Number System Class 9 Notes Mathematics

Number System Class 9 Notes

INTRODUCTION

                We know that all counting numbers are called natural numbers. If we include 0 also in the set of natural numbers then we get set of whole numbers and if we include –1, – 2, – 3 … then we get set of integers. If 1 and 2 are any two consecutive natural numbers then between these two, we have infinite numbers in the decimal form or fraction form. These numbers are divided into two parts:

                (1) Rational numbers and

                (2) Irrational numbers

                In this section, we will learn about the rational and irrational numbers.

Rational Numbers :

                The numbers which can be expressed in the form of \frac{p}{q}, q ≠ 0, p and q are integers, are called rational numbers. For example \frac{2}{3},\frac{{37}}{{15}},\frac{{ - 17}}{{19}} etc. are rational numbers.

                Remark

                All natural numbers, whole numbers and integers are rational numbers.

Decimal representation of a rational number :

                If the decimal representation of a number is either terminating or non-terminating and repeating (i.e. recurring decimal), then it is called a rational number.

(A)          Terminating Decimal

                A decimal number in which after the decimal point, the number of digits are finite, is called a terminating decimal.

                For example 0.5, 0.6875, 0.15 etc. are terminating decimals.

(B)          Recurring decimals are of two types

                   (a)   Pure recurring decimal

                A decimal number in which all the digits after decimal point are repeated, is known as a pure recurring decimal. For example

                (i) 0.\bar 3 = 0.3333 …,    

                (ii) 3.\overline {16} = 3.161616 …

                    (b)   Mixed recurring decimal :

                A decimal number in which after the decimal point, at least one figure is not repeated and some figures are repeated is called a mixed recurring decimal.

                For example. 0.1\,\overline 9 = 0.1999 …, 0.57\overline {32} =  0.573232 …

Conversion of Recurring decimal into a fraction :

 (a)           Long Method

 Step 1 : Take the mixed recurring decimal and put it equal to x.

Step 2 : Count the number of non-recurring digits after the decimal point, let it be n.

Step 3 : Multiply both sides of x by 10ⁿ so that only the repeating digits are on the right hand side of the decimal point.

Step 4 : Multiply both sides of the equation obtained in step 3 by 10^m where m is the number of repeating digits.

Step 5 : Subtract the equation obtained in step 3 from equation obtained in step 4.

Step 6 : Divide both sides of the resulting equation by the coefficient of x.

Step 7 : Write the rational number thus obtained in the simplest p/q form.

 Ex 1.      Express 0.\overline 6 to p/q form.

Sol          Let,         x = 0.\overline 6

                i.e.,        x = 0.6666 …..                     …(i)

                multiply both sides of (i) by 10.

                             10x = 6.666                          …(ii)

                Subtract (i) from (ii)

                             10x = 6.666

                             – x = 0.666

                             9x = 6   

                \Rightarrow       x = \frac{6}{9} = \frac{2}{3}

 Ex. 2.     Express 0.\overline {47} to p/q form.

Sol.         Let,         x = 0.\overline {47}

                i.e.,        x = 0.474747…                  …(i)

                multiply both sides of  (i) by 100.  ∴ m = 2

                             100 x = 47.4747                 …(ii)

                Subtract (i) from (ii)

                             100 x = 47.474747…

                             –  x = 0.474747…

                             99x = 47

              \Rightarrow x = \frac{{47}}{{99}}

 Ex. 3.     Express0.12\overline 3   to p/q form.

Sol.         Let,         x = 0.12\overline 3

                i.e.,        x = 0.12333…                    …(i)

                multiply both sides of (i) by 100.

                             100x = 12.333…                 …(ii)

                multiply both sides of (ii) by 10.

                             1000x = 123.333…             …(iii)

                Subtract (ii) from (iii)

                             1000x = 123.333 ……

                             – 100x = 12.333 …….

                             900x = 111.0

                \Rightarrow   x = \frac{{111}}{{900}} = \frac{{3 \times 37}}{{900}} = \frac{{37}}{{300}}

(b)           Direct Method

Step 1 :  To obtain numerator subtract the number formed by non-repeating digits from the complete number without decimal.

Step 2 :  To obtain denominator take number of nine equal to number of repeating digits and after that put number of zeros equal to number of non-repeating digits.

