Rs Aggarwal 2020 2021 for Class 8 Maths Chapter 1 - Rational Numbers

Answer:

If $\frac{a}{b}$ is a fraction and $m$ is a non-zero integer, then $\frac{a}{b}=\frac{a\times m}{b\times m}$.

Now,

(i) $\frac{-3}{5}=\frac{-3\times 4}{5\times 4}=\frac{-12}{20}$

(ii) $\frac{-3}{5}=\frac{-3\times -6}{5\times -6}=\frac{18}{-30}$

(iii)$\frac{-3}{5}=\frac{-3\times 7}{5\times 7}=\frac{-21}{35}$

(iv)$\frac{-3}{5}=\frac{-3\times -8}{5\times -8}=\frac{24}{-40}$

Answer:

If $\frac{a}{b}$ is a rational number and $m$ is a common divisor of $a \mathrm{and} b$, then $\frac{a}{b}=\frac{a÷m}{b÷m}$.

∴ $\frac{-42}{98}=\frac{-42÷14}{98÷14}=\frac{-3}{7}$

Answer:

If $\frac{a}{b}$ is a rational integer and $m$ is a common divisor of $a \mathrm{and} b$, then $\frac{a}{b}=\frac{a÷m}{b÷m}$.

∴​ $\frac{-48}{60}=\frac{-48÷12}{60÷12}=\frac{-4}{5}$

Answer:

A rational number $\frac{a}{b}$ is said to be in the standard form if $a$ and $b$ have no common divisor other than unity and $b>0$.
Thus,

(i) The greatest common divisor of 12 and 30 is 6.
   
     ∴ $\frac{-12}{30}=\frac{-12÷6}{30÷6}=\frac{-2}{5}$ (In the standard form)

(ii)The greatest common divisor of 14 and 49 is 7.    
     
    ∴ $\frac{-14}{49}=\frac{-14÷7}{49÷7}=\frac{-2}{7}$ (In the standard form)

(iii) $\frac{24}{-64}=\frac{24\times (-1)}{-64\times -1}=\frac{-24}{64}$
   
     The greatest common divisor of 24 and 64 is 8.          
    
     ∴ $\frac{-24}{64}=\frac{-24÷8}{64÷8}=\frac{-3}{8}$ (In the standard form)

(iv) $\frac{-36}{-63}=\frac{-36\times (-1)}{-63\times -1}=\frac{36}{63}$
 
     The greatest common divisor of 36 and 63 is 9.     
  
      ∴ $\frac{36}{63}=\frac{36÷9}{63÷9}=\frac{4}{7}$ (In the standard form)

Answer:

We know:
(i) Every positive rational number is greater than 0.
(ii) Every negative rational number is less than 0.

Thus, we have:

(i)$\frac{3}{8}$ is a positive rational number.
    ∴ $\frac{3}{8}>0$

(ii)$\frac{-2}{9}$ is a negative rational number.
    ∴ $\frac{-2}{9}<0$

(iii) $\frac{-3}{4}$ is a negative rational number.
    ∴ $\frac{-3}{4}<0$
    Also,
    $\frac{1}{4}$ is a positive rational number.
    ∴ $\frac{1}{4}>0$
    Combining the two inequalities, we get:
   $\frac{-3}{4}<\frac{1}{4}$

(iv)Both $\frac{-5}{7}$ and $\frac{-4}{7}$ have the same denominator, that is, 7.
    So, we can directly compare the numerators.

    ∴ $\frac{-5}{7}<\frac{-4}{7}$

(v)The two rational numbers are $\frac{2}{3}$ and $\frac{3}{4}$.
    The LCM of the denominators 3 and 4 is 12.
    Now,
   $\frac{2}{3}=\frac{2\times 4}{3\times 4}=\frac{8}{12}$
    Also, 
   $\frac{3}{4}=\frac{3\times 3}{4\times 3}=\frac{9}{12}$
    Further
   $\frac{8}{12}<\frac{9}{12}$

    ∴$\frac{2}{3}<\frac{3}{4}$

(vi)The two rational numbers are $\frac{-1}{2}$ and $-1$.
    We can write $-1=\frac{-1}{1}$.
    The LCM of the denominators 2 and 1 is 2.
    Now,
    $\frac{-1}{2}=\frac{-1\times 1}{2\times 1}=\frac{-1}{2}$
    Also,
    $\frac{-1}{1}=\frac{-1\times 2}{1\times 2}=\frac{-2}{2}$
    ∵ $\frac{-2}{1}<\frac{-1}{1}$
    ∴ $-1<\frac{-1}{2}$

Answer:

1. The two rational numbers are $\frac{-4}{3}\mathrm{and}\frac{-8}{7}$.

The LCM of the denominators 3 and 7 is 21.

Now,
 
$\frac{-4}{3}=\frac{-4\times 7}{3\times 7}=\frac{-28}{21}$

Also,

$\frac{-8}{7}=\frac{-8\times 3}{7\times 3}=\frac{-24}{21}$

Further,
 
$\frac{-28}{21}<\frac{-24}{21}$

∴ $\frac{-4}{3}<\frac{-8}{7}$

2. ​The two rational numbers are $\frac{7}{-9}\mathrm{and}\frac{-5}{8}$.

The first fraction can be expressed as $\frac{7}{-9}=\frac{7\times -1}{-9\times -1}=\frac{-7}{9}$.

The LCM of the denominators 9 and 8 is 72.

Now, 

$\frac{-7}{9}=\frac{-7\times 8}{9\times 8}=\frac{-56}{72}$

Also,

$\frac{-5}{8}=\frac{-5\times 9}{8\times 9}=\frac{-45}{72}$

Further,
 
$\frac{-56}{72}<\frac{-45}{72}$

∴​ $\frac{7}{-9}<\frac{-5}{8}$

3. ​The two rational numbers are $\frac{-1}{3}\mathrm{and}\frac{4}{-5}$.

$\frac{4}{-5}=\frac{4\times -1}{-5\times -1}=\frac{-4}{5}$

The LCM of the denominators 3 and 5 is 15.

