ALGEBRAIC EXPRESSIONS AND IDENTITIES

by maths on October 28, 2011

INTRODUCTION
In the previous class, we have learnt about algebraic expressions and their addition and subtraction. Most of the expressions that we worked with had integer coefficients. In this chapter, we shall study multiplication of algebraic expressions in the form of monomials and binomials etc. Also, we shall learn to work with algebraic expressions that contain both integer and fractional coefficients. In other words, we shall work with algebraic expressions containing rational numbers as the coefficients of various terms. We shall also learn how to factorize algebraic expressions. But before all these things, we review here what we have learnt earlier.
6.2 REVIEW OF CONCEPTS AND DEFINITIONS
In algebra, we generally come across two types of symbols, namely constants and variables.
CONSTANT A symbol having a fixed numerical value is called a constant.
VARIABLE A symbol which takes various numerical values is called a variable.
ILLUSTRATION 1 We know that the perimeter P of a square of side s is given by P=4 x s. Here, 4 is a constant and P and s are

variables.

ILLUSTRATION 2

The perimeter P of a rectangle of sides 1 and b is given by P = 2(1 + b). Here, 2 is a constant and l and b are variables.

ALGEBRAIC EXPRESSIONS

A combination of constants and variables connected by the signs of fundamental operations of addition, subtraction, multiplication and division is called an algebraic expression.

TERMS

Various parts of an algebraic expression which are separated by the signs of + or — are called the ‘terms’ of the expression.
ILLUSTRATION 3 2x^2 — 3xy + 5y^2 is an algebraic expression consisting of three terms, namely, 2x^2,—3xy and 5y^2.
ILLUSTRATION 4 The expression 2x^3-3x^2+ 4x —7 is an algebraic expression consisting of four terms, namely, 2x^3,-3x^2,4x and —7.
MONOMIAL An algebraic expression containing only one term is called a monomial.
ILLUSTRATION 5 —5, 3y, 7xy, x2yz, a2bc3 etc. are all monomials. Two monomials containing unlike terms when added give a binomial as defined below.
BINOMIAL An algebraic expression containing two terms is called a binomial.
ILLUSTRATION 6 The expressions 2x —3, 3x + 2y, xyz —5 etc. are all binomials.
Note that 3x + 7x is not a binomial, because 3x + 7x = lOx, which is a monomial.
TRINOMIAL An algebraic expression containing three terms is called a trinomial.
In other words, if three monomials are such that no two contain like terms, then their sum is a trinomial.
ILLUSTRATION 7 The expressions a—b+2x^2+y^2—xy,x^3—2y^3-2x^2 y^2z etc. are trinomials.
FACTORS Each term man algebraic expression is a product of one or more numbers (s) and / or literal(s). These number(s) and / or literal(s) are known as the factors of that term.
A constant factor is called a numerical factor, while a variable factor is known as a literal
factor.
COEFFICIENT In a term of an algebraic expression any of the factors with the sign of the term is called the coefficient of the product of the other factors.
ILLUSTRATION 8 In —5xy, the coefficient of x is —5y; the coefficient of y is —5x and the coefficient of xy is —5.
ILLUSTRATION 9 In —x , the coefficient of x is —1.
ILLUSTRATION 10 In 3a^2bc, the coefficient of a^2 is 3bc, the coefficient of b is x3a^2c and the coefficient of c is 3a^2b.
CONSTANT TERM A term of the expression having no literal factor is called a constant term,
ILLUSTRATION 11 In the algebraic expression x^2 — xy + yz —4, the constant term is —4.
LIKE AND UNLIKE TERMS The terms having the same literal factors are called like or similar terms, otherwise they are called unlike terms.
ILLUSTRATION 12 In the algebraic expression 2a^2b + 3ab^2 — 7ab — 4ba^2, we have 2 2 a^2b and
—4ba^2 as like terms, whereas 3ab^2 and —7ab are unlike terms.
EXERCISE 6.1
1. Identify the terms, their coefficients for each of the following expressions:
(i) 7x^2 yz—5xy (ii) x^2 +x+1 (iii) 3x^2 y^2 —5x^2 y^2 z^2 +z^2
(iv) 9—ab+bc—ca (v) \frac{a}{2}+\frac{b}{2} – ab (vi) O.2x-0.3xy+0.5y
2. Classify the following polynomials as monomials, binomials, trinomials. Which
polynomials do not fit in any category?
(i) x + y (ii) 1000 (iii) x + x^2 + x^3 + x^4
(iv) 7 + a + 5b (v) 2b — 3b^2 (vi) 2y — 3y^2 + 4y^3
(vii) 5x—4y+3x (viii) 4a-15a^2 (ix) xy+yz+zt+tx
(x) pqr (xi) p^2q+pq^2 (xii) 2p+2q
ANSWERS
1. Terms Coefficients Terms Coefficients
(i) 7x^2yz 7 (ii) x^2 1
-5xy -5 x 1
x 1
(iii) 3x^2 y^2 3 (iv) -ab -1
– 5x^2 y^2 z^2 -5 bc 1
z^2 1 9 9

(v) \frac{a}{2} \frac{1}{2}
\frac{b}{2} \frac{1}{2}
-ab -1
(vi) 0.2x 0.2
-0.3xy -0.3
0.5y 0.5
2. Monomial \ Binomial Trmomial None of these
(ii), (x) (i), (v), (viii) (xi), (xii) (iv), (vi), (vii) (iii), (ix)
6.2.1 ADDITiON OF ALGEBRAIC EXPRESSIONS
In adding algebraic expressions, we collect different groups of like terms and find the sum of like terms in each group. Note that the sum of several like terms is another like term. whose coefficient is the sum of the coefficients of those like terms.
Following examples will illustrate the same.
ILLUSTRATIVE EXAMPLES
Examplel Add:7×2 —4x+5,—3×2 +2x—land 5×2 —x+9.
Solution We have,
Required sum

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by maths on October 25, 2011

