{3 \times 5}} = \dfrac{{10}}{{15}}$$

The rational numbers between $$\dfrac{{12}}{{15}}$$ and $$\dfrac{{10}}{{15}}$$ is,

$$\dfrac{{\dfrac{{12}}{{15}} + \dfrac{{10}}{{15}}}}{2} = \dfrac{{22}}{{30}}$$

So, $$\dfrac{{10}}{{15}} < \dfrac{{22}}{{30}} < \dfrac{{12}}{{15}}$$

The rational number between $$\dfrac{{10}}{{15}}$$ and $$\dfrac{{22}}{{30}}$$ is,

$$\dfrac{{\dfrac{{10}}{{15}} + \dfrac{{22}}{{30}}}}{2} = \dfrac{{42}}{{30 \times 2}}$$

$$ = \dfrac{{21}}{{30}}$$

The rational number between $$\dfrac{{12}}{{15}}$$ and $$\dfrac{{22}}{{30}}$$ is,

$$\dfrac{{\dfrac{{12}}{{15}} + \dfrac{{22}}{{30}}}}{2} = \dfrac{{46}}{{30 \times 2}}$$

$$ = \dfrac{{23}}{{30}}$$

Thus, $$\dfrac{{10}}{{15}} < \dfrac{{21}}{{30}} < \dfrac{{22}}{{30}} < \dfrac{{23}}{{30}} < \dfrac{{12}}{{15}}$$.

Therefore the three rational numbers are $$\dfrac{{21}}{{30}},\dfrac{{22}}{{30}},\dfrac{{23}}{{30}}$$.

$${\textbf{Step - 1: Let us consider the rational number as x.}} $$

                  $${\text{The formula for finding rational number is,}}$$ $$x = \dfrac{1}{2}\left( {\dfrac{a}{b} + \dfrac{c}{d}} \right)$$

$${\textbf{Step -2: So, using this formula find the rational number between}}$$$$ \mathbf{- 3}$$ $${\textbf{and}}$$ $$\mathbf{1.}$$

                  $$ \Rightarrow x = \dfrac{1}{2}\left( { - 3 + 1} \right)$$

                  $$ \Rightarrow x = \dfrac{1}{2}\left( { - 2} \right)$$

                  $$ \Rightarrow x =  - 1$$

$${\textbf{Hence, the rational number between }}$$$$\mathbf{ - 3}$$  $${\textbf{and}}$$ $$\mathbf{1}$$ $${\textbf{is}}$$ $$ \mathbf{- 1.}$$

We know that the multiplicative inverse or reciprocal for a number $$x$$, denoted by $$\dfrac{1}{x}$$ or $$x^{-1},$$ is a number which when multiplied by $$x$$ yields the multiplicative identity $$1.$$

Therefore the multiplicative inverse of rational numbers $$\displaystyle \dfrac { 8 }{ 13 } ,-\dfrac { 13 }{ 11 } ,\dfrac { 12 }{ 17 } ,-\dfrac { 101 }{ 100 } ,\dfrac { 26 }{ 23 } $$ 

is $$\dfrac { 13 }{ 8 } ,-\dfrac { 11 }{ 13 } ,\dfrac { 17 }{ 12 } ,-\dfrac { 100 }{ 101 } ,\dfrac { 23 }{ 26 } $$ respectively.