Now, 

$\frac{-1}{3}=\frac{-1\times 5}{3\times 5}=\frac{-5}{15}$

Also,

$\frac{-4}{5}=\frac{-4\times 3}{5\times 3}=\frac{-12}{15}$

Further,
 
$\frac{-12}{15}<\frac{-5}{15}$

∴ $\frac{4}{-5}<\frac{-1}{3}$

4. The two rational numbers are $\frac{9}{-13}\mathrm{and}\frac{7}{-12}$.

$\mathrm{Now}, \frac{ 9}{-13}=\frac{9\times -1}{-13\times -1}=\frac{-9}{13} \mathrm{and} \frac{7}{-12}=\frac{7\times -1}{-12\times -1}=\frac{-7}{12}$

The LCM of the denominators 13 and 12 is 156.

Now, 

$\frac{-9}{13}=\frac{-9\times 12}{13\times 12}=\frac{-108}{156}$

Also,

$\frac{-7}{12}=\frac{-7\times 13}{12\times 13}=\frac{-91}{156}$

Further,
 
$\frac{-108}{156}<\frac{-91}{156}$

∴ $\frac{9}{-13}<\frac{7}{-12}$

5. The two rational numbers are $\frac{4}{-5} \mathrm{and} \frac{-7}{10}$.

∴​ $\frac{4}{-5}=\frac{4\times -1}{-5\times -1}=\frac{-4}{5}$

The LCM of the denominators 5 and 10 is 10.

Now,
 
$\frac{-4}{5}=\frac{-4\times 2}{5\times 2}=\frac{-8}{10}$

Also,

$\frac{-7}{10}=\frac{-7\times 1}{10\times 1}=\frac{-7}{10}$

Further,
 
$\frac{-8}{10}<\frac{-7}{10}$

∴ $\frac{-4}{5}<\frac{-7}{10}, or, \frac{4}{-5}<\frac{-7}{10}$

6. The two rational numbers are $\frac{-12}{5}\mathrm{and} -3.$
 $-3 \mathrm{can} \mathrm{be} \mathrm{written} \mathrm{as} \frac{-3}{1}$.

The LCM of the denominators is 5.

Now,
 
$\frac{-3}{1}=\frac{-3\times 5}{1\times 5}=\frac{-15}{5}$

Because $\frac{-15}{5}<\frac{-12}{5}$, we can conclude that $-3<\frac{-12}{5}$.

Answer:

(i)We will write each of the given numbers with positive denominators.

One number = $\frac{-3}{7}$ 
Other number =$\frac{6}{-13}=\frac{6\times (-1)}{-13\times (-1)}=\frac{-6}{13}$

 LCM of 7 and 13 = 91

 ∴ $\frac{-3}{7}=\frac{-3\times 13}{7\times 13}=\frac{-39}{91}$ 

And,

$\frac{-6}{13}=\frac{-6\times 7}{13\times 7}=\frac{-42}{91}$$\frac{-6}{13}=\frac{-6\times 7}{13\times 7}=\frac{-42}{91}$$\frac{-6}{13}=\frac{-6\times 7}{13\times 7}=\frac{-42}{91}$

Clearly,

$-39>-41$

∴ ​$\frac{-39}{91 }>\frac{-42}{91}$

Thus,

$\frac{-3}{7}>\frac{6}{-13}$

(ii) We will write each of the given numbers with positive denominators.

One number = $\frac{5}{-13}=\frac{5\times (-1)}{-13\times (-1)}=\frac{-5}{13}$ 

Other number =$\frac{-35}{91}$

 LCM of 13 and 91 = 91

 ∴ $\frac{-5}{13}=\frac{-5\times 7}{13\times 7}=\frac{-35}{91}$ and $\frac{-35}{91}$

Clearly,
 
$-35=-35$

 ∴ $\frac{-35}{91 }=\frac{-35}{91}$

Thus,
 
$\frac{-5}{13}=\frac{-35}{91}$ 
 

(iii) We will write each of the given numbers with positive denominators.

One number = $-2$
 
We can write -2 as$\frac{-2}{1}$.
Other number =$\frac{-13}{5}$

 LCM of 1 and 5 = 5

 ∴​ $\frac{-2}{1}=\frac{-2\times 5}{1\times 5}=\frac{-10}{5}$ and $\frac{-13}{5}=\frac{-13\times 1}{5\times 1}=\frac{-13}{5}$

Clearly,

$-10>-13$

  ∴ $\frac{-10}{5}>\frac{-13}{5}$

Thus,
 
$\frac{-2}{1}>\frac{-13}{5}$ 
 
$-2>\frac{-13}{5}$

(iv) We will write each of the given numbers with positive denominators.

One number = $\frac{-2}{3}$ 
Other number =$\frac{5}{-8}=\frac{5\times (-1)}{-8\times (-1)}=\frac{-5}{8}$

 LCM of 3 and 8 = 24

∴ ​$\frac{-2}{3}=\frac{-2\times 8}{3\times 8}=\frac{-16}{24}$ and $\frac{-5}{8}=\frac{-5\times 3}{8\times 3}=\frac{-15}{24}$

Clearly,

$-16<-15$
 
∴ $\frac{-16}{24}<\frac{-15}{24}$

Thus,
 
$\frac{-2}{3}<\frac{-5}{8}$ 
 
$\frac{-2}{3}<\frac{5}{-8}$

(v) $\frac{-3}{-5}=\frac{-3\times -1}{-5\times -1}=\frac{3}{5}$

$\frac{3}{5}$ is a positive number.

Because every positive rational number is greater than 0, $\frac{3}{5}>0\Rightarrow 0<\frac{3}{5}$.

(vi) We will write each of the given numbers with positive denominators.