11. What number should be subtracted from \frac{3}{7} to get \frac{5}{4}
12. What shold be added to \LARGE{(} \frac{2}{3} + \frac{3}{5} \LARGE{)} to get \frac{-2}{15}
13. What should be added to \LARGE{(} \frac{1}{2} + \frac{1}{3} + \frac{1}{5} \LARGE{)} to get 3?
14. What should be subtracted from \LARGE{(} \frac{3}{4} – \frac{2}{3} \LARGE{)} to get \frac{-1}{6}?
15. Fill in the blanks:
(i) \frac{-4}{13} – \frac{-3}{26} = …. (ii) \frac{-9}{14} + ….= -1 (iii) \frac{-7}{9} + …. = 3 (iv) …. + \frac{15}{23} = 4
ANSWERS
1. (i) \frac{1}{4} (ii) \frac{11}{9} (iii) \frac{-7}{11} (iv) \frac{-15}{13} (v) \frac{-5}{8} (vi) \frac{3}{2} (vii) \frac{-1}{14} (viii) \frac{-5}{66}
2. (i) \frac{1}{15} (ii) \frac{2}{21} (iii) \frac{-1}{7} (iv) \frac{-23}{9} (v) \frac{37}{56} (vi) \frac{-3}{26} (vii) \frac{-1}{14} (viii) \frac{29}{75} (ix) \frac{1}{13} (x) \frac{-17}{72} (xi) \frac{29}{63}
3. \frac{2}{9} 4. \frac{11}{3} 5. \frac{11}{3} 6. \frac{-41}{7} 7. \frac{103}{72} 8. \frac{41}{33} 9. \frac{1}{21} 10. \frac{-5}{2} 11. \frac{-23}{28} 12. \frac{-7}{5} 13. \frac{59}{30} 14. \frac{1}{4} 15. (i) \frac{-5}{26} (ii) \frac{-5}{14} (iii) \frac{34}{9} (iv) \frac{77}{23}
1.7 SIMPLIRCATION OF EXPRESSIONS INVOLVING ADDITION AND SUBTRACTION OF RATIONAL NUMBERS
Uptill now, we have been simplifying rational expressions involving addition and subtraction of more than two rational numbers by making groups of pairs of rational numbers having either the same denominator or having some common factor in their denominators, by using commutativity and asociativity of addition (see illustrations 1, 2 and 3 on pages 1.8 — 1.9). We may simplify expressions involving addition and subtraction more easily by using the following algorithm.
ALGORITHM
Step I Find the LCM of the denominators of all the numbers involved.
Step II Divide the LCM by the deno minator of the first rational number and get a quotient.
Step III Multiply the first numerator by the quotient obtained in step II and get an integer.
Step IV Repeat steps II and III for the remaining rational numbers in the sum and obtain integers.
Step V Retain the given symbols addition and subtraction between the given rationals and get an expression involving integers. Simplify this expression and get
an integer.
Step VI Obtain the required sum equal to the rational number whose
numerator is equal to the integer obtained in step V and deno- minator equal to the LCM obtained in step I. Reduce this number to the lowest form if it is not already so.
ILLUSTRATION
Suppose, we wish to simplify
\frac{2}{5} + \frac{8}{3} – \frac{12}{15} + \frac{4}{5} – \frac{2}{3}
Step I LCM of denominators 5,3,15,5 and 3 is 15
Step II 15 \div 5 =3 (Quotient)
Step III First numerator = Numerator of \frac{2}{3} = 2
\therefore First numerator X Quotient = 2×3=6
Step IV Integers obtained are :
8×5, 12×1, 4×3, 2×5
Step V 2×3+8×5-12×1+4×3-2×5
= 6+40-12+12-10
= 58-22=36
Step VI Required sum = \frac{36}{15} = \frac{12}{5}
i.e., \frac{2}{5} + \frac{8}{3} – \frac{12}{15} + \frac{4}{5} – \frac{2}{3} = \frac{12}{5}
The above procedure, in shorter form, may be written as follows:
\frac{2}{5} + \frac{8}{3} – \frac{12}{15} + \frac{4}{5} – \frac{2}{3}
= \frac{(2X3)+(8X5)-(12X1)+(4X3)-(2X5)}{15}
= \frac{6+40-12+12-10}{15} = \frac{58-22}{15} = \frac{36}{15} = \frac{12}{5}
In order to understand the above procedure properly, let us discuss some more examples.
ILLUSTRATIVE EXAMPLES
Example 1 Find:\frac{3}{7} + \LARGE{(} – \frac{6}{11} \LARGE{)} + \frac{8}{21} + \LARGE{(} \frac{-5}{22} \LARGE{)}
Solution We have,
\frac{3}{7} + \LARGE{(} – \frac{6}{11} \LARGE{)} + \frac{8}{21} + \LARGE{(} \frac{-5}{22} \LARGE{)}

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by maths on October 25, 2011