Ex. 4.       (i)    0.\overline {45}  = \frac{{45--0}}{{99}} = \frac{{45}}{{99}} = \frac{5}{{11}}

                (ii)   0.7\overline {37}  = \frac{{737 - 7}}{{990}} = \frac{{730}}{{990}} = \frac{{73}}{{99}}

                (iii)  0.46\overline {573}  = \frac{{46573 - 46}}{{99900}} = \frac{{46527}}{{99900}}

Ex. 5.     Reduce 4 \frac{1}{8} into decimals (By long division method.)

Sol.         4\frac{1}{8} = \frac{{33}}{8}                                                            

                \Rightarrow        4\frac{1}{8}= 4.125 is terminating as the remainder is zero.

Ex. 6      What can be the maximum number of digit be in the repeating block of digits in the decimal expansion of \frac{1}{{17}}? Perform the division to check your answer.

Sol.      

  \frac{{\rm{1}}}{{17}}{\rm{ = }}{\rm{.}}\overline {{\rm{0588235294117647}}}

                Hence the maximum number of digits in the repeating block of the decimal representation of the fraction  \frac{1}{{17}} is (17 – 1) = 16

                Which may be verified by long division.

Ex.7.      You know that \frac{1}{7} = 0.\overline {142857} . Can you predict what will the decimal  expansion of \frac{2}{7},{\rm{ }}\frac{3}{7},{\rm{ }}\frac{4}{7},{\rm{ }}\frac{5}{7},{\rm{ }}\frac{6}{7} without actually doing the long division? If so, how.

Sol.          \frac{1}{7} = 0.\overline {142857} \Rightarrow \,\,\, = 2 \times \frac{1}{7} = 2 \times 0.\overline {142857}  = 0.\overline {285714} ,\, \,\frac{3}{7} = 3 \times \frac{1}{7} = 3 \times 0.\overline {142857}  = 0.\overline {428571}, \frac{4}{7} = 4 \times \frac{1}{7} = 4 \times 0.\overline {142857}  = 0.\overline {571428} ,\, \,\frac{5}{7} = 5 \times \frac{1}{7} = 5 \times 0.\overline {142857}  = 0 {714285} \frac{6}{7} = 6 \times \frac{1}{7} = 6 \times 0.\overline {142857}  = 0.\overline {857142}

Properties of rational numbers

        If a, b, c are three rational numbers. Then

(i)    Commutative property of addition:  For any two rational numbers a and b.

        a + b = b + a

(ii)   Associative property of addition : For any three rational numbers a, b and c.

        (a + b) + c = a + (b + c)

(iii)  Additive inverse :

        (a + (–a)) = 0 = (–a) + a

        0 is identity element, –a is called additive inverse of a.

(iv)  Commutative property of multiplication. For any two rational numbers a and b.

        a.b = b.a               

(v)   Associative property of multiplication. For any three rational numbers a, b and c.

        (a.b).c = a.(b.c)

(vi)  Multiplicative inverse

a \times \frac{1}{a} = 1 = \frac{1}{a} \times a

1 is called multiplicative identity and \frac{1}{a} is called multiplicative inverse of a or reciprocal of a. (a ≠ 0)

(vii) Distributive property :

        a.(b + c) = a.b + a.c

Rational Numbers between given two rational numbers :

                To find the rational numbers between given two rational numbers, we have two methods.

                Method – I

                Let a and b are two given rational numbers such that a < b

                then,

                           a < b

                \Rightarrow       a + a < b + a       [adding ‘a’ on both sides]

                \Rightarrow       2a < a + b

                \Rightarrow       a < \frac{{a + b}}{2}

                Again,

                           a < b

                \Rightarrow       a + b < b + b.      [adding ‘b’ on both sides]

                \Rightarrow       a + b < 2b    \Rightarrow \frac{{a + b}}{2} < b.

                ∴ a < \frac{{a + b}}{2} < b.\,\,\,\,i.e.\,\frac{{a + b}}{2} lies between a and b.

                Hence 1^st rational number between a and b is \frac{{a + b}}{2} 

                For next rational number 

\frac{{a + \frac{{a + b}}{2}}}{2} = \frac{{\frac{{2a + a + b}}{2}}}{2} = \frac{{3a + b}}{4}

                ∴           a < \frac{{3a + b}}{4} < \frac{{a + b}}{2} < b.