One number = $\frac{-8}{9}$

Other number = $\frac{-9}{10}$

 LCM of 9 and 10 = 90

∴​$\frac{-8}{9}=\frac{-8\times 10}{9\times 10}=\frac{-80}{90}$ and $\frac{-9}{10}=\frac{-9\times 9}{10\times 9}=\frac{-81}{90}$

Clearly,

$-81<-80$

∴​$\frac{-81}{90}<\frac{-80}{90}$

Thus,
 
$\frac{-9}{10}<\frac{-8}{9}$ 

Answer:

(i) We will write each of the given numbers with positive denominators.

We have:

$\frac{4}{-9}=\frac{4\times (-1)}{-9\times (-1)}=\frac{-4}{9}$ and$\frac{7}{-18}=\frac{7\times (-1)}{-18\times (-1)}=\frac{-7}{18}$

Thus, the given numbers are $\frac{-4}{9}, \frac{-5}{12}, \frac{-7}{18} \mathrm{and} \frac{-2}{3}.$

LCM of 9, 12, 18 and 3 is 36.

Now, 

$\frac{-4}{9}=\frac{-4\times 4}{9\times 4}=\frac{-16}{36}$

$\frac{-5}{12}=\frac{-5\times 3}{12\times 3}=\frac{-15}{36}$

$\frac{-7}{18}=\frac{-7\times 2}{18\times 2}=\frac{-14}{36}$

$\frac{-2}{3}=\frac{-2\times 12}{3\times 12}=\frac{-24}{36}$

Clearly, 

$\frac{-24}{36}<\frac{-16}{36}<\frac{-15}{36}<\frac{-14}{36}$

∴ ​$\frac{-2}{3}<\frac{-4}{9}<\frac{-5}{12}<\frac{-7}{18}$ 
 
That is

$\frac{-2}{3}<\frac{4}{-9}<\frac{-5}{12}<\frac{7}{-18}$

(ii) We  will write each of the given numbers with positive denominators.

We have:

$\frac{5}{-12}=\frac{5\times (-1)}{-12\times (-1)}=\frac{-5}{12}$ and$\frac{9}{-24}=\frac{9\times (-1)}{-24\times (-1)}=\frac{-9}{24}$

Thus, the given numbers are $\frac{-3}{4}, \frac{-5}{12}, \frac{-7}{16} \mathrm{and} \frac{-9}{24}.$

LCM of  4, 12, 16 and 24 is 48.

Now,
 
$\frac{-3}{4}=\frac{-3\times 12}{4\times 12}=\frac{-36}{48}$

$\frac{-5}{12}=\frac{-5\times 4}{12\times 4}=\frac{-20}{48}$

$\frac{-7}{16}=\frac{-7\times 3}{16\times 3}=\frac{-21}{48}$

$\frac{-9}{24}=\frac{-9\times 2}{24\times 2}=\frac{-18}{48}$

Clearly, 

$\frac{-36}{48}<\frac{-21}{48}<\frac{-20}{48}<\frac{-18}{48}$

∴​ $\frac{-3}{4}<\frac{-7}{16}<\frac{-5}{12}<\frac{-9}{24}$ 

That is

$\frac{-3}{4}<\frac{-7}{16}<\frac{5}{-12}<\frac{9}{-24}$

(iii) We will write each of the given numbers with positive denominators.

We have:

$\frac{3}{-5}=\frac{3\times (-1)}{-5\times (-1)}=\frac{-3}{5}$

Thus, the given numbers are $\frac{-3}{5}, \frac{-7}{10}, \frac{-11}{15} \mathrm{and} \frac{-13}{20}.$

LCM of 5, 10, 15 and 20 is 60.

Now, 

$\frac{-3}{5}=\frac{-3\times 12}{5\times 12}=\frac{-36}{60}$

$\frac{-7}{10}=\frac{-7\times 6}{10\times 6}=\frac{-42}{60}$

$\frac{-11}{15}=\frac{-11\times 4}{15\times 4}=\frac{-44}{60}$

$\frac{-13}{20}=\frac{-13\times 3}{20\times 3}=\frac{-39}{60}$

Clearly,
 
$\frac{-44}{60}<\frac{-42}{60}<\frac{-39}{60}<\frac{-36}{60}$

∴​ $\frac{-11}{15}<\frac{-7}{10}<\frac{-13}{20}<\frac{-3}{5}$.

That is 

$\frac{-11}{15}<\frac{-7}{10}<\frac{-13}{20}<\frac{3}{-5}$

(iv) We will write each of the given numbers with positive denominators.

We have:

$\frac{13}{-28}=\frac{13\times (-1)}{-28\times (-1)}=\frac{-13}{28}$

Thus, the given numbers are $\frac{-4}{7}, \frac{-9}{14}, \frac{-13}{28} \mathrm{and} \frac{-23}{42}.$

LCM of 7, 14, 28 and 42 is 84.

Now, 

$\frac{-4}{7}=\frac{-4\times 12}{7\times 12}=\frac{-48}{84}$

$\frac{-9}{14}=\frac{-9\times 6}{14\times 6}=\frac{-54}{84}$

$\frac{-13}{28}=\frac{-13\times 3}{28\times 3}=\frac{-39}{84}$

$\frac{-23}{42}=\frac{-23\times 2}{42\times 2}=\frac{-46}{84}$

Clearly, 

$\frac{-54}{84}<\frac{-48}{84}<\frac{-46}{84}<\frac{-39}{84}$

∴​ $\frac{-9}{14}<\frac{-4}{7}<\frac{-23}{42}<\frac{-13}{28}$.
 