= \LARGE{(} \frac{3}{5}+\frac{2}{5} \LARGE{)} + \LARGE{[} \frac{(-7)}{6}+\frac{(-5)}{6} \LARGE{]}
= \frac{3+2}{5} + \frac{(-7)+(-5)}{6} = \frac{5}{5} + \frac{(-12)}{6} = 1+(-2)=1-2=-1
(ii) Re-arranging and grouping the numbers in pairs in such a way that each group contains a pair of rational numbers with equal denominators, we have
\frac{4}{3} + \frac{-4}{5} + \frac{-2}{3} + \frac{7}{5} – 2
= \LARGE{(} \frac{4}{3}+\frac{-2}{3} \LARGE{)} + \LARGE{(} \frac{-4}{5}+\frac{7}{5} \LARGE{)} -2
= \LARGE{(} \frac{4}{3}+\frac{(-2)}{3} \LARGE{)} + \LARGE{(} \frac{(-4)}{5}+\frac{7}{5} \LARGE{)} -2
= \frac{4+(-2)}{3} + \frac{(-4)+7}{5}-2
= \frac{2}{3} + \frac{3}{5} -2
= \LARGE{(} \frac{2}{3} + \frac{3}{5} \LARGE{)} -2
= \frac{2X5+3X3}{15} + (-2)
= \frac{10+9}{15} + (-2) = \frac{19}{15} + \frac{(-2)}{1} = \frac{19+(-2)X15}{15} = \frac{19+(-30)}{15} = \frac{-11}{15}
Example 3 Re-arrange suitably and find the sum in each of the following:
(i) \frac{5}{3} + \frac{11}{2} + \frac{-9}{4} + \frac{-8}{3} + \frac{-7}{2} (ii) \frac{-4}{7} + \frac{7}{6} + \frac{2}{7} + 3 + \frac{-11}{6}
Solution (i) Re-arranging and grouping the numbers in pairs in such a way that each group contains a pair of rational numbers with equal denominators, we have
\frac{5}{3} + \frac{11}{2} + \frac{-9}{4} + \frac{-8}{3} + \frac{-7}{2}
= \LARGE{(} \frac{5}{3} + \frac{-8}{3} \LARGE{)} + \LARGE{(} \frac{11}{2} + \frac{-7}{2} \LARGE{)} + \frac{-9}{4}
= \frac{5-(-8)}{3} + \frac{11+(-7)}{2} + \frac{(-9)}{4}
= \frac{-3}{3} +\frac{4}{2} + \frac{(-9)}{4}
= -1+2+ \frac{(-9)}{4}
= (-1+2) + \frac{(-9)}{4}
= 1 + \frac{(-9)}{4} = \frac{1}{1} + \frac{(-9)}{4} = \frac{1X4+(-9)}{4} = \frac{4+(-9)}{4} = \frac{(-5)}{4}
(ii) Re-arranging and grouping the numbers in pairs in such a way that each
group contains a pair of rational numbers with equal denominators, we have
\frac{-4}{7} + \frac{7}{6} + \frac{2}{7} + 3 + \frac{-11}{6}
= \LARGE{(} \frac{-4}{7} + \frac{2}{7} \LARGE{)} + \LARGE{(} \frac{7}{6} + \frac{-11}{6} \LARGE{)} + 3
= \frac{(-4)+2}{7} + \frac{7+(-11)}{6}+3
= \frac{-2}{7} + \frac{(-4)}{6} + 3
= \frac{(-2)}{7} + \frac{(-2)}{3} + 3
= \LARGE{(} \frac{(-2)}{7} + \frac{(-2)}{3} \LARGE{)} +3
= \frac{(-2)X3+(-2)X7}{21} + 3
= \frac{(-20)}{21} + \frac{3}{1} = \frac{(-20)+3X21}{21} = \frac{(-20)+63}{21} = \frac{43}{21}
EXERCISE 1.2
1. Verifr commutativity of addition of rational numbers for each of the following pairs of rational numbers:
(i) \frac{-11}{5} and \frac{4}{7} (ii) \frac{4}{9} and \frac{7}{-12} (iii) \frac{-3}{5} and \frac{-2}{-15} (iv) \frac{2}{-7} and \frac{12}{-35} (v) 4 and \frac{-3}{5} (vi) -4 and \frac{4}{-7}
2. Verify associativity of addition of rational numbers i.e., (x + y) + z = x ± (y + z), when:
(i) x = \frac{1}{2},y=\frac{2}{3},z=- \frac{1}{5} (ii) x= \frac{-2}{5},y=\frac{4}{3},z=\frac{-7}{10} (iii) x = \frac{-7}{11},y = \frac{2}{-5}, z = \frac{-3}{22} (iv) X=-2,Y = \frac{3}{5},z = \frac{-4}{3}
3. Write the additive inverse of each of the following rational numbers:
(i) \frac{-2}{17} (ii) \frac{3}{-11} (iii) \frac{-17}{5} (iv) \frac{-11}{-25}
4. Write the negative (additive inverse) of each of the following:
(i) \frac{-2}{5} (ii) \frac{7}{-9} (iii) \frac{-16}{13} (iv) \frac{-5}{1} (v) 0 (vi) 1 (vii) -1
5. Using commutativity and associativity of addition of rational numbers, express each of the following as a rational number:
(i) \frac{2}{3} + \frac{7}{3} + \frac{-4}{5} + \frac{-1}{3} (ii) \frac{3}{7} + \frac{-4}{9} + \frac{-11}{7} + \frac{7}{9} (iii) \frac{2}{5} + \frac{8}{3} + \frac{-11}{15} + \frac{4}{5} + \frac{-2}{3} (iv) \frac{4}{7} + 0 \frac{-8}{9} + \frac{-13}{7} + \frac{17}{21}
6. Re-arrange suitably and find the sum in each of the following:
(i) \frac{11}{12} + \frac{-17}{3} + \frac{11}{2} + \frac{-25}{2} (ii) \frac{-6}{7} + \frac{-5}{6} + \frac{-4}{9} + \frac{-15}{7} (iii) \frac{3}{5} + \frac{7}{3} + \frac{9}{5} + \frac{-13}{15} + \frac{-7}{3} (iv) \frac{4}{13} + \frac{-5}{8} + \frac{-8}{13} + \frac{9}{13} (v) \frac{2}{3} + \frac{-4}{5} + \frac{1}{3} + \frac{2}{5} (vi) \frac{1}{8} + \frac{5}{12} + \frac{2}{7} + \frac{7}{12} + \frac{9}{7} + \frac{-5}{16}
ANSWERS
3. (i) \frac{2}{17} (ii) \frac{3}{11} (iii) \frac{17}{5} (iv) \frac{-11}{25}
4. (i) \frac{2}{5} (ii) \frac{7}{9} (iii) \frac{16}{13} (iv) 5 (v) 0 (vi) -1 (vii) 1
5. (i) \frac{8}{5} (ii) \frac{-17}{21} (iii) \frac{37}{15} (iv) \frac{-86}{63}
6. (i) \frac{-141}{12} (ii) \frac{-77}{18} (iii) \frac{23}{15} (iv) \frac{-25}{104} (v) \frac{3}{5} (vi) \frac{267}{112}
1.5 SUBTRACTION OF RATIONAL NUMBERS
If \frac{a}{b} and \frac{c}{d} are two rational numbers, then subtracting \frac{c}{d} from \frac{a}{b} means adding additive
inverse (negative) of \frac{c}{d} to \frac{a}{b}
The subtraction of \frac{c}{d} from \frac{a}{b} is written as \frac{a}{b} – \frac{c}{d}
Thus, we have
\frac{a}{b} – \frac{c}{d} = \frac{a}{b} + \LARGE{(} \frac{-c}{d} \LARGE{)}
ILLUSTRATIVE EXAMPLES
Example 1 Subtract \frac{3}{4} from \frac{5}{6}
Soiution The additive inverse of \frac{3}{4} is\frac{-3}{4}
\therefore \frac{5}{6} – \frac{3}{4} = \frac{5}{6} + \frac{-3}{4} = \frac{5X2}{6X2} + \frac{-3X3}{4X3} = \frac{10}{12} + \frac{-9}{12} = \frac{10+(-9)}{12} = \frac{1}{12}
Example 2 Subtract \frac{-3}{8} from \frac{-5}{7}.
Solution The additive inverse of \frac{-3}{8} is \frac{3}{8}
\therefore \frac{-5}{7} – \LARGE{(} \frac{-3}{8} \LARGE{)} = \frac{-5}{7} + \frac{3}{8}
= \frac{(-5)X8+3X7}{56} = \frac{-40+21}{56} = – \frac{19}{56}
Example 3 Subtract \frac{-3}{5} from \frac{9}{10}
SOlutIOn The additive inverse of \frac{-3}{5} is – \LARGE{(} \frac{-3}{5} \LARGE{)} = \frac{3}{5}
\therefore \frac{9}{10} – \LARGE{(} \frac{-3}{5} \LARGE{)} = \frac{9}{10} + \frac{3}{5} = \frac{9+3X2}{10} = \frac{9+6}{10} = \frac{15}{10} = \frac{3}{2}
Example 4 The sum of two rational numbers is \frac{-3}{5}. If one of the number is \frac{-9}{20}, find the other.
Solution It is given that
Sum of the numbers = \frac{-3}{5} and, One of the numbers = \frac{-9}{20}
Suppose the other rational number is x. Since the sum is \frac{-3}{5}
\therefore X + \LARGE{(} \frac{-9}{20} \LARGE{)} = \frac{-3}{5}
\Rightarrow x = \frac{-3}{5} – \LARGE{(} \frac{-9}{20} \LARGE{)}
\Rightarrow x = \frac{-3}{5} + \frac{9}{20} \LARGE{[} \because – \LARGE{(} \frac{-9}{20} \LARGE{)} = \frac{9}{20} \LARGE{]}
\Rightarrow x = \frac{(-3)X4+9X1}{20}
\Rightarrow x = \frac{(-3)X4+9X1}{20} \frac{-12+9}{20} = \frac{-3}{20}
Example 5 What number should be added to \frac{-5}{8} so as to get \frac{5}{9}?
Solution Suppose x is the rational number to be added to \frac{-5}{8} to get \frac{5}{9} . Then,
\frac{-5}{8} + x = \frac{5}{9}
\Rightarrow x = \frac{5}{9} – \LARGE{(} \frac{-5}{8} \LARGE{)} \LARGE{[} Transposing \frac{-5}{8} to RHS \LARGE{]}
\Rightarrow x = \frac{5}{9} +\frac{5}{8}
\Rightarrow x = \frac{5X8+5X9}{72} = \frac{40+45}{72} = \frac{85}{72}
EXISTENCE OF RIGHT IDENTITY The rational number 0 is the right identity. That is, for any rational number \frac{a}{b} we have
EXERCISE 1.3
1. Subtract the first rational number from the second in each of the following:
(i) \frac{3}{8},\frac{5}{8} (ii) \frac{-7}{9},\frac{4}{9} (iii) \frac{-2}{11},\frac{-9}{11} (iv) \frac{11}{13},\frac{-4}{13} (v) \frac{1}{4},\frac{-3}{8} (vi) \frac{-2}{3},\frac{5}{6b} (vii) \frac{-6}{7},\frac{-13}{14} (viii) \frac{-8}{33},\frac{-7}{22}
2. Evaluate each of the following:
(i) \frac{2}{3} – \frac{3}{5} (ii) – \frac{4}{7} – \frac{2}{-3} (iii) \frac{4}{7} – \frac{-5}{-7} (iv) -2 – \frac{5}{9} (v) \frac{-3}{-8} – \frac{-2}{7} (vi) \frac{-4}{13} – \frac{-5}{26} (vii) \frac{-5}{14} – \frac{-2}{7} (viii) \frac{13}{15} – \frac{12}{25} (ix) \frac{-6}{13} – \frac{-7}{13} (x) \frac{7}{24} – \frac{19}{36} (xi) \frac{5}{63} – \frac{-8}{21}
3. The sum of the two numbers is \frac{5}{9} . If one of the numbers is \frac{1}{3},find the other.
4. The sum of two numbers is \frac{-1}{3}. If one of the numbers is \frac{-12}{3}, find the other.
5. The sum of two numbers is \frac{-4}{3}. If one of the numbers is —5, find the other.
6. The sum of two rational numbers is —8. If one of the numbers is \frac{-15}{7} find the other.
7. What should be added to \frac{-7}{8} so as to get \frac{5}{9}
8. What number should be added to \frac{-5}{11} so as to get \frac{26}{33}?
9. What number should be added to \frac{-5}{7} to get \frac{-2}{3}?
10. What number should be subtracted from \frac{-5}{3} to get \frac{5}{6}?

