Question

SOLVE THE FOLLOWING :-



1. What are the multiplicative and additive identities of rational numbers?



2. Write the additive inverse of 19/-6 and -⅔.



3. Write the multiplicative inverse of -13/19 and -7.



4. Mention any 4 rational numbers which are less than 5.



5. Find any five rational numbers between -3/5 and 1/3.



6. Use appropriate property and find

(-2/3 × 1/5) + ( 2/3 × 1/5 )



7. Which property you use to compute 1/4 × (3 × 4/7) as (1/4 × 3 ) × 4/7 ? Justify.



8. Represent -2/11, -5/11, and -9/11 on the number line.​

Answer 1

Solution : 1

The rational number 0 is the additive identity for rational numbers. The rational number 1 is the multiplicative identity for rational numbers

______________________________________

Solution - 2

$\sf \: (a). \: \: \dfrac{19}{ - 6} \: = \: \dfrac{19}{6}$

$\sf \: (b). \: \: \dfrac{ - 2}{3} \: = \: \dfrac{2}{3}$

______________________________________

Solution - 3

$\sf \: (a). \: \: \dfrac{ - 13}{ 19} \: = \: \dfrac{ - 19}{13}$

$\sf \: (b). \: \: - 7 \: = \: \dfrac{ - 1}{7}$

______________________________________

Solution - 4

$\sf \: The \: rational \: number \: that \: are \: less \: than \: 5 \: are \: \underline{4, \: 3, \: 2, \: 1}.$

______________________________________

Solution - 5

$\bf \: To \: find :$ ↴

$\sf \: We \: have \: to \: find \: five \: rational \: number \: between \: \dfrac{ - 3}{5} \: and \: \dfrac{1}{3}$

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

$\sf \: First \: of \: all \: we \: have \: to \: make \: the \: denominators \: as \: same.$

$\sf \: Taking \: LCM \: of \: 5 \: and \: 3$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{3}}}&{\underline{\sf{\:\:5 - 3\:\:\:}}}\\ {\underline{\sf{5}}}& \underline{\sf{\:\:5 - 1\:\:\:}} \\\underline{\sf{}}&{\sf{\:\: 1 - 1\:\:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

$\sf \: LCM = 3 \times 5 = 15$

$\sf \longrightarrow \: \dfrac{ - 3}{5} \times \dfrac{3}{3} = \dfrac{ - 9}{15}$

$\sf \longrightarrow \: \dfrac{1}{3} \times \dfrac{5}{5} = \dfrac{5}{15}$

$\sf \: Now,$

$\dfrac{ { - 9}}{15}, \: \boxed{ \dfrac{ - 8}{15}, \: \dfrac{ - 7}{15}, \: \dfrac{ - 6}{15}, \: \dfrac{ - 5}{15}, \: \dfrac{ - 4}{15}, } \: \dfrac{5}{15}$

$\sf \: Answer = \dfrac{ - 8}{15}, \: \dfrac{ - 7}{15}, \: \dfrac{ - 6}{15}, \: \dfrac{ - 5}{15}, \: \dfrac{ - 4}{15}$

______________________________________

Solution - 6

$\sf \: \bigg( \dfrac{ - 2}{3} \times\dfrac{1}{5} \bigg) + \bigg( \dfrac{2}{3} \times \dfrac{1}{5} \bigg)$

$\sf \: \bigg( \dfrac{ 2}{3} \bigg) \times\bigg(\dfrac{ - 1}{5} \bigg) + \bigg( \dfrac{2}{3} \bigg) \times \bigg(\dfrac{1}{5} \bigg)$

$\sf \: \bigg( \dfrac{ 2}{3} \bigg) \times \bigg[ \bigg(\dfrac{ - 1}{5} + \dfrac{1}{5} \bigg) \bigg]$

$\sf \: \bigg( \dfrac{ 2}{3} \bigg) \times \bigg[ \bigg(\dfrac{ - 1 + 1}{5} \bigg) \bigg]$

$\sf \: \bigg( \dfrac{ 2}{3} \bigg) \times \bigg(\dfrac{ 0}{5} \bigg)$

$\sf \: \dfrac{ 2}{3} \times 0$

$\sf \: Answer = 0$

______________________________________

Solution - 7

$\sf \dfrac{1}{4} \: \bigg(3 \times \dfrac{4}{7} \bigg) \: as \: \bigg( \dfrac{1}{4} \times 3 \bigg) \times \dfrac{4}{7}$

$\sf \dfrac{1}{4} \: \bigg(3 \times \dfrac{4}{7} \bigg)$

$\sf \: = \dfrac{12}{28}$

$\sf \: = \dfrac{3}{7} - - - - - - \boxed{1}$

$\sf \bigg( \dfrac{1}{4} \times 3 \bigg) \times \dfrac{4}{7}$

$\bigg( \dfrac{3}{4} \bigg) \times\bigg( \dfrac{4}{7} \bigg)$

$\sf \: = \dfrac{3}{7} - - - - - - \boxed{2}$

$\underline{ \sf \: Now, \: from \: 1 \: and \: 2 }$

$\sf \dfrac{1}{4} \: \bigg(3 \times \dfrac{4}{7} \bigg) \: = \: \bigg( \dfrac{1}{4} \times 3 \bigg) \times \dfrac{4}{7}$

$\sf \: Associative \: property \: of \: Rational \: no. \: is \: used \: here.$

______________________________________

Solution - 8

$\sf \: In \: this \: question \: we \: have \: to \: represent \: \dfrac{ - 2}{11} , \: \dfrac{ - 5}{11} \: and \: \dfrac{ - 9}{11} \: on \: number \: line.$

$\sf \: See \: the \: attached \: image...$