That is

$\frac{-9}{14}<\frac{-4}{7}<\frac{-23}{42}<\frac{13}{-28}$


 

Answer:

(i) We will first write each of the given numbers with positive denominators. We have:

   $\frac{8}{-3}=\frac{8\times (-1)}{-3\times (-1)}=\frac{-8}{3}$

Thus, the given numbers are $-2, \frac{-13}{6}, \frac{-8}{3} \mathrm{and} \frac{1}{3}$

LCM of 1, 6, 3 and 3 is 6

Now,
 
$\frac{-2}{1}=\frac{-2\times 6}{1\times 6}=\frac{-12}{6}$

$\frac{-13}{6}=\frac{-13\times 1}{6\times 1}=\frac{-13}{6}$

$\frac{-8}{3}=\frac{-8\times 2}{3\times 2}=\frac{-16}{6}$

and 

$\frac{1}{3}=\frac{1\times 2}{3\times 2}=\frac{2}{6}$

Clearly,Thus,
 
$\frac{2}{6}>\frac{-12}{6}>\frac{-13}{6}>\frac{-16}{6}$

∴​ $\frac{1}{3}>-2>\frac{-13}{6}>\frac{-8}{3}$. i.e $\frac{1}{3}>-2>\frac{-13}{6}>\frac{8}{-3}$


(ii) We will first write each of the given numbers with positive denominators. We have:

   $\frac{7}{-15}=\frac{7\times (-1)}{-15\times (-1)}=\frac{-7}{15}$ and $\frac{17}{-30}=\frac{17\times (-1)}{-30\times (-1)}=\frac{-17}{30}$ 

Thus, the given numbers are $\frac{-3}{10}, \frac{-7}{15}, \frac{-11}{20} \mathrm{and} \frac{-17}{30}$

LCM of 10, 15, 20 and 30 is 60

Now,
 
$\frac{-3}{10}=\frac{-3\times 6}{10\times 6}=\frac{-18}{60}$ 

$\frac{-7}{15}=\frac{-7\times 4}{15\times 4}=\frac{-28}{60}$

$\frac{-11}{20}=\frac{-11\times 3}{20\times 3}=\frac{-33}{60}$

and 

$\frac{-17}{30}=\frac{-17\times 2}{30\times 2}=\frac{-34}{60}$

Clearly,
 
$\frac{-18}{60}>\frac{-28}{60}>\frac{-33}{60}>\frac{-34}{60}$

∴ $\frac{-3}{10}>\frac{-7}{15}>\frac{-11}{20}>\frac{-17}{30}$. i.e $\frac{-3}{10}>\frac{7}{-15}>\frac{-11}{20}>\frac{17}{-30}$

(iii) We will first write each of the given numbers with positive denominators. We have:

   $\frac{23}{-24}=\frac{23\times (-1)}{-24\times (-1)}=\frac{-23}{24}$ 

Thus, the given numbers are $\frac{-5}{6}, \frac{-7}{12}, \frac{-13}{18}\mathrm{and}\frac{-23}{24}$

LCM of 6, 12, 18 and 24 is 72

Now, 

$\frac{-5}{6}=\frac{-5\times 12}{6\times 12}=\frac{-60}{72}$

$\frac{-7}{12}=\frac{-7\times 6}{12\times 6}=\frac{-42}{72}$

$\frac{-13}{18}=\frac{-13\times 4}{18\times 4}=\frac{-52}{72}$

and 

$\frac{-23}{24}=\frac{-23\times 3}{24\times 3}=\frac{-69}{72}$

Clearly,
 
$\frac{-42}{72}>\frac{-52}{72}>\frac{-60}{72}>\frac{-69}{72}$

∴​ $\frac{-7}{12}>\frac{-13}{18}>\frac{-5}{6}>\frac{-23}{24}$. i.e $\frac{-7}{12}>\frac{-13}{18}>\frac{-5}{6}>\frac{23}{-24}$

(iv) The given numbers are $\frac{-10}{11}, \frac{-19}{22}, \frac{-23}{33} \mathrm{and} \frac{-39}{44}$

LCM of 11, 22, 33 and 44 is 132

Now, 

$\frac{-10}{11}=\frac{-10\times 12}{11\times 12}=\frac{-120}{132}$

$\frac{-19}{22}=\frac{-19\times 6}{22\times 6}=\frac{-114}{132}$

$\frac{-23}{33}=\frac{-23\times 4}{33\times 4}=\frac{-92}{132}$

and 

$\frac{-39}{44}=\frac{-39\times 3}{44\times 3}=\frac{-117}{132}$

Clearly,
 
$\frac{-92}{132}>\frac{-114}{132}>\frac{-117}{132}>\frac{-120}{132}$

∴ $\frac{-23}{33}>\frac{-19}{22}>\frac{-39}{44}>\frac{-10}{11}$

Answer:

1. True
A whole number can be expressed as $\frac{a}{b}, \mathrm{with} b=1 \mathrm{and} a\ge 0$. Thus, every whole number is rational.

2. True
Every integer is a rational number because any integer can be expressed as $\frac{a}{b}, \mathrm{with} b=1 \mathrm{and} 0>a\ge 0$. Thus, every integer is a rational number.

3. False
$0=\frac{a}{b}, \mathrm{for} a=0 \mathrm{and} b\ne 0.$ Thus, 0 is a rational and whole number.

Answer:

(i)


(ii)

 
(iii)


(iv)


(v)

(vi)


(vii)


(viii)

Answer:

(i)

(ii)


(iii)


(iv)


(v)


(vi)


(vii)


(viii)

Answer:

(i) True
A negative number always lies to the left of 0 on the number line.

(ii) False
A negative number always lies to the left of 0 on the number line.

(iii) True
Negative and positive numbers always lie on the opposite sides of 0 on the number line.

(iv) False
The negative sign cancels off and the number becomes $\frac{18}{13}$; it lies to the right of 0 on the number line.

Answer:

1. $\frac{-2}{5 }+\frac{4}{5}=\frac{-2+4}{5}=\frac{2}{5}$


2. $\frac{-6}{11}+\frac{-4}{11}=\frac{-6+(-4)}{11}=\frac{-6-4}{11}=\frac{-10}{11}$


3. $\frac{-11}{8}+\frac{5}{8}=\frac{-11+5}{8}=\frac{-6}{8}=\frac{-3\times 2}{4\times 2}=\frac{-3}{4}$


4. $\frac{-7}{3}+\frac{1}{3}=\frac{-7+1}{3}=\frac{-6}{3}=\frac{-3\times 2}{3}=-2$


5. $\frac{5}{6}+\frac{-1}{6}=\frac{5+(-1)}{6}=\frac{4}{6}=\frac{2\times 2}{3\times 2}=\frac{2}{3}$


6. $\frac{-17}{15}+\frac{-1}{15}=\frac{-17+(-1)}{15}=\frac{-17-1}{15}=\frac{-18}{15}=\frac{-6\times 3}{5\times 3}=\frac{-6}{5}$

Answer:

1. The denominators of the given rational numbers are 4 and 5.

LCM of 4 and 5 is 20.

