Rational Numbers

1.0Introduction

Natural numbers

Counting numbers $1,2,3,4,5$,.....are known as natural numbers. The set of all natural numbers can be represented by $N=\{1,2,3,4,5,\dots ..\}$

Whole numbers

If we include 0 among the natural numbers, then the numbers $0,1,2,3,4,5\dots$....are called whole numbers. The set of whole numbers can be represented by $W=\{0,1,2,3,4,5,\dots ..$

Clearly, every natural number is a whole number but 0 is a whole number which is not a natural number. So, every whole number is not a natural number.

Integers

All counting numbers and their negatives including zero are known as integers. The set of integers can be represented by $Z\text{ or }I=\{\dots ..,-4,-3,-2,-1,0,1,2,3,4,\dots ...\}$

Positive integers

The set ${\mathrm{I}}^{+}=\{1,2,3,4,\dots$.$\}isthesetofallpositiveintegers.Clearly,positiveintegersand$ natural numbers are synonyms.

Negative integers

The set $I^{-}=\{\dots ..,-3,-2,-1\}$ is the set of all negative integers. 0 is neither positive nor negative.

Fractions

A number which can be expressed in the form $\frac{p}{q}$, where $p,q$ are whole numbers and $q\ne 0$ is called a fraction. E.g. $\frac{3}{4},\frac{1}{2},\frac{7}{19}$ etc.

• All whole numbers are integers but not all the integers are whole numbers.
• All natural numbers are whole numbers but not all the whole numbers are natural numbers.

2.0Definition:

A number which can be expressed in the form $\frac{p}{q}$, where $\mathrm{p},\mathrm{q}$ are integers and $\mathrm{q}\ne 0$ is called a rational number. E.g. $\frac{2}{3},\frac{-7}{9},-3,0,1$ etc.

3.0Standard form of rational number

A rational number is said to be in standard form if its denominator is positive and it is in the lowest terms.

Steps to express given rational number in standard form : Step 1 : If not already so, make the denominator of the given rational number positive. Step 2 : Divide both the numerator and the denominator by their HCF.

• Q. Write the following rational numbers in their standard form. (i) $\frac{-6}{2}$ (ii) $\frac{-3}{-5}$ (iii) $2\frac{3}{7}$ (iv) $-8\frac{3}{7}$ (v) 0.4 Solution: (i) $\frac{-6}{2}=-3=\frac{-3}{1}$ (ii) $\frac{-3}{-5}=\frac{(-3)÷(-1)}{(-5)÷(-1)}=\frac{3}{5}$ (iii) $2\frac{3}{7}=\frac{17}{7}$ (iv) $-8\frac{3}{7}=\frac{-59}{7}$ (v) $0.4=\frac{4}{10}=\frac{2}{5}$
• If denominators of two rational numbers are same, then lesser the numerator, lesser the rational number will be and greater the numerator, greater the rational number will be. E.g., $\frac{5}{11}<\frac{7}{11}$ and $\frac{-5}{11}>\frac{-7}{11}$
• If numerators of two rational numbers are same then after writing the rational numbers in standard form:

• If both the rational numbers are -ve then; rational numbers with smaller denominators will be smaller and rational numbers with greater denominator will be greater. E.g., $\frac{-6}{2}<\frac{-6}{5}$
• If both the rational numbers are +ve then, rational number with smaller denominator will be greater and rational number with greater denominator will be smaller. E.g., $\frac{6}{2}>\frac{6}{5}$

4.0Properties of rational numbers

Closure property

(i) The sum of two rational numbers is again a rational number. E.g. $\frac{-2}{9}+\frac{7}{9}=\frac{(-2)+7}{9}=\frac{5}{9}$, a rational number. Thus, rational numbers are closed under addition.

(ii) The difference of two rational numbers is again a rational number. E.g., $\frac{2}{3}-\frac{7}{8}=\frac{16-21}{24}=\frac{-5}{24}$, a rational number. Thus, rational numbers are closed under subtraction.

(iii) The product of two rational numbers is again a rational number, E.g., $\frac{5}{6}\times \left(\frac{-3}{4}\right)=\frac{-15}{24}=\frac{-5}{8}$, a rational number. Thus, rational numbers are closed under multiplication.