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by maths on October 21, 2011

Similarly, we have
- \LARGE{(} – \frac{5}{7} \LARGE{)} = \frac{5}{7}, – \LARGE{(} – \frac{15}{13} \LARGE{)} = \frac{15}{13},- \LARGE{(} – \frac{7}{12} \LARGE{)} and so on
Thus for any ‘rational number \frac{a}{b} , we have
- \LARGE{(} – \frac{a}{b} \LARGE{)} = \frac{a}{b}
Remark Wehave, 0+0=0=0+0
0 is the additive inverse of itself, that is, 0 = 0.
NOTE: It should be noted that 0 is the only rational number which is its own negative.
ILLUSTRATION 4 Write the additive inverse of each of the following rational numbers:
(i) \frac{4}{9} (ii) \frac{-13}{7} (iii) \frac{5}{-11} (iv) \frac{-11}{-14}
Solution (i) The additive inverse of \frac{4}{9} is – \frac{4}{9} = \frac{-4}{9}
(ii) Te additive inverse of \frac{-13}{7} iS – \LARGE{(} \frac{-13}{7} \LARGE{)} = – \LARGE{(} – \frac{13}{7} \LARGE{)} = \frac{13}{7}
(iii) We have,\frac{5}{-11} = \frac{-5}{11}
The additive inverse of \frac{-5}{11} is – \LARGE{(} \frac{-5}{11} \LARGE{)} = – \LARGE{(} – \frac{5}{11} \LARGE{)} = \frac{5}{11}
(iv) We have,\frac{-11}{-14} = \frac{11}{14}
The additive inverse of \frac{11}{14} is – \LARGE{(} \frac{11}{14} \LARGE{)} = \frac{-11}{14}
ILLUSTRATIVE EXAMPLES
Example l Verify: \LARGE{(} \frac{a}{b} + \frac{c}{d} \LARGE{)} + \frac{e}{f} = \frac{a}{b} + \LARGE{(} \frac{c}{d} + \frac{e}{f} \LARGE{)} for \frac{a}{b} = \frac{-2}{3}, \frac{c}{d} = \frac{5}{7} and \frac{e}{f} = \frac{-1}{6}
Solution We have,
= \LARGE{(} \frac{a}{b} + \frac{c}{d} \LARGE{)} + \frac{e}{f}
= \LARGE{(} \frac{-2}{3} + \frac{5}{7} \LARGE{)} + \frac{-1}{6}
= \frac{(-2X7+3X5)}{21} + \frac{-1}{6}
= \frac{(-14)+15}{21} + \frac{-1}{6}
= \frac{1}{21} + \frac{(-1)}{6} = \frac{1X2+(-1)X7}{42} = \frac{2+(-7)}{42} = \frac{(-5)}{42}=\frac{-5}{42}
and \frac{a}{b} + \LARGE{(} \frac{c}{d} + \frac{e}{f} \LARGE{)}
= \frac{-2}{3} + \LARGE{(} \frac{5}{7} + \frac{-1}{6} \LARGE{)}
= \frac{-2}{3} + \frac{5X6+7X(-1)}{42}
= \frac{-2}{3} + \frac{30+(-7)}{42}
= \frac{(-2)}{3} + \frac{23}{42} = \frac{(-2)X14+23X1}{42} = \frac{(-28)+(23)}{42} = \frac{(-5)}{42} = \frac{-5}{42}
\therefore \LARGE{(} \frac{a}{b} + \frac{c}{d} \LARGE{)} + \frac{e}{f} = \frac{a}{b} + \LARGE{(} \frac{c}{d} + \frac{e}{f} \LARGE{)}
Exampe 2 Using commutativity and associativity of addition of rational numbers, express each of the following as a rational number:
(i) \frac{3}{5}+ \frac{-7}{6} + \frac{2}{5} + \frac{-5}{6} (ii) \frac{4}{3} + \frac{-4}{5} + \frac{-2}{3} + \frac{7}{5} – 2
Solution (i) Re-arranging and grouping the numbers in pairs in such a way that each group contains a pair of rational numbers with equal denominators, we have
\frac{3}{5}+ \frac{-7}{6} + \frac{2}{5} + \frac{-5}{6}
= \LARGE {(} \frac{3}{5} + \frac{2}{5} \LARGE {)} + \LARGE {(} \frac{-7}{6} + \frac{-5}{6} \LARGE {)}