Now, 

$\frac{3}{4}=\frac{3\times 5}{4\times 5}=\frac{15}{20}$ and $\frac{-3}{5}=\frac{-3\times 4}{5\times 4}=\frac{-12}{20}$

∴ $\frac{3}{4}+\frac{-3}{5}=\frac{15}{20}+\frac{-12}{20}=\frac{15+(-12)}{20}=\frac{15-12}{20}=\frac{3}{20}$

2.​ The denominators of the given rational numbers are 8 and 12.

LCM of 8 and 12 is 24.

Now, 

$\frac{5}{8}=\frac{5\times 3}{8\times 3}=\frac{15}{24}$ and $\frac{-7}{12}=\frac{-7\times 2}{12\times 2}=\frac{-14}{24}$

∴​ $\frac{5}{8}+\frac{-7}{12}=\frac{15}{24}+\frac{-14}{24}=\frac{15+(-14)}{24}=\frac{15-14}{24}=\frac{1}{24}$

3. ​The denominators of the given rational numbers are 9 and 6.

LCM of 9 and 6 is 18.

Now, 

$\frac{-8}{9}=\frac{-8\times 2}{9\times 2}=\frac{-16}{18}$ and $\frac{11}{6}=\frac{11\times 3}{6\times 3}=\frac{33}{18}$

∴​ $\frac{-8}{9}+\frac{11}{6}=\frac{-16}{18}+\frac{33}{18}=\frac{-16+33}{18}=\frac{-16+33}{18}=\frac{17}{18}$

4.​ The denominators of the given rational numbers are 16 and 24.

LCM of 16 and 24 is 48.

Now, 

$\frac{-5}{16}=\frac{-5\times 3}{16\times 3}=\frac{-15}{48}$ and $\frac{7}{24}=\frac{7\times 2}{24\times 2}=\frac{14}{48}$

∴​ $\frac{-5}{16}+\frac{7}{24}=\frac{-15}{48}+\frac{14}{48}=\frac{-15+14}{48}=\frac{-1}{48}$

5. We will first write each of the given numbers with positive denominators.

$\frac{7}{-18}=\frac{7\times (-1)}{-18\times (-1)}=\frac{-7}{18}$

​The denominators of the given rational numbers are 18 and 27.

LCM of 18 and 27 is 54.

Now, 

$\frac{-7}{18}=\frac{-7\times 3}{18\times 3}=\frac{-21}{54}$ and $\frac{8}{27}=\frac{8\times 2}{27\times 2}=\frac{16}{54}$

∴ $\frac{7}{-18}+\frac{8}{27}=\frac{-21}{54}+\frac{16}{54}=\frac{-21+16}{54}=\frac{-5}{54}$

6. ​We will first write each of the given numbers with positive denominators.

$\frac{1}{-12}=\frac{1\times (-1)}{-12\times (-1)}=\frac{-1}{12}$ and $\frac{2}{-15}=\frac{2\times (-1)}{-15\times (-1)}=\frac{-2}{15}$

​The denominators of the given rational numbers are 12 and 15.

LCM of 12 and 15 is 60.

Now, 

$\frac{-1}{12}=\frac{-1\times 5}{12\times 5}=\frac{-5}{60}$ and $\frac{-2}{15}=\frac{-2\times 4}{15\times 4}=\frac{-8}{60}$

∴ $\frac{1}{-12}+\frac{2}{-15}=\frac{-5}{60}+\frac{-8}{60}=\frac{-5+(-8)}{60}=\frac{-5-8}{60}=\frac{-13}{60}$

7. We can write -1 as$\frac{-1}{1}$.

The denominators of the given rational numbers are 1 and 4.

LCM of 1 and 4 is 4.

Now, 

$\frac{-1}{1}=\frac{-1\times 4}{1\times 4}=\frac{-4}{4}$ and $\frac{3}{4}=\frac{3\times 1}{4\times 1}=\frac{3}{4}$

∴ $-1+\frac{3}{4}=\frac{-4}{4}+\frac{3}{4}=\frac{-4+3}{4}=\frac{-1}{4}$

8. ​We can write 2 as$\frac{2}{1}$.

The denominators of the given rational numbers are 1 and 4.

LCM of 1 and 4 is 4.

Now, 

$\frac{2}{1}=\frac{2\times 4}{1\times 4}=\frac{8}{4}$ and $\frac{-5}{4}=\frac{-5\times 1}{4\times 1}=\frac{-5}{4}$

∴ $2+\frac{(-5)}{4}=\frac{8}{4}+\frac{(-5)}{4}=\frac{8+(-5)}{4}=\frac{8-5}{4}=\frac{3}{4}$

9. ​We can write 0 as$\frac{0}{1}$.

The denominators of the given rational numbers are 1 and 5.

LCM of 1 and 5 is 5, that is, (1 $\times$ 5).

Now,
 
$\frac{0}{1}=\frac{0\times 5}{1\times 5}=\frac{0}{5}=0$ and $\frac{-2}{5}=\frac{-2\times 1}{5\times 1}=\frac{-2}{5}$

∴ $0+\frac{(-2)}{5}=\frac{0}{5}+\frac{(-2)}{5}=\frac{0+(-2)}{5}=\frac{0-2}{5}=\frac{-2}{5}$

Answer:

1. LHS = $\frac{-12}{5}+\frac{2}{7}$

LCM of 5 and 7 is 35.