(iv) When a rational number is divided by a non-zero rational number, a rational number is obtained. E.g., $\frac{3}{5}÷\left(\frac{-4}{7}\right)=\frac{3}{5}\times \left(\frac{-7}{4}\right)=\frac{-21}{20}$, a rational number. But, division of a rational number by zero is not defined. Thus, rational numbers are not closed under division.

Commutative Property

(i) The sum of two rational numbers remains the same even if the order in which they are added is changed. E.g., $\frac{5}{7}+\left(\frac{-9}{10}\right)=\frac{50-63}{70}=\frac{-13}{70}$ $\left(\frac{-9}{10}\right)+\frac{5}{7}=\frac{-63+50}{70}=\frac{-13}{70}$ So, $\frac{5}{7}+\left(\frac{-9}{10}\right)=\left(\frac{-9}{10}\right)+\frac{5}{7}$ Thus, the addition of rational numbers is commutative.

(ii) The difference of two rational numbers is not the same if the order in which they are subtracted is changed. E.g., $\frac{3}{5}-\frac{4}{7}=\frac{21-20}{35}=\frac{1}{35}$ $\frac{4}{7}-\frac{3}{5}=\frac{20-21}{35}=\frac{-1}{35}$ So, $\frac{3}{5}-\frac{4}{7}\ne \frac{4}{7}-\frac{3}{5}$ Thus, the subtraction of rational numbers is not commutative.

(iii) The product of two rational numbers remains the same even if the order in which they are multiplied is changed. E.g., $\frac{2}{9}\times \left(\frac{-3}{7}\right)=\frac{-6}{63}=\frac{-2}{21}$ $\left(\frac{-3}{7}\right)\times \frac{2}{9}=\frac{-6}{63}=\frac{-2}{21}$ So, $\frac{2}{9}\times \left(\frac{-3}{7}\right)=\left(\frac{-3}{7}\right)\times \frac{2}{9}$ Thus, the product of rational numbers is commutative.

(iv) The quotient of two rational numbers is not the same if the order in which they are divided is changed. E.g., $\frac{3}{5}÷\left(\frac{-4}{7}\right)=\frac{3}{5}\times \left(\frac{-7}{4}\right)=\frac{-21}{20}$ $\left(\frac{-4}{7}\right)÷\left(\frac{3}{5}\right)=\frac{-4}{7}\times \frac{5}{3}=\frac{-20}{21}$ So, $\frac{3}{5}÷\left(\frac{-4}{7}\right)\ne \left(\frac{-4}{7}\right)÷\left(\frac{3}{5}\right)$ Thus, the division of rational numbers is not commutative. (i) Yes, (ii) Yes, Yes (iii) Yes, No, No (iv) Yes, Yes

Associative Property

(i) The sum of three or more rational numbers remains the same even if the order in which they are grouped is changed. E.g., $\left(\frac{-3}{4}+\frac{5}{7}\right)+\left(\frac{-9}{14}\right)=\left(\frac{-21+20}{28}\right)+\left(\frac{-9}{14}\right)$ $=\frac{-1}{28}+\left(\frac{-9}{14}\right)=\frac{-1+(-18)}{28}=\frac{-19}{28}$ $=\frac{-3}{4}+\left[\frac{5}{7}+\left(\frac{-9}{14}\right)\right]=\frac{-3}{4}+\left[\frac{10+(-9)}{14}\right]$ $=\frac{-3}{4}+\frac{1}{14}=\frac{-21+2}{28}=\frac{-19}{28}$ So, $\left(\frac{-3}{4}+\frac{5}{7}\right)+\left(\frac{-9}{14}\right)=\frac{-3}{4}+\left[\frac{5}{7}+\left(\frac{-9}{14}\right)\right]$ Thus, the addition of rational numbers is associative.