                Next

\frac{{\frac{{a + b}}{2} + b}}{2} = \frac{{a + b + 2b}}{{2 \times 2}} = \frac{{a + 3b}}{4}

                 ∴           a < \frac{{3a + b}}{4} < \frac{{a + b}}{2} < \frac{{a + 3b}}{4} < b and continues like this

Ex.8.      Find 3 rational numbers between \frac{1}{3} and \frac{1}{2}

                1^st           = \frac{{\frac{1}{3} + \frac{1}{2}}}{2} = \frac{{\frac{{2 + 3}}{6}}}{2} = \frac{5}{{12}}   ∴ \frac{1}{3} < \frac{5}{{12}} < \frac{1}{2}

                2^nd          = \frac{{\frac{1}{3} + \frac{5}{{12}}}}{2} = \frac{{\frac{{4 + 5}}{{12}}}}{2} = \frac{9}{{24}}  ∴   \frac{1}{3} < \frac{9}{{24}} < \frac{5}{{12}} < \frac{1}{2}

                3^rd           = \frac{{\frac{5}{{12}} + \frac{1}{2}}}{2} = \frac{{\frac{5}{{12}} + \frac{6}{{12}}}}{2} = \frac{{11}}{{24}}  ∴ \frac{1}{3} < \frac{9}{{24}} < \frac{5}{{12}} < \frac{{11}}{{24}} < \frac{1}{2}

                Verification  \frac{8}{{24}} < \frac{9}{{24}} < \frac{{10}}{{24}} < \frac{{11}}{{34}} < \frac{{12}}{{24}}. \left( {{\rm{as }}\,\frac{8}{{24}} = \frac{1}{3}\,{\rm{ and }}\,\frac{{12}}{{24}} = \frac{1}{2}} \right)                 

                Method – II

                To find n rational numbers between a and b (a < b), we use following algorithm.

                (i)            Find d = \frac{{b - a}}{{n + 1}}.

                (ii)           1^st rational number will be a + d.

                                2^nd rational number will be a + 2d.

                                3^rd rational number will be a + 3d and so on………..

                                n^th rational number is a + nd.

 Ex. 9.     Find 5 rational numbers between \frac{3}{5} and \frac{4}{5}

                Here       a = \frac{3}{5},\,\,\,b = \frac{4}{5}{\rm{ and }} n = 5

                ∴           d = \frac{{b - a}}{{n + 1}} = \frac{{\frac{4}{5} - \frac{3}{5}}}{{5 + 1}} = \frac{1}{{5 \times 6}} = \frac{1}{{30}}

                1^st = a + d = \frac{3}{5} + \frac{1}{{30}} = \frac{{19}}{{30}},             

                2^nd  = a + 2d = \frac{3}{5} + \frac{2}{{30}} = \frac{{20}}{{30}},         

                3^rd   = a + 3d = \frac{3}{5} + \frac{{23}}{{30}} = \frac{{21}}{{30}},

                4^th  = a + 4d = \frac{3}{5} + \frac{4}{{30}} = \frac{{22}}{{30}},          

                5^th a + 5d = \frac{3}{5} + \frac{5}{{30}} = \frac{{23}}{{30}}

Representation of Rational numbers on the number line by means of magnifying glass

                 The process of visualization of number on the number line through a magnifying glass in known as successive magnification.

                Sometimes, we are unable to check the numbers like 3.765 and 4.\overline {26} on the number line, we seek the help of magnifying glass by dividing the part into subparts and subparts into again equal subparts to ensure the accuracy of the given number.

Method to Find  Such Numbers on the Number Line :

1. Choose the two consecutive integral numbers in which the given number lies.
2. Choose the two consecutive decimal points in which the given decimal part lies by dividing the two given decimal parts into required equal parts.
3. Visualise the required number through magnifying glass.

Ex.10.    Visualise 3.765 on the number line.

Sol.         Since 3 < 3.765 < 4

                3.7 < 3.765 < 3.8

                and 3.76 < 3.765 < 3.77

                Step 1.                                       

                Step 2.                                   

                Step 3.                                   

Ex. 11.   Visualise 4.\overline {26} on the number line by magnifying glass.

Sol.         Since       4 < 4.\overline {26} < 5

                                4.2 < 4.\overline {26} < 4.3

                and         4.262 < 4.\overline {26} < 4.263

                Step 1.                   