$\frac{-12\times 7}{5\times 7}+\frac{2\times 5}{7\times 5}=\frac{-84}{35}+\frac{10}{35}=\frac{-84+10}{35}=\frac{-74}{35}$

RHS = $\frac{2}{7}+\frac{-12}{5}$

LCM of 5 and 7 is 35.

$\frac{2\times 5}{7\times 5} +\frac{-12\times 7}{5\times 7}=\frac{10}{35}+\frac{-84}{35}=\frac{10-84}{35}=\frac{-74}{35}$

∴ $\frac{-12}{5}+\frac{2}{7}=\frac{2}{7}+\frac{-12}{5}$

2. ​LHS = $\frac{-5}{8}+\frac{-9}{13}$

LCM of 8 and 13 is 104.

$\frac{-5\times 13}{8\times 13}+\frac{-9\times 8}{13\times 8}=\frac{-65}{104}+\frac{-72}{104}=\frac{-65+(-72)}{104}=\frac{-65-72}{104}=\frac{-137}{104}$

RHS = $\frac{-9}{13}+\frac{-5}{8}$

LCM of 13 and 8 is 104.

$\frac{-9\times 8}{13\times 8} +\frac{-5\times 13}{8\times 13}=\frac{-72}{104}+\frac{-65}{104}=\frac{-72-65}{104}=\frac{-137}{104}$

∴ $\frac{-5}{8}+\frac{-9}{13}=\frac{-9}{13}+\frac{-5}{8}$

3. ​LHS = $\frac{3}{1}+\frac{-7}{12}$

LCM of 1 and 12 is 12.

$\frac{3\times 12}{1\times 12}+\frac{-7\times 1}{12\times 1}=\frac{36}{12}+\frac{-7}{12}=\frac{36+(-7)}{12}=\frac{36-7}{12}=\frac{29}{12}$

RHS = $\frac{-7}{12}+\frac{3}{1}$

LCM of 12 and 1 is 12.

$\frac{-7\times 1}{12\times 1} +\frac{3\times 12}{1\times 12}=\frac{-7}{12}+\frac{36}{12}=\frac{-7+36}{12}=\frac{29}{12}$

∴ $3+\frac{-7}{12}=\frac{-7}{12}+3$

4. LHS = ​$\frac{2}{-7}+\frac{12}{-35}$

We will write the given numbers with positive denominators.

$\frac{2}{-7}=\frac{2\times (-1)}{-7\times (-1)}=\frac{-2}{7}$ and $\frac{12}{-35}=\frac{12\times (-1)}{-35\times (-1)}=\frac{-12}{35}$

LCM of 7 and 35 is 35.

$\frac{-2\times 5}{7\times 5}+\frac{-12\times 1}{35\times 1}=\frac{-10}{35}+\frac{-12}{35}=\frac{-10+(-12)}{35}=\frac{-10-12}{35}=\frac{-22}{35}$

RHS = $\frac{12}{-35}+\frac{2}{-7}$

We will write the given numbers with positive denominators.

$\frac{12}{-35}=\frac{12\times (-1)}{-35\times (-1)}=\frac{-12}{35}$ and $\frac{2}{-7}=\frac{2\times (-1)}{-7\times (-1)}=\frac{-2}{7}$

LCM of 35 and 7 is 35.

$\frac{-2\times 5}{7\times 5} +\frac{-12\times 1}{35\times 1}=\frac{-10}{35}+\frac{-12}{35}=\frac{-10+(-12)}{35}=\frac{-10-12}{35}=\frac{-22}{35}$

∴​ $\frac{2}{-7}+\frac{12}{-35}=\frac{12}{-35}+\frac{2}{-7}$

Answer:

1.
LHS =  $\left\{\left(\frac{3}{4}+\frac{-2}{5}\right)+\frac{-7}{10}\right\}$

$\left\{\left(\frac{15-8}{20}\right)+\frac{-7}{10}\right\}=\left(\frac{7}{20}+\frac{-7}{10}\right)=\left(\frac{7}{20}+\frac{-14}{20}\right)=\left(\frac{7+(-14)}{20}\right)=\frac{-7}{20}$

RHS =  $\left\{\frac{3}{4}+\left(\frac{-2}{5}+{\displaystyle \frac{-7}{10}}\right)\right\}$

$\left\{\frac{3}{4}+\left(\frac{-4}{10}+{\displaystyle \frac{-7}{10}}\right)\right\}=\left\{{\displaystyle \frac{3}{4}}+\left({\displaystyle \frac{-4-7}{10}}\right)\right\}=\left\{{\displaystyle \frac{3}{4}}+\left({\displaystyle \frac{-11}{10}}\right)\right\}=\left({\displaystyle \frac{3}{4}}+{\displaystyle \frac{-11}{10}}\right)=\left({\displaystyle \frac{15}{20}}+{\displaystyle \frac{-22}{20}}\right)=\left({\displaystyle \frac{15-22}{20}}\right)=\frac{-7}{20}$

∴​ $\left(\frac{3}{4}+\frac{-2}{5}\right)+\frac{-7}{10}=\frac{3}{4}+\left({\displaystyle \frac{-2}{5}}+{\displaystyle \frac{-7}{10}}\right)$


2.
LHS =  $\left\{\left(\frac{-7}{11}+\frac{2}{-5}\right)+\frac{-13}{22}\right\}$

We will first make the denominator positive.

$\left\{\left(\frac{-7}{11}+\frac{2\times (-1)}{-5\times (-1)}\right)+\frac{-13}{22}\right\}=\left\{\left(\frac{-7}{11}+\frac{-2}{5}\right)+\frac{-13}{22}\right\}$

$\left\{\left(\frac{-7}{11}+\frac{-2}{5}\right)+\frac{-13}{22}\right\}=\left\{\left({\displaystyle \frac{-35}{55}}+{\displaystyle \frac{-22}{55}}\right)+{\displaystyle \frac{-13}{22}}\right\}=\left\{\left({\displaystyle \frac{-35-22}{55}}\right)+{\displaystyle \frac{-13}{22}}\right\}=\left({\displaystyle \frac{-57}{55}}+{\displaystyle \frac{-13}{22}}\right)=\frac{-114}{110}+\frac{-65}{110}=\frac{-114-65}{110}=\frac{-179}{110}$

RHS = $\left\{\frac{-7}{11}+\left(\frac{2}{-5}+{\displaystyle \frac{-13}{22}}\right)\right\}$

We will first make the denominator positive.