(ii) The difference of three or more rational numbers is not the same if the order in which they are grouped is changed. E.g., $\left(\frac{3}{8}-\frac{2}{7}\right)-\frac{5}{6}=\left(\frac{21-16}{56}\right)-\frac{5}{6}$ $=\frac{5}{56}-\frac{5}{6}=\frac{15-140}{168}=-\frac{125}{168}$ $=\frac{3}{8}-\left(\frac{2}{7}-\frac{5}{6}\right)=\frac{3}{8}-\left(\frac{12-35}{42}\right)$ $=\frac{3}{8}-\left(-\frac{23}{42}\right)=\frac{63-(-92)}{168}=\frac{155}{168}$ So, $\left(\frac{3}{8}-\frac{2}{7}\right)-\frac{5}{6}\ne \frac{3}{8}-\left(\frac{2}{7}-\frac{5}{6}\right)$ Thus, the subtraction of rational numbers is not associative.

(iii) The product of three or more rational numbers remains the same even if the order in which they are grouped is changed. E.g., $\left[\frac{2}{5}\times \left(\frac{-1}{3}\right)\right]\times \left(\frac{-7}{11}\right)=\left(\frac{-2}{15}\right)\times \left(\frac{-7}{11}\right)=\frac{14}{165}$ $\frac{2}{5}\times \left[\frac{-1}{3}\times \left(\frac{-7}{11}\right)\right]=\frac{2}{5}\times \frac{7}{33}=\frac{14}{165}$ So, $\left[\frac{2}{5}\times \left(\frac{-1}{3}\right)\right]\times \left(\frac{-7}{11}\right)=\frac{2}{5}\times \left[\frac{-1}{3}\times \left(\frac{-7}{11}\right)\right]$ Thus, the product of rational numbers is associative.

(iv) The quotient of three or more rational numbers is not the same if the order in which they are grouped is changed. E.g., $\left(\frac{3}{8}÷\frac{2}{7}\right)÷\frac{5}{6}=\left(\frac{3}{8}\times \frac{7}{2}\right)÷\frac{5}{6}=\frac{21}{16}÷\frac{5}{6}=\frac{21}{16}\times \frac{6}{5}=\frac{63}{40}$ $=\frac{3}{8}÷\left(\frac{2}{7}÷\frac{5}{6}\right)=\frac{3}{8}÷\left(\frac{2}{7}\times \frac{6}{5}\right)=\frac{3}{8}÷\frac{12}{35}=\frac{3}{8}\times \frac{35}{12}=\frac{35}{32}$ So, $\left(\frac{3}{8}÷\frac{2}{7}\right)÷\frac{5}{6}\ne \frac{3}{8}÷\left(\frac{2}{7}÷\frac{5}{6}\right)$ Thus, the division of rational numbers is not associative.

Note

• Q. Rearrange suitably and add: $\frac{6}{11}+\frac{-2}{7}+\frac{3}{7}+\frac{9}{11}+\frac{5}{7}$ Solution: Let us rearrange the numbers suitably to make addition easier. $\frac{6}{11}+\frac{9}{11}=\frac{6+9}{11}=\frac{15}{11}$ $\frac{-2}{7}+\frac{3}{7}+\frac{5}{7}=\frac{-2+3+5}{7}=\frac{6}{7}$ Now, $\frac{15}{11}+\frac{6}{7}=\frac{105+66}{77}=\frac{171}{77}$
  
• Q. Rearrange suitably and multiply: $\frac{-16}{7}\times \frac{5}{11}\times \frac{21}{-16}\times \frac{22}{40}$ Solution: $=\frac{-16}{7}\times \frac{5}{11}\times \frac{21}{-16}\times \frac{22}{40}$ $=\left(\frac{-16}{7}\times \frac{21}{-16}\right)\times \left(\frac{5}{11}\times \frac{22}{40}\right)$ $=\left(\frac{3}{1}\right)\times \left(\frac{1}{4}\right)=\frac{3}{1}\times \frac{1}{4}=\frac{3}{4}$

Additive inverse and Multiplicative inverse

Special numbers 0 and 1

If we multiply a rational number by 1 , then the product is the number itself. E.g., $\frac{-9}{11}\times 1=1\times \left(\frac{-9}{11}\right)=\frac{-9}{11}$ One is called the multiplicative identity of rational numbers.