                Step 2.                   

                Step 3.                   

 Exercise

1.            Find three rational numbers between 5 and 6.

Answer

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2.           Insert 4 rational numbers between –5 and 2.

Answer

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 3.            Write four rational numbers between 2 and 3. 

Answer

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4.            Find a rational number between \frac{{ - 2}}{3} and \frac{1}{4}.

Answer

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5.            Find three rational numbers between 1/3 and 1/2.                                      

Answer

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 6.            Find five rational numbers between –1 and –2 .

Answer

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7.            Find five rational numbers between 3/5 and 4/5.                        

Answer

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8.            Find three rational numbers between \frac{{ - 3}}{7} and \frac{{ - 1}}{7}.

Answer

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 9.            Express the following in the form of p/q.

                (A) 0.\overline 3              

                (B) 0.\overline {37}          

                (C) 0.\overline {54}

                (D) 0.\overline {05}

                (E) 1.\overline 3                                

                (F) 0.\overline {621} Answer

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 10.         Convert the following into rational numbers of form p/q

                (A) 0.\overline 6                                               

                (B) 0.222…          

                (C) 0.31818…

                (D) 0.571428571428…

Answer

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11.          Change each of the given decimals in the form p/q by long division method

                (i) 3.7222 …..                         (ii) 2.56767 …..

Answer

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12.          Change each of the given decimals in the form of p/q by direct method

                (i) {\rm{3}}{\rm{.7}}\overline {{\rm{237}}}                            

                (ii) {\rm{4}}{\rm{.25}}\overline {\rm{6}} Answer

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Irrational Numbers

Definition :

            All real numbers which are not rational are irrational numbers. These are non-recurring as well as non-terminating type of decimal numbers

            For example \sqrt 2 ,\sqrt 3 , \sqrt[3]{4} , 2+\sqrt 3 , \sqrt {2 + \sqrt 3 }

              \sqrt 3 = 1.73205080756 ……        and   \sqrt 2 = 1.4142123562 …..

Properties of Irrational Numbers

(a)        Negative of an irrational number is an irrational number.

            Eg:-  –\sqrt 3 , – \sqrt[4]{5} are irrational.

(b)        Sum and difference of a rational and an irrational number is an irrational number.

            For example :

(i)         Two numbers are 2 and\sqrt 3

            Sum = 2 +\sqrt 3 , is an irrational number.

             Difference = 2 –\sqrt 3 , is an irrational number.

             Also \sqrt 3 – 2 is an irrational number.

(ii)        Two numbers are 4 and \sqrt[3]{3}

            Sum = 4 + \sqrt[3]{3}, is an irrational number.

            Difference = 4 – \sqrt[3]{3}, is an irrational number.

(c)        Sum and difference of two irrational numbers is not necessarily an irrational number.

            For example :

(i)         Two irrational numbers are \sqrt 3 , 2\sqrt 3

            Sum = \sqrt 3 + 2\sqrt 3 = 3\sqrt 3 , is an irrational.

            Difference = 2\sqrt 3 –\sqrt 3 = \sqrt 3 , is an irrational.

(ii)        Two irrational numbers are 2 +\sqrt 3 and 2 – \sqrt 3

            Sum = (2 +\sqrt 3 ) + (2 +\sqrt 3 ) = 4, a rational number

            Two irrational numbers are  \sqrt 3 + 3,  \sqrt 3 – 3

            Difference =  \sqrt 3 + 3 –  \sqrt 3 + 3 = 6, a rational number

(d)        Product of a rational number with an irrational number is not always irrational.

            For example :

(i)         2 is a rational number and\sqrt 3 is an irrational number.

            2 ×  \sqrt 3 = 2\sqrt 3 , an irrational number.

(ii)        0 a rational number and \sqrt 3 an irrational number

            0 ×\sqrt 3   = 0, a rational number

(e)        Product of a non-zero rational number with an irrational number is always irrational.

            For example :

(i)         \frac{4}{3}× \sqrt 3 =  \frac{4}{3}\sqrt 3 =\frac{4}{3}  is an irrational number.

(f)         Product of an irrational with an irrational is not always irrational.