$\left\{\frac{-7}{11}+\left({\displaystyle \frac{2\times (-1)}{-5\times (-1)}}+{\displaystyle \frac{-13}{22}}\right)\right\}=\left\{{\displaystyle \frac{-7}{11}}+\left({\displaystyle \frac{-2}{5}}+{\displaystyle \frac{-13}{22}}\right)\right\}$

$\left\{{\displaystyle \frac{-7}{11}}+\left({\displaystyle \frac{-2}{5}}+{\displaystyle \frac{-13}{22}}\right)\right\}=\left\{{\displaystyle \frac{-7}{11}}+\left({\displaystyle \frac{-44}{110}}+{\displaystyle \frac{-65}{110}}\right)\right\}=\left\{{\displaystyle \frac{-7}{11}}+\left({\displaystyle \frac{-44+(-65)}{110}}\right)\right\}=\frac{-7}{11}+\frac{-109}{110}=\frac{-70}{110}+\frac{-109}{110}=\frac{-70-109}{110}=\frac{-179}{110}$

∴​ $\left(\frac{-7}{11}+\frac{2}{-5}\right)+\frac{-13}{22}=\frac{-7}{11}+\left({\displaystyle \frac{2}{-5}}+{\displaystyle \frac{-13}{22}}\right)$


3.
LHS = $-1+\left({\displaystyle \frac{-2}{3}}+{\displaystyle \frac{-3}{4}}\right)$

$\left\{\frac{-1}{1}+\left({\displaystyle \frac{-2}{3}}+{\displaystyle \frac{-3}{4}}\right)\right\}=\left\{{\displaystyle \frac{-1}{1}}+\left({\displaystyle \frac{-8}{12}}+{\displaystyle \frac{-9}{12}}\right)\right\}=\left\{{\displaystyle \frac{-1}{1}}+\left({\displaystyle \frac{-8-9}{12}}\right)\right\}=\left\{{\displaystyle \frac{-1}{1}}+\left({\displaystyle \frac{-17}{12}}\right)\right\}=\left({\displaystyle \frac{-1}{1}}+{\displaystyle \frac{-17}{12}}\right)=\left({\displaystyle \frac{-1\times 12}{1\times 12}}+{\displaystyle \frac{-17\times 1}{12\times 1}}\right)=\left(\frac{-12+(-17)}{12}\right)=\left(\frac{-12-17}{12}\right)=\frac{-29}{12}$

RHS = $\left\{\left(-1+{\displaystyle \frac{-2}{3}}\right)+\frac{-3}{4}\right\}$

$\left\{\left({\displaystyle \frac{-1}{1}}+{\displaystyle \frac{-2}{3}}\right)+\frac{-3}{4}\right\}=\left\{\left({\displaystyle \frac{-3}{3}}+{\displaystyle \frac{-2}{3}}\right)+{\displaystyle \frac{-3}{4}}\right\}=\left\{\left({\displaystyle \frac{-3-2}{3}}\right)+{\displaystyle \frac{-3}{4}}\right\}=\left\{\left({\displaystyle \frac{-5}{3}}\right)+{\displaystyle \frac{-3}{4}}\right\}=\left({\displaystyle \frac{-5}{3}}+{\displaystyle \frac{-3}{4}}\right)=\left({\displaystyle \frac{-20}{12}}+{\displaystyle \frac{-9}{12}}\right)=\left(\frac{-20-9}{12}\right)=\frac{-29}{12}$

∴ $-1+\left({\displaystyle \frac{-2}{3}}+{\displaystyle \frac{-3}{4}}\right)=\left(-1+{\displaystyle \frac{-2}{3}}\right)+\frac{-3}{4}$

Answer:

(i) Addition is commutative, that is, $a+b=b+a$.

Hence, the required solution is $\left({\displaystyle \frac{-3}{17}}\right)+\left({\displaystyle \frac{-12}{5}}\right)=\left({\displaystyle \frac{-12}{5}}\right)+\boxed{\left({\displaystyle \frac{-3}{7}}\right)}$.
 
(ii) Addition is commutative, that is, $a+b=b+a$.

Hence, the required solution is $-9+\frac{-21}{8}=\frac{-21}{8}+\boxed{-9}$.

(iii) Addition is associative, that is, $\left(a+b\right)+c=a+\left(b+c\right)$.

Hence, the required solution is $\left(\frac{-8}{13}+\frac{3}{7}\right)+\left(\frac{-13}{4}\right)=\boxed{\left({\displaystyle \frac{-8}{13}}\right)}+\left[{\displaystyle \frac{3}{7}}+\left({\displaystyle \frac{-13}{4}}\right)\right]$.

(iv) Addition is associative, that is, $\left(a+b\right)+c=a+\left(b+c\right)$.

Hence, the required solution is $-12+\left(\frac{7}{12}+\frac{-9}{11}\right)=\left(-12+{\displaystyle \frac{7}{12}}\right)+\frac{-9}{11}$.

(v) Addition is associative, that is, $\left(a+b\right)+c=a+\left(b+c\right)$.

Hence, the required solution is$\frac{19}{-5}+\left({\displaystyle \frac{-3}{11}}+{\displaystyle \frac{-7}{8}}\right)=\left\{\frac{19}{-5}+\boxed{\left({\displaystyle \frac{-3}{11}}\right)}\right\}+\frac{-7}{8}$.

(vi) 0 is the additive identity, that is, $0+a=a$.

Hence, the required solution is $\frac{-16}{7}+\boxed{0}=\boxed{0}+\frac{-16}{7}=\frac{-16}{7}$.