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+

by maths on October 21, 2011

= \frac{1}{12}+\frac{-4}{9}=\frac{3}{36}=\frac{3+(-16)}{36}=\frac{-13}{36}
\therefore \frac{-3}{4}+\LARGE{(} \frac{5}{6}+\frac{-4}{9} \LARGE{)}=\LARGE{(} \frac{-3}{4}+\frac{5}{6} \LARGE{)}+\frac{-4}{9}
Similarly, it can be verified for other rational numbers.
Remark 1. From the associativity of addition of rational numbers, we find that three or more rational numbers can be added without using parentheses and from the commutativity of addition of rational numbers we observe that the order in which the -numbers are arranged before addition does not affect the sum. It follows from these two properties that while adding three or more rational numbers, we can re-arrange the numbers in any order before adding them and the use of paraentheses is not essential.
Remark 2. While adding three or more rational numbers we re-arrange and group them in such a way that each group contains a pair of numbers either with a common denominator or their denominators have a common divisor. The following illustrations will illustrate the sum.
ILLUSTRATION 1 Simplify: \frac{4}{3}+\frac{3}{5}+\frac{-2}{3}+\frac{-11}{5}
Solution We find that out of the four rational numbers to be added, two have the same denominator 3 and the remaining two have the same denominator 5. So, we re-arrange and group them in such a way that each group contains a pair of numbers with a common denomiator,
\therefore \frac{4}{3} + \frac{3}{5} + \frac{-2}{3} + \frac{-11}{5} = \LARGE{(} \frac{4}{3}+\frac{-2}{3} \LARGE{)} + \LARGE{(} \frac{3}{5}+\frac{-11}{5} \LARGE{)}
= \frac{4+(-2)}{3} + \frac{3+(-11)}{5}
= \frac{2}{3} + \frac{-8}{5}
= \frac{2X5}{3X5} + \frac{(-8)X3}{5X3} \LARGE{[} \therefore LCM of 3 and 5 is 15
\because Each rational number is
expressed with denominator 15 \LARGE{]}
\frac{10}{15} + \frac{-24}{15} = \frac{-10+(-24)}{15} + \frac{-14}{15}
ILLUSTRATION 2 Simplify \frac{3}{8} + \frac{7}{2} + \frac{-3}{5} + \frac{-9}{8} + \frac{-3}{2} + \frac{6}{5}
Solution Re-arranging and grouping the numbers iii pairs in such a way that each group contains a pair of rational numbers with a common denominator, we have
\frac{3}{8} + \frac{7}{2} + \frac{-3}{5} + \frac{-9}{8} + \frac{-3}{2} + \frac{6}{5}
= \LARGE{(} \frac{3}{8}+\frac{9}{8} \LARGE{)} + \LARGE{(} \frac{7}{2}+\frac{-3}{2} \LARGE{)} + \LARGE{(} \frac{-3}{5} + \frac{6}{5} \LARGE{)}
= \frac{3+9}{8} + \frac{7+(-3)}{2} + \frac{(-3)+6}{5}
= \frac{12}{8} + \frac{4}{2} + \frac{3}{5}
= \frac{3}{2} + 2+ \frac{3}{5}
= \frac{3X5}{2X5} + \frac{2X10}{1X10} + \frac{3X2}{5X2} = \frac{15}{10} + \frac{20}{10} + \frac{6}{10} + \frac{15+20+6}{10} = \frac{41}{10}
ILLUSTRA1ION 3 Simplify: \frac{-3}{10} + \frac{7}{15} + \frac{3}{-20} + \frac{-9}{10} + \frac{13}{15} + \frac{13}{-20}
Solution Re-arranging and grouping the numbers in pairs such that each group
contains a pair of rational numbers with equal denominators, we have
\frac{-3}{10} + \frac{7}{15} + \frac{3}{-20} + \frac{-9}{10} + \frac{13}{15} + \frac{13}{-20}
= \LARGE{(} \frac{-3}{10}+\frac{-9}{10} \LARGE{)} + \LARGE{(} \frac{7}{15}+\frac{13}{15} \LARGE{)} + \LARGE{(} \frac{3}{-20}+\frac{13}{-20} \LARGE{)}
= \LARGE{(} \frac{-3}{10}+\frac{-9}{10} \LARGE{)} + \LARGE{(} \frac{7}{15}+\frac{13}{15} \LARGE{)} + \LARGE{(} \frac{-3}{20}+\frac{-13}{20} \LARGE{)}
= \frac{(-3)+(-9)}{10} + \frac{7+13}{15} + \frac{(-3)+(-13)}{20}
= \frac{-12}{10} + \frac{20}{15} + \frac{-16}{20}
= \frac{-6}{5} + \frac{4}{3} + \frac{-4}{5}
= \LARGE{(} \frac{-6}{5}+\frac{-4}{5} \LARGE{)} + \frac{4}{3}
= \frac{(-6)+(-4)}{5} + \frac{4}{3} = \frac{-10}{5} + \frac{4}{3} = \frac{-10}{5} + \frac{4}{3} = -2 + \frac{4}{3} = \frac{-6+4}{3} = \frac{-2}{3}
EXISTENCE OF AODVE IDENTITY (ZERO) The sum of any rational number and zero (O) is the national number itself.
In other words, if \frac{a}{b} is any rational number; then
\frac{a}{b} + 0 = \frac{a}{b} = 0 + \frac{a}{b}
Verification We have
(i) \frac{2}{3} + 0 = \frac{2}{3} + \frac{0}{3} = \frac{2+0}{3} = \frac{2}{3} and, 0 + \frac{2}{3} = \frac{0}{3} + \frac{2}{3} = \frac{0+2}{3} = \frac{2}{3}
\therefore \frac{2}{3} + 0 = \frac{2}{3} = 0 + \frac{2}{3}
(ii) \frac{-4}{5} + 0 = \frac{-4}{5} + \frac{0}{5} = \frac{-4+0}{5} = \frac{-4}{5} and 0 + \frac{-4}{5} = \frac{0}{5} + \frac{-4}{5} = \frac{0+(-4)}{5} \frac{-4}{5}
\frac{-4}{5} + 0 = \frac{-4}{5} = 0 + \frac{-4}{5}
Similarly, it can be verified for other rational numbers.
Remark The rational number 0 is called the identity element for the addition of rational numbers.
EXISTENCE OF NEGATIVE (ADDITIVE INVERSE) OF A RATIONAL NUMBER For every rational number \frac{a}{b} there is a rational number \frac{c}{d} such that
\frac{a}{b} + \frac{c}{d} = 0 = \frac{c}{d} + \frac{a}{b}
The rational numbers \frac{a}{b} and \frac{c}{d} satisfying the above property are called additive inverse or negative of each other. The additive inverse of \frac{a}{b} is written as – \frac{a}{b}
Verification: We have,
\frac{3}{5} + \frac{-3}{5} = \frac{3+(-3)}{5} = \frac{0}{5} = 0
and, \frac{-3}{5}+\frac{3}{5} = \frac{(-3)+3}{5} = \frac{0}{5} = 0
\therefore \frac{3}{5} + \frac{-3}{5} = 0 = \frac{-3}{5} + \frac{3}{5}
It follows from this that the additive inverse (negative) of \frac{3}{5} is \frac{-3}{5}and the additive inverse
of \frac{-3}{5} is \frac{3}{5}.But, the additive inverse (negative) of \frac{3}{5} is written as – \frac{3}{5} and that of \frac{-3}{5} is
Written as -\LARGE{(} \frac{-3}{5} \LARGE{)} Therefore, \frac{-3}{5} and – \frac{3}{5} represent the same rational number, that is,
\frac{-3}{5} = – \frac{3}{5}
Also, \frac{3}{5} = \frac{-3}{5} represent the same rational number,that is
-\LARGE{(} \frac{-3}{5} \LARGE{)} = \frac{3}{5} \Rightarrow – \LARGE{(} – \frac{3}{5} \LARGE{)} = \frac{3}{5} \LARGE{[} \because \frac{-3}{5} = – \frac{3}{5}