Additive inverse of a rational number

Multiplicative inverse of a rational number

• Multiplicative inverse of +ve rational number is always the +ve.
• Multiplicative inverse of -ve rational number is always the -ve.
• 0 has no multiplicative inverse.
• The multiplicative inverse of 1 and -1 are the numbers themselves.

Note: Multiplicative inverse of $(-5)$ is not ' 5 ', it is $\left(\frac{1}{-5}\right)$, and Additive inverse of $(-5)$ is not $\left(\frac{1}{-5}\right)$, it is ' 5 '.

• Q. Write the additive inverse of (i) $\frac{6}{8}$ (ii) $\frac{-7}{11}$ (iii) $\frac{8}{-17}$ (iv) $\frac{-5}{-9}$ (v) $\frac{17}{1}$ Solution: Additive inverse is a number with the same magnitude but opposite sign. So, the additive inverse are (i) $\frac{-6}{8}$ (ii) $\frac{7}{11}$ (iii) $\frac{8}{17}$ (iv) $\frac{-5}{9}$ (v) $\frac{-17}{1}$
• Q. Is 0.7 the multiplicative inverse of $1\frac{3}{7}$ ? Explanation: $1\frac{3}{7}=\frac{10}{7}$ and $0.7=\frac{7}{10}$ $=\frac{10}{7}\times \frac{7}{10}=1$ Yes, 0.7 is the multiplicative inverse of $1\frac{3}{7}$.

Distributive property of multiplication

In general, for rational number $a$, $b$ and $c$, $\mathrm{a}(\mathrm{b}\pm \mathrm{c})=\mathrm{ab}\pm \mathrm{a}$. This property is known as distributive property of multiplication.

• Q. Solve using distributive property of multiplication. $\left(\frac{-2}{3}\times \frac{5}{6}\right)+\left(\frac{-2}{3}\times \frac{7}{6}\right)$ Solution: $=\left(\frac{-2}{3}\times \frac{5}{6}\right)+\left(\frac{-2}{3}\times \frac{7}{6}\right)=\frac{-2}{3}\times \left(\frac{5}{6}+\frac{7}{6}\right)$ $=\frac{-2}{3}\times \left(\frac{5+7}{6}\right)=\frac{-2}{3}\times \frac{12}{6}=\frac{-2}{3}\times \frac{2}{1}=\frac{-4}{3}$

5.0Rational numbers on the number line

To represent a rational number on the number line, divide each unit length on the number line into as many parts as the denominator of the rational number and move as many steps starting from 0 , on the number line as the numerator (towards the right for positive rational numbers and towards the left for negative rational numbers). Every rational number can be represented by a unique point on the number line. E.g., Look at the following number line

• Q. Which rational number do the letters $X,Y$ and $Z$ represent on the following number line?
  
  Explanation: The points X and Y lie between 0 and -1 . The distance between 0 and -1 is divided into 5 equal parts. So, Y represent $\frac{-1}{5}$ and X represent $\frac{-3}{5}$. The point $Z$ lies between 0 and 1 . The distance between 0 and 1 is divided into 5 equals parts. So, Z represents $\frac{3}{5}$.
• Q. Represent the following on the number line : (i) $\frac{3}{7}$ (ii) $\frac{8}{-5}$ Solution: (i) $\frac{3}{7}$ will lie between 0 and 1 on the number line.
  
  For $\frac{3}{7}$, the denominator is 7 so, divide the distance between 0 and 1 into 7 equal parts. The points A, B, C, D, E and F do this. The point C represents $\frac{3}{7}$. (ii) $\frac{8}{-5}=\frac{-8}{5}=-1\frac{3}{5}$ lies between -1 and -2 on the number line.
  
  As the denominator is 5 , we will divide the distance between -1 and -2 into 5 equal parts. The points A, B, C and D do this. The point B represents $\frac{-8}{5},-1$ is same as $\frac{-5}{5}$ and -2 is $\frac{-10}{5}$.

6.0Rational numbers between two rational numbers

You already know that between two whole numbers or two integers, only a definite number of whole numbers or integer numbers exist. But, it isn't the same for two rational numbers. There are innumerable rational numbers between any two rational numbers.