Answer:

The additive inverse of $\frac{a}{b}$ is $\frac{-a}{b}$. Therefore, $\frac{a}{b}+\left(\frac{-a}{b}\right)=0$
(i) Additive inverse of $\frac{1}{3}\mathrm{is}\frac{-1}{3}.$

(ii) Additive inverse of  $\frac{23}{9}\mathrm{is}\frac{-23}{9}.$

(iii) Additive inverse of -18 is 18.

(iv) Additive inverse of $\frac{-17}{8}\mathrm{is}\frac{17}{8}.$

(v) In the standard form, we write $\frac{15}{-4}\mathrm{as}\frac{-15}{4}.$

    Hence, its additive inverse is $\frac{15}{4}$.

(vi) We can write:
 
$\frac{-16}{-5}=\frac{-16\times (-1)}{-5\times (-1)}=\frac{16}{5}$

    Hence, its additive inverse is $\frac{-16}{5}$.

(vii) Additive inverse of $\frac{-3}{11}\mathrm{is}\frac{3}{11}.$

(viii) Additive inverse of 0 is 0.

(ix) In the standard form, we write $\frac{19}{-6}\mathrm{as}\frac{-19}{6}.$

     Hence, its additive inverse is $\frac{19}{6}$.

(x) We can write:
 
$\frac{-8}{-7}=\frac{-8\times (-1)}{-7\times (-1)}=\frac{8}{7}$

Hence, its additive inverse is $\frac{-8}{7}$.

Answer:

(i) $\left(\frac{1}{3}-\frac{3}{4}\right)=\frac{1}{3}+\left(\mathrm{Additive} \mathrm{inverse} \mathrm{of}\frac{3}{4}\right)$

                     = $\left({\displaystyle \frac{1}{3}}+{\displaystyle \frac{-3}{4}}\right)=\left({\displaystyle \frac{4}{12}}+{\displaystyle \frac{-9}{12}}\right)=\left(\frac{4-9}{12}\right)=\frac{-5}{12}$


(ii)  $\left(\frac{1}{3}-\frac{-5}{6}\right)=\frac{1}{3}+\left(\mathrm{Additive} \mathrm{inverse} \mathrm{of}\frac{-5}{6}\right)$

                       = $\left({\displaystyle \frac{1}{3}}+{\displaystyle \frac{5}{6}}\right)$ (Because the additive inverse of $\frac{-5}{6}\mathrm{is}\frac{5}{6}$)

                       =$\left({\displaystyle \frac{2}{6}}+{\displaystyle \frac{5}{6}}\right)=\left(\frac{2+5}{6}\right)=\frac{7}{6}$


(iii) $\left(\frac{-3}{5}-\frac{-8}{9}\right)=\frac{-3}{5}+\left(\mathrm{Additive} \mathrm{inverse} \mathrm{of}\frac{-8}{9}\right)$

                          = $\left({\displaystyle \frac{-3}{5}}+{\displaystyle \frac{8}{9}}\right)$ (Because the additive inverse of $\frac{-8}{9}\mathrm{is}\frac{8}{9}$)

                          =$\left({\displaystyle \frac{-27}{45}}+{\displaystyle \frac{40}{45}}\right)=\left(\frac{-27+40}{45}\right)=\frac{13}{45}$


(iv) $\left(-1-\frac{-9}{7}\right)=-1+\left(\mathrm{Additive} \mathrm{inverse} \mathrm{of}\frac{-9}{7}\right)$

                        =$\left({\displaystyle \frac{-1}{1}}+{\displaystyle \frac{9}{7}}\right)$ (Because the additive inverse of $\frac{-9}{7}\mathrm{is}\frac{9}{7}$)

                        =$\left({\displaystyle \frac{-7}{7}}+{\displaystyle \frac{9}{7}}\right)=\left(\frac{-7+9}{7}\right)=\frac{2}{7}$


(v) $\left(1-\frac{-18}{11}\right)=1+\left(\mathrm{Additive} \mathrm{inverse} \mathrm{of}\frac{-18}{11}\right)$

                       =$\left({\displaystyle \frac{1}{1}}+{\displaystyle \frac{18}{11}}\right)$ (Because the additive inverse of $\frac{-18}{11}\mathrm{is}\frac{18}{11}$)

                       = $\left({\displaystyle \frac{11}{11}}+{\displaystyle \frac{18}{11}}\right)=\left(\frac{11+18}{11}\right)=\frac{29}{11}$


(vi) $\left(0-\frac{-13}{9}\right)=0+\left(\mathrm{Additive} \mathrm{inverse} \mathrm{of}\frac{-13}{9}\right)$

                       =$\left(0+{\displaystyle \frac{13}{9}}\right)$ (Because the additive inverse of $\frac{-13}{9}\mathrm{is}\frac{13}{9}$)

                       =$\frac{13}{9}$

(vii) $\left(\frac{-6}{5}-\frac{-32}{13}\right)=\frac{-6}{5}+\left(\mathrm{Additive} \mathrm{inverse} \mathrm{of}\frac{-32}{13}\right)$

                             =$\left({\displaystyle \frac{-6}{5}}+{\displaystyle \frac{32}{13}}\right)$ (Because the additive inverse of $\frac{-32}{13}\mathrm{is}\frac{32}{13}$)

                             =  $\left({\displaystyle \frac{-78}{65}}+{\displaystyle \frac{160}{65}}\right)=\left(\frac{-78+160}{65}\right)=\frac{82}{65}$


(viii) $\left(\frac{-4}{7}-\frac{-7}{1}\right)=\frac{-4}{7}+\left(\mathrm{Additive} \mathrm{inverse} \mathrm{of}\frac{-7}{1}\right)$

                           = $\left({\displaystyle \frac{-4}{7}}+{\displaystyle \frac{7}{1}}\right)$ (Because the additive inverse of $\frac{-7}{1}\mathrm{is}\frac{7}{1}$)

                           = $\left({\displaystyle \frac{-4}{7}}+{\displaystyle \frac{49}{7}}\right)=\left(\frac{-4+49}{7}\right)=\frac{45}{7}$