Question

Find a rational number between the following pair of rational numbers.
(i)$ \dfrac{4}{3}$ and $ \dfrac{2}{5}$
(ii)$ - \dfrac{2}{7}$ and $ \dfrac{5}{6}$
(iii)$ \dfrac{5}{{11}}$ and $ \dfrac{7}{8}$
(iv)$ \dfrac{7}{4}$ and $ \dfrac{8}{3}$

Answer 1

Hint: the base of the number should be made the same by taking out the LCM of the numbers in the denominator and then multiplying by a required number to make the denominator equal to the LCM.

Complete step-by-step answer:
(i) $ \dfrac{4}{3}$ and $ \dfrac{2}{5} $
Find the LCM for 3 and 5.
LCM for 3 and 5, $ = 3 \times 5 = 15 $
Multiply the first rational number by 5 and second by 3, both in numerator and denominator to make the base the same.
The number becomes
$ \dfrac{{4 \times 5}}{{3 \times 5}} = \dfrac{{20}}{{15}} $ and $ \dfrac{{2 \times 3}}{{5 \times 3}} = \dfrac{6}{{15}} $
Choose any number between $ \dfrac{{20}}{{15}}$ and $ \dfrac{6}{{15}} $
The number is $ \dfrac{7}{{15}} $ (can be any number with base 15 and lying between 20 and 6)
Hence, the required rational number is $ \dfrac{7}{{15}}$ .

(ii) $ - \dfrac{2}{7}$ and $ \dfrac{5}{6} $
Find the LCM for 3 and 5.
LCM for 7 and 6 $ = 7 \times 6 = 42$
Multiply the first rational number by 6 and second by 7, both in numerator and denominator to make the base the same.
The number becomes
$ - \dfrac{{2 \times 6}}{{7 \times 6}} = - \dfrac{{12}}{{42}}$ and $ \dfrac{{5 \times 7}}{{6 \times 7}} = \dfrac{{35}}{{42}} $
Choose any number between $ - \dfrac{{12}}{{42}}$ and $ \dfrac{{35}}{{42}} $
The number is $ \dfrac{{31}}{{42}}$ (can be any number with base 42 and lying between -12 and 35)
Hence, the required rational number is $ \dfrac{{31}}{{42}} $ .

(iii) $ \dfrac{5}{{11}}$ and $ \dfrac{7}{8} $
Find the LCM for 11 and 8.
LCM for 11 and 8 $ = 11 \times 8 = 88 $

Multiply the first rational number by 8 and second by 11, both in numerator and denominator to make the base the same.

The number becomes
$ \dfrac{{5 \times 8}}{{11 \times 8}} = \dfrac{{40}}{{88}}$ and $ \dfrac{{7 \times 11}}{{8 \times 11}} = \dfrac{{77}}{{88}} $

Choose any number between $ \dfrac{{40}}{{88}}$ and $ \dfrac{{77}}{{88}} $
The number is $ \dfrac{{50}}{{88}}$ (can be any number with base 88 and lying between 40 and 77)
Hence, the required rational number is $ \dfrac{50}{{88}} $ .

(iv)$ \dfrac{7}{4}$ and $ \dfrac{8}{3} $
Find the LCM for 4 and 3.
LCM for 4 and 3 $ = 4 \times 3 = 12 $
Multiply the first rational number by 3 and second by 4, both in numerator and denominator to make the base the same.
The number becomes
$ \dfrac{{7 \times 3}}{{4 \times 3}} = \dfrac{{21}}{{12}} $ and $ \dfrac{{8 \times 4}}{{3 \times 4}} = \dfrac{{24}}{{12}} $
Choose any number between $ \dfrac{{21}}{{12}}$ and $ \dfrac{{24}}{{12}} $
The number is $ \dfrac{{22}}{{12}}$ (can be any number with base 12 and lying between 21 and 24)

Hence, the required rational number is $ \dfrac{{22}}{{12}} $ .

Note: LCM should be taken out for the terms in the denominator and multiplied by a suitable factor in the numerator and denominator.
Any number can be chosen after making the base the same, provided the difference of the terms in the numerator should be more than 2. For instance, rational numbers between $ \dfrac{2}{3}$ and $ \dfrac{5}{3}$ can be $ \dfrac{3}{3},\dfrac{4}{3}$ .Out of them any one can be chosen.