Question

Find 8 rational number between -1\3and-2\3

Answer 1

SOLUTION :

Finding 8 rational number between $\frac{-1}{3} \:and \: \frac{-2}{3}$

So,

$\frac{-1}{3}\times \frac{9}{9} \: \: \: \:and \: \: \: \: \frac{-2}{3} \times \frac{9}{9}$

$\frac{-9}{27} \: \: \: \: and \: \: \: \: \frac{-18}{27}$

Now,

We can find rational between them

$\frac{-9}{27} \: , \: \frac{-10}{27} \: , \: \frac{-11}{27} \: , \frac{-12}{27} \: , : \frac{-13}{27}\: , \: \frac{-14}{27}\:, \:\frac{-15}{27}\:, \:\frac{-16}{27}\:, \:\frac{-17}{27}\:, \:\frac{-18}{27}$

Answer 2

$ist \: rational \: number \\ \\ = \frac{ \frac{ - 1}{3} + ( \frac{ - 2}{3} ) }{2} \\ \\ = \frac{ - 3}{6} \\ \\ = \frac{ - 1}{2} \\ \\ second \: rational \: number \: \: \\ \\ = \frac{ \frac{ - 1}{3} + ( \ \frac{ - 1}{2} )}{2} \\ \\ = \frac{ \frac{ - 2 - 3}{6} }{2} \\ \\ = \frac{ - 5}{12} \\ \\ 3 {}^{rd} \: rational \: number \\ \\ = \frac{ \frac{ - 1}{2} + ( \frac{ - 2}{3}) }{2} \\ \\ = \frac{ \frac{ - 3 - 4}{6} }{2} \\ \\ = \frac{ - 7}{12} \\ \\ \\ 4 {}^{th} rational \: number \\ \\ = \frac{ \frac{ - 1}{2} + ( \frac{ - 5}{12}) }{2} \\ \\ = \frac{ \frac{ - 6 - 5}{12} }{2} \\ \\ = \frac{ - 11}{24} \\ \\ \\ 5 {}^{th} \: rational \: number \\ \\ = \frac{ \frac{ - 7}{12} + ( \frac{ - 11}{24} )}{2} \\ \\ = \frac{ \frac{ - 14 - 11}{24} }{2} \\ \\ = \frac{ - 25}{48} \\ \\ \\ 6 {}^{th} \: rational \: number \\ \\ = \frac{ \frac{ - 11}{24} + ( \frac{ - 25}{48} ) }{2} \\ \\ = \frac{ \frac{ - 22 - 25}{48} }{2} \\ \\ = \frac{ - 47}{96} \\ \\ 7 {}^{th} \: rational \: number \\ \\ = \frac{ \frac{ - 25}{48} + ( \frac{ - 47}{96} ) }{2} \\ \\ = \frac{ \frac{ - 50 - 47}{96} }{2} \\ \\ = \frac{ - 97}{ 192} \\ \\ 8 {}^{th} \: rational \: number \\ \\ = \frac{ \frac{ - 47}{96} + ( \frac{ - 97}{192} )}{2} \\ \\ = \frac{ \frac{ - 94 - 97}{192} }{2} \\ \\ = \frac{ - 191}{192 \times 2} \\ \\ = \frac{ - 191}{384}$


hope it helps..