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by maths on October 21, 2011

\frac{a}{b}+\LARGE{(} \frac{c}{d}+ \frac{e}{f} \LARGE{)}= \LARGE{(} \frac{a}{b}+ \frac{c}{d} \LARGE{)}+ \frac{e}{f}
\frac{-3}{4}+ \LARGE{(} \frac{5}{6}+ \frac{-4}{9} \LARGE{)} and \LARGE{(} \frac{-3}{4}+ \frac{5}{6} \LARGE{)}+ \frac{-4}{9}
We have,
\frac{-3}{4}+ \LARGE{(} \frac{5}{6}+ \frac{-4}{9}[latex]\LARGE{)}[/latex]= \frac{-3}{4}+ \LARGE{(} \frac{15}{18}+ \frac{-8}{18} \LARGE{)}
= \frac{-3}{4}+ \frac{15-8}{18}
= \frac{-3}{4}+ \frac{7}{18}=\frac{-27}{36}+\frac{14}{36} \frac{-27+14}{36}= \frac{-13}{36}
and \LARGE{(} \frac{-3}{4}+\frac{5}{6} \LARGE{)}+\frac{-4}{9}=\LARGE{(} \frac{-9}{12}+\frac{10}{12} \LARGE{)}+\frac{-4}{9}
= \frac{-9+10}{12}+\frac{-4}{9}

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by maths on October 20, 2011

(i) \frac{1}{4}+\frac{2}{3}=\frac{3}{12}+\frac{8}{12}=\frac{3+8}{12}=\frac{11}{12} which is a rational number.
(ii) \frac{-3}{4}+\frac{5}{6}=\frac{-9}{12}+\frac{10}{12}=\frac{-9+10}{12}=\frac{1}{12} which is a rational number.
(iii) \frac{-5}{4}+\frac{-1}{24}=\frac{-30}{24}+\frac{-1}{24}=\frac{-30-1}{24}=\frac{-31}{24} ,which is a rational number.
COMMUTATIVITY The addition of rational numbers is commutative i.e., if \frac{a}{b} and \frac{c}{d} are any two rational numbers, then
\frac{a}{b}+\frac{c}{d}=\frac{c}{d}+\frac{a}{b}
Verification: In order to verify this property, let us consider two expressions
\frac{5}{6}+\frac{-4}{9} and \frac{-4}{9}+\frac{5}{6}
We have
\frac{-4}{9}+\frac{5}{6}=\frac{-8}{18}+\frac{15}{18}=\frac{-8+15}{18}=\frac{7}{18} and, \frac{5}{6}+\frac{-4}{9}=latex]\frac{-8+15}{18}[/latex]+\frac{-8}{18}=\frac{15+(-8)}{18} = \frac{7}{18}
\therefore \frac{5}{6}=\frac{-4}{9}=\frac{-4}{9}=\frac{-4}{9}+\frac{5}{6}
Similarly, it can be verified for other pairs of rational numbers.
ASSOCIATIVITY The addition of rational numbers is associative i.e. if \frac{a}{b}, – and – are any three rational numbers, then