            For example :

(i)         \sqrt 3 × \sqrt 3 = \sqrt {3 \times 3}   =\sqrt 9 = 3 a rational number

(ii)        2 \sqrt 3 × 3 \sqrt 2 = 2 × 3 \sqrt {3 \times 2} = 6 \sqrt 6 an irrational number

(iii)        \sqrt[3]{3}× \sqrt[3]{{{3^2}}} = \sqrt[3]{{3 \times {3^2}}} =  \sqrt[3]{{{3^3}}}= 3 a rational number

(iv)       (2 +\sqrt 3 ) (2 –\sqrt 3 ) = (2)² – (\sqrt 3 )² = 4 – 3 = 1 a rational number

(v)        (2 +\sqrt 3 ) (2 +\sqrt 3 ) = (2 +\sqrt 3 )² = (2)² + (\sqrt 3 )2 + 2(2) × (\sqrt 3 ) = 4 + 3 + 4 \sqrt 3 = 7 + 4 \sqrt 3 an irrational number.

Important Notes :

            (i)         \sqrt {-2} ≠  -\sqrt 2 i.e., -\sqrt 2 is not an irrational number.

            (ii)         -\sqrt 2 ×  \sqrt {-3} ≠ \left( {\sqrt {-2 \times -3}  = \sqrt 6 } \right)

            (iii)       The set of rational numbers together with all irrational numbers form the set of real numbers. It is denoted  by R.

Ex. 1.     Prove that \sqrt 2 is not a rational number

Sol.         Method – I

                Let us find the square root of 2 by long division method as shown below :

⇒           \sqrt 2 = 1.414215…

                Clearly, the decimal representation of \sqrt 2 is neither terminating nor repeating. Hence, it is not a rational number.

            Method II         

            We shall prove this by the method of contradiction. If possible, let us assume that \sqrt 2 is a rational number. Then,

                        \sqrt 2 = \frac{p}{q}, where p and q are integers having no common factor and q ≠ 0.

⇒        2 = \frac{{{p^2}}}{{{q^2}}}          [Squaring both sides]

⇒        p² = 2q²                       … (i)

⇒        p² is an even integer

⇒        p is an even integer       

⇒        p = 2m, where m is an integer

⇒        p² = 4m²

⇒        2q² = 4m²                          [Using (i)]

⇒        q² = 2m²

⇒        q² is an even integer

⇒        q is an even integer

            So, both p and q are even integers and therefore have a common factor 2. But, this contradicts that p and q have no common factor.

            Hence, \sqrt 2 is not a rational number.

Insertion of irrational numbers between two real numbers

Let a and b are two given real numbers, then irrational number between a and b is           

Example :

            2.101001000100001…………. is between 2 & 5 and it is non-recurring as well as non-terminating.

            Also 2.201001000100001………….

Example :

            As  \sqrt 2 = 1.414213562……………

            and  \sqrt 3 = 1.732050808……………

            As \sqrt 3   >  \sqrt 2 and \sqrt 2 has 4 in the 1st place of decimal while \sqrt 3 has 7 in the 1st place of decimal.

            ∴         1.501001000100001…………., 1.601001000100001………….

Representation of irrational number on a number line

Ex. 7.     Represent  \sqrt 2 on the number line.

Sol.         For representing \sqrt 2 on the number line we denote 0 (zero) on the line by the point O and 1 Unit Length by the point A such that OA = 1 Unit Length.

            Steps of Construction :

            (i)         Draw the line segment OA = 1 unit.

            (ii)        Draw AB = 1 unit, perpendicular to OA. (by using compass)

            (iii)       Join O to B.

                        Now, also by Pythagoras Theorem, we              

                        have

                                                OB²  = OA² + AB²

                                                OB  = \sqrt {O{A^2} + A{B^2}}

                                                OB = \sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}}

                                                      =\sqrt {1 + 1}   = \sqrt 2

            (iv)       With O as centre and OB as a radius, draw an arc to intersect the number line l at P.

                        Hence, it is clear that OP = OB = \sqrt 2

                        Thus, the point P represent  the number \sqrt 2 on the number line l.

Representation of Irrational numbers like \sqrt x

            To represent \sqrt x on number line we follow the following steps :

Step I        Obtain the positive real number x (say)

Step II       Draw a line and mark a point A on it.

Step III      Mark a point B on the line such that AB = x units.

Step IV      From point B mark a distance of 1 unit and mark the new point as C.

Step V       Draw a perpendicular bisector of line segment AC to find the midpoint of AC. Let it be ‘O’.

