Question

please answer correctly .​

Answer 1

To find : ↴

$\sf \: We \: have \: to \: find \: four \: rational \: number \: between \: \: \dfrac{1}{5} \: and \: \dfrac{1}{8}$

$\bf \: \underline{ So, \: Let's \: find \: it} :$ ↴

$\sf \: So, \: first \: of \: all \: We \: will \: find \: the \: LCM \: of \: 5 \: and \: 8.$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:5 \: - \: 8\:\:\:}}}\\ {\underline{\sf{2}}}& \underline{\sf{\:\:5 \: - \: 4\:\:\:}} \\\underline{\sf{2}}&\underline{\sf{\:\:5 \: - \: 2\: \:\:}}\\{ \underline{\sf{5}}}&{\underline{\sf{\:5 \: - \: 1\:\:\:}}} \\ {\underline{\sf{}}}&{{\sf{\:\:1\:\:\:}}} \underline{\sf{}}&{\sf{\:\:\:\:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

$\bf \: LCM \: = \: 2 \: \times \: 2 \times \: 2 \: \times \: 5$

$\sf \:So, \: The \: LCM \: = \: 40$

$\sf \: Now, \: We \: will \: make \: the \: denominator \: same.$

$\sf \dfrac{1}{5} \: \times \: \dfrac{8}{8} \: \rightarrow \: \dfrac{8}{40}$

$\sf \dfrac{1}{8} \: \times \: \dfrac{5}{5} \: \rightarrow \: \dfrac{5}{40}$

$\sf \: Now, \: We \: have \: to \: multiply \: each \: of \: them \: by \: \: \dfrac{10}{10}$

$\sf \dfrac{8}{40} \: \times \: \dfrac{10}{10} \: \rightarrow \: \dfrac{80}{400}$

$\sf \dfrac{5}{40} \: \times \: \dfrac{10}{10} \: \rightarrow \: \dfrac{50}{400}$

$\sf \: So, \: the \: numbers \: between \: \dfrac{80}{100} \: and \: \dfrac{50}{100} \: are \: the \: answer.$

$\sf \red{Answer \: } = \: \dfrac{ - 72}{100} \: ,\: \dfrac{ - 71}{10} \: ,\: \dfrac{ - 70}{100} \: ,\: \dfrac{ - 69}{100}$

More to Know : ↴

The numbers which can be expressed in the form of $\frac{P}{q}$, where P and q both are Integers but q is not equal to 0 are known as Rational numbers.

$\sf Some \:Rational \:numbers \:are\: \: \dfrac{8}{7} , \dfrac{-8}{14} , \dfrac{-5}{-5}$

Every natural number, whole number, integer and fractions are rational but the converse may not possible.

Note - 0 is a rational number which is neither Positive nor nagitive.