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by maths on October 19, 2011

Step III Find-the LCM of the denominators obtained in step II.
Step IV Express each one of the rational numbers in step I so that the LCM obtained in step III becomes their common denominator.
Step V Write a rational number whose numerator is equal to the sum of the numerators of rational numbers obtained in step IV and denominator as the LCM obtained in step III.
Step VI The rational number obtained in step V is the required sum.
Following examples will illustrate the above procedure.
ILLUSTRATIVE EXAMPLES
Example 1 Add \frac{5}{12} and \frac{3}{8}.
Solution Clearly, denominators of the given numbers are positive.
The LCM of denominators 12 and 8 is 24.
Now, we express \frac{5}{12} and \frac{3}{8} into forms in which both of them have the same
denominator 24.
We have,
\frac{5}{12}=\frac{5X2}{12X2} = \frac{10}{24} and, \frac{3}{18}= \frac{3X3}{8X3}=\frac{9}{24}
\therefore \frac{5}{12}+\frac{3}{8}=\frac{10}{24}+\frac{9}{24}=\frac{10+9}{24}=\frac{19}{24}
Example 2 Add \frac{7}{9} and 4.
Solution We have, 4=\frac{4}{1}
Clearly, denominators of the two rational numbers are positive. We now rewrite them so that they have a common denominator equal to the LCM of the denominators.
LCM of 9 and 1 is 9.
We have, \frac{4}{1} = \frac{4X9}{1X9} = \frac{36}{9}
\therefore \frac{7}{9}+4= \frac{7}{9} + \frac{4}{1}=\frac{7}{9}+\frac{36}{9}=\frac{7+36}{9}=\frac{43}{9}
Example 3 Add \frac{3}{8} and \frac{-5}{12}
Solution The denominators of the given rational numbers’ are 8 and 12 respectively. The LCM of 8 and 12 is 24.
Now we re-write the given rational numbers into forms in which both of them have the same denominator.
\frac{3}{8}=\frac{3X3}{8X3}=\frac{9}{24} and \frac{-5}{12}=\frac{-5X2}{12X2}=\frac{-10}{24}
\therefore \frac{3}{8}+\frac{-5}{12}=\frac{9}{24}+\frac{(-10)}{24}=\frac{9-10}{24}=\frac{-1}{24}
Example 4 Simplify:\frac{8}{-15} + \frac{4}{-3}
Solution We have,
\frac{8}{-15}+\frac{4}{-3}=\frac{8}{-15}+\frac{-4}{3} \LARGE{[} \therefore \frac{8}{-15} =\frac{8X-1}{(-15)X(-1)}=\frac{-8}{15} and \frac{4}{-3}=\frac{4x-1}{(-3)x(-1)}=\frac{-4}{3} \LARGE{]}
Re-writing \frac{-4}{3} in the form in which it has denominator 15, we get
\frac{-4}{3}=\frac{-4X5}{3X5}=\frac{-20}{15}
\therefore \frac{8}{-15}+\frac{4}{-3}=\frac{-8}{15}=\frac{-4}{3}
= \frac{-8}{15}+\frac{-20}{15} \LARGE{[} \because c=\frac{-20}{15} \LARGE{]}
= \frac{(-8)+(-20)}{15}=\frac{-28}{15}
Example 5 Simplify:\frac{7}{-26}+\frac{16}{39}
Solution We have,
\frac{7}{-26}+\frac{16}{39}=\frac{-7}{26}+\frac{16}{39} \LARGE{[} \because \frac{7}{-26}=\frac{7X-1}{(-26)X(-1)}=\frac{-7}{26} \LARGE{]}
LCM of26and39is 78.
Re-writing \frac{7}{-26} and \frac{16}{39} in forms having the same denominator 78, we get
\frac{-7}{26}=\frac{-7X3}{26X3}=\frac{-21}{78},\frac{16}{39}=\frac{16X2}{39X2}=\frac{32}{78}
\therefore \frac{7}{-26}+\frac{16}{39}=\frac{-7}{26}+\frac{16}{39}=\frac{-21}{78}+\frac{32}{78}=\frac{11}{78}
EXERCISE 1.1
1. Add the following rational numbers:
(i) \frac{-5}{7} and \frac{3}{7}(ii) \frac{-15}{4} and \frac{7}{4} (iii) \frac{-8}{11} and \frac{-4}{11} (iv) \frac{6}{13} and \frac{-9}{13}
2. Add the following rational numbers:
(ii) \frac{3}{4} and \frac{-5}{8} (ii) \frac{5}{-9} and \frac{7}{3} (iii) -3 and \frac{3}{5} (iv) \frac{-7}{27} and \frac{11}{18} (v) \frac{31}{-4} and \frac{-5}{8} (vi) \frac{5}{36} and \frac{-7}{12} (vii) \frac{-5}{16} and \frac{7}{24} (viii) \frac{7}{-18} and \frac{8}{27}
3. Simplify:
(i) \frac{8}{9}+\frac{-11}{6} (ii) \frac{-5}{16}+\frac{7}{24} (iii) \frac{1}{-12}+\frac{2}{-15} (iv) \frac{-8}{-19}+\frac{-4}{57} (v) \frac{7}{9}+\frac{3}{-4} (vi) \frac{5}{26}+\frac{11}{-39} (vii) \frac{-16}{9}+\frac{-5}{12} (viii) \frac{-13}{8}+\frac{5}{36} (ix) 0+\frac{-3}{5} (x) 1+\frac{-4}{5} (xi) 3+\frac{5}{-7}
4. Add and express the sum as a mixed fraction:
(i) \frac{-12}{5} and \frac{43}{10} (ii) \frac{24}{7} and \frac{-11}{4} (iii) \frac{-31}{6} and \frac{-27}{8} (iv) \frac{101}{6} and \frac{7}{8}
ANSWERS
1. (i) \frac{-2}{7} (ii) —2 (iii) \frac{-12}{11} (iv) \frac{-3}{13}
2. (i) \frac{1}{8} (ii) \frac{16}{9} (iii) \frac{-12}{5} (iv)\frac{19}{54} (v) \frac{-67}{8} (vi) \frac{-4}{9} (vii) \frac{-1}{48} (viii) \frac{-5}{54}
3. (i) \frac{-17}{18} (ii) \frac{-1}{48} (iii) \frac{-13}{60} (iv) \frac{-28}{57} (v) \frac{1}{36} (vi) \frac{-7}{78} (vii) \frac{-79}{36} (viii) \frac{-107}{72} (ix) \frac{-3}{5} (x) \frac{1}{5} (xi) \frac{-16}{7}
4 (i) 1\frac{9}{10} (ii) \frac{19}{28} (iii) —8 \frac{13}{24} (iv) 17 \frac{17}{24}
1.4 PROPERTIES OF ADDITION OF RATIONAL NUMBERS
In this section, we shall learn some properties of addition of rational numbers. These properties are similar to those of addition of integers which we have learnt in the previous class.
CLOSURE PROPERTY The sum of any two rational numbers is always a rational number.
Thus, if \frac{a}{b} and \frac{c}{d} are any two rational numbers, then\LARGE{(} \frac{a}{b}+\frac{c}{d} \LARGE{)} is also a rational number.
Verification: In order to verify this property. Let us consider the following:

