Question

(i). Five rational numbers between $2$ and $1$ are

(ii). Four rational numbers between $\frac{2}{3}$ and $\frac{3}{5}$ are

Answer 1

(i) We have to five rational numbers between the given numbers, first we will convert the given nos. its equivalent rational number, by multiplying same number in the numerator and denominator. There exist infinite numbers between any two rationals.

For the given first numbers.we multiply the number by $6$.

i.e

$1\times \frac{6}{6}=\frac{6}{6}$ and $2\times \frac{6}{6}=\frac{12}{6}$

$\frac{6}{6}<\frac{7}{6}<\frac{8}{6}<\frac{9}{6}<\frac{10}{6}<\frac{11}{6}<\frac{12}{6}$

Thus $5$ rational number between $1$ and $2$ are $\frac{7}{6},\frac{8}{6},\frac{9}{6},\frac{10}{6},\frac{11}{6}$

(ii) As in $\frac{2}{3},\frac{3}{5}$ denominator isn't same, to make denominators same take LCM of $3$ and $5.$

i.e LCM $(3,5)=15$

Now, we have to make the denominators same of the given fractions.

$\frac{2}{3}\times \frac{5}{5}=\frac{10}{15}$ and $\frac{3}{5}\times \frac{3}{3}=\frac{9}{15}$

fractions are $\frac{10}{15}$ and $\frac{9}{15}$

multiply $\frac{10}{15}$ and $\frac{9}{15}$ by $\frac{5}{5}$

$\frac{10}{15}\times \frac{5}{5}=\frac{50}{75}$ and $\frac{9}{15}\times \frac{5}{5}=\frac{45}{75}$

$\Rightarrow \frac{50}{75},\frac{45}{75}$

$\frac{45}{75}<\frac{46}{75}<\frac{47}{75}<\frac{48}{75}<\frac{49}{75}<\frac{50}{75}$

Thus $4$ rational number between $\frac{2}{3}$and $\frac{3}{5}$ are $\frac{46}{75},\frac{47}{75},\frac{48}{75},\frac{49}{75}$