Step VI      Draw a circle with centre O and radius OC.

Step VII    Draw a line perpendicular to AC passing through B and intersecting the semi circle at D. Length BD is equal to \sqrt x .

            Explanation

            We have,

            AB = x units and BC = 1 unit.

∴        AC = (x + 1) units

⇒        OA = OC = \frac{{x + 1}}{2} units

⇒        OD = \frac{{x + 1}}{2}units                   [ OA = OC = OD]

            Now,     OB  = AB – OA = x - \frac{{x + 1}}{2} = \frac{{x - 1}}{2}

            Using Pythagoras Theorem in ΔOBD, we have

                        OD² = OB² + BD²

⇒        BD² = OD² – OB²

⇒        BD² = {\left( {\frac{{x + 1}}{2}} \right)^2} - {\left( {\frac{{x - 1}}{2}} \right)^2}

⇒        BD =    \sqrt {\frac{{\left( {{x^2} + 2x + 1} \right)\, - \,\left( {{x^2} - 2x + 1} \right)}}{2}} =  \sqrt {\frac{{4x}}{4}} = \sqrt x

            This shows that \sqrt x exists for all real numbers x > 0.

INTRODUCTION

            In previous section we have studied about rational and irrational numbers and their representation on number line and some properties of rational and irrational numbers.

            In this section we shall study the laws of rational powers (exponents) of real numbers. First we shall define the principal n^th root of a positive real number.

Principal n^th Root of a Positive Real Number

            If ‘a’ is a positive real number and ‘n’ is a positive integer, then the principal nth root of ‘a’ is

            a^1/n    or  \sqrt[n]{a}

            i.e.,  a^1/n  = \sqrt[n]{a}

Principal n^th root of  a negative real number

            If x is a negative number and n is odd positive integer, then the principal nth root of x is defined as –|x|^1/n 

            Note – Principal nth root of negative number is not defined in real numbers if n is even.

            i.e. it is not real number.

            For example

            (– 4)^1/2 = \sqrt { - 4} is not a  real number.

Rational Exponent

            For any positive real number ‘a’ and a rational number p/q

            where,  q ≠ 0, we define,

            a^p/q = (a^p)^1/q =\sqrt[q]{{{a^p}}}

⇒        a^p/q = (a^1/q)^p =  {\left( {\sqrt[q]{a}} \right)^{\,\,p}}

            i.e.,  a^p/q is the principal q^th root of a^p 

            or,  a^p/q is q^th root of a raised to the power of p.

Laws of Rational Exponents

            Let a, b are two positive real numbers and m, n are two positive rational numbers then,

(i)         a^m × aⁿ  = a^m+n 

(ii)        a^m ÷ aⁿ = a^m–ⁿ 

(iii)       (a^m)ⁿ  =  a^mn 

(iv)       a^–n   = \frac{1}{{{a^n}}}

(v)        (a.b)^m  =  a^m.b^m 

(vi)       {\left( {\frac{a}{b}} \right)^m} = \frac{{{a^m}}}{{{b^m}}}

(vi)       a⁰= 1

Surds

            Surds :  An irrational number of the form \sqrt[n]{a} is given a special name Surd, where ‘a’ is called radicand, it should always be a rational number and cannot be expressed as n^th power of any other rational number, also the symbol \sqrt[n]{{}} is called the radical sign and the index n is called order of the surd.

            \sqrt[n]{a} is read as nth root of ‘a’ and can also be written as {a^{\frac{1}{n}}}.

How to identify a surd ?

(i)         \sqrt[3]{4} is a surd as radicand is a rational number.

            Similar examples \sqrt[3]{5},\,\,\sqrt[4]{{12}},\,\,\sqrt[5]{7},\,\,\sqrt {12} ,.............

(ii)        2 + \sqrt 3 is a surd (as surd + rational number will give a surd)

                        Similar examples \sqrt 3 -\sqrt 2 ,\,\,\sqrt 3  + 1,\,\,\sqrt[3]{3} + 1,.............

(iii)       \sqrt {7-4\sqrt 3 } is a surd as 7 – 4 \sqrt 3 is a perfect square of \left( {2-\sqrt 3 } \right)

            Similar examples \sqrt {7 + 4\sqrt 3 } ,\,\,\sqrt {9-4\sqrt 5 } ,\,\,\sqrt {9 + 4\sqrt 5 } ,..............