• Q. How many rational numbers lie between 0 and 1? Explanation:
  
  Divide the number line into 10 equal parts between the points 0 and 1 . We can easily see that, there are 9 points between 0 and 1, i.e., $\frac{1}{10},\frac{2}{10},\frac{3}{10},\frac{4}{10},\frac{5}{10},\frac{6}{10},\frac{7}{10},\frac{8}{10},\frac{9}{10}$
  
  Again, divide the number line between $\frac{1}{10}$ and $\frac{2}{10}$ into 10 equal parts, we get. $\frac{11}{100},\frac{12}{100},\frac{13}{100},\frac{14}{100},\frac{15}{100},\frac{16}{100},\frac{17}{100},\frac{18}{100},\frac{19}{100}$

We can further divide the number line between the points $\frac{11}{100}$ and $\frac{12}{100}$ into 10 equal parts. We get $\frac{111}{1000},\frac{112}{1000},\frac{113}{1000},\frac{114}{1000},\frac{115}{1000},\frac{116}{1000},\frac{117}{1000},\frac{118}{1000},\frac{119}{1000}$, If we go on increasing the divisions between two rational numbers, we can accommodate an infinite number of rational numbers.

• Q. Find any 5 rational numbers between $\frac{-1}{2}$ and $\frac{1}{2}$. Solution: Let us write $\frac{-1}{2}$ as $\frac{-5}{10}$ and $\frac{1}{2}$ as $\frac{5}{10}$. The rational numbers between these are $\frac{-4}{10},\frac{-3}{10},\frac{-2}{10},\frac{-1}{10},\frac{0}{10},\frac{1}{10},\frac{2}{10},\frac{3}{10}$ and $\frac{4}{10}$. Write any 5 rational numbers for the answer. The numbers are $\frac{-4}{10},\frac{-3}{10},\frac{-2}{10},\frac{-1}{10},\frac{2}{10}$.
• Q. Find any two rational numbers between $\frac{1}{4}$ and $\frac{3}{5}$. Solution: First, find the equivalent rational numbers of both the given rational numbers by converting their denominators into a common denominator. LCM of 4 and 5 is 20 . $\frac{1}{4}=\frac{1\times 5}{4\times 5}=\frac{5}{20}$ and $\frac{3}{5}=\frac{3\times 4}{5\times 4}=\frac{12}{20}$ Two rational numbers between $\frac{1}{4}\left(=\frac{5}{20}\right)$ and $\frac{3}{5}\left(=\frac{12}{20}\right)$ are $\frac{6}{20}$ and $\frac{7}{20}$ Note : "The rational numbers do not have a unique representation in the form $\frac{p}{q}$, where $p$, q are integers and $\mathrm{q}\ne 0$. eg. $\frac{1}{2}=\frac{2}{4}=\frac{3}{6}=\frac{4}{8}$, these are called equivalent rational numbers."

Alternate method

We can also use the idea of mean to find rational numbers between any two given rational numbers. The number that is midway between the two given rational numbers, a and $\mathrm{b}(\mathrm{a}<\mathrm{b})$ is the mean of $a$ and $b$, i.e. $x=\left[\frac{a+b}{2}\right]$