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RECAPITULATION

by maths on October 18, 2011

RATIONAL NUMBER A number of the form – or a number which can be expressed in the

form \frac{p}{q}, where p and q are integers and {q}\neq{0}, is called a rational number.
In other words, a rational number is any number that can be expressed as the quotient of
two integers with the condition that the divisor is not zero.
Each of the numbers \frac{7}{9},\frac{-2}{11},\frac{-3}{-2},\frac{15}{-7} is a rational number.
It should be noted that every integer is a rational number.
In the rational number \frac{p}{q} , integer p is known as the numerator and non-zero integer q is
called the denominator,
If the numerator and denominator of a rational number are of the same sign, then it is said to be positive. Otherwise, it is negative.
Rational numbers \frac{7}{11} and \frac{-25}{-9} are positive whereas \frac{-11}{7} and \frac{4}{-15} are negative rational numbers.
If \frac{p}{q} is a rational number and m is a non-zero integer, then
\frac{p}{q}=\frac{pxm}{qxm}
Rational number \frac{pxm}{qxm} is a rational number equivalent to \frac{p}{q}
If \frac{p}{q} is a rational number and m is a common divisor of p and q, then
\frac{p}{q}=\frac{p \div m}{q \div m}
LOWEST FORM OF A RATIONAL NUMBER A rational number \frac{p}{q} is said to be in the lowest
form or simplest form if p and q have no common factor other than1.
Rational number \frac{5}{7} is in the lowest form, but \frac{15}{21} is not in the lowest form. In order to
reduce a rational number \frac{p}{q} in the lowest form, we divide its numerator p and denominator q by the HCF of p and q.
A rational number \frac{p}{q} is said to be in standard form, if its denominator q is a positive integer and p and q have no common divisor other than 1.
EQUALITY OF RATIONAL NUMBERS Two rational numbers \frac{p}{q} and \frac{r}{s} are equal if p x s = q x r.
\frac{p}{q}=\frac{r}{s}
\Leftrightarrow pxs=qxr
\LeftrightarrowNumerator of first x Denominator of second
= Numerator of second x Denominator of first
For example ,\frac{-7}{21} =\frac{3}{-9} and \frac{5}{7}=\frac{20}{28} because (—7)x(—9)=21x3and5x28=7×20.
1.3 ADDITION OF RATIONAL NUMBERS
In class VII, we have defined the operation of addition of rational numbers. The addition
of rational numbers is carried out in the same way as that of addition of fractions which
we have learnt in earlier classes. If two rational numbers are to be added, we first express
each one of them as rational numbers with positive denominator. For addition purpose,
we divide the rational numbers into the following two categories:
1.3.1 RATIONAL NUMBERS WITH SAME DENOMINATORS
In order to add two rational numbers having the same denominator, we follow the
following steps:
Step I Obtain the numerators of the two given rational numbers and their common denominator.
Step II Add the numerators obtained in step I
Step III Write a rational number whose numerator is the sum obtained in step II, and whose denominator is the common denominator of the given rational numbers.
It follows from the above steps that if \frac{p}{q} and \frac{r}{q}. are two rational numbers with the same
denominator, then
\frac{p}{q}+\frac{r}{q}=\frac{p+r}{q}
Following examples will illustrate the above procedure for the addition of two rational numbers with the same denominator
ILLUSTRATIVE EXAMPLES
3 13
Example 1 Add \frac{3}{5} and \frac{13}{5}.
Solution We have,
\frac{3}{5} + \frac{13}{5} = \frac{3+13}{5} = \frac{16}{5}
Example 2 Add \frac{7}{9} and\frac{-12}{9}
Solution We have,
\frac{7}{9} + \frac{-12}{9} = \frac{7+(-12)}{9} = \frac{-5}{9}
Example 3 Add \frac{-5}{9} and
\frac{-17}{9}
Solution We have,
\frac{-5}{9}+\frac{-17}{9}=\frac{(-5)+(-17)}{9}=\frac{-22}{9}
Example4 Add \frac{4}{-11} and \frac{7}{11}.
Solution We first express \frac{4}{-11} as a rational number with positive denominator.
We have,\frac{4}{-11} = \frac{4x(-1)}{(-11)x(-1)} = \frac{-4}{11}
.’. \frac{4}{-11} = \frac{7}{11} = \frac{-4}{11}+\frac{7}{11} = \frac{(-4)+7}{-1} = \frac{3}{11}
1.3.2 NUMBERS WITH DISTINCT DENOMINATORS
To find the sum of two rational numbers which do not have the same denominator, we follow the following steps:
Step I Obtain the rational numbers and see whether their denominators are positive or not. If the denominator of one (or both) of the numbers is negative, re-write it so that the denominator becomes positive.
Step II Obtain the denominators of the rational numbers in step I.

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RATIONAL NUMBERS

by maths on October 18, 2011

INTRODUCTION
In class VI, we began our study of numbers with counting numbers or natural numbers i.e.,
1, 2, 3, 4,…. By including 0 to natural numbers, we got whole numbers i.e., 0, 1, 2, 3, 4,….
The negative of natural numbers were put together with whole numbers to get integers
i.e., .,. —4, —3, —2, —1, 0, 1, 2, 3, 4,…. Addition, subtraction, multiplication and division operations were defined on integers and various properties (closure property, commuta
4.36 tivity, associativity, existence of identity, existence of inverse) of these operations were
discussed. In class VII, the concept of rational numbers was introduced and addition,
subtraction, multiplication and division operations on rational numbers were defined. In this chapter, we will learn about various properties of these operations on rational
numbers.
Let us first recall, in brief, what we have learn about rational numbers in class VII.

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