(iv)        \sqrt[3]{{\sqrt 3 }}is a surd as \sqrt[3]{{\sqrt 3 }} = {\left( {{3^{\frac{1}{2}}}} \right)^{\frac{1}{3}}} = {3^{\frac{1}{6}}} = \sqrt[6]{3}

            Similar examples \sqrt[3]{{\sqrt[3]{5}}},\,\,\sqrt[4]{{\sqrt[5]{6}}},...........           

            These are not surds.

(v)         \sqrt[3]{8}because \sqrt[3]{8} = \sqrt[3]{{{2^3}}}  = 2 which is a rational number.

(vi)        \sqrt {2 + \sqrt 3 } because 2 +\sqrt 3 is not a perfect square or 2 + \sqrt 3 is not rational.

(vii)      \sqrt[3]{{1 + \sqrt 3 }} because radicand is an irrational number.

Laws of Surds

(a)        {\left( {\sqrt[n]{a}} \right)^n} = \sqrt[n]{{{a^n}}} = a          

(b)        \sqrt[n]{a} \times \sqrt[n]{b} = \sqrt[n]{{ab}}            [Here order should be same]

            For example

            (i) \sqrt[3]{2} \times \sqrt[3]{6} = \sqrt[3]{{2 \times 6}} = \sqrt[3]{{12}}

            But  \sqrt[3]{3} \times \sqrt[4]{6} \ne \sqrt[3]{{3 \times 6}} \ne \sqrt[4]{{3 \times 6}}          (because order is not same)

            First make their order same & then we can multiply.

(c)        \sqrt[n]{a} \div \sqrt[n]{b} = \sqrt[n]{{\frac{a}{b}}}

(d)        \sqrt[n]{{\sqrt[m]{a}}} = \sqrt[{nm}]{a} = \sqrt[m]{{\sqrt[n]{a}}},          For example \sqrt {\sqrt {\sqrt 2 } }  = \sqrt[8]{2}

(e)        \sqrt[n]{a} = \sqrt[{n \times p}]{{{a^p}}}  [important for changing order of surds]

            or \sqrt[n]{{{a^m}}} = \sqrt[{n \times p}]{{{a^{m \times p}}}}

            For example

            (i)         \sqrt[3]{{{6^2}}} , make its order 6

                        then \sqrt[3]{{{6^2}}} = \sqrt[{3 \times 2}]{{{6^{2 \times 2}}}} = \sqrt[6]{{{6^4}}}

            (ii)        \sqrt[3]{6}, make its order 15  then \sqrt[3]{6} = \sqrt[{3 \times 5}]{{{6^{1 \times 5}}}} = \sqrt[{15}]{{{6^5}}}

Types of Surds

(i)         Pure Surd : A surd which has unity only as its rational factor, the other factor being irrational, is called pure surd.

            For example  \sqrt 3 ,\,\,\sqrt {15} ,\,\,\sqrt 2 ,\,\,\sqrt[4]{3},\,\,\sqrt[3]{{135}},\,\,\sqrt[4]{{1875}},\,\,\sqrt[6]{8}

(ii)        Mixed Surd : A surd which has a rational factor other than unity, the other factor being irrational, is called a mixed surd.

            For example  2\sqrt 3 ,\,\,2\sqrt[3]{5},\,\,3\sqrt[3]{5},\,\,5\sqrt[4]{3}

Comparison of Surds

            It is clear that if x > y > 0 and n > 1 is a positive integer then \sqrt[n]{x} > \sqrt[n]{y}.

            For example

            \sqrt[3]{{16}} >\sqrt[3]{{12}}  and  \sqrt[5]{{36}} >\sqrt[5]{{25}}  and so on.

Rationalization of Surds

             If the product of two surds is a rational number then each of them is called the rationalizing factor (R.F.) of the other.

            The process of converting a surd to a rational number by using an appropriate multiplier is known as rationalization.

Remarks :

            (i)         R.F. of \sqrt a is \sqrt a

            (ii)        R.F. of \sqrt[3]{a} is  \sqrt[3]{{{a^2}}}

            (iii)       R.F. of \sqrt a  + \sqrt b is \sqrt a -\sqrt b & vice versa

            (iv)       R.F. of a + \sqrt b is a-\sqrt b & vice versa

  

 

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