• Q. Find a rational number that is midway between $\frac{1}{5}$ and $\frac{1}{4}$. Solution: We find the mean of the given rational numbers. $\frac{\left(\frac{1}{5}+\frac{1}{4}\right)}{2}=\frac{\left(\frac{4+5}{20}\right)}{2}=\frac{\frac{9}{20}}{2}=\frac{9}{20}\times \frac{1}{2}=\frac{9}{40}$
• Q. Find 3 rational numbers between $\frac{1}{8}$ and $\frac{1}{2}$. Solution: Let $r_1$ be a rational number between $\frac{1}{8}$ and $\frac{1}{2}$ $\therefore r_1=\frac{\left(\frac{1}{8}+\frac{1}{2}\right)}{2}=\frac{\left(\frac{1+4}{8}\right)}{2}=\frac{5}{8}\times \frac{1}{2}=\frac{5}{16}$ So, we have $\frac{1}{8}<\frac{5}{16}<\frac{1}{2}$. Let $r_2$ be a rational number between $\frac{5}{16}$ and $\frac{1}{2}$. $\therefore r_2=\frac{\left(\frac{5}{16}+\frac{1}{2}\right)}{2}=\frac{\left(\frac{5+8}{16}\right)}{2}=\frac{13}{16}\times \frac{1}{2}=\frac{13}{32}$ So, we have $\frac{1}{8}<\frac{5}{16}<\frac{13}{32}<\frac{1}{2}$. Let $r_3$ be a rational number between $\frac{5}{16}$ and $\frac{13}{32}$. $\therefore r_3=\frac{\left(\frac{5}{16}+\frac{13}{32}\right)}{2}=\frac{\left(\frac{10+13}{32}\right)}{2}=\frac{23}{32}\times \frac{1}{2}=\frac{23}{64}$ So, we have, $\frac{1}{8}<\frac{5}{16}<\frac{23}{64}<\frac{13}{32}<\frac{1}{2}$ Thus, 3 rational numbers between $\frac{1}{8}$ and $\frac{1}{2}$ are $\frac{5}{16},\frac{23}{64}$ and $\frac{13}{32}$.
• Q. John bought 25 kg of rice and he used $1\frac{3}{4}\mathrm{kg}$ on the first day, $4\frac{1}{2}\mathrm{kg}$ on the second day. Find the quantity of rice left. Explanation: Total quantity of rice $=25\mathrm{kg}$. Rice used on first day $=1\frac{3}{4}\mathrm{kg}=\frac{7}{4}\mathrm{kg}$. Rice used on second day $=4\frac{1}{2}\mathrm{kg}=\frac{9}{2}\mathrm{kg}$ Remaining quantity of rice left $=\frac{25}{1}-\frac{7}{4}-\frac{9}{2}$ $=\frac{100-7-18}{4}=\frac{75}{4}=18\frac{3}{4}\mathrm{kg}$.
• Q. Kriyaan reads $\frac{1}{3}$ part of a book in 1 hour. How much part of the book will he read in $2\frac{1}{5}$ hours? Solution: The part of the book read by Kriyaan in 1 hour $=\frac{1}{3}$ So, the part of the book read by him in $2\frac{1}{5}$ hours $=2\frac{1}{5}\times \frac{1}{3}$ $=\frac{11}{5}\times \frac{1}{3}=\frac{11}{15}$
• Q. John bought 25 kg of Rice and he used $1\frac{3}{4}\mathrm{kg}$ on the first day, $4\frac{1}{2}\mathrm{kg}$ on the second day. Find the remaining quantity of rice left. Solution: Rice used on first day $=1\frac{3}{4}\mathrm{kg}$ Rice used on second day $=4\frac{1}{2}\mathrm{kg}$ Quantity of rice used on first day and second day: $=1\frac{3}{4}+4\frac{1}{2}=\frac{7}{4}+\frac{9}{2}=\frac{7}{4}+\frac{18}{4}=\frac{(7+18)}{4}=\frac{25}{4}$ Remaining quantity of rice left : $=\frac{25}{1}-\frac{25}{4}=\frac{100-25}{4}=\frac{75}{4}=18\frac{3}{4}$ Remaining quantity of rice is $18\frac{3}{4}\mathrm{kg}$.
• Q. There are 15000 books in a library. One-fifth of the book are English, two-third of the books are French and the rest are Spanish. Find the number of Spanish books in the library? Solution: Fraction of books which are English and French $=\frac{1}{3}+\frac{2}{5}$ Least common multiple of $(3,5)=15$. Make each denominator as 15 by multiplying each fraction by an appropriate number. $=\frac{5+6}{15}=\frac{11}{15}$ Fraction of books which are Spanish $=\frac{4}{15}$ Number of Spanish books in the library $=\frac{4}{15}\times 15000$ $=4\times 1000=4000$
• Q. By what rational number should $\frac{-7}{12}$ be multiplied to get the product as $\frac{5}{14}$ ? Solution: Let the required number be ' $x$ '. So, $\frac{-7}{12}\times x=\frac{5}{14}$ $\mathrm{x}==\frac{5}{14}\times \frac{12}{-7}$ $x=\frac{30}{-49}$ Hence, the required number is $-\frac{